View Full Version : Today Special Relativity dies...
ram1024
Jun20-04, 02:09 AM
Mark it on your calendars, people. Let us begin.
Case #1:
|(-> (o) <-)|
man standing on a movable platform bed. at the EXACT center between two photon emitters. SR concludes that the simultaneously emitted photons from the two emitters will be detected by the observer at the same exact time.
ADDENDUM:the photon emitters are tied to precise atomic clocks
these clocks are perfectly aligned and synchronized and in all cases they move within the same inertial frame so they can stay calibrated.
(True / False) ?
Case #2:
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Platform is moving. SR concludes the photons (still emitted simultaneously) will be detected by the observer at the exact same time.
(True / False) ?
Case #3:
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Man is moving on platform towards an emitter. SR concludes that photons are NOT detected at the exact same time.
(True / False) ?
once we square these we'll move on to stage 2.
Simultaneously, but simultaneously in which frame? You cant' omit that.
Case #1 there is only frame, so assuming simultaneously in that frame, true.
Case #2 there is one frame (not counting the background), so assuming simultaneously in that one frame, true.
Case #3, there are two frames (man and emitters). So in which frame are the photons emitted simultaneously? If man frame, false (detected at the same time). If emitter frame, true (not detected at the same time).
That's my answer according to my understanding of SR.
Chronos
Jun20-04, 04:28 AM
huh? if both frames are moving relative to each other, they must obey SR to measure one to the other. what is so difficult about that? it has been proven in particle accelerators many times.
In all cases we have 2 refrence frames.
In case 1 these are the same so in both reference frames the phtons are emitted and detectd simultaneously
In case 2 these refrnce frames are different, in our reference frame i.e. the 'stationary' frame we conclude the phtons are emitted and detected simulatenously. In the refrence frame of the man THIS IS NOT THE CASE; the man does not think the phtons are emitted simultaneously, but he does conclude that they are detected simultaneously.
In case 3 in our refernce frame we conclude that the phtons are emitted but not detected simulatenously. In the refernce frame of the man, the photons are not emitted simulataneously and they are not detected simulataneously.
russ_watters
Jun20-04, 09:54 AM
We've had this discussion before: you can't prove/disprove a theory with a thought experiment, only with evidence.
In case 2 these refrnce frames are different, in our reference frame i.e. the 'stationary' frame we conclude the phtons are emitted and detected simulatenously. In the refrence frame of the man THIS IS NOT THE CASE; the man does not think the phtons are emitted simultaneously, but he does conclude that they are detected simultaneously.
For Case#2 I didn't consider the stationary background frame in my reply. If the photons are emitted simultaneously in the stationary background frame, the photons are not emitted simultaneously in the moving platform frame, therefore the man does NOT detect them at the same time. There I disagree with jcsd.
ram, will you please specify the frames in which the photons are emitted simultaneously? Is it the stationary background frame in all cases? You can't expect too many replies for a poorly specified scenario.
For Case#2 I didn't consider the stationary background frame in my reply. If the photons are emitted simultaneously in the stationary background frame, the photons are not emitted simultaneously in the moving platform frame, therefore the man does NOT detect them at the same time. There I disagree with jcsd.
ram, will you please specify the frames in which the photons are emitted simultaneously? Is it the stationary background frame in all cases? You can't expect too many replies for a poorly specified scenario.
There is ceratinly plenty of ambiguity as ram failed to define reference frames, I just assumed that ram's drawings refer to the point of view of the 'stationary frame' and hence it is in this frame that the photons are always emitted simulatenously.
However you are incorrect still as simulatenity fails only at distance, so if the photons are detected simulataneously in one frame, they're detected simulatenously in all frames.
edited for about 100 typos
However you are incorrect still as simulatenity fails only at distance, so if the photons are detected simulataneously in one frame, they're detected simulatenously in all frames.
Yes I know simultaneity fails only at distance. And, the photons cannot be detected at the same time by the man [for case#2, if photons are emitted simultaneously in background frame] They would be detected at the same time by a midpoint observer in the stationary frame. But the man does not remain at that midpoint, so he does not detect them at the same time.
Just to add my voice to what has already been said. You must state which frame the lights emit simultaneously in. Is it according to us, the background observer or is it according to the emitters themselves.
In the first case, they are one in the same.
In the second case they aren't
In the third case they are the same again.
Also, in the third case, it depends on where the man is on the platform when he detects the photons. If he is exactly between the two emitters when he does, he will detect the photons simultaneously. But this doe not mean that the photons were emitted at the same time in his frame (in fact, from his frame, they can't have been emitted simultaneously.) That is the whole point of Relativity of Simultaneity, that distance separated events that are simultaneous in one frame are not simultaneous in a frame moving relative to the first.
It is a consequence of the invarience of the speed of light for all observers.
And as also already pointed out, you can not kill a theory with a thought experiment. All a thought experiment can do is show you the consequences of any given postulates. It takes a physical experiment to show whether or not the results of the thought experiment agree with reality or not. So far, every physical experiment designed to test the predictions of Relativity has supported the conclusions of Relativity.
(NOTE: This was in response to the original posting)
In all cases, I will assume by "simultaneously emitted photons" you mean that emission was simultaneous in the frame in which the picture is drawn.
Case #1: True.
Case #2: False.
Case #3: True.
Since there is some disagreement in case #2, so I'll explain my answers.
disagreement with wespe:
Given the presentation of the problem, I'm assuming that "simultaneous" means that they're simultaneous according to the "background". However, you are correct if we assume that ram2048 means that emission was simultaneous in the frame of the platform.
disagreement with jcsd:
In our frame, the guy is at the midpoint of the emitters when they flash, but he's moving towards the right emitter; thus he has to detect the right photon first. In the moving frame, they are indeed not emitted simultaneously, but the guy will still detect the right photon first.
Yes I know simultaneity fails only at distance. And, the photons cannot be detected at the same time by the man [for case#2, if photons are emitted simultaneously in background frame] They would be detected at the same time by a midpoint observer in the stationary frame. But the man does not remain at that midpoint, so he does not detect them at the same time.
Yep, sorry you're correct. I what I incorrecty did was take the refernce frame of the platform which is the same as the staionary frame in the other two case.
In all cases, I will assume by "simultaneously emitted photons" you mean that emission was simultaneous in the frame in which the picture is drawn.
Case #1: True.
Case #2: False.
Case #3: False.
Hurkyl, by "Case #3: False.", do you mean "detected at the same time" or "not detected at the same time"?
Ram, look at the mess you put us in. Clean this up! LOL
Good catch; it's been fixed.
ram1024
Jun20-04, 11:25 AM
the clocks in all cases emit photon simultaneously RELATIVE TO EACH OTHER.
these clocks are perfectly aligned and synchronized and in all cases they move within the same inertial frame so they can stay calibrated.
the clocks in all cases emit photon simultaneously RELATIVE TO EACH OTHER.
these clocks are perfectly aligned and synchronized and in all cases they move within the same inertial frame so they can stay calibrated.
Do you realize you wasted people's time by not specifying this in the first place? OK, in that case, #1 and #2 are true.
But, as Janus says, #3 is still unclear: where is the man when he detects the photons?
ram1024
Jun20-04, 11:40 AM
does it matter where the man is?
he's obviously NOT in the center...
you can put real values to it if you like and work it out from there. however you want to make it easier for you.
and as far as wasting people's time, we've got ALL DAY, you in a hurry? :D
does it matter where the man is?
he's obviously NOT in the center...
Then, he does not detect them at the same time. #3: true
and as far as wasting people's time, we've got ALL DAY, you in a hurry? :D
I might have better things to do if you'll excuse me
Take care.
ram1024
Jun20-04, 11:58 AM
so do we have (True, True, True) yet?
need couple more opinions/verifications then we can move to the next stage
ram1024
Jun20-04, 11:58 AM
editting main post to include the clock bit, since it confuses you guys. I thought it was self explanatory, my bad.
editting main post to include the clock bit, since it confuses you guys. I thought it was self explanatory, my bad.
You need to define which refernce frame the [phtons are emitted simultaneously, the stationery observer's, the platform's or the man's.
edited to add so the clocks are in the rest frame of the platform?
ram1024
Jun20-04, 01:14 PM
hehehe that's part of my hook.
what i can GIVE you, if the clocks are perfectly synchronized together in the stationary frame, they are locked onto the platform so they cannot move in relation to each other, and therefore cannot become unsynchronized, moving OR not.
they emit photons simultaneously in both the moving frame AND the stationary frame, basically.
hehehe that's part of my hook.
what i can GIVE you, if the clocks are perfectly synchronized together in the stationary frame, they are locked onto the platform so they cannot move in relation to each other, and therefore cannot become unsynchronized, moving OR not.
they emit photons simultaneously in both the moving frame AND the stationary frame, basically.
No they're not, you see they 'don't become unsychronized they either are or aren't scynorised in a partcular inertial frame, as in method used to synchronize them in one partcular frame only makes them unscychronised in another frame.
ram1024
Jun20-04, 01:23 PM
how so?
stationary they shoot light at the exact same moment. on the platform they shoot light at the exact same moment. MOVING the platform they STILL shoot light at the exact same moment.
they are never in a position where there exists ANY relative motion between them to "unsynch" them in respect to each other. these two clocks define a precise INSTANT in time where two events ARE perfectly simultaneous.
any unsimultaneity by the observers can only be concluded as perceptual errors.
(True / False) ?
does it matter where the man is?
he's obviously NOT in the center...
you can put real values to it if you like and work it out from there. however you want to make it easier for you.
and as far as wasting people's time, we've got ALL DAY, you in a hurry? :D
Yes it does matter, if he is at the midpoint he will detect both simultaneously, if he's not he won't.
Example: the two following gifs show what happens according to the frame the emitters are in and the frame the man is in for the same situation. (the actual animation deals with a railway car moving along a track, but we will just assume that the track represents the platform. In this situation we have to men, one that stays stationary to the platform and one that moves with it.
We further stipulate that the moving man is next to the stationary man when the light is first detected by either.
The first animation shows things from the perspective of the emitters and the stationary man:
http://home.teleport.com/~parvey/train1.gif
The photons expand at c in two spherical fronts that reach the mid point at the same time by the observation of both men (both men see the photons arrive at the same time.)
The second animation shows what happens according to the man movig relative to the platform/track.
Since the speed of light is invarient for all observers, he also must see photons exapand outward from the point of emmission as a sphere in his frame. But from his perspective, the emitters do not stay at the the point of emmission. Therefore, in order for him to detect the emissions from both emitters at the same time, and at the same time as the man stationary to the platform, in his frame, the emiiters do not emit simultaneously, but one emits after the other.
http://home.teleport.com/~parvey/train2.gif
The reason things have to be this way is that the two men, being at the same point at the same time according to both of them, and both seeing the flash from the emitters arrive simultaneously is a spacetime event that is invarient and must be agreed upon by everyone.
They 'shoot light' simulateously only in their own rest frame, this is not the case in other rest frames. There is no perceptual error as there is nothing in particular that ties the photon to the rest frame of it's emitter.
Today Special Relativity dies...
Mark it on your calendars, people. Let us begin.
...
I guess I should throw out my GPS receiver then. Shame about that...
ram1024
Jun20-04, 01:49 PM
Since the speed of light is invarient for all observers
that's an assumption. I'll elaborate when we get to the second stage. need hurkyl russ or tom to confirm (True, True, True) on the first 3 cases.
I guess I should throw out my GPS receiver then. Shame about that...
nah they'll probably just reprogram the service. should be simple enough to do with all the computers they have
nah they'll probably just reprogram the service. should be simple enough to do with all the computers they have
:rolleyes:
You are aware that GPS receivers use SR to determine their exact location relative to the satellites? That's why you need to have 4 sat's in view. You triangulate your position with three, but you need the fourth to solve for the SR clock bias.
that's an assumption. I'll elaborate when we get to the second stage. need hurkyl russ or tom to confirm (True, True, True) on the first 3 cases.
nah they'll probably just reprogram the service. should be simple enough to do with all the computers they have
You won't get an answer until you specify which frame the clocks are synchronised in and whether the man is in the middle of the two emitters, otherwise there is no sensible answer.
ram1024
Jun20-04, 02:14 PM
they are synchronized IN the rest frame and REMAIN synchronized in all inertial frames for the clocks and the platform
it's already been specified :D
But they're not and that's whole point; you can't use the assumptions of Galilean relativity to disprove special relatvity.
ram1024
Jun20-04, 02:59 PM
that's the point being made. by ANY standard used they are synchronized. if you can't accept that you can't get any further into the experiment.
if they're NOT simultaneous you MUST be able to explain why?
they're synchronized in the rest frame, and remain synchronized no matter what inertial frame they're in because THEY'RE THE SAME CLOCK. they're attached, see? :D
Tom decided not to join us. i think he already knows where i'm going with this...
that's an assumption. I'll elaborate when we get to the second stage. need hurkyl russ or tom to confirm (True, True, True) on the first 3 cases.
No, it is a postulate. A postulate that has been experimentally confirmed in numerous ways.
Tom Mattson
Jun20-04, 03:40 PM
Tom decided not to join us. i think he already knows where i'm going with this...
I didn't say that, I said that the others were "kicking your butt just fine without me". Boy oh boy, talk about problems with reading comprehension! :biggrin:
The point that you aren't getting is that you can't simply declare that the clocks are synchronized in every frame. In fact, they won't be, because simultaneity is relative. That is, if the ticks from the two clocks are simultaneous in one frame, then if they are spatially separated then they cannot be simultaneous in any other frame.
they're synchronized in the rest frame, and remain synchronized no matter what inertial frame they're in because THEY'RE THE SAME CLOCK.
No, they aren't. You quite clearly stipulated that there are two different clocks. The fact that they are of identical construction and are perfectly synchronized in their own rest frame does not negate the fact that the two clocks are not one and the same.
ram1024
Jun20-04, 03:46 PM
fine, the emitter/clocks are on a geared track with exactly the same amount of cogs.
they are synchronized together in the center of the platform then wheeled to their locations at either end across the cogged tracks at exactly the same rate.
at any given point in the experiment at ANY inertial frame, the emitters can be wheeled back to the center to verify they are still synchronized.
attacking the synch of the clocks gets you nowhere.
a vbiew from one rest frame:
|> o <|
|-------->o<--------|
A view from another:
|> 0 |
----|-------->0<--------|
It should be clear that in the second rest frame in order for them to be detected simulatenously the two phtons cannot of been emitted simulateously.
lets do some maths for this
Let x^{\mu}_1 and x^{\mu}_2 (where x^0 = t)be the postion 4 -vector of the emitters as it emits the light where the orgin is half way between the emitters in the the rest frame of the emitters, where c is the speed of light and l the diatnce between the the two emitters:
x^{\nu}_1 = \left(\begin{array}{c}0\\{\frac{l}{2}}\\0\\0\end{a rray}\right)
x^{\nu}_2 = \left(\begin{array}{c}0\\{\frac{-l}{2}}\\0\\0\end{array}\right)
Let x'^{\mu}_1 and x'^{\mu}_2 be the postion 4 vectors of the emitters as they emit light in rest frame of someone who is moving with velocity u relative to the emitters:
the following realtionships apply:
x'^{\mu}_1 = {\Lambda^{\mu}}_{\nu}x^{\nu}_1
x'^{\mu}_2 = {\Lambda^{\mu}}_{\nu}x^{\nu}_2
therefore:
x'^{\mu}_1 = \left(\begin{array}{c}{\frac{-\gamma l\beta}{2c}}\\{\frac{\gamma l}{2}}\\0\\0\end{array}\right)
x'^{\mu}_2 = \left(\begin{array}{c}{\frac{\gamma l\beta}{2c}}\\{\frac{-\gamma l}{2}}\\0\\0\end{array}\right)
So in the first frame the photons are emitted both at t = 0
but in the second frame the're emitted at \frac{-\gamma l\beta}{2c} and \frac{\gamma l\beta}{2c} respectively which are only equal if u = 0.
They're not the same clock as they're not local.
Tom Mattson
Jun20-04, 03:52 PM
fine, the emitter/clocks are on a geared track with exactly the same amount of cogs.
they are synchronized together in the center of the platform then wheeled to their locations at either end across the cogged tracks at exactly the same rate.
This doesn't address the point. You needn't specify the method of synchronization, you need to specify the (one!) frame in which they are synchronized, and you need to not assume that that condition holds in every other frame. From the context, it seems clear enough to me that you mean that the clocks are synchronized in their rest frame. Yes?
at any given point in the experiment at ANY inertial frame, the emitters can be wheeled back to the center to verify they are still synchronized.
On this basis, all you can say for certain is that the clocks will be in synch when they are at rest at the center of the track (provided of course that their motion to the center was perfectly symmetric).
attacking the synch of the clocks gets you nowhere.
No one is "attacking the synch of the clocks" (whatever the heck that means). People are trying to get you to explicitly state which frame the clocks are synchronized in, and to get you to stop insisting that the clocks will be synchronized according to all inertial observers. Unless you do that, you are posing a problem in Galilean relativity. That sort of problem is well-defined enough, but it has the unfortunate difficulty of not being descriptive of the actual universe. :frown:
that's the point being made. by ANY standard used they are synchronized. if you can't accept that you can't get any further into the experiment.
The problem is that you are trying to get people to accept something that they know isn't acceptable.
Your inability to grasp the concept of the Relativity of Simultanity can in no way be used as proof against SR. It is a failure on your part, not on the part of the theory.
The Relativity of simultaneity is a direct consequence of the two postulates of Relativity. To disprove it, you have to show one or both postulates to be in error by direct physical experiment.
All you are doing is showing that SR and Galilean relativity are incompatible. But we already know that. It takes a real physical experiment to determine which of the two is correct for our universe. And every real experiment performed to date has come down on the side of SR.
ram1024
Jun20-04, 04:01 PM
you mean it doesn't work according to SR's universe
yes that's quite the point, since SR is what i'm disproving :D
in any case explain to me how they could POSSIBLY become unsynchronized.
what reason is there to assume they AREN'T synchronized, in other words.
do tell
Tom Mattson
Jun20-04, 04:04 PM
you mean it doesn't work according to SR's universe
No, I meant what I said: It doesn't work according the real universe.
yes that's quite the point, since SR is what i'm disproving :D
No, you aren't disproving it. You are simply denying it. Try to understand the difference.
in any case explain to me how they could POSSIBLY become unsynchronized.
what reason is there to assume they AREN'T synchronized, in other words.
do tell
That question is equivalent to asking, "Why are the laws of physics the same in every frame, and why is the speed of light the same in every frame." It is equivalent because the relativity of simultaneity is derived from those premises.
And the answer to both sets of questions is the same: Because that's the way it is.
you mean it doesn't work according to SR's universe
yes that's quite the point, since SR is what i'm disproving :D
in any case explain to me how they could POSSIBLY become unsynchronized.
what reason is there to assume they AREN'T synchronized, in other words.
do tell
They don't BECOME unscyhronised, they ARE unsychronised in different inertial reference frames.
ram1024
Jun20-04, 04:06 PM
sure i can win any argument with "because i said so" too :|
that tells us nothing
Tom Mattson
Jun20-04, 04:09 PM
sure i can win any argument with "because i said so" too :|
!!!
That's exaclty what you are doing! You are simply declaring something about the universe that is known to be false, and then saying that it disproves SR.
that tells us nothing
You sure are a dense little fella.
I'm not saying that it's true "because I said so", I'm saying it's true "because the universe says so". It says so when we ask it, via experimientation.
Relatvity shows that if the clocks are sychronised in all refrence frames you get results that cause irresovable paradoxes and/or do not fit in with empirical obsrevrations.
ram1024
Jun20-04, 04:15 PM
lemme get this straight.
nothing in the known universe is being done to CAUSE the clocks to become unsynchronized, yet you want me to believe that they ARE because you say so.
don't give me this "the universe says so" because the universe didn't do anything to the clocks. we can discern NO universal change that would CAUSE them to be changed.
the clocks in all cases emit photon simultaneously RELATIVE TO EACH OTHER.
In this case, then, indeed, (true, true, true).
However, I would like to comment on other things you have said:
what i can GIVE you, if the clocks are perfectly synchronized together in the stationary frame, they are locked onto the platform so they cannot move in relation to each other, and therefore cannot become unsynchronized, moving OR not.
This will not work. There is no such thing as a rigid body, so you cannot use that as a way to keep things in sync.
In fact, if you sync the clocks while they're stationary (in the background frame), then accelerate the platform, the clocks cannot be synchronized in their rest frame, nor in the background frame.
This is easy to see in the background frame; due to length contraction, one of the clocks must have been displaced more than the other.
they emit photons simultaneously in both the moving frame AND the stationary frame, basically.
Which cannot happen, according to SR, unless the situation is trivial; either the two clocks are at the same place, or two frames are the same.
(for simplicity, I'm speaking in one spatial dimension)
I'm lagging somewhat behind the course of the conversation (darned TV!) but I'm going to post this anyways.
Tom Mattson
Jun20-04, 04:25 PM
nothing in the known universe is being done to CAUSE the clocks to become unsynchronized,
A better way to look at it is the way JCSD described. Nothing "causes" them to be unsynchronized in other frames, they simply are unsynchronized. There is no way that the postulates of SR can hold and for simultaneity to be absolute. So the question is, Do the postulates hold?
Experimentation has answered with an emphatic "YES".
yet you want me to believe that they ARE because you say so.
No, you are expected to accept it because the evidence says so. And even if you don't accept it, surely you must be able to accept that simply assuming that SR is false does not disprove it.
don't give me this "the universe says so"
And why not? Experimentation is the final court of appeals in science. If you aren't open to that, then there is no hope for you.
because the universe didn't do anything to the clocks.
That's correct, because nothing was "done" to the clocks, period. It's not as though some invisible agent resets the clocks so that they are out of synch when a moving observer passes by. It is just a simple consequence of the fact that, in our universe, the laws of physics and the speed of light are the same for everyone.
we can discern NO universal change that would CAUSE them to be changed.
But we can discern that the clocks don't tick at the same rate for all observers.
ram1024
Jun20-04, 04:45 PM
which brings the final conclusion that perceptions are WRONG in moving frames. not that time/speed/measurement etc CHANGE <ludicrous i KNOW>.
let's move onto stage 2. i think i have enough of an idea of where we're at to continue now.
Case #4:
observer1
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
observer2
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Two trains side by side (4 clocks now, wheee!). both containing the exact same set up. in the first train, the whole train moves forward at 5ft/s. it starts moving the instant the photons are released. at the same moment on the other train, the platform is stationary and the OBSERVER runs ahead at 5ft/s.
who receives the photon from the right emitter first? who receives the photon from the left first? if they both receive photons at the SAME time, how can you justify previous responses that observer 2 on his own would receive photons NOT simultaneously but observer 1 WOULD.
please make answers detailed :D
we'll move to stage 3 <conclusion> soon
ram1024
Jun20-04, 04:50 PM
stupid "code" screwing up my pretty trains
Tom Mattson
Jun20-04, 05:14 PM
which brings the final conclusion that perceptions are WRONG in moving frames.
It has nothing to do with "perceptions". It is what his instruments would record. Sorry, but you can't argue with experimental evidence just because it doesn't fit your view of how the universe should work.
not that time/speed/measurement etc CHANGE <ludicrous i KNOW>.
It is ludicrous to the moving observer. You have absolutely no basis for telling him that what he measures isn't real.
who receives the photon from the right emitter first?
In which frame?
who receives the photon from the left first?
In which frame?
if they both receive photons at the SAME time,
In which frame?
ram1024
Jun20-04, 05:17 PM
why are you saying "in which frame" ?
they are both happening at the same time. if they WEREN'T happening at the same time there would be no point in asking which observer receives a photon first, now would there?
Tom Mattson
Jun20-04, 05:19 PM
why are you saying "in which frame" ?
Why do you think? I just got through explaining that simultaneity is relative! For some reason, you saw fit to put 4 trains in this problem, with observers in different states of motion. When you ask about when something is observed, you have to state the frame from which the measurements are made.
they are both happening at the same time. if they WEREN'T happening at the same time there would be no point in asking which observer receives a photon first, now would there?
In which frame?
ram1024
Jun20-04, 05:23 PM
there's no 4 trains.. there's 2 trains.
each train progressing exactly as depicted in the diagrams presented.
train1:
step1
step2
step3
step4
train2:
step1
step2
step3
step4
you know that's what i meant, we've been discussing this very thing for 4 pages, don't feign ignorance now in order to wear down my patience, friend :D
You say that the moving frame measurements are "WRONG" because they disagree with the measurements in a different frame.
Why is that?
The difference should not be surprising, since they're different measurements. Is there a reason you think they should be the same?
ram1024
Jun20-04, 05:32 PM
You say that the moving frame measurements are "WRONG" because they disagree with the measurements in a different frame. Why is that?
because, EVERYTHING cannot be relative. something has to be real. it makes no sense to depict the universe as a place where two people can disagree on something and BOTH be correct. 1=2 ? yes=no ?
The difference should not be surprising, since thy're different measurements. Is there a reason you think they should be the same?
even with shrunken rulers (or whatever the heck zany contraptions you people measure with), the distance to be measured is also shrunken so "relatively" you should measure the same lengths if this were true
Tom Mattson
Jun20-04, 05:33 PM
there's no 4 trains.. there's 2 trains.
OK, my mistake.
I also see now that you don't have different inertial frames. The clocks, the observer on the train, and the observer running on the platform are all in the same frame. However, it is still true that you did not specify the frame in which the measurements are taken. I am going to assume that you mean the frame of the clocks and the observers.
Now the question is: Is the observer on the platform running in such a way that he is always at the midpoint of the train, as determined by him?
ram1024
Jun20-04, 05:41 PM
guy on train 1 is standing stationary on moving train 5 ft/s to the right
guy on train 2 is running on stationary train 5 ft/s to the right.
the only thing you need to tell me is which of the two observers detects a photon first. i don't think there's any "measurable" frame where one could claim he got hit first and it NOT be true from any other reference frame.
i don't think so anyways, perhaps if that's arguable it will be brought up
loseyourname
Jun20-04, 05:42 PM
because, EVERYTHING cannot be relative. something has to be real. it makes no sense to depict the universe as a place where two people can disagree on something and BOTH be correct. 1=2 ? yes=no ?
Ah yes, the argument from personal incredulity. We see this all the time in debates with creationists. They also don't seem to understand that refusing to believe the evidence does nothing to discredit a theory with an overwhelming amount of evidential support. Maybe this thread belongs in the philosophy of science forum instead of theory development. That would be the proper place to debate the validity of accepting evidence that prima facie defies common sense.
because, EVERYTHING cannot be relative. something has to be real. it makes no sense to depict the universe as a place where two people can disagree on something and BOTH be correct. 1=2 ? yes=no ?
If it's the concept of relativity that's bothering then take a look at the spacetime continuum, in that a different inertial rst frames are analogous to defing different co-ordiante systems in Euclidian space, a velocity is like a rotation.
ram1024
Jun20-04, 05:44 PM
Now the question is: Is the observer on the platform running in such a way that he is always at the midpoint of the train, as determined by him?
that's worded funny so i'm going to make an assumption that you're talking about the runner on train 2 keeping up with the midpoint of train one relative to him.
yes
Tom Mattson
Jun20-04, 05:44 PM
because, EVERYTHING cannot be relative.
Have you absorbed nothing of what we write???
The laws of physics and the speed of light are not relative. There you have it! There is your anchor of sanity to keep your ship from being tossed about in the sea of ambiguity.
something has to be real.
If you think that "relative" means the same as "unreal", then you are mistaken. Measured quantities that are subject to change under changes in inertial frames are real for the one who measures them.
it makes no sense to depict the universe as a place where two people can disagree on something and BOTH be correct.
No one is saying that it does make sense. What we are saying is that you have to be very specific about the statements you make. This can be easily illustrated, by the following example. Notice that I am taking an example that would be true even in a Galilean universe!.
If ship takes off along the +x-axis at 0.5c relative to the Earth, then an Earth bound observer would say, "The velocity of the ship is 0.5c in the positive x direction." However, an observer on a ship that is moving along the same axis with the same speed, but behind the ship, notices that he isn't gaining any ground on the first ship. He then rightly concludes that, "The velocity of the ship is zero."
Observer 1: "The velocity of the ship is 0.5c in the positive x direction."
Observer 2: "The velocity of the ship is zero."
Now, one might say: Oh, my, how can these two statements both possibly be correct? They directly contradict each other. Which velocity is "real"???
The problem here is that the statements are both ill-defined. We have to be more specific:
Observer 1: "The velocity of the ship relative to Observer 1 is 0.5c in the positive x direction."
Observer 2: "The velocity of the ship relative to Observer 2 is zero."
Specifying the parts in blue and red are necessary to make the statements well-defined. Notice that they also differentiate the statements, so that both can be correct simultaneously.
1=2 ? yes=no ?
No.
even with shrunken rulers (or whatever the heck zany contraptions you people measure with), the distance to be measured is also shrunken so "relatively" you should measure the same lengths if this were true
No, I never measure with a "shrunken ruler". It's always the other guy's rulers that are shrunken, as measured by me. Likewise, mine are shrunken as measured by him.
ram1024
Jun20-04, 05:48 PM
Ah yes, the argument from personal incredulity. We see this all the time in debates with creationists. They also don't seem to understand that refusing to believe the evidence does nothing to discredit a theory with an overwhelming amount of evidential support. Maybe this thread belongs in the philosophy of science forum instead of theory development. That would be the proper place to debate the validity of accepting evidence that prima facie defies common sense.
which is why i'm using the evidence to contradict itself. completely relying on the observations and analysis of people who understand the theory and making them justify things that i believe to be paradoxial in the theory.
i wouldn't trust myself to supply my own experimental or mathematical evidence any more than you would. that is not my forte' :D
Tom Mattson
Jun20-04, 05:49 PM
that's worded funny so i'm going to make an assumption that you're talking about the runner on train 2 keeping up with the midpoint of train one relative to him.
In that case, then both observers would report that they receive the light simultaneously.
ram1024
Jun20-04, 05:52 PM
k good. now we move to conclusion... i think
because, EVERYTHING cannot be relative. something has to be real.
What makes you think that relative and real are mutually exclusive?
it makes no sense to depict the universe as a place where two people can disagree on something and BOTH be correct. 1=2 ? yes=no ?
The universe in under no obligation to behave by rules which you deem as making sense.
it makes no sense to depict the universe as a place where two people can disagree on something and BOTH be correct.
Why not?
One common example is when two people face each other, they disagree on what is "left" and what is "right". Yet, they're both correct.
Another example is the pitch of the motor of a race car; people in front depict it as a high pitch, people behind depict it as a low pitch. Yet they're both correct.
The only reason you have problem with SR is that you refuse to accept that measurements of length and time are yet another thing that is relative.
ram1024
Jun20-04, 06:10 PM
In that case, then both observers would report that they receive the light simultaneously
so here we have emitted simultaneously, and DETECTED simultaneously, despite difference in relative position to the emitters of the photons.
i thought SR was supposed to claim non-simultaneity for moving towards a light source vis-avis Case#3 post 1?
ram1024
Jun20-04, 06:13 PM
The only reason you have problem with SR is that you refuse to accept that measurements of length and time are yet another thing that is relative.
i'm fine with people's perceptions being relative, i'm not going to divulge my master plan just yet, though
'm fine with people's perceptions being relative
Ok, let me rephrase.
In one measurement system, the direction considered "left" can be exactly what another measurement system considers "right", yet both answers are correct.
If I put a sound detector in front of a race car, it will measure a sound higher pitched than an identical sound detector placed behind the race car, yet both are correct.
I specifically chose measurements that aren't psychological; they can be unambiguously performed by measuring devices.
HallsofIvy
Jun20-04, 06:25 PM
Yet they do become unsychronized- the experiment you describe has been done (in a much more sophisticated form using cosmic rays) the experimental results have consistently supported relativity- events that were synchornized in one frame of reference are unsychronized in different frames of reference- that's the one sure thing in science: the experimental evidence.
Tom Mattson
Jun20-04, 06:29 PM
so here we have emitted simultaneously, and DETECTED simultaneously, despite difference in relative position to the emitters of the photons.
Hold on: I thought both observers were located at the midpoint of the train. In other words, I thought that both of them were equidistant from the sources. If not, then they certainly will not both report having received the light pulses simultaneously. This has nothing to do with Einstein vs. Galileao, it's simple geometry.
i thought SR was supposed to claim non-simultaneity for moving towards a light source vis-avis Case#3 post 1?
It is nonsimultaneous for observers in a different inertial frame. But you set this up so that the clocks, the emitters, and the observers are all in the same frame, so SR effects don't even show up. They won't show up until a third party, who is moving relative to the frame of the original observers, enters the scene.
ram1024
Jun20-04, 06:29 PM
or "flaws in the experiments"
might be more apt. we'll see as this discussion continues :D
Tom Mattson
Jun20-04, 06:34 PM
or "flaws in the experiments"
No, that's not it. Professional physicists perform experiments to the most exacting standards possible. Each part of an experiment is worked out in great detail by a whole team of scientists. For some internet bozo to come along and suggest that they don't do their job correctly is just plain laughable. Really, what other profession is subject to this indignity? What would you say if someone who obviously has no clue about your profession (assuming you have one) were to come along and tell you that you don't really understand the thing that you go in and do all day, every day?
might be more apt. we'll see as this discussion continues :D
No, we won't see it in this discussion. It's not possible, even in principle, to determine if the experiments are at fault, because your thought experiments don't make any connection to the experiments.
ram1024
Jun20-04, 06:46 PM
so do me a favor and go back to case #1 2, and 3 for me Tom and state your "true / false" predictions
i'm getting conflicting data from two people who i know to be good at this SR thing.
Tom Mattson
Jun20-04, 06:50 PM
First I need you to answer my comments from Post #71. Are both observers at the midpoint of the train, or aren't they?
ram1024
Jun20-04, 06:56 PM
Case #2:
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Platform is moving. SR concludes the photons (still emitted simultaneously) will be detected by the observer at the exact same time. (observer is standing exactly in the center of the platform, equidistant from both sources)
(True / False) ?
Case #3:
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Man is moving on platform towards an emitter. SR concludes that photons are NOT detected at the exact same time.(observer starts at the exact center of the platform, equidistant from the two sources and moves towards the right emitter)
(True / False) ?
As case #2 is now stated, I revert to my original response (false). Since you don't specify the frame in which "simultaneously" is measured, the natural assumption is that you mean it is the frame in which the pictures are drawn.
Case #3 is similarly ambiguous; my true was, again, under the assumption that "simultaneously" is measured in the frame in which the pictures are drawn.
ram1024, I see that you still waste other's time. Please see that post#76 itself does not contain enough information to give answers.
If you didn't change the conditions, my understanding is:
The emitters are stationary wrt the platform in both cases. The emitters emit light simultaneously according to the clocks synchronized in the platform frame.
Case#2: Does the midpoint observer (co-moving with the emitters on the platform) detect the photons at the same time? (my answer is yes, so, True)
Case#3: Does the moving observer (moving towards one of the emitters) detect the photons at the same time, given that the observer was at the midpoint (according to clocks on the platform) when the photons were emitted ? (my answer is no, so, True)
ram1024, I must add that you don't deserve any attention if you insist on asking ambiguous questions.
ram1024
Jun20-04, 07:51 PM
the emitters are welded TO the train and NEVER move relative to each other.
the conditions set forth are NOT ambiguous. you have answered as i expected SR to answer. Tom and Hurkyl are putting forth different answers though, but they're not really detailing their answers so it's hard for me to piece together where the difference in understanding is.
i'm pretty sure my outline of the experiment is not vague.
two synchronous clocks set a specified distance apart emitting a photon at what would be to THEM synchronous times (no inertial effects happen to one that do not also happen to the other). drop a moving observer in the center in one case, and in the other case a stationary observer but the whole train is moving.
i can continue to keep explaining it over and over using different words until you understand it exactly if you want, but in the interest of progress let's keep that to a minimum :D
Tom Mattson
Jun20-04, 07:58 PM
the emitters are welded TO the train and NEVER move relative to each other.
That part is clear.
the conditions set forth are NOT ambiguous.
It's not the conditions, it's the questions. When you asked in the first 3 cases about simultaneity, you never specified which frame. As Hurkyl says, he was assuming that it was the frame of the train. But since you didn't say so, there's no way to tell.
Tom and Hurkyl are putting forth different answers though, but they're not really detailing their answers so it's hard for me to piece together where the difference in understanding is.
Where do we do that? I have yet to answer Cases 1-3, because the questions aren't well posed.
Yea, OK, I gave you your (true,true) answer. Now what is it you don't understand?
ram1024
Jun20-04, 08:04 PM
i guess i'm failing to see how something detected simultaneously by one observer is not the same in every frame. therefore when i pose a question that says "does the observer detect the photons simultaneously" and you answer back "what frame" it makes no sense to me because no matter how i look at it if he does then he DOES if he doesn't then he DOESN'T. there's no frame that will turn one into the other.
so my response to "what frame" would probably be "pick one"
i guess i'm failing to see how something detected simultaneously by one observer is not the same in every frame. therefore when i pose a question that says "does the observer detect the photons simultaneously" and you answer back "what frame" it makes no sense to me because no matter how i look at it if he does then he DOES if he doesn't then he DOESN'T. there's no frame that will turn one into the other.
so my response to "what frame" would probably be "pick one"
Are you replying to me?
When you said "detected simultaneously", I did not ask "in which frame", because the detection occurs at one point if they are detected at the same time. When you said "emitted simultanously", I did ask "in which frame", because the emissions are separated by distance. OK?
So, since I gave you the answers you expected, assume that I'm interpreting your question as you intended. So again, what is it you don't understand?
ram1024
Jun20-04, 08:24 PM
move to case #4 <stage 2>, wespe. it's not true or false it's more of an essay question :D
O Great One
Jun20-04, 08:27 PM
Man, I can't believe everybody is giving you such a hard time ram1024. The man in all three cases does detect the photons simultaneously.
case 1: The man is exactly in the middle and the photons are emitted simultaneously in the frame of reference of the platform, therefore the man detects the photons simultaneously.
case 2: The man must receive the photons at the same time, otherwise there would be an absolute frame of reference, which is forbidden.
case 3: The man is equally distant from each emitter at the time of emission, therefore the man will detect each photon at the same time.
Tom Mattson
Jun20-04, 08:36 PM
i guess i'm failing to see how something detected simultaneously by one observer is not the same in every frame. therefore when i pose a question that says "does the observer detect the photons simultaneously" and you answer back "what frame" it makes no sense to me because no matter how i look at it if he does then he DOES if he doesn't then he DOESN'T. there's no frame that will turn one into the other.
so my response to "what frame" would probably be "pick one"
I asked you "in which frame" with regards to the detection when I thought there were 4 trains, with different observers in different states of motion. The picture confused me a bit.
But now, as regards case 4, I am still confused because you originallly said that both observers are at the midpoint of the train. In response to that, I said that the detection is simultaneous for both. But then you said that the two have different positions relative to the emitters. So, I have to ask, which one is it? Are you considering displacement in another dimension, perpendicular to the axis of the train?
ram1024
Jun20-04, 08:39 PM
actually if they're giving me a hard time it's probably because i deserve it :D
but i'm ridiculously patient and not really functioning under a deadline so meh, it doesn't matter :D
you have pointed out something interesting though with
case 3: The man is equally distant from each emitter at the time of emission, therefore the man will detect each photon at the same time.
that. say a man standing on a platform halfway to halfway (crap i better make a picture)
|(-> (o) (o) <-)|
observer1 observer2
at the time of emission, observer 2 is at the exact center. but at the exact moment the photons are emitted he starts running left and ends up at the position where observer one is. according to what you said he gets hit by photons from both sides simultaneously. but observer 1 just standing there doesn't. he gets hit by photon to the left first and then by photon to the right.
case 3: The man is equally distant from each emitter at the time of emission, therefore the man will detect each photon at the same time.
No, the photons meet at the midpoint of the stationary platform. The observer was at the midpoint, but moved to a further location, so he cannot detect them at the same time. I stand by my answer.
ram1024, I'm checking your case #4.
But there's a problem. Case#4:observer1 is not the same experiment as case#2. Because now there is acceleration involved. In case#2, we assumed the platform was inertial. I will think on this and post my reply.
ram1024
Jun20-04, 08:46 PM
here's case #4 again
Case #4:
observer1
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
observer2
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Two trains side by side (4 clocks now, wheee!). both containing the exact same set up. in the first train, the whole train moves forward at 5ft/s. it starts moving the instant the photons are released. at the same moment on the other train, the platform is stationary and the OBSERVER runs ahead at 5ft/s.
who receives the photon from the right emitter first? who receives the photon from the left first? if they both receive photons at the SAME time, how can you justify previous responses that observer 2 on his own would receive photons NOT simultaneously but observer 1 WOULD.
some re-explaining (not changing the problem any), in train 1 the observer has his feet welded to the floor of the train. in train 2, the train has its wheels welded to the tracks. i'm dead serious :D.
okay that being said train 1 moves at 5ft/s to the right
and on TOP of train 2, observer 2 moves at 5ft/s to the right.
ram1024
Jun20-04, 08:50 PM
Case#4:observer1 is not the same experiment as case#2. Because now there is acceleration involved. In case#2, we assumed the platform was inertial. I will think on this and post my reply.
we can ditch the acceleration and say train 1 was running along the track at 5ft/s the whole time and when it hit the point where both trains were aligned nose-to-nose, THEN the experiment begins
(it changes the outcome for my relativity but i don't think it does for yours (sr))
we can ditch the acceleration and say train 1 was running along the track at 5ft/s the whole time and when it hit the point where both trains were aligned nose-to-nose, THEN the experiment begins
No, you can't do that. Because, there is length contraction. The ends of two trains will not meet at the same time, each one will see the other contracted. (If you make one of the trains longer, still no, becaue length contraction is mutual)
I have to admit I may not be able to answer your question if it involves acceleration. I don't know much GR. Maybe this is a good time for me to start GR.
ram1024
Jun20-04, 09:09 PM
damn, bummer then
teach me while you're at it, wespe. contractions perplex me
As has already come up, #4 has not been formulated in a way consistent with SR.
In other words, you're assuming the answer before you "do" the experiment.
If you accelerate the train, you run into the problem of no rigid bodies, and if both trains had been moving with different constant velocitys (possibly zero), then it cannot be the case that the two setups are identical (particularly the same proper length) and that the endpoints can coincide simultaneously in either of their two rest frames.
Furthermore, yet again I can only assume that when you say "simultaneous" you mean in the frame you draw pictures (which I'll now call the "picture frame")
Anyways, if I assume you do something like accept there's a rigid body problem, or allow one train to be longer than the other, or something or other, I deduce that both observers will observe the right photon before the left photon. If you can get the two observers along the same worldline, then they will detect the right photon simultaneously, and similarly for the left photon.
This is consistent with my previous answers where, in both case #2 and #3 (under the assumption that "simultaneously" was measured in the picture frame), I stated that the observer would not detect both photons simultaneously.
damn, bummer then
teach me while you're at it, wespe. contractions perplex me
However, if you consider that after the acceleration is over, train1 will be inertial again, and length contraction must have already occured. (you can't reject length contraction at this point. If you do, you have to also reject the given SR answers). That would explain some things. Anyway, if expert people here consider your question worthy, you will get your answer quickly. My answer will be delayed since acceleration is involved.
Take care.
geistkiesel
Jun20-04, 09:21 PM
Mark it on your calendars, people. Let us begin.
Case #1:
|(-> (o) <-)|
man standing on a movable platform bed. at the EXACT center between two photon emitters. SR concludes that the simultaneously emitted photons from the two emitters will be detected by the observer at the same exact time.
ADDENDUM:the photon emitters are tied to precise atomic clocks
these clocks are perfectly aligned and synchronized and in all cases they move within the same inertial frame so they can stay calibrated.
(True / False) ?
Case #2:
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Platform is moving. SR concludes the photons (still emitted simultaneously) will be detected by the observer at the exact same time.
(True / False) ?
Case #3:
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
|(-> (o) <-)|
Man is moving on platform towards an emitter. SR concludes that photons are NOT detected at the exact same time.
(True / False) ?
once we square these we'll move on to stage 2.
!. True, the photon detected simultaneously.
2. False. The photons are detected sequentially. Light is measured at c in all inertial frames. The fact that the photon source is moving or not is irelevant. The observer is moving to the right toward a photon moving to the left. The left moving photon has a shorter distance to travel than does the left photon, therfore the times of flight of both photons is different. SR should predict sequential detection and that the photons were not emitted simulaneously in both stationary and moving frames.
3. same as 2). The photons will move at the speed of light. The left and right moving photons move at the same speed (from whatever frame measured). As the photon moving left has a shorter disance to cover befor meeting the observer moving right this photon will be detected first.
In both 2 and 3 SR will predict the photons were not emitted simultaneoulsy in the moving frame.
(under the assumption that "simultaneously" was measured in the picture frame)
I was under the impression that "simultaneously" was measured in the platform/emitters frame. (from post #14). Maybe ram1024 would like to make that clear.
O Great One
Jun20-04, 09:30 PM
actually if they're giving me a hard time it's probably because i deserve it :D
but i'm ridiculously patient and not really functioning under a deadline so meh, it doesn't matter :D
you have pointed out something interesting though with
that. say a man standing on a platform halfway to halfway (crap i better make a picture)
|(-> (o) (o) <-)|
observer1 observer2
at the time of emission, observer 2 is at the exact center. but at the exact moment the photons are emitted he starts running left and ends up at the position where observer one is. according to what you said he gets hit by photons from both sides simultaneously. but observer 1 just standing there doesn't. he gets hit by photon to the left first and then by photon to the right.
This is correct. Let's say that I'm 10 miles from somebody who then runs directly towards me at a speed always relative to me at 5 m.p.h. No matter what I do after he starts running towards me, he will reach me in 2 hours. If I start moving away at 20 m.p.h. relative to the earth, he will then be moving towards me at 25 m.p.h. relative to the earth, so that he maintains his constant speed of 5 m.p.h. relative to me. If light always moves towards us at the same constant speed relative to us, then the time it takes for light to reach us depends on the distance at the time of emission.
If light always moves towards us at the same constant speed relative to us, then the time it takes for light to reach us depends on the distance at the time of emission.
Yes, light always moves towards us at the same constant speed relative to us. And the distance is the same when lights were emitted. So the time it takes for light to reach us are the same. BUT, according to the moving observer, they were not emitted at the same time, so he detects them at different times. [or you can say: since he detects them at different times, they were not emitted at the same time according to him] That's the whole point of relative simultaneity.
geistkiesel
Jun20-04, 09:42 PM
Do you realize you wasted people's time by not specifying this in the first place? OK, in that case, #1 and #2 are true.
But, as Janus says, #3 is still unclear: where is the man when he detects the photons?
Number 3. is as clear as the rest. Once tghe photons have been emitted, whether from a moving platform or not, the motion of the sources is irelevant. The moving source may measure the speed of light, or he can simply look at his photon detectors. He is moving tot he right towarda photon moving tot he left. The time of flight of both photons is different.The left moving photon meet the observer before the right moving photon.
AS long as some v > 0 is specified the moving observer will always detect the photons in sequence. The measured delta t between measurements will surely differ from those measured by a stationary observer, and the distance traveled by the photons will be measured differently between the two observers, but this does not change the reality that the time of flight of both photons is different. The distance from L to Midpont is the same as R to the midpoint in both frames, though the frame to frame measurement number comparison may differ.
SR does not say that the right moving photon will move faster than the left moving photon in any of the cases. This would have to be the conclusion if the observer ever saw the detected photons simultaneously in 2 and 3. SR merely says the photons were not emitted simultaneously in both frames.
As long as some v > 0 is specified the moving observer will always detect the photons in sequence. It seems we have an absolute frame implied. In Grounded's examples, the measured photons are emasured against a stationary source, as the speed of light is invariant with the speed of the source, correct? Therefore meassuring the observers velocity wrt the wave lengths ppasing through the observers position, the observers speed wrt the photon source speed = 0 can be determined.
O Great One
Jun20-04, 09:51 PM
Yes, light always moves towards us at the same constant speed relative to us. And the distance is the same when lights were emitted. So the time it takes for light to reach us are the same. BUT, according to the moving observer, they were not emitted at the same time, so he detects them at different times. [or you can say: since he detects them at different times, they were not emitted at the same time according to him] That's the whole point of relative simultaneity.
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
russ_watters
Jun20-04, 09:57 PM
:rolleyes:
You are aware that GPS receivers use SR to determine their exact location relative to the satellites? That's why you need to have 4 sat's in view. You triangulate your position with three, but you need the fourth to solve for the SR clock bias. Now that, I did not know. Thanks.
And hey, don't throw out the reciever quite yet - I'm sure SR will be replaced with another theory that is mathematically exactly the same, if any of the anti-relativity guys ever get around to working out the math. :rofl:
ram1024
Jun20-04, 09:57 PM
EEK :surprise:
was trying to stay on target and only get opinions using SR in this thread... i guess that's all out the window now
russ_watters
Jun20-04, 10:02 PM
that's an assumption. I'll elaborate when we get to the second stage. need hurkyl russ or tom to confirm (True, True, True) on the first 3 cases. Sorry - today was Father's Day, which meant I was having lunch with my parents/grandparents, then playing golf with my dad. I'm not quite at my computer 24/7.
ram1024
Jun20-04, 10:06 PM
oh crap today was father's day?
ram1024
Jun20-04, 10:19 PM
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
hehe yeh it's things like this that make me a skeptic :D
good work spotting it though, Great One. apparantly it's not easy to spot :|
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
Both must be true. Given the constant speed of light, there is no other explanation (well, if you have one...). Look at it this way: simultaneity is defined by what the midpoint observer concludes in this setup. Since there can be any number or midpoint observers moving at different speeds, each one concludes differently, and each of them are correct according to themselves. That's relative simultaneity, as opposed to absolute (one that everyone agrees on) simultaneity. If you have a better definition of simultaneity, I'd like to hear that. Of course, defining simultaneity has consequences. It must be consistent with what you define as time, and with casuality. Note that all observers still agree that events occuring at a single point in space are absolutely simultaneous. Only events separated by distance causes disagreement. But this disagreement does not yield any casuality paradoxes, since SR also limits all speeds to be below c. Considering the speeds involved in our daily lives, it has little impact and not a big deal really (except when you need very high precision and stuff)..
Take care.
hehe yeh it's things like this that make me a skeptic :D
good work spotting it though, Great One. apparantly it's not easy to spot :|
Ram1024, that's funny, but, did you read #93 and #96? Hurkyl still thinks simultaneity was measured in "picture frame". Maybe you should make it clear before wasting any more of his time.
ram1024
Jun21-04, 12:13 AM
the consequences are way too far-fetched to believe in.
Length Contractions, Time Dialation, No Absolute Time, No Absolute Space...
it's like alice in wonderland on your guys' side of the fence :D
i'll define my version some time tomorrow i think. sure would like more people to toss in their opinions first of all, though.
it's intriguing that people who know SR still can't agree what the answers are to these.
russ_watters
Jun21-04, 12:18 AM
the consequences are way too far-fetched to believe in.
Length Contractions, Time Dialation, No Absolute Time, No Absolute Space...
it's like alice in wonderland on your guys' side of the fence :D
i'll define my version some time tomorrow i think. sure would like more people to toss in their opinions first of all, though.
it's intriguing that people who know SR still can't agree what the answers are to these. Clearly, this approach isn't working. Perhaps you'd be more inclined to accept SR if you started looking at what the evidence actually shows. Real experiments showing these "alice in wonderland" concepts like time dilation and length contractions.
How about GPS satellites? Do you know how they keep their clocks synchronized?
How about speed of light measurements: do you know that they all give the same answer?
ram1024
Jun21-04, 12:21 AM
Hurkyl has unfortunately come up with the "correct" answers for #2 #3 and #4, but i'm not sure he acquired them through SR since he didn't really elaborate.
if he has indeed come to that conclusion based on SR then i am pretty much SOL for refuting it since that's what my relativity would say as well
was trying to hide him under the mattress or something so i could get a bit further along in the discussion, but you just HAD to go and point him out :wink:
geistkiesel
Jun21-04, 12:45 AM
Hurkyl has unfortunately come up with the "correct" answers for #2 #3 and #4, but i'm not sure he acquired them through SR since he didn't really elaborate.
if he has indeed come to that conclusion based on SR then i am pretty much SOL for refuting it since that's what my relativity would say as well
was trying to hide him under the mattress or something so i could get a bit further along in the discussion, but you just HAD to go and point him out :wink:
you gave up too much., too soon. try this one out for size.
All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.
At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.
To demonstrate the following:
Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.
Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.
At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1
vtx = -vtx - vt1 – vt1
2tx = -2t1
tx = -t1
Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.
O Great One
Jun21-04, 12:58 AM
OK. So the guy moving to the left concludes that the light on the left was emitted before the light on the right. Let's say that there was a guy running to the right then he would have to conclude that the light on the right was emitted before the light on the left. BOTH CAN'T BE TRUE.
Both must be true. Given the constant speed of light, there is no other explanation (well, if you have one...). Look at it this way: simultaneity is defined by what the midpoint observer concludes in this setup. Since there can be any number or midpoint observers moving at different speeds, each one concludes differently, and each of them are correct according to themselves. That's relative simultaneity, as opposed to absolute (one that everyone agrees on) simultaneity. If you have a better definition of simultaneity, I'd like to hear that. Of course, defining simultaneity has consequences. It must be consistent with what you define as time, and with casuality. Note that all observers still agree that events occuring at a single point in space are absolutely simultaneous. Only events separated by distance causes disagreement. But this disagreement does not yield any casuality paradoxes, since SR also limits all speeds to be below c. Considering the speeds involved in our daily lives, it has little impact and not a big deal really (except when you need very high precision and stuff)..
Take care.
It is logically impossible that both statements are true. Let's imagine that the platform is balanced like a teeter-totter and when the light is emitted the platform tips in that direction. So the guy running towards the left thinks that the platform tips to the left and then to the right, and the guy running towards the right thinks the platform tips to the right and then to the left. Try to explain how they are both correct!
ram1024
Jun21-04, 01:01 AM
actually it all suits my purposes. i don't care to retute SR as long as it sees things MY way ;D
geistkiesel
Jun21-04, 01:05 AM
As has already come up, #4 has not been formulated in a way consistent with SR.
In other words, you're assuming the answer before you "do" the experiment.
If you accelerate the train, you run into the problem of no rigid bodies, and if both trains had been moving with different constant velocitys (possibly zero), then it cannot be the case that the two setups are identical (particularly the same proper length) and that the endpoints can coincide simultaneously in either of their two rest frames.
Furthermore, yet again I can only assume that when you say "simultaneous" you mean in the frame you draw pictures (which I'll now call the "picture frame")
Anyways, if I assume you do something like accept there's a rigid body problem, or allow one train to be longer than the other, or something or other, I deduce that both observers will observe the right photon before the left photon. If you can get the two observers along the same worldline, then they will detect the right photon simultaneously, and similarly for the left photon.
This is consistent with my previous answers where, in both case #2 and #3 (under the assumption that "simultaneously" was measured in the picture frame), I stated that the observer would not detect both photons simultaneously.
Hurkyl, the following is similar to the problem you had me solving a while back. This should be slam dunk easy for you.
All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.
At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.
To demonstrate the following:
Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.
Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.
At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1
vtx = -vtx - vt1 – vt1
2tx = -2t1
tx = -t1
Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.
It is logically impossible that both statements are true. Let's imagine that the platform is balanced like a teeter-totter and when the light is emitted the platform tips in that direction. So the guy running towards the left thinks that the platform tips to the left and then to the right, and the guy running towards the right thinks the platform tips to the right and then to the left. Try to explain how they are both correct!
That's because there is no totally rigid objects in reality. Imagine that the platform is several light years long. When one side goes down, this event cannot have an immediate effect on the midpoint or the other side. Even if the change propagates at light speed, it has to travel for several years to reach the midpoint or to make the other side go up. Therefore, what the observer sees is more up-to-date than what he feels under his feet. Was this what you meant?
O Great One
Jun21-04, 02:19 AM
OK. Let's try again. Let's say that when the light is emitted, the emitter blows up in a huge explosion. Somebody is standing off in the distance observing this. He either observes the explosion on the left and then the explosion on the right or he observes the explosion on the right and then on the left.
ahrkron
Jun21-04, 02:32 AM
That will depend on how that person is moving. It is the original situation all over again. If he is moving to the left, he will get first the light from the left, and since he measures the speed of that light to be the same as that from the right, he is entitled to deduce that the blast from the left occured first.
Same happens if the person in the distance is moving to the right.
ram1024
Jun21-04, 03:02 AM
the emitter exploding is a good one, Great. i was tempted to use explosion as emitters but was afraid someone would say something crazy like "the explosion imparts and inertial vectored thrust quotient based on non-linear cohesian and sub temporal disjunction, and therefore the light particles get immersed in an endothermic tesla-radiation that elevates their states to 15th dimensional particles. SR predicts that these particles exist everywhere at once. Good Game, ram1024, YOU LOSE"
geistkiesel
Jun21-04, 05:40 AM
In this case, then, indeed, (true, true, true).
However, I would like to comment on other things you have said:
This will not work. There is no such thing as a rigid body, so you cannot use that as a way to keep things in sync.
In fact, if you sync the clocks while they're stationary (in the background frame), then accelerate the platform, the clocks cannot be synchronized in their rest frame, nor in the background frame.
This is easy to see in the background frame; due to length contraction, one of the clocks must have been displaced more than the other.
Which cannot happen, according to SR, unless the situation is trivial; either the two clocks are at the same place, or two frames are the same.
(for simplicity, I'm speaking in one spatial dimension)
I'm lagging somewhat behind the course of the conversation (darned TV!) but I'm going to post this anyways.
Can't you rotate the clocks 90 degrees, accelerate and then rotate back to align as before acceleration?
geistkiesel
Jun21-04, 05:46 AM
A better way to look at it is the way JCSD described. Nothing "causes" them to be unsynchronized in other frames, they simply are unsynchronized. There is no way that the postulates of SR can hold and for simultaneity to be absolute. So the question is, Do the postulates hold?
Experimentation has answered with an emphatic "YES".
No, you are expected to accept it because the evidence says so. And even if you don't accept it, surely you must be able to accept that simply assuming that SR is false does not disprove it.
And why not? Experimentation is the final court of appeals in science. If you aren't open to that, then there is no hope for you.
That's correct, because nothing was "done" to the clocks, period. It's not as though some invisible agent resets the clocks so that they are out of synch when a moving observer passes by. It is just a simple consequence of the fact that, in our universe, the laws of physics and the speed of light are the same for everyone.
But we can discern that the clocks don't tick at the same rate for all observers.
Tom, Can't you rotate the clocks 90 degrees transverse to the accelerated direction and when velocity is achieved rotate the clocks -90 degrees?
but i'm not sure he acquired them through SR since he didn't really elaborate.
As Tom said, it's just geometry.
The method is consistent with SR, but in this problem, it is also consistent with Newton. No switching of frames of reference was involved, and that seems to be the particular disagreement you have with SR.
A lot of people detracting from SR forget that it has all the "ordinary" tools for analyzing a single reference frame too! (e.g. geometry, vectors, calculus)
Can't you rotate the clocks 90 degrees, accelerate and then rotate back to align as before acceleration?
That's an interesting question; I hadn't pondered if there was any simple 2-D way to do this.
the emitter exploding is a good one, Great. i was tempted to use explosion as emitters but was afraid someone would say something crazy like "the explosion imparts and inertial vectored thrust quotient based on non-linear cohesian and sub temporal disjunction, and therefore the light particles get immersed in an endothermic tesla-radiation that elevates their states to 15th dimensional particles. SR predicts that these particles exist everywhere at once. Good Game, ram1024, YOU LOSE"
Bah! That's silly!
They're only 14th dimension particles. :biggrin:
OK. Let's try again. Let's say that when the light is emitted, the emitter blows up in a huge explosion. Somebody is standing off in the distance observing this. He either observes the explosion on the left and then the explosion on the right or he observes the explosion on the right and then on the left.
Do you mean: "those simultaneity conclusions were harmless illusions, but let's add something real like explosions". It doesn't matter. The order of events can be really different for different observers. None of them is more correct or real than the other. You can't be at both places to see which event "really" occured when. As long as there is no paradox, what is your objection?
Actually, O Great One, I think you can ask such questions in the relativity forum and get better answers there. [edit: provided that you will ask questions and you are willing to learn relativity's answer. Otherwise this forum is the proper place, but I suggest you create a new thread]
This thread is in a strange state.. Ram1024 seems content with Hurkyl's answers, even though I suggested there were a misunderstanding between them, Ram1024 doesn't try to clarify and Hurkyl doesn't try to investigate.. On the other hand I don't want more time wasted, so I don't mind if this thread dies.
ram1024
Jun21-04, 01:44 PM
well i can't really disprove SR if it's giving answers that i deem to be "correct" :D
the whole idea for this was to try and point out a paradox of reference frames using simultaneity of one observer (which i believe to be immutable)
maybe if Tom can chime in and dispute Hurkyl's answers we can try again.
well i can't really disprove SR if it's giving answers that i deem to be "correct" :D
I see problems, ram1024:
You don't know enough about SR to find the answers yourself. So you have to rely on others for those answers. Then you fail to specify the scenario clearly and you get different answers. One of the answers happen to match your version and you give up. You were trying to disprove a theory you don't know enough about in the first place. How will it be next time? Don't you see the problem, even after 100 replies I'm not sure what the scenario is anymore. Next time, don't omit "obvious" details, use some math if possible.
ram1024
Jun21-04, 02:20 PM
well if SR wasn't so counter intuitive it'd be alot easier for me to describe situations that can be easily viewed by it :|
SR contains too many things that go against what i believe to be reality, all to accomodate this "constant relative light speed".
i haven't given up
well if SR wasn't so counter intuitive it'd be alot easier for me to describe situations that can be easily viewed by it :|
SR contains too many things that go against what i believe to be reality, all to accomodate this "constant relative light speed".
Exactly. All those counter intuitive things were invented to satisfy constant relative light speed. So, all you have to do is disprove constant relative light speed. Or, find an alternate theory that satisfies constant relative light speed. Good luck!
ram1024
Jun21-04, 03:01 PM
constant relative itself makes no sense. that means it's not constant NOT relative. think about THAT one :D
constant relative itself makes no sense. that means it's not constant NOT relative. think about THAT one :D
Makes no sense? You have the option of disproving it. You should be able to explain how it was measured constant in experiments.
Yes, relative, but still constant. Speed is a ratio. Due to SR effects this ratio remains constant. Or, you could even say it's absolute, in the sense that all observers agree it's constant.
ram1024
Jun21-04, 05:40 PM
but ONLY if you give all observers their own reality :|
but ONLY if you give all observers their own reality :|
All observers can have their own reality as long as these realities don't create a paradox. I kind of liked your idea about a common reality created by superimposing these individual realities (or sth like that). But that common reality is of no use to us mere mortals :)
geistkiesel
Jun21-04, 07:21 PM
but ONLY if you give all observers their own reality :|
Ram: Some gross physics might help. When discussing a moving frame relative to a stationary frame or a frame < your frame there are practical limits. Trains and stationary platforms are a slam dunk open obvious. The train moves, the platform is a stationary platorm. Who is going to realistically sta3e that the platform can be considered moving while the train is stationar.? The train started in the stationary frame and it is that entity that is moviing. Because mathematics says you can do it doesn't mean you have to do it.
What about relative motion between two bodies. If there is a time acceleration history then the frames can be distinguished such that V(frame1) < V(frame2) and while the frames are inertial, constant velocity, one is always moving slower than the other even though the math lets you swap out the frames "because you can".
This is illogical to do it once much less more than once, but taking the digression v1>v2>v3>v4> to extremes, there has to be zero velocity as a lower limit. So why swap out frames, just because you can?
I suppose that c would be the max velocity until that barrier is overcome.
Your day is up, but ram1024 I think you won after all: can't you sense the anguish, the anticipated high pitched moans of panicked SR theorists, contemplating the demise of their cherished theory. ..?
ram1024
Jun21-04, 09:16 PM
if i had some real experimental data to work with i could take it further.
or a better grounding in maths and the formulas used :(
what i had was logic, and it appears i come to the same conclusions that SR does. <shrug>
ram1024
Jun21-04, 09:17 PM
albeit with a different system... so maybe i need to polish that :D
ram1024
Jun21-04, 09:21 PM
on a different note, since SR confirms the same thing that i was thinking about "reality" that means my Spacebuoy HAS to work.
and now since civilians are making space flights, maybe i can sneak one of my spacebuoys on-board and prove an universal stationary relative frame :D
HallsofIvy
Jun21-04, 10:22 PM
ram1024: What you "deem" to be correct is irrelevant to whether relativity is correct or not. The only thing that is relevant is the experimental evidence- which is strongly in favor of relativity.
By the way "logic" is useless without facts.
ram1024
Jun21-04, 10:26 PM
if what i deem to be correct leads to the same results as relativity then i don't care, either way like i said before, my purposes are realized by them.
i think i'll port the space-buoy over to this thread since it's agreed that it will work now :D
ram1024
Jun21-04, 10:28 PM
URF (Universal Rest Frame) Space-Buoy© <- ke ke
the machine consists OF:
1 heavy duty computing device
4 photon receptors
4 equal length long *** poles and 4 photon emitters connected to sychronized clocks
some thrusters for movement.
now the idea for this consists of detecting the difference in arrival times of the photons from different directions.
if the photons from any source arrive SOONER than other sources, we can conclude we are moving towards that emitter. we merely perform some calculations, use our thrusters to slow us down, and take another measurement.
when we finally receive the light from all sources at the same time we will be at complete universal rest, and objects functioning in this relativistic inertial frame will bear true space/time
but ONLY if you give all observers their own reality
Reality, in SR, is worldlines living in space-time.
Observers don't have their "own reality"; they just measure reality differently.
ram1024
Jun21-04, 11:02 PM
well ya, but "their measurement" are also perfectly valid, according to einstein assertions, so ya they DO have their own reality :D
ArmoSkater87
Jun22-04, 02:45 AM
ok...i read the first section of arguements, and jesus u all make it way more complicated than it really is. It might seem this way to me because i'm a high school student going into 12th grade, and i have yet to take BC calc and AP physics when school starts. Even though i cant go into the detail that u all go into all the frames and ****, but from the very biginning i saw that the answers were true, true, true. I'll come some time to see what else is up, well, have fun argueing about SR. By the way, u just cant try to disprove the theory of one of the greatest geniuses of all time, SR has been proven correct in several experiments. (-_-)
geistkiesel
Jun22-04, 04:02 AM
ok...i read the first section of arguements, and jesus u all make it way more complicated than it really is. It might seem this way to me because i'm a high school student going into 12th grade, and i have yet to take BC calc and AP physics when school starts. Even though i cant go into the detail that u all go into all the frames and ****, but from the very biginning i saw that the answers were true, true, true. I'll come some time to see what else is up, well, have fun argueing about SR. By the way, u just cant try to disprove the theory of one of the greatest geniuses of all time, SR has been proven correct in several experiments. (-_-)
Hey, ArmoSkater87, I gotta question for ya. How do you know who is and who is not the "greatest geniuses of all time" ? I mean if you aren't up there with the greatest, how do you comprehend those that are the greatest? I suspect because soemone convinced you they were "the greatest".
So sad to see someone so rigidly robotic at such a tender young age. You start believing in the complexities of scientific understanding as rigidly established as you stated in your post then you are missing the dynamic exchanges that proves the oppostite situation as the rule, the dynamics of change and reformation of structure and form. Relativity theory is a temporary state of scientific discussion, nothing more, nothing less.
well ya, but "their measurement" are also perfectly valid, according to einstein assertions, so ya they DO have their own reality :D
Don't forget Gallileo and Newton too!
URF (Universal Rest Frame) Space-Buoy© <- ke ke
the machine consists OF:
1 heavy duty computing device
4 photon receptors
4 equal length long *** poles and 4 photon emitters connected to sychronized clocks
some thrusters for movement.
now the idea for this consists of detecting the difference in arrival times of the photons from different directions.
if the photons from any source arrive SOONER than other sources, we can conclude we are moving towards that emitter. we merely perform some calculations, use our thrusters to slow us down, and take another measurement.
when we finally receive the light from all sources at the same time we will be at complete universal rest, and objects functioning in this relativistic inertial frame will bear true space/time
Sorry, but this device will not detect uniform motion. If you had two of these devices, and they were moving relative to each other, all the photons for each device would still arrive all at the same time.
ram1024
Jun22-04, 12:31 PM
Janus, do me a flavor and go back to page 1.
answer True/False for case 1,2 and 3
then hit page 4 for case #4 which is an essay type question
ram1024
Jun22-04, 01:52 PM
and yes, the UFR Spacebuoy does NOT work, but i'm not going to tell you why... not yet :D
and yes, the UFR Spacebuoy does NOT work, but i'm not going to tell you why... not yet :D
How old are you? :)
Anyway, UFR Spacebuoy imparts and inertial vectored thrust quotient based on non-linear cohesian and sub temporal disjunction, and therefore the light particles get immersed in an endothermic tesla-radiation that elevates their states to 15th dimensional particles. SR predicts that these particles exist everywhere at once. Good Game, ram1024, YOU LOSE
ok...i read the first section of arguements, and jesus u all make it way more complicated than it really is. It might seem this way to me because i'm a high school student going into 12th grade, and i have yet to take BC calc and AP physics when school starts. Even though i cant go into the detail that u all go into all the frames and ****, but from the very biginning i saw that the answers were true, true, true. I'll come some time to see what else is up, well, have fun argueing about SR. By the way, u just cant try to disprove the theory of one of the greatest geniuses of all time, SR has been proven correct in several experiments. (-_-)
Actually the only one that is true is the last statement, the first statemnt is defintely false as it implies the clocks will stay synchronized in all inertial frames which is not the case and the second one suggests the same thing.
ram1024
Jun22-04, 03:17 PM
How old are you? :)
Anyway, UFR Spacebuoy imparts and inertial vectored thrust quotient based on non-linear cohesian and sub temporal disjunction, and therefore the light particles get immersed in an endothermic tesla-radiation that elevates their states to 15th dimensional particles. SR predicts that these particles exist everywhere at once. Good Game, ram1024, YOU LOSE
heh i'm 28, and that's not as far from the truth as you might suppose ;D
15th dimensional particles! holy hell! they're everywhen!
ArmoSkater87
Jun22-04, 05:19 PM
Hey geistkiesel, your right about me not being anywhere up there, but in my opinion i dont have to be to know who has a briliant mind and who doesnt. Einstein follwed Galileo and Newton, who were the first to explain motion mathematically. Altough many more followed, it seems to me that Einstein has incorperated much more into his theories than earlier minds had before him. I'm not saying that Einstein is the greatest genius that ever lived, im just saying that he has ONE of the most briliant minds that anyone has ever seen. I can simply say this because of the fact that there have been so few in that field like him.
quote:
"Actually the only one that is true is the last statement, the first statemnt is defintely false as it implies the clocks will stay synchronized in all inertial frames which is not the case and the second one suggests the same thing."
haha, way over my head jcsd
haha, way over my head jcsd
It isn't/doesn't have to be though, relativistic kinematics are suprisingly easy to understand.
ArmoSkater87
Jun22-04, 05:41 PM
jcsd, I though pretty much everyone agreed on true, true, true for the first stage. So why are u saying only the last statement is true.
There's ambiguities in the statements:
The first statement purpisely suggetss that the clocks will appear synchronized to all onsrevers this is not the case.
the second statement is true except for the bit in the brackets which is only true in one inertial frame.
ArmoSkater87
Jun22-04, 05:51 PM
oh i thought we asume they are synchronized in every case and every frame
oh i thought we asume they are synchronized in every case and every frame
But this is the crux, they can't be synchronised in every frame as they are seperated by distance.
ram1024
Jun22-04, 07:23 PM
all the clocks are synchronous in the stationary "picture-frame" as hurkyl would call it.
we're determining simultaneity/detection times for the observers which i believe to be immutable (meaning that no matter what frame you choose to look at there will never be a frame where events happening to one observer will change order)
in case 1 we're assuming the train is stationary (i know nothing can be truly stationary in SR, calm down let's not get bogged down)
in case 2 we're moving the train, but because it's a relativistic inertial frame we should get the same results as in case 1 right?
in case 3 we're moving the observer, we SHOULD get a different result than in step 2, right?
that was the main focus of this "gedankenexperiment". Hurkyl caught on a bit too quickly and foiled my plans, and now i can't get anyone to admit where they stand on any of the cases :D
ArmoSkater87
Jun22-04, 07:53 PM
thanks for the clarification! :P
russ_watters
Jun22-04, 09:53 PM
thanks for the clarification! :P To simplify a little for your sake, picture two clocks that are 300,000km apart and synchronized from Hurkyl's "picture frame." Each clock can send a signal to the other saying what time it is: since it takes 1sec for the signal to reach the opposing clock, each clock compares its own time to the signal it recieves and concludes the other clock is a second behind. The paradox can obviously be resolved easily enough knowing the speed of light and the distance between the clocks and calculating the transmission delay.
This is true in Galilean Relativity, ie, before you even start to consider Einstein's SR/GR. The only thing added by SR is that the speed of light is constant.
ram1024
Jun22-04, 09:54 PM
if you're gonna tell him it's constant at least tell him HOW it's constant.
Alkatran
Jun22-04, 10:29 PM
actually it all suits my purposes. i don't care to retute SR as long as it sees things MY way ;D
I find that quote really amusin, RAM.
Anyways, I just though I'd come in and tell you why all of this time dilation and non-simultaneity is popping up. Consider this: You have two objects in a two-dimensional universe. One is standing still in the reference frame, while the other is moving to the right at a high speed. As soon as the second object is at the same position as the first light is emitted from where they are positioned.
Now, the un-moving object perceives the light as expanding uniformely around it (in a circle) and SO DOES THE MOVING ONE. But how is that possible?? Well, check the attacked picture for what Einstein was thinking for relativity.
Now, remember this is a 2d universe, the up/down part of the picture represents the passage of time. (Notice that the light is expanding as it goes up? That's light moving away. Now, notice how the moving object (the blue one) has a "reality" that is warped in relation to time? Notice how it's actual passage through time is warped (it moves slightly faster upwards). Finally, notice that in it's reality the light is in a circle around it.
Do you understand now? That really for us to have a concept of simultaneous we could define a certain speed as "0" and say only things in that frame are simultaneous? (Think about this, if the observers know of the effects or relativity, shouldn't they compensate for it when deciding wether or not something was simultaneous? :devil: )
all the clocks are synchronous in the stationary "picture-frame" as hurkyl would call it.
Earlier in this thread you said
the clocks in all cases emit photon simultaneously RELATIVE TO EACH OTHER.
these clocks are perfectly aligned and synchronized and in all cases they move within the same inertial frame so they can stay calibrated.
So which is it? Now, for cases 1 & 3 it doesn't matter, because the "picture frame" and the "clock frame" are one in the same.
But in case 2, the clock frame and the "picture frame move relative to each other, and thus the clocks can't emit their photons simultaneously in both frames. They can emit simultaneously in one or the other, but not both.
By the way, u just cant try to disprove the theory of one of the greatest geniuses of all time, SR has been proven correct in several experiments. (-_-)
No, that’s a myth. Lorentz theory has been proven correct in several experiments. Einstein stole many of his SR theory ideas from Lorentz and modified Lorentz’s ideas.
The “time dilation” of atomic clocks was invented by Lorentz in the 1890s. So was “length contraction”, the “speed limit of c”, “mass increase due to motion”. See Lorentz's, “Versuch Einer Theorie Der Elektrischen Und Optischen Erscheinungen In Bewegten Körpern,” published in 1895 when Einstein was just 16 years old.
”To fill this gap, I introduced the principle of the constancy of the velocity of light, which I borrowed from H.A. Lorentz’s theory of the stationary luminiferous ether, and which, like the principle of relativity, contains a physical assumption that seemed to be justified only by the relevant experiments (experiments by Fizeau, Rowland, etc.).” A. Einstein, 1912
It isn't/doesn't have to be though, relativistic kinematics are suprisingly easy to understand.
Straight line non-accelerated relative motion can’t slow down the rate of any clock. As Lorentz pointed out, a “force” has to be placed on a clock timing mechanism, or removed from it, to change the clock’s rates. “Kinematics” without force changes no clock rates.
ram1024
Jun22-04, 11:24 PM
So which is it? Now, for cases 1 & 3 it doesn't matter, because the "picture frame" and the "clock frame" are one in the same.
But in case 2, the clock frame and the "picture frame move relative to each other, and thus the clocks can't emit their photons simultaneously in both frames. They can emit simultaneously in one or the other, but not both.
no one's yet given me a reason to believe they'd emit photons non-simultaneously in EITHER frame, moving or not.
does the phrase "relative to each other" actually have a meaning in this case? i mean i just threw it out there without thinking of consequences because i cannot fathom the difference it would make.
so if you can, tell me what the difference would be if they emitted photons simultaneously in the "picture frame" and in the "relative to each other frame)
thanks in advance
no one's yet given me a reason to believe they'd emit photons non-simultaneously in EITHER frame, moving or not.
does the phrase "relative to each other" actually have a meaning in this case? i mean i just threw it out there without thinking of consequences because i cannot fathom the difference it would make.
so if you can, tell me what the difference would be if they emitted photons simultaneously in the "picture frame" and in the "relative to each other frame)
thanks in advance
Please read http://www.bartleby.com/173/9.html
ArmoSkater87
Jun22-04, 11:43 PM
thanks to whoever helped me out :P
ram1024
Jun23-04, 12:13 AM
Please read http://www.bartleby.com/173/9.html
Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result
this case depicted is actually case 3 of my list, not case 2. note the sources are "lightning flashes" which are not in any way tied to the train.
the train moving towards B)flash is equivalent to my guy running towards B)emitter on top of the train in case 3.
case 2 is the crux of the argument. if "stationary" doesn't exist how can case 1 yield that light hits the observer simultaneously, yet in case 2 if you take that frame as being stationary (inertially relative frame) you DON'T get hit by light simultaneously?
i'm not sure i explained that well enough for you to understand :( sorry if not, i'll do better
Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result
this case depicted is actually case 3 of my list, not case 2. note the sources are "lightning flashes" which are not in any way tied to the train.
the train moving towards B)flash is equivalent to my guy running towards B)emitter on top of the train in case 3.
case 2 is the crux of the argument. if "stationary" doesn't exist how can case 1 yield that light hits the observer simultaneously, yet in case 2 if you take that frame as being stationary (inertially relative frame) you DON'T get hit by light simultaneously?
i'm not sure i explained that well enough for you to understand :( sorry if not, i'll do better
Yes, it doesn't matter whether the sources are tied to the train or not, because speed of light is independent of its source.
So, what is the difference between your case #2 and #3?
The difference is: in case #2, the photons are emitted simultaneously relative to the emitters (which means also relative to the man). In case #3, the photons are emitted simultaneously relative to the emitters (as you said this is always the case), therefore they cannot be emitted simultaneously relative to the man. See, it matters in which frame they are emitted simultanously. So how do we know in which frame? It is the frame that the clocks (tied to the emitters) were synchronized in. And how were they synchronized? By sending two light signals to the emitters from the midpoint in that frame. Naturally it follows that: after this synchronization, simultaneous light signals will be received at the same time only at the midpoint in that frame (not some other frame where the midpoint moves [is somewhere else when the photons meet]).
ram1024
Jun23-04, 12:57 AM
Yes, it doesn't matter whether the sources are tied to the train or not, because speed of light is independent of its source
The difference is: in case #2, the photons are emitted simultaneously relative to the emitters (which means also relative to the man). In case #3, the photons are emitted simultaneously relative to the emitters (as you said this is always the case), therefore they cannot be emitted simultaneously relative to the man
i think you're confusing the emissions with the intercepts... if the sources do not matter in speed, remove them from the picture and view the motions of the observers in case 2 and case 3. they're identical. the photons from both frames were emitted at exactly the same time and place. the only difference is, one guy is running and one guy is riding the train.
are you telling me runners and train riders have different perspectives at the same speed?
<we're getting close to the point where you're going to have the same revelation that i had i think. good stuff>
i think you're confusing the emissions with the intercepts... if the sources do not matter in speed, remove them from the picture and view the motions of the observers in case 2 and case 3. they're identical. the photons from both frames were emitted at exactly the same time and place.
No I'm not confusing anything. Did you read carefully what I wrote? The sources can only emit light according to how they were synchronized (before the experiment started). How do you suppose they know when to emit the photons? It is this synchronization that makes the cases 2 and 3 different. Actually, change the sources with mirrors, and you can think of the emitted photons as reflected photons sent from the midpoint (they were sent to synchronize the clocks). In case 2, the man sent these photons to synchronize the clocks, so he receives them back at the same time. In case 3, the stationary midpoint sent them, so it receives them back at the same time, and the man does not. Yes it is this simple.
ram1024
Jun23-04, 01:49 AM
i can see where you would get this confused.
go back to the link on the page. the lightning is synchronized simultaneous ACCORDING to the simultaneous intercepts of the embankment.
it is deriving simultaneity FROM that frame. and then comparing it to all others.
the clock/emitters in my experiment are NOT deriving synchronization or simultaneity FROM any frame. they are synchronized together (let's say zero-distance) and then methodically placed into position using exacting methods to ensure they are never subjected to anything that would "unsynch" them
in this case simultaneous MEANS true simultaneity not "according to how you look at the picture".
i'm going to take a wild stab in the dark that SR doesn't allow ANYTHING to be truly simultaneous unless it emanates from ONE location... right?
i can see where you would get this confused.
go back to the link on the page. the lightning is synchronized simultaneous ACCORDING to the simultaneous intercepts of the embankment.
it is deriving simultaneity FROM that frame. and then comparing it to all others.
Yes.
the clock/emitters in my experiment are NOT deriving synchronization or simultaneity FROM any frame. they are synchronized together (let's say zero-distance) and then methodically placed into position using exacting methods to ensure they are never subjected to anything that would "unsynch" them
No, after the acceleration they go out of synch, even if they both go under the same acceleration. They may be still in synch in their final rest frame, but in all other frames they will look out of synch. The end result is the same as if they were synchronized by a signal from the midpoint in their final rest frame.
in this case simultaneous MEANS true simultaneity not "according to how you look at the picture".
i'm going to take a wild stab in the dark that SR doesn't allow ANYTHING to be truly simultaneous unless it emanates from ONE location... right?
There is no such thing as true simultaneity or simultaneity in all frames [when events separated by distance]. That Enistein gedanken proves [edit:no, discusses] just that. Seems you didn't get the point from that page.
editing to add: The point is, if two distant clocks are synchronized in their frame, they can not be [look] synchronized in other frames.
ram1024
Jun23-04, 02:37 AM
There is no such thing as true simultaneity or simultaneity in all frames [when events separated by distance]. That Enistein gedanken proves just that. Seems you didn't get the point from that page
that's because the page assumes simultaneity is relative to begin with.
i see i'm going to have to employ stricter methods to convince you :D
ONE EMITTER.
TWO OBSERVERS.
Case #5
(o) <-)|(-> (o)
One emitter simultaneously shoots 2 photons towards two ovservers equal distance from the center (where the emitter is). the observers carry synchronized clocks to time their photon receptions
SR predicts that since this is a "inertial frame" light will hit both observers at the same time. (True / False) ?
ram1024
Jun23-04, 02:55 AM
i realize case 5 is a loaded question.
so i'm going to answer it myself so we can move on with no hard feelings :D
SR cannot make predictions as to the simultaneity of reception BECAUSE it does not believe in sychronous clocks at a distance.
fine. moving on to case #6
ram1024
Jun23-04, 03:01 AM
Case #6
(o) <-)|(-> (o)
Same thing as case #5 except this time the two clocks use laser light and the distance between them to synchronize. They synchronize in such a way that the light from the center hits them both at what appears to them to be the same "time"
the train is them sped up FASTER in the direction it was travelling (let's say to the right) and another light is pulsed.
SR predicts they receive light non-simultaneous now (because of clocks getting messed up) even though nothing happened that changed clock synch relative to the other clock. (True / False)
i see i'm going to have to employ stricter methods to convince you :D
You will either make me go insane, or I will give up on you. I'm not sure if you are reading what I'm saying. If you are thinking I don't qualify or something, just say so and I will leave you alone. Otherwise please try a bit harder and hopefully you will be able to answer your own questions.
Case #5
(o) <-)|(-> (o)
One emitter simultaneously shoots 2 photons towards two ovservers equal distance from the center (where the emitter is). the observers carry synchronized clocks to time their photon receptions
SR predicts that since this is a "inertial frame" light will hit both observers at the same time. (True / False) ?
"At the same time" means "simultaneously". Simultaneity is meaningless unless you specify the frame it is measured in. So your question is meaningless. SR will predict that light will hit both observers at the same time in the frame of the emitter/observers/picture. There is no other frame seen here, but no doubt there will be in the next case. All this confusion is due to your omitting the frame bit.
Case #6
(o) <-)|(-> (o)
Same thing as case #5 except this time the two clocks use laser light and the distance between them to synchronize. They synchronize in such a way that the light from the center hits them both at what appears to them to be the same "time"
the train is them sped up FASTER in the direction it was travelling (let's say to the right) and another light is pulsed.
SR predicts they receive light non-simultaneous now (because of clocks getting messed up) even though nothing happened that changed clock synch relative to the other clock. (True / False)
No, they will receive the light at the same time in the train frame. But will the clocks show the same time when they do? If the clocks remain synchronized according to the train frame*, yes. In any case, they will not look like receiving them at the same time in the picture frame. And even if the clocks remained synchronized in the train frame, they will not look synchronized in the picture frame.
*OK I can't talk so confidently when acceleration is involved. But I can say it depends on how the train is accelerated. Is it pulled from the front, or pushed from the back, or both "at the same time" (wrt what? If wrt themselves, then I -think- they will remain synchronized in the train frame)
Alkatran
Jun23-04, 07:05 AM
Ram are you just ignoring what people have been trying to tell you?! It's been said time and time again that if something is "at the same time" in one frame, it doesn't mean it is in another! (most likely no others!)
This is where your flaw in thinking is: Just because your clock are undergoing the same speed changes doesn't mean they are in synch! They are only in synch in the "clock" frame ("train" frame)
If you are moving, your reality is distorted (see my first post, which you seem to have conveniently ignored) through time, so things that are NOW for me are LATER for you, or maybe BEFORE. This is how we can both be moving slow as seen by the other person!
Straight line non-accelerated relative motion can’t slow down the rate of any clock. As Lorentz pointed out, a “force” has to be placed on a clock timing mechanism, or removed from it, to change the clock’s rates. “Kinematics” without force changes no clock rates.
The clocks don't 'slow down' there is no mystery force, they simply run at different rates.
no one's yet given me a reason to believe they'd emit photons non-simultaneously in EITHER frame, moving or not.
does the phrase "relative to each other" actually have a meaning in this case? i mean i just threw it out there without thinking of consequences because i cannot fathom the difference it would make.
so if you can, tell me what the difference would be if they emitted photons simultaneously in the "picture frame" and in the "relative to each other frame)
thanks in advance
Go back to the animations I gave earlier. In that example, if both observers do not agree that both flashes struck each other at the same time, we would have a paradox.
For instance: Suppose that there is an explosive device at the point where they pass each other that has to be triggered by both. Each observer will only trigger the device if he sees the flashes arrive simultaneously. It is obvious that in the frame of the embankment observer that the flashes arrive at the same instant as the observer on the railway car. Thus both should trigger the device and set off the explosive. If the Railway car observer did not see the flash arrive at the same instant, he would not trigger the device from his side and the device would not go off. This would set up a paradox where for one observer the track is destroyed,and for the other it isn't.
Now we include the postulates of Relativity.
1. The laws of physics remain the the same for all observers in all frames regardlesss of their relative motion.
I don't think we have any argument there because if this weren't true then we'd really have a case of different realities in different frames.
2. The speed of light(in a vacuum) is invarient for all observers regardless of their relative moton.
IOW, the speed of light is a constant when measured relative to an observer by that observer.
Now this is a sticking point for some. In fact, earlier you said that is was just an assumption. But it is much more than that. It in itself is a consequence of the first postulate.
The speed of light is determined by two universal constants : permeability and permitivity. These two constants also are what determine the strengths of the magnetic and electrostatic fields. Thus in order for the speed of light to have different values relative to an observer as measured by that observer, these constants would have to have different values for this observer also(the strength of magnets would change, etc.). The laws of physics would have to change for every observer depending on his relative motion to others.
It is actually worse than that. For if two beams of light were directed at the observer from two sources each moving at different velocities from the observer, in order for that observer to see the light from each be dependant on the velocity of each source (Galileian addition of velocities), these two constants would each have to have two different values each at the same time.
In fact, it is possible to show that if the speed of light wasn't invarient for every observer, light produced in one frame could not even be detected by a frame that was moving relative to that frame.
Add to this the fact that the invarience of the speed of light has been confirmed by multiple experiments.
The animations I provided show what happens according to the frames of both observers in order to uphold the principles above. As a result, simultanity is found to fail at distance between frames.
But this is a small price to pay, since at least this doesn't lead to any real paradoxes.
russ_watters
Jun23-04, 10:39 AM
Straight line non-accelerated relative motion can’t slow down the rate of any clock. As Lorentz pointed out, a “force” has to be placed on a clock timing mechanism, or removed from it, to change the clock’s rates. “Kinematics” without force changes no clock rates.[emphasis added] Its sloppy grammar to say clocks slow down in SR, though I admit I do it too. In actuality (and of course, you already know this, as you've been part of this same discussion many, many times), time itself passes at different rates for clocks in different frames, while the clocks themselves are, to a local observer, unaffected.
russ_watters
Jun23-04, 10:49 AM
Ram, just to hammer this in a little more, please please remember that most of the simultenaity issues we're dealing with here are the same for Galilean Relativity as they are for Einstein's. You seem to be trying to argue against SR, but in actuality, you're arguing against the original basis of physics.
One more little thought experiment:
-Suppose you have three identical clocks next to each other and synchronized.
-Move two of the clocks 300,000km apart and keep the third (ignore SR effects).
-At a pre-determined time (call it T0), both clocks emit a signal.
-If you are 200,000km from one clock and 100,000 from the other, at what time according to your clock do you recieve each signal (T0+X, T0+Y)?
-Are the clocks synchronized in your frame?
ram1024
Jun23-04, 11:24 AM
of course they are. I know the speed of light and i know the distance from the emitters. they won't hit me simultaneously, but i can easily deduce the time of emission from distance and arrival times.
ram1024
Jun23-04, 11:36 AM
http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif
i'll have you look at your own animations.
look at the first animation and then look at the second one.
what looks fishy about the second animation?
that's right, the speed of the bubble expansion from the left emitter is MUCH faster than the speed of the bubble on the right. in order to satisfy "constant relative to all viewers". SOMEONE made up length contraction. but wait. if we contract the distance to the right of the train it just leads to the light getting to the train even FASTER. we need to contract the distance BEHIND the train so that light arriving from that direction will arrive simultaneously. OH NO! protect our precious light speed!
ram1024
Jun23-04, 11:47 AM
Ram are you just ignoring what people have been trying to tell you?! It's been said time and time again that if something is "at the same time" in one frame, it doesn't mean it is in another! (most likely no others!)
This is where your flaw in thinking is: Just because your clock are undergoing the same speed changes doesn't mean they are in synch! They are only in synch in the "clock" frame ("train" frame)
that's why i made the point in case #6 that we're determining "simultaneous synchro" by the setup in case #5.
give me your analysis for case #6 if you will be so kind :D
ArmoSkater87
Jun23-04, 01:15 PM
no comment, for the most part...WOOSHHHHHHHHHHHH :P
Why I'm jumping in I have no idea. Masochism?
ONE EMITTER.
TWO OBSERVERS.
Case #5
(o) <-)|(-> (o)
One emitter simultaneously shoots 2 photons towards two ovservers equal distance from the center (where the emitter is). the observers carry synchronized clocks to time their photon receptions
SR predicts that since this is a "inertial frame" light will hit both observers at the same time. (True / False) ?
I presume that the observer's clocks are synchronized in their own frame. If so, then the answer is: When the two observers detect the light, their clocks will read the same time. (Of course a third observer in relative motion to these two guys will disagree that the clocks were ever synchronized according to his frame.) So what?
I have a suggestion.
Restate this problem using spacetime diagrams. ram's snapshot diagrams may seem unambiguous to him, but it's obvious that they don't seem that way to everyone else.
There's a reason that physicists discuss and explain relativity in terms of spacetime: it works!
And nothing has to be conceded on either side of the argument; spacetime is just as valid a concept in Galilean relativity as it is in SR.
ram1024
Jun23-04, 02:10 PM
I presume that the observer's clocks are synchronized in their own frame. If so, then the answer is: When the two observers detect the light, their clocks will read the same time. (Of course a third observer in relative motion to these two guys will disagree that the clocks were ever synchronized according to his frame.) So what?
don't jump in and do one case and expect a revelation :D the "paradox" only comes after you realize the issue from multiple vantage points.
russ_watters
Jun23-04, 02:51 PM
of course they are. I know the speed of light and i know the distance from the emitters. they won't hit me simultaneously, but i can easily deduce the time of emission from distance and arrival times. Well herein lies the basic problem you are having with these thought experiments: you don't know what we're talking about when we say "simultaneous." You essentially answered 'yes, they are simultaneous in my frame' and then provided an explanation that says 'no, they are not simultaneous in my frame, but you can calculate that they are simultaneous to an outside observer.'
Frame, frame, frame, frame, frame, frame, frame!!!!!!
In your frame, the events are not simultaneous - and thats the question I asked. That's what we mean when we say two events are not simultaneous according to a specific observer. You need to get on board with that concept. Its the root of the misunderstanding here.
russ_watters
Jun23-04, 02:56 PM
in order to satisfy "constant relative to all viewers". SOMEONE made up length contraction. but wait. if we contract the distance to the right of the train it just leads to the light getting to the train even FASTER. we need to contract the distance BEHIND the train so that light arriving from that direction will arrive simultaneously. OH NO! protect our precious light speed! While it may be easy enough to ignore length contraction for a train moving at 100km/hr, it is not easy to ignore for a train moving at a significant fraction of the speed of light as shown in the animations.
I mentioned before that all these thought experiments may be counterproductive: since thought experiments exist only in your head, you may start thinking the data exists only in your head too. It doesn't. The data has been collected from real experiments and demonstrates that C is constant. Maybe we should start looking at real experiments instead of thought experiments.
ram1024
Jun23-04, 03:18 PM
see that's the thing, if simultaneity can be real at a single point, then simultaneity MUST be able to be real at a distance. not "according to an observer" but according to "reality".
to say it doesn't happen is like saying "no two things in the universe EVER happen at the same time"
whether or not they happen "at the same time to you" is merely a matter of perception and is NOT reality
ram1024
Jun23-04, 03:20 PM
if you have any real experimental data handy you can share it, but i unfortunately do not have access to any
what looks fishy about the second animation?
that's right, the speed of the bubble expansion from the left emitter is MUCH faster than the speed of the bubble on the right. in order to
No, it doesn't look much faster to me, it looks very close. In fact, they should be the same, if the animation is accurate. Maybe you are fixing your eyes on the red dots? Remember this is the train's perspective. The speeds must be the same wrt the train.
whether or not they happen "at the same time to you" is merely a matter of perception and is NOT reality
Why can't that perception be your reality? Nothing can interact faster than light. And aren't you one of those observers? Will you prefer someone else's perception as your reality?
I mentioned before that all these thought experiments may be counterproductive: since thought experiments exist only in your head, you may start thinking the data exists only in your head too. It doesn't. The data has been collected from real experiments and demonstrates that C is constant. Maybe we should start looking at real experiments instead of thought experiments.
Russ makes an excellent point. While these thought experiments may serve to illustrate what relativity says in various situations and to show that it's perfectly self-consistent, they cannot prove that relativity is in fact how the world really works. Only experiment can do that.
what looks fishy about the second animation?
that's right, the speed of the bubble expansion from the left emitter is MUCH faster than the speed of the bubble on the right.
I had a look at them again; they look the same speed to me.
The motion of the track kind of plays an optical illusion, though; if you still think the left bubble expands faster than the one on the right, crop the image so you can't see the track (maybe by shrinking your browser window and scrolling the image partially off the window) and see if you still think one expands faster than the other.
Alkatran
Jun23-04, 05:19 PM
http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif
i'll have you look at your own animations.
look at the first animation and then look at the second one.
what looks fishy about the second animation?
that's right, the speed of the bubble expansion from the left emitter is MUCH faster than the speed of the bubble on the right. in order to satisfy "constant relative to all viewers". SOMEONE made up length contraction. but wait. if we contract the distance to the right of the train it just leads to the light getting to the train even FASTER. we need to contract the distance BEHIND the train so that light arriving from that direction will arrive simultaneously. OH NO! protect our precious light speed!
They're expanding at the same speed. The train track gives the illusion that one is moving more quickly than the other (remember that this is lgiht speed relative to the train and NOT the track!)
Watch the bubbles (the left sides of them) at each frame they move approximately one "track" (little brown line). They are moving at the same speed.
no one's yet given me a reason to believe they'd emit photons non-simultaneously in EITHER frame, moving or not.
Take case two. Suppose each clock is reset to zero when the photons are emitted simultaneously in the picture frame.
Now, note the times at which each clock receives the others photon; they will be different. (as can easily be shown in the picture frame)
If the photons were emitted simultaneously in the clock frame, then the clocks would be synchronized in the clock frame. Furthermore, it takes the same time for the photon to get from A to B as it does from B to A. (Remember that the clocks are stationary in their rest frame!) Thus, the clocks must read the same time when they receive the other's photon.
Since the clocks, in fact, do not read the same time when they receive the other's photon, we conclude that the photons were not emitted simultaneously in the clock frame.
Or just imagine a device to sychronize them:
at the half way point we have a device that emits a photon to each clock, when that photon arives at the clocks they tick. This synchronises the clocks in the staionary frame, but in a moving frame one photon will have further to travel than the other so they CANNOT tick at the same time in the moving frame. As our photon device sychronises the clocks perfectly in the rest frame this must hold true for all clocks that are synchronised in their rest frame whether we use this device or not.
ram1024
Jun23-04, 05:44 PM
They're expanding at the same speed. The train track gives the illusion that one is moving more quickly than the other (remember that this is lgiht speed relative to the train and NOT the track!)
Watch the bubbles (the left sides of them) at each frame they move approximately one "track" (little brown line). They are moving at the same speed.
wondering if you guys have eyes...
look at example 2 yet again:
http://home.teleport.com/~parvey/train2.gif
now halfway through the animation, the light from emitter(L) starts. it covers a FULL distance in that time while light from emitter(R) travels half the remaining distance.
add that to the fact that the picture is skewed (look at the bubbles in the last frame they're not even centered to the sources anymore)
the light on the left is travelling about 4x faster than the light on the right.
now halfway through the animation, the light from emitter(L) starts. it covers a FULL distance in that time while light from emitter(R) travels half the remaining distance.
So? The left emitter was much closer to the meeting point when it fires than the right emitter was.
add that to the fact that the picture is skewed (look at the bubbles in the last frame they're not even centered to the sources anymore)
The sources moved. What else would you expect?
wondering if you guys have eyes...
no wonder we are having communication problems.
we can't even agree on something as simple as this.
ram1024
Jun23-04, 06:21 PM
so speed is distance over time
left light travels 1 full distance in 1/2 the time the right light.
left light is comparatively twice as fast.
what happened to relatively constant?
The two light "spheres" expand at the same rate. You are "measuring" distance (I suppose) by counting the moving rails: incorrect. The speed of the light is invariant with respect to the observer.
so speed is distance over time
Pull out your ruler and measure the distance on your screen between the point where the right emitter fires and the point where the bubbles meet.
Pull out your stopwatch and measure how long it takes.
Divide.
Repeat for the left emitter.
Compare.
My measurements:
7.5 cm distance for the right emitter.
I average 1.83 seconds over three trials.
Velocity is 4.10 cm/sec
3.75 cm for the left emitter.
0.83 seconds
Velocity is 4.52 cm/sec
That's a 10% relative error; well within my confidence in my time measurements.
Certainly nowhere near twice as fast.
ram1024
Jun23-04, 06:50 PM
is that picture based on experimental data?
because it's way way off. :yuck:
What's off about it?
Light travels at the same speed in all directions from the point of emission.
What else should happen?
ram1024
Jun23-04, 07:01 PM
well in the second picture nothing's moving.
picture 1. the train is making relative progress towards the emitter (source) of the photon to the right.
picture 2. the train is making NO relative progress towards the emitter (source)
poke a pen on the center of the emissions in picture 1. movement is measured by the train towards emitter to the right and away from emitter to the left.
poke a pen on the center of emissions in picture 2. no relative motion is made towards the locations of the sources. indeed the locations of the sources seem to be travelling down the tracks. what's with that?
poke a pen on the center of emissions in picture 2. no relative motion is made towards the locations of the sources. indeed the locations of the sources seem to be travelling down the tracks. what's with that?
In picture 2, the sources are the red dots, the same as in picture one, and they are very clearly drifting leftwards with the tracks.
And because the sources are moving, they cannot occupy the center of emissions for the entire animation!
ram1024
Jun23-04, 07:11 PM
i'm not talking about the red dots. ignore the red dots completely.
poke your pen in the centers of the expanding spheres.
picture 1 = movement towards expanding sphere to the right
picture 2 = no movement towards expanding sphere to the right.
so what's going on?
i'm not talking about the red dots. ignore the red dots completely.
poke your pen in the centers of the expanding spheres.
picture 1 = movement towards expanding sphere to the right
picture 2 = no movement towards expanding sphere to the right.
so what's going on?
The centers of the expanding spheres should not be moving, do you agree? If one thing goes left at c and one thing goes right at c, then their midpoint should be stationary, right?
Why is this not a contradiction? Because "The center of the sphere of light" is not an object; it is a geometric calculation.
The two animations demonstrate how an object can satisfy this geometric description in one frame and not the other. The emitters, which are stationary in the first animation, remain in the center of the sphere. The mitters, which are moving in the second animation, leave the center of the sphere.
But in any given frame, the center of a sphere of light cannot move; the light going left has the same speed as the light going right, so the point midway between them must remain stationary.
ram1024
Jun23-04, 07:32 PM
it SHOULD not be moving. which means MOTION of the train should bring the train closer to it.
it IS a contradition, because in picture 1, the train is without a doubt making relative motion towards it.
in the second image the train and the two sources/centerpoints move together. why are the centerpoints moving? they shouldn't be.
poke your pen in the centers of the expanding spheres.
picture 1 = movement towards expanding sphere to the right
I don't see center of spheres moving in any pictures.
If the author of the animation is the same person who posted the link, maybe he can create a version showing the center of spheres in the second picture.
it SHOULD not be moving. which means MOTION of the train should bring the train closer to it.
And if the train is NOT MOVING (as in the second picture), then the train should not get closer.
in the second image the train and the two sources/centerpoints move together.
The train is not moving. The sources are. The centerpoints are not.
ram1024
Jun23-04, 07:34 PM
not moving relative to the picture frame, i'm talking about moving relative to the tracks, just like the train is moving.
i'm talking about moving relative to the tracks, just like the train is moving.
Yet the train is not moving in picture #2; the tracks are.
ram1024
Jun23-04, 07:38 PM
no no no hurkyl. the second picture is not the train stationary, it's supposed to be the vantage point from the train IF the train was stationary.
what would be the point of comparing a moving train to a stationary one?
they both move in both pictures, that's the exercise, they're both supposed to get relative views of "what happens" to the same even viewed from two vantages.
the problem is. in picture 1, the train makes definate relative movement towards the emitter to the right.
in picture 2 there is no movement towards it.
i can only conclude that the picture is somehow interpretting the data incorrectly OR, someone fudged the calculations intentionally.
no no no hurkyl. the second picture is not the train stationary
Sure it is; it remains at exactly the same point in the picture at all times, right? That is the very definition of stationary.
what would be the point of comparing a moving train to a stationary one?
It's the same train; the appropriate choice adjective "moving" or "stationary" depends on which reference frame you are using.
they both move in both pictures,
No they don't; the train remains at exactly the same point in picture #2. It's velocity is zero. It's stationary.
they're both supposed to get relative views of "what happens"
And, by golly, the train is stationary relative to itself.
the problem is. in picture 1, the train makes definate relative movement towards the emitter to the right.
in picture 2 there is no movement towards it.
In both pictures, the distance between the train and the emitter is very clearly decreasing.
I point out, once again, that if the emitter is moving, then it does not remain at the center of the sphere light.
i can only conclude that the picture is somehow interpretting the data incorrectly OR, someone fudged the calculations intentionally.
That person is you. You are mentally converting picture #2 back into picture #1 before you do any interpretation.
russ_watters
Jun23-04, 07:46 PM
see that's the thing, if simultaneity can be real at a single point, then simultaneity MUST be able to be real at a distance. not "according to an observer" but according to "reality".
to say it doesn't happen is like saying "no two things in the universe EVER happen at the same time"
whether or not they happen "at the same time to you" is merely a matter of perception and is NOT reality Well here's the thing, and its a catch-22 for you: your definitions of words like "simultaneous," "reality," and "perception" aren't the definitions science uses. So even if you are correct that perception does not equal reality (you're not), you still have to stipulate to it for the purpose of examining what the theory says. Otherwise, you're arguing that the sky is orange and defining orange to be the color between green and indigo. You won't get very far in science with that approach.
Regarding perception vs reality: what is perception? Its measurements, observations, data collected from an observer's reality. If these perceptions aren't reality, then there is no way to know what reality is and no use for science. Science has no choice but to assume that our perceptions are real. Otherwise, there isn't anything that science can "know."
Basically (not surprisingly), your objections to science are philosophical in nature. And I'm sorry, but you can't ever hope to understand science if you can't accept its philosophy.
From my thought experiment, ram - what if you don't know your distance between the two distant clocks. In that case, if the signals reach you at different times, how can you figure out if the signals are simultaneous in another reference frame? (Hint: you can't.)
ram1024
Jun23-04, 07:56 PM
what if you don't know your distance between the two distant clocks. In that case, if the signals reach you at different times, how can you figure out if the signals are simultaneous in another reference frame? (Hint: you can't.)
but if you DO know the distance you CAN. and simultaneity would be the same to everyone else who did their own calculations.
i'll point out the problem in the pictures in detail as soon as i can figure out how to work my paint program :surprise:
no no no hurkyl. the second picture is not the train stationary, it's supposed to be the vantage point from the train IF the train was stationary.
what would be the point of comparing a moving train to a stationary one?
they both move in both pictures, that's the exercise, they're both supposed to get relative views of "what happens" to the same even viewed from two vantages.
the problem is. in picture 1, the train makes definate relative movement towards the emitter to the right.
in picture 2 there is no movement towards it.
i can only conclude that the picture is somehow interpretting the data incorrectly OR, someone fudged the calculations intentionally.
The emitters are the red dots, which it is assumed in this case emitted one brief flash, represented by the expanding rings. Once that flash is emitted, there is no reason for there to be any connection between the center of emission and the source that intially emitted it.
The point is that there is no prefered frame of reference. You can not absolutely say whether it is the train or tracks that are "moving". Both observers have equal claim that it is they that are stationary and that it is the other that is moving. Thus each observer will measure events as if they are the in stationary frame. In this case, the train observer sees the flash expand outward at the speed of light as a sphere from the point of emission. But the initial emitters move away from that point.
The observer beside the tracks has no motion relative to the emitters, thus for him, the center of expanision and the emitters remain together.
Alkatran
Jun23-04, 08:19 PM
so speed is distance over time
left light travels 1 full distance in 1/2 the time the right light.
left light is comparatively twice as fast.
what happened to relatively constant?
I think I understand why you don't understand now. You REFUSE to believe that light can be at c for more than one observer at the same time. I am willing to accept there is a better solution than relativity out there, but it better pass all the tests relativity has before I accept it.
Consider this: In picture two the track is being shoved sideways by some massive force at a constant speed. The train is standing still because it is rolling on the track at just the right speed to be at 0. The light is emitted and you get what you're seeing!
Does that makes more sense to you?
ram1024
Jun23-04, 08:52 PM
stupid paint program is worthless for animated gifs.
oh well now i have the file done but my webhosting is down
:mad:
ram1024
Jun23-04, 09:18 PM
http://www.imagedump.com/index.cgi?pick=get&tp=87816
temporary host :|
You're still confusing "source of sphere of light" with "center of sphere of light".
The center is a geometric concept. Its definition depends on the (simultaneous!) measurement of the extent of the sphere of light at a given time in the given frame.
The source is a physical object. It is not required to remain at the center of the sphere of light in a given frame.
ram1024
Jun24-04, 03:57 AM
that's why source is in "parenthesis"
in picture two we're NOT making relative motion towards "the source (geometric center"
in picture one we are.
let's assume that "the source" was an explosion and the physical source no longer exists.
how would you perform your calculations in picture 2?
that's why source is in "parenthesis"
in picture two we're NOT making relative motion towards "the source (geometric center"
in picture one we are.
The two animations are different views of the same thing. In both animations, the sources are fixed (attached) to the track. Animation #2 takes the view of someone on the train (thats why the train doesn't move). The sources (red dots) certainly move. The centers of the light spheres never move in either animation: that's what "invariant speed of light" means--any observer will see the light moving in a perfect sphere from wherever and whenever it started according to him. The speed of the light has nothing to do with the speed of the source.
let's assume that "the source" was an explosion and the physical source no longer exists.
how would you perform your calculations in picture 2?
I'm not sure what calculation you are trying to do. If you are calculating the speed of the light: remember the speed of light is constant relative to you the observer. The animation shows this pretty well.
ram1024
Jun24-04, 04:35 AM
remember the speed of light is constant relative to you the observer. The animation shows this pretty well.
not if you take out the physical sources. basically you're saying that sources moving towards a stationary body and emitting photons simultaneously is exactly the same thing as sources stationary from a stationary body but emitting light non simultaneously.
the difference is where reality would agree with it. I didn't want to resort to it, but play the "where's the photon" game with both pictures. unless you're going to tell me the location of the photon is "relative" as well it's pretty closed case.
not if you take out the physical sources. basically you're saying that sources moving towards a stationary body and emitting photons simultaneously is exactly the same thing as sources stationary from a stationary body but emitting light non simultaneously.
Think of the source as emitting a single pulse of light and then shutting off. So, from an observer's view in describing how the light emanates from the source--all I care about is where the source was at the instant it flashed (according to me).
The "reality" is the fact that the photons from both sources arrive at the train together. That has nothing to do with who's viewpoint you take. The animations just show you that what one observer views as two simultaneous flashes of light, another observer views as two non-simultaneous flashes. But everyone agrees that the two pulses hit the train together.
the difference is where reality would agree with it. I didn't want to resort to it, but play the "where's the photon" game with both pictures. unless you're going to tell me the location of the photon is "relative" as well it's pretty closed case.
Please do play "where's the photon"! It's very instructive. Just be aware that relativity (and reality) insist that photons always travel at speed c with respect to the observer.
One trap to be wary of is that nothing prevents an observer from measuring that the apparent rate* at which light approaches another (moving) object may differ from c. For example in animation #1 the train is moving, so the rate at which the train and second light pulse come together is greater that the speed of light! But that's no problem. (But from the train's point of view (animation #2), the light speed is just c as usual.) It's tricky stuff.
* (I would refer to this rate as a "third party separation velocity" to distinguish it from a velocity with respect to the observer.)
You may be interested in checking out some of the other threads in TD. We've recently analyzed the crap out of this "train paradox" thing.
And what about Lorenz transformation?
the difference is where reality would agree with it. I didn't want to resort to it, but play the "where's the photon" game with both pictures. unless you're going to tell me the location of the photon is "relative" as well it's pretty closed case.
Ram, I understand what you mean: you want to run the two animations in parallel (you would have to move one of them manually to keep the trains aligned), then you want to see the egde of the spheres at the same positions all the time. But they won't be. If that was the case, there would be a common reality agreed in both frames. But, do you remember there was an animation I posted the link for. There were two frames skewed relative to each other in the time dimension. If we could run these animations skewed like that (instead of just parallel), you would see it makes sense. Maybe I could make such animations in flash.. But.. I'm not sure you would be convinced anyway.. In post#204, Hurkyl unbelievably bothered to make measurements on those animations for you, but you just ignored.. Nothing we say will convince you if you don't want to learn how this works..
Alkatran
Jun24-04, 06:38 AM
Ram, I understand what you mean: you want to run the two animations in parallel (you would have to move one of them manually to keep the trains aligned), then you want to see the egde of the spheres at the same positions all the time. But they won't be. If that was the case, there would be a common reality agreed in both frames. But, do you remember there was an animation I posted the link for. There were two frames skewed relative to each other in the time dimension. If we could run these animations skewed like that (instead of just parallel), you would see it makes sense. Maybe I could make such animations in flash.. But.. I'm not sure you would be convinced anyway.. In post#204, Hurkyl unbelievably bothered to make measurements on those animations for you, but you just ignored.. Nothing we say will convince you if you don't want to learn how this works..
I agree completely. Any anyone bull-headed enough to go through 12 pages of people telling them basic things about relativity, such as the fact that c is relative, and still come out and say:
ot if you take out the physical sources. basically you're saying that sources moving towards a stationary body and emitting photons simultaneously is exactly the same thing as sources stationary from a stationary body but emitting light non simultaneously.
the difference is where reality would agree with it. I didn't want to resort to it, but play the "where's the photon" game with both pictures. unless you're going to tell me the location of the photon is "relative" as well it's pretty closed case.
is CLEARLY not listening to what we're saying. He's just glazing over the posts to find errors so he can get HIS point across. Then he wonders why all of our answers are different depending on the frame we are in! :eek:
Alkatran
Jun24-04, 06:41 AM
Alright Ram, you disagree with the pictures because of the non-simultanity. But remember that picture with space distorted through time? (It was UP (ahead) in the direction of movement, and DOWN (back) the other way) Guess what that means? The train experiences certain "spaces" before an unmoving observer (the ones ahead of the train), but it also experiences certain ones AFTER (the events behind the train). Do you get how that can both perceive them as different YET?!?! :mad:
This hasn't been said... maybe it needs to be?
If the green fuzzy dots drawn in animation #2 are objects, then in animation #1 they would be moving to the right with the same speed as the train.
Where can we find Lorenz transformation in this thread?
I did not find it here.
russ_watters
Jun24-04, 10:41 AM
but if you DO know the distance you CAN. Yes! Now put the two together: the events are simultaneous when viewed from one frame, but are not simultaneous when viewed from another.
Again, "simultaneous" means "happening at the same time." If two signals reach you at different times, then you did not receive them simultaneously. Why is that such a difficult concept to accept? ...and simultaneity would be the same to everyone else who did their own calculations. But only if they agreed on who'se frame would be considered the universal reference rame from which to do them. Remember, by any form of relativity, there isn't one. In fact, the top-down/outside-in reference frame we're using in our thought experiments does not exist in our hypothetical 2d universe. You could call it a virtual reference frame, one in which communication is instantaneous. But regardless, it still doesn't change the fact that you received the signals at different times.
Maybe something more physical: if two people throw baseballs at you and one hits you in the head a second before the second one hits you in the head, then they didn't hit you simultaneously.
ram1024
Jun25-04, 01:53 AM
so answer case #6 already :|
ram1024
Jun25-04, 03:15 AM
just in case you couldn't find the post here it is again
Case #6
(o) <-)|(-> (o)
Same thing as case #5 except this time the two clocks use laser light and the distance between them to synchronize. They synchronize in such a way that the light from the center hits them both at what appears to them to be the same "time"
the train is them sped up FASTER in the direction it was travelling (let's say to the right) and another light is pulsed.
SR predicts they receive light non-simultaneous now (because of clocks getting messed up) even though nothing happened that changed clock synch relative to the other clock. (True / False)
explaining your answer in this case helps too. so try to :D
Didn't I post an explanation of this a couple of pages back:
Or just imagine a device to sychronize them:
at the half way point we have a device that emits a photon to each clock, when that photon arives at the clocks they tick. This synchronises the clocks in the staionary frame, but in a moving frame one photon will have further to travel than the other so they CANNOT tick at the same time in the moving frame. As our photon device sychronises the clocks perfectly in the rest frame this must hold true for all clocks that are synchronised in their rest frame whether we use this device or not.
Let us add Lorenz transformation to our "story".
observer1
|(->__.__.__.__.__(o)__.__.__.__.__<-)|
|(->__.__.__.__.__(o)__.__.__.__.__<-)|
|(->__.__.__.__.__(o)__.__.__.__.__<-)|
|(->__.__.__.__.__(o)__.__.__.__.__<-)|
observer2
|(->__.__.__.__.__(o)__.__.__.__.__<-)|
|(->___.___.___.___._(o)->.__.__.__<-)|
|(->____.____.____.____.(o)->_._._.<-)|
|(->______.______.______.__(o)->...<-)|
So there is no difference between observer1 and observer2 because space scale shrinks like a rubber sheet without losing its density ( exactly as (x/1)*(1/x) = 1 ) in the direction of the movement of observer2, and expends in the opposite side of observer2.
These opposite states can be observed as Doppler Effect.
So in both cases the speed of light is invariant in both directions.
just in case you couldn't find the post here it is again
Case #6
(o) <-)|(-> (o)
Same thing as case #5 except this time the two clocks use laser light and the distance between them to synchronize. They synchronize in such a way that the light from the center hits them both at what appears to them to be the same "time"
Using the laser pulse to synchronized their clocks (call them A and B) in their own frame is a perfectly good method, as jcsd explained. No problem there.
the train is them sped up FASTER in the direction it was travelling (let's say to the right) and another light is pulsed.
Rather than introduce acceleration, and endless arguments about how it would affect the answer, just introduce another observer (call him Joe) who happens to be moving by. Now your once "stationary" arrangement of laser plus clocks A and B is moving according to this new observer Joe. No one will claim that anything was done to "mess up" those clocks--Joe does nothing but pass by. No one will claim that clocks A and B were accelerated: pure canonical SR. (Note that acceleration can certainly be handled, but why complicate matters?)
SR predicts they receive light non-simultaneous now (because of clocks getting messed up) even though nothing happened that changed clock synch relative to the other clock. (True / False)
In my version, it easy to see that SR predicts that clocks A and B will do exactly what they always did. Why shouldn't they? No one did anything to mess with them.
SR further predicts that measurements made by moving observer Joe will show that those laser pulses arrive at A and B at different times according to Joe.
ram1024
Jun25-04, 07:13 PM
SR further predicts that measurements made by moving observer Joe will show that those laser pulses arrive at A and B at different times according to Joe.
exactly why i didn't want to introduce more observers :|
thanks for not following the rules :grumpy:
drawing up case #7. i'll get through to you this time... :frown:
exactly why i didn't want to introduce more observers :|
thanks for not following the rules :grumpy:
drawing up case #7. i'll get through to you this time... :frown:
I was trying to make things easy for you. But... if you insist on accelerating the train, just be sure to do it right. Accelerate each piece of the train uniformly (according to observers on the train) so that each piece is always moving together according to observers on the train. Do this right and the train will be accelerated and the clocks will not be affected (as far as folks on the train can tell). Once you get it moving to the speed you want (with respect to something else of course), fire off that laser again. SR predicts that the light will hit the clocks A and B at the same time according to observers on the train.
Of course, an observer sitting on the tracks watching that train speed up will see those clocks slowly get out of synch. And slow down.
I'm sure you'll make everything clear in case #7. :smile:
ram1024
Jun26-04, 12:21 AM
Case #7
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
in this setup, we have but one emitter and one observer. keeping it simple-like. In all cases the emitter is going to emit a pulse of light on the first "frame" of the setup. assume uniform motion (no acceleration).
step1: emitter stays the same place towards the observer.
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves towards the observer.
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves away from observer.
This is simply a demonstration of what you're saying that light doesn't care what its source does, right? In all 3 cases light would reach the observer at the same time if the first "frame" were synchronized.
now we're going to do what you guys do to things...
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
we're going to take the same set ups from above and simply CHANGE the relative motion so that the emitters are stationary and the observer is the one that's moving. this shouldn't change ANYTHING as far as you guys see it right? these cases should be EXACTLY the same as the ones above, we just changed perspective....
Discuss.
Discuss.
What's there to discuss? Yes, changing the perspective will not change measurements. Maybe you should tell what you think can be wrong.
ram1024
Jun26-04, 03:03 AM
Original Quote by JanusThe point is that there is no prefered frame of reference. You can not absolutely say whether it is the train or tracks that are "moving". Both observers have equal claim that it is they that are stationary and that it is the other that is moving. Thus each observer will measure events as if they are the in stationary frame. In this case, the train observer sees the flash expand outward at the speed of light as a sphere from the point of emission. But the initial emitters move away from that point.
Original Quote by wespe What's there to discuss? Yes, changing the perspective will not change measurements. Maybe you should tell what you think can be wrong.
Step 1, 2, and 3. doesn't matter where the emitters move, the light will hit the observer at the same time.
Step 4 and 5. DOES matter where the OBSERVER moves. the light will NOT hit the observer at the same time.
HENCE. the transposition of reference frames for pictures 1 and 2 for the "trains" would NOT result in the same situation.
you cannot create the same situation by merely "changing' who is moving. for reasons outlined previously
Alkatran
Jun26-04, 04:24 AM
Case #7
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Alright, let's do this then.
Case #7
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
We are watching from a frame at rest. The emitter is moving. The time it takes for light to reach the observer is distance/c (since we are at "rest" compared to them) You could say we are in the observer's frame.
Case #7
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Same thing as moving towards.
Case #7
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Alright, so now the observer is moving and we have SWITCHED FRAMES. We are now observering from the emitters frame and not the observers! This is why you get the same results.
Step 4 and 5. DOES matter where the OBSERVER moves. the light will NOT hit the observer at the same time.
I admit you had me confused for a while.
Ok, combine step 2 and 3. There are actually two emitters. They emit light when they meet. The two emitted light beams go side by side, they can be considered one. Note that the emitters are separating.
(o)___________________<-a.b->
But you can't combine step 4 and 5. If you do, the emitters would not be separating and there would be two observers. In short, they are two different perspectives so you can't combine them like that.
Step 1, 2, and 3. doesn't matter where the emitters move, the light will hit the observer at the same time.
Step 4 and 5. DOES matter where the OBSERVER moves. the light will NOT hit the observer at the same time.
I'm not sure what your point is, since in your last two steps you change to a different observer! Of course different frames measure different times. :smile:
Call your observer A. If the light flashes from a distance L (as measured by A) then A will observe that the light takes the same time to reach him, regardless of the relative motion of A and the light source.
Your steps 1, 2, and 3 seem to take a view from A's frame. But steps 4 and 5 take a view from a frame in which A is moving. Of course that frame will measure different times. It should be no surprize to you by now that time measurements are frame dependent.
ram1024
Jun26-04, 05:08 AM
Case #7
Step: 1
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
in this setup, we have but one emitter and one observer. keeping it simple-like. In all cases the emitter is going to emit a pulse of light on the first "frame" of the setup. assume uniform motion (no acceleration).
emitter stays the same place towards the observer.
Step: 2
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves towards the observer.
Step: 3
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves away from observer.
This is simply a demonstration of what you're saying that light doesn't care what its source does, right? In all 3 cases light would reach the observer at the same time if the first "frame" were synchronized.
now we're going to do what you guys do to things...
Step 4:
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Step 5:
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
we're going to take the same set ups from above and simply CHANGE the relative motion so that the emitters are stationary and the observer is the one that's moving. this shouldn't change ANYTHING as far as you guys see it right? these cases should be EXACTLY the same as the ones above, we just changed perspective....
http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif
Your steps 1, 2, and 3 seem to take a view from A's frame. But steps 4 and 5 take a view from a frame in which A is moving. Of course that frame will measure different times. It should be no surprize to you by now that time measurements are frame dependent.
that is exactly why picture 2 is NOT the same situation as picture 1. :surprise:
geistkiesel
Jun26-04, 05:32 AM
Alright Ram, you disagree with the pictures because of the non-simultanity. But remember that picture with space distorted through time? (It was UP (ahead) in the direction of movement, and DOWN (back) the other way) Guess what that means? The train experiences certain "spaces" before an unmoving observer (the ones ahead of the train), but it also experiences certain ones AFTER (the events behind the train). Do you get how that can both perceive them as different YET?!?! :mad:
There is simple resolution to this. If you say that the moving observer will detect photons emitted simultaneously in a stationary platform as not being emitted simultaneously in the moving platform. This is a definition of lost simultaneity.
Here is your physical problem.In the stationary frame the photons were emitetd simultaeously. There is no ionstance of time where there were not two photons moving fronm their respective sources.
SR says, that photon B is emitted before photon A, which means the photons did not exist as a pair for a t > 0. There is no SR postuilates that will surpress the existence of the photons emitted simultaneously in the stationary by virtue of measuring the arrival times on the moving frame. Nor does SR provide for 'gost emitters' that would allow for the delayed emission of photons in the moving frame that have already been emitted. simultabneously in he stationary frame. The mere fact that SR predicts the photons were not emitted simultaneously is proof of the intrinsic error and fault and uselessness of SR.
Another problem: The arrival times of the photons will always be sequential in the moving frame, but this has nothing to do with SR. Use red ants or Camarros moving at a constant speed and you will always get sequential arrival times. It is silly beyond conmprehension to have any simultabneous arrival of photons, red ants or Camarros once the midpoint of the photons has been moved. Silly. Silly. Silly.
However, one may work backward from the arrival times and determine if the photons were emitted simultaneously using time and velocity data of the moving frame.
geistkiesel
Jun26-04, 05:38 AM
I'm not sure what your point is, since in your last two steps you change to a different observer! Of course different frames measure different times. :smile:
Call your observer A. If the light flashes from a distance L (as measured by A) then A will observe that the light takes the same time to reach him, regardless of the relative motion of A and the light source.
Your steps 1, 2, and 3 seem to take a view from A's frame. But steps 4 and 5 take a view from a frame in which A is moving. Of course that frame will measure different times. It should be no surprize to you by now that time measurements are frame dependent.
You are assuming time dilation and SR. You also assume frame switching and claim moving frames and stationary frames can b e exchanged for mahematical purposes. Hiowever, trains and stationary platform are not pohysically switchable. It is the trains that are seen to accelerate and move, never the stationary platforms. You people are confused.
How are you going to handle the fact that photons emitted simultaneously in a stationary frame are determined to have been emitted sequentially in the moving frame. The latter tells us that there was some t > 0 when both photons had not been emitted, yet the photons had been emitted in the stationary frame simultaneously. Where was the non-emitted photon when the emitted photon was existing by itself?
You are assuming time dilation and SR. You also assume frame switching and claim moving frames and stationary frames can b e exchanged for mahematical purposes. Hiowever, trains and stationary platform are not pohysically switchable. It is the trains that are seen to accelerate and move, never the stationary platforms. You people are confused.
You have a long road to travel before you understand Galilean relativity, never mind Einsteinian relativity. :smile: Baby steps!
How are you going to handle the fact that photons emitted simultaneously in a stationary frame are determined to have been emitted sequentially in the moving frame. The latter tells us that there was some t > 0 when both photons had not been emitted, yet the photons had been emitted in the stationary frame simultaneously. Where was the non-emitted photon when the emitted photon was existing by itself?
That's only a problem for those old-fashioned folks who still think that simultaneity is absolute. But it's not!
geistkiesel
Jun26-04, 05:58 AM
Take case two. Suppose each clock is reset to zero when the photons are emitted simultaneously in the picture frame.
Now, note the times at which each clock receives the others photon; they will be different. (as can easily be shown in the picture frame)
If the photons were emitted simultaneously in the clock frame, then the clocks would be synchronized in the clock frame. Furthermore, it takes the same time for the photon to get from A to B as it does from B to A. (Remember that the clocks are stationary in their rest frame!) Thus, the clocks must read the same time when they receive the other's photon.
Since the clocks, in fact, do not read the same time when they receive the other's photon, we conclude that the photons were not emitted simultaneously in the clock frame.
You are clock smart and physics crazy.
If photons are emitted simultaneously in the stationary plartform and are ultimately determined to have been emitted sequentially in the moving frame then there is some time t > 0 that one of gthe photons did not exist. The stationary frame proves the photons were emitted simultabeously there. The stationary frame observed two photons emitted into the universe at all times.
You would have us believe that measuring from a moving platform allows the supression of the existence of photons already emitted?
The moving observer comes along and determines that no, first there was one emitted photon, then later there was another emitted photobn. You just don't have a physical basis to make the assumption of a loss of simultaneity when you require the suppression and later emission of photons that were already emitted in the stationary frame.
Where was the non emitted photon in the moving frame when the emitted photon existed by itself?
As Grounded's posts show, time
dilation and frame shrinking is due to an error of emitting the addition of the velocity of the observer wrt the photon light source. This occurs as when counting the passing "wave lengths" the observer will count fewer/time than if she had included her own velocity, assuming she is moving against the stream form the source. So no wonder one always measures c from any inertial frame.
geistkiesel
Jun26-04, 06:04 AM
Reality, in SR, is worldlines living in space-time.
Observers don't have their "own reality"; they just measure reality differently.
You mean the reality of suppressing and reviving photons that had already been emitted in the stationary frame? This is the result is it not when you say what was simultaneous in the stationary platform is not simultaneous in the moving frame? B emitted befoe A?
You keep forgetting Grounded's posts that show the intrinsic error in Sr by the lack of adding the observers velocity wrt the source of the photons. So easy to forget isn't it?
ram1024
Jun26-04, 06:05 AM
omg don't distract MY Doc Al! :D
geistkiesel
Jun26-04, 06:13 AM
all the clocks are synchronous in the stationary "picture-frame" as hurkyl would call it.
we're determining simultaneity/detection times for the observers which i believe to be immutable (meaning that no matter what frame you choose to look at there will never be a frame where events happening to one observer will change order)
in case 1 we're assuming the train is stationary (i know nothing can be truly stationary in SR, calm down let's not get bogged down)
in case 2 we're moving the train, but because it's a relativistic inertial frame we should get the same results as in case 1 right?
in case 3 we're moving the observer, we SHOULD get a different result than in step 2, right?
that was the main focus of this "gedankenexperiment". Hurkyl caught on a bit too quickly and foiled my plans, and now i can't get anyone to admit where they stand on any of the cases :D
You shouldn't listen to Hurkyl. He is a professional confuser. He will keep you off guard as much as possible.
If and observer and a midpoint in a stationary frame are colocated for an ionstant, it is ludicrous to think the photon heading to th emidpoint fopr a simultabneous arrival there will also arrive simultabneously at the point that had been colocated bu that has now moved. Yet this is how SR defines loss of simultabeity. There is no "speed of light " implications in the scenario just discussed.Absolutely none.
You have him right here Ram1024. Ask how can two photons emitted simultaneously suddenly change to one photon emitted, then later, another photon emitted? First there were two photons in the universe, then by some fancy mathematics, one photon gets suppressed in a ghost state until the moving observer is ready to let the photon into her reality.
All this time the photons are speeding toward the midpoint of the sources.
we're going to take the same set ups from above and simply CHANGE the relative motion so that the emitters are stationary and the observer is the one that's moving. this shouldn't change ANYTHING as far as you guys see it right? these cases should be EXACTLY the same as the ones above, we just changed perspective....
Still not getting your point. What do you mean "shouldn't change anything"? If you change frames, you have changed the time measurement.
Of course, what I THINK you are struggling with is this: If all we did was change perspectives, then nothing should change right? Well, for that observer (A?) in your diagrams NOTHING HAS CHANGED. It's only us, the outside observers who have changed our perspective and thus our time measurements.
Time measurements ARE a matter of perspective (in a sense). So if you are talking time, then perspective matters. Just like if you are talking about what you see, then it matters where you stand.
Of course if you are thinking: Something's funny here. How can REALITY change if all I'm doing is changing perspective? Ah... now you're beginning to see the light? ( :smile: ) In relativity (and in the real world) time and distance are frame dependent--but there are quantities that are not frame dependent! In some sense those quantities are more "real" that our frame-dependent times and distances. The "real" physical laws governing the world can't be frame dependent, right? Right! That's the entire point of SR. That's why physical laws must be "Lorentz invariant".
that is exactly why picture 2 is NOT the same situation as picture 1. :surprise:
If by "picture" you mean those animations, then OF COURSE they are different--they are views from different frames. But note that the physical reality--the fact that the lights meet the train at the same time & place--is still true in both views!
ram1024
Jun26-04, 06:22 AM
if in order to make light constant you need to break other "Reality" laws such as Simultaneity, Absolute time, Absolute space, and True universal rest, then i don't need it.
show me the data. you guys messed up your calculations somewhere down the line
geistkiesel
Jun26-04, 06:22 AM
Yes, it doesn't matter whether the sources are tied to the train or not, because speed of light is independent of its source.
So, what is the difference between your case #2 and #3?
The difference is: in case #2, the photons are emitted simultaneously relative to the emitters (which means also relative to the man). In case #3, the photons are emitted simultaneously relative to the emitters (as you said this is always the case), therefore they cannot be emitted simultaneously relative to the man. See, it matters in which frame they are emitted simultanously. So how do we know in which frame? It is the frame that the clocks (tied to the emitters) were synchronized in. And how were they synchronized? By sending two light signals to the emitters from the midpoint in that frame. Naturally it follows that: after this synchronization, simultaneous light signals will be received at the same time only at the midpoint in that frame (not some other frame where the midpoint moves [is somewhere else when the photons meet]).
Pure SR scientific smog. How can two photons emitted simultaneously in one frame, now turn into photons emitted sequentially when observed in a moving frame? What law of physics are you using to say that a photon once emitted into the universe, is now suppressed from existience because the moving observer's mathematics tells her that is the reality? There must be a time in the moving frame, some t > 0 when there was only one photon in the universe if Sr is true. A photon in the universe is what is meant by "emitted", as in "emitted simultaneously".
You've still got the trap of mathematics tied aound your neck I see. You shuld try studying physics.
Grounded's posts have negated all your assumptions regarding time and space dilations. Why do you persist?
ram1024
Jun26-04, 06:24 AM
keep in mind that light speed is also governed by that Maxwell equation you brought up earlier Permitivity and Permeability or whatever.
nowhere in that equation does it say "perspective". . .
if in order to make light constant you need to break other "Reality" laws such as Simultaneity, Absolute time, Absolute space, and True universal rest, then i don't need it.
Whose "laws" are those? You're living in the past, man. :smile:
show me the data. you guys messed up your calculations somewhere down the line
Are you joking? We have libraries of data! All of modern physics is based upon SR. Entire fields of physics would make no sense without it. It's been tested so many times that it's hard to take these threads seriously.
keep in mind that light speed is also governed by that Maxwell equation you brought up earlier Permitivity and Permeability or whatever.
nowhere in that equation does it say "perspective". . .
Right! Light speed is something that is NOT a matter of "perspective". It's an invariant.
You shouldn't listen to Hurkyl. He is a professional confuser.
I wouldn't want to take anything away from Hurkyl's professional rep, but you must admit that you are pretty easy to confuse. :rofl:
ram1024
Jun26-04, 07:00 AM
Right! Light speed is something that is NOT a matter of "perspective". It's an invariant
it's completely perspective.
take for a moment everything we KNOW to be real, distances, time, simultaneity. these are things we use every day.
how often do we travel at significant "relativistic speeds" ?
never.
so postulating that doing so would make distance/time/simultaneity UNreal is completely unreasonable.
let's work from reality for a moment and move in the right direction this time. from what we know of reality, what can we deduce from experiments calculating light speed as constant.
A. people don't know how to calculate it correctly
B. their perceptions get screwed up
C. they have no idea what "frame" they're working in <a laughable excuse, as the rest of the world works in the "reality frame">
D. the data is wrong
E. probably all of the above.
put lightly, what good does it do ANY of us to say something is 50 light years away if when we go to travel that distance our years aren't the same, nor is the distance that light travels in a year the same BECAUSE our years are different?
take for a moment everything we KNOW to be real, distances, time, simultaneity. these are things we use every day.
how often do we travel at significant "relativistic speeds" ?
never.
so postulating that doing so would make distance/time/simultaneity UNreal is completely unreasonable.
So we're back to stating that "relativity can't be true because it contradicts what I know is true from common sense". (Of course, relativity is perfectly consistent with your actual everyday experience--since our speeds are so low, the "strange" SR effects aren't evident.)
The most we can do in a thread like this is explain what relativity says and show why it's not contradictory with itself or anything you've actually observed. So if you want to understand relativity, then fine. But if all you want to do is deny relativity, then ... :zzz:
I was trying to make things easy for you. But... if you insist on accelerating the train, just be sure to do it right. Accelerate each piece of the train uniformly (according to observers on the train) so that each piece is always moving together according to observers on the train. Do this right and the train will be accelerated and the clocks will not be affected (as far as folks on the train can tell).
That's not quite correct; the observers on the train will still observe the front clock running fast and the back clock running slow. Heuristically, recall that in an accelerated frame, clocks in the direction of acceleration run fast, and clocks on the other side run slow.
As a concrete example, consider this case.
The (x,t) coordinates of the left edge of the train is given by:
x = 1s * c * cosh (tau / 1s)
t = 1s * sinh (tau / 1s)
And the right edge is given by
x = 2s * c * cosh (tau / 2s)
t = 2s * sinh (tau / 2s)
You can check that:
tau is the proper time along these worldlines.
The lines of simultaneity for the observers at each end are lines that pass through the origin.
The train always has length (1s * c) according to the observers. (as measured along a line of simultaneity)
The right clock always reads twice the left clock. (again, according to the lines of simultaneity)
In general, analyzing from a stationary frame, if the train is accelerated towards the right, and you choose two pairs of events that train observers think occur at the same time, then the left edge will have a greater space displacement and lesser time displacement than the right edge; the proper time experienced by the left edge cannot be as much as that of the right edge.
take for a moment everything we KNOW to be real, distances, time, simultaneity. these are things we use every day.
how often do we travel at significant "relativistic speeds" ?
never.
so postulating that doing so would make distance/time/simultaneity UNreal is completely unreasonable.
This seems exactly backwards to me. We never travel at significant "relativistic speeds", so we most certainly don't "KNOW" anything about it.
The only reason we assumed what we "KNOW" about slow speeds carries over to fast speeds was because there was no reason to think otherwise. With the advent of Maxwell's equations and Special Relativity, now there is reason to think otherwise.
At the current juncture in time, there has never been any experimental evidence that what we "KNOW" about slow carries over to fast... and overwhelming evidence that SR is correct. The unreasonable thing is to cling to prerelativistic assumptions.
put lightly, what good does it do ANY of us to say something is 50 light years away if when we go to travel that distance our years aren't the same, nor is the distance that light travels in a year the same BECAUSE our years are different?
It doesn't. But it does do good to say something is 50 light-years away according to a specified coordinate chart.
That's not quite correct; the observers on the train will still observe the front clock running fast and the back clock running slow. Heuristically, recall that in an accelerated frame, clocks in the direction of acceleration run fast, and clocks on the other side run slow.
This is why I didn't want to introduce acceleration, since the final reading of those clocks will depend on how you do it. I was envisioning a step-by-step process where a small thrust is given to each part of the train simultaneously (according to the rest frame of the train). Each small thrust must be given at the same time in the current rest frame of the train. If you do that, wouldn't you agree that observers on the train would always measure the proper length of the train to be unchanged and the clocks would still be synched? Meaning: fire off that light in the middle and both clocks would still read the same time when the photons hit. Am I wrong?
In general, analyzing from a stationary frame, if the train is accelerated towards the right, and you choose two pairs of events that train observers think occur at the same time, then the left edge will have a greater space displacement and lesser time displacement than the right edge; the proper time experienced by the left edge cannot be as much as that of the right edge.
I think in your example you've accelerated the train uniformly according to the rest frame of the track. In which case, yes, the clocks would lose synch even in their own final rest frame.
geistkiesel
Jun26-04, 09:54 AM
So we're back to stating that "relativity can't be true because it contradicts what I know is true from common sense". (Of course, relativity is perfectly consistent with your actual everyday experience--since our speeds are so low, the "strange" SR effects aren't evident.)
The most we can do in a thread like this is explain what relativity says and show why it's not contradictory with itself or anything you've actually observed. So if you want to understand relativity, then fine. But if all you want to do is deny relativity, then ... :zzz:
Prove it. Do you understand the request? Prove it? Where is the missing photon?
Show from the postulates of SR that where two photons are emitted simultaneously in a stationary frame that a moving frame observer will observe or determine, the photons were emitted sequentially, non-simultaneously, and explain where the photon that was second to appear was wrt the stationary frame, the moving frame the universe?
Are the photons in the stationary frame the same photons in the moving frame?
DOc Al why don't you stop your clowing and posturing, it isn't close to physics and it isn't even in the same zip code as funny.
geistkiesel
Jun26-04, 09:58 AM
keep in mind that light speed is also governed by that Maxwell equation you brought up earlier Permitivity and Permeability or whatever.
nowhere in that equation does it say "perspective". . .
Keep in mind that using Maxwell's equations and disregarding the observers speed wrt the source of the EM radiation will result is the sillinness that the speed of light is constant when measured in all inertial framess. Light slows down mass shrinkles, loss of simultaneity, siliness in the universe. RAM1024 you must read grounded's papers again, and learn them. Everything you have been saying will glow like a warm ray of peace at last.
geistkiesel
Jun26-04, 10:02 AM
Whose "laws" are those? You're living in the past, man. :smile:
Are you joking? We have libraries of data! All of modern physics is based upon SR. Entire fields of physics would make no sense without it. It's been tested so many times that it's hard to take these threads seriously.
When did you decide that including the observers velocity was not required? Include the observer's velocity and your precious constant speed of light will remain subject to the obvious that the relative velocity of the speed of light is easily measured.
Doc Al's physics is measured in the width of his smirk.
geistkiesel
Jun26-04, 10:13 AM
Still not getting your point. What do you mean "shouldn't change anything"? If you change frames, you have changed the time measurement.
Of course, what I THINK you are struggling with is this: If all we did was change perspectives, then nothing should change right? Well, for that observer (A?) in your diagrams NOTHING HAS CHANGED. It's only us, the outside observers who have changed our perspective and thus our time measurements.
Time measurements ARE a matter of perspective (in a sense). So if you are talking time, then perspective matters. Just like if you are talking about what you see, then it matters where you stand.
Of course if you are thinking: Something's funny here. How can REALITY change if all I'm doing is changing perspective? Ah... now you're beginning to see the light? ( :smile: ) In relativity (and in the real world) time and distance are frame dependent--but there are quantities that are not frame dependent! In some sense those quantities are more "real" that our frame-dependent times and distances. The "real" physical laws governing the world can't be frame dependent, right? Right! That's the entire point of SR. That's why physical laws must be "Lorentz invariant".
If by "picture" you mean those animations, then OF COURSE they are different--they are views from different frames. But note that the physical reality--the fact that the lights meet the train at the same time & place--is still true in both views!
Your changing perspectives is a ruse. You switch views when it would be a physical impossibility to do that in a real physical situation. You are dummer than a fence post if you believe that your precious mathematical models are not subject to limitations. Just because your math says it can be done does not prove it is physically possible. Trains move, stationaary platforms are stationary throughout the course of events. Change you perspective and the physics will follow? Is this the secret of SR?
You people are going to be chewing on Grounded's thesis for quite a spell.
When a moving frame observer determines simultanously emitted photons in a stationary frame were sequentially emitted in the moving frame, then where did the nonemitted photon go, or where was it when the photons were sequentially emitted in the moving frame? You must simply signore the stationary platform as nonexistent and the only reality left for consideration is the one your moving observer perceives?
geistkiesel
Jun26-04, 10:17 AM
Right! Light speed is something that is NOT a matter of "perspective". It's an invariant.
Regardless of ME measuring the relative speed of light in all inertial frames will provide just that: A Constant speed of light and an observer moving wrt the photon source velocity. The speed of light is invariant but the measurement of the speed of light is limited only by the imagination of experimental physicists.
Prove it. Do you understand the request? Prove it? Where is the missing photon?
You are the one claiming to have lost a photon. What did you do with it? And where were the photons before you turned on the light? :smile:
Show from the postulates of SR that where two photons are emitted simultaneously in a stationary frame that a moving frame observer will observe or determine, the photons were emitted sequentially, non-simultaneously, and explain where the photon that was second to appear was wrt the stationary frame, the moving frame the universe?
Again? Do you even bother to read what I've written in all the threads we've been through? (I know you don't understand it, but at least make an effort to read it.) Read through them again. You'll see that I've calculated exactly where everything is according to everyone. I'm not going to do it again, until YOU give it a try.
Are the photons in the stationary frame the same photons in the moving frame?
Not sure what you're asking here. The photons just are; they don't belong to a frame. Instead of photons, we'd be better off thinking of pulses of light (containing many photons). These pulses emanate from the light sources and can be detected in either frame. (Obviously, M' detects different photons than M. But don't get hung up on that detail. Consider them as both viewing the same pulses of light.)
DOc Al why don't you stop your clowing and posturing, it isn't close to physics and it isn't even in the same zip code as funny.
You know nothing about physics, poor geistkiesel.
If you do that, wouldn't you agree that observers on the train would always measure the proper length of the train to be unchanged and the clocks would still be synched? Meaning: fire off that light in the middle and both clocks would still read the same time when the photons hit. Am I wrong?
Yah... I started considering the situation earlier when Ram considered synching the clocks then accelerating the train. I could quickly see that it was obvious that the answer depends on how you do it, so I wondered just how you would have to do it to keep the clocks synched in their own frame.
Doing all of the analysis in the stationary frame, if the train is accelerated towards the right, then no matter how the train gets acceelrated, if I consider two points that correspond to the same time in the train's frame, the right point has a greater temporal displacement and a lesser spatial displacement than the left point. Thus, if compute the proper time difference for each of the ends, the right end must have experienced more proper time than the left end.
For my example, if we switch to units where c = 1 (for simplicity), we consider the case where the left edge of the train lies along the worldline:
r(tau) = < cosh tau, sinh tau >
(exercise: check that tau is, indeed, the proper time along the worldline)
The velocity is given by
v(tau) = < sinh tau, cosh tau >
Thus, for the observer on the left edge his line of simultaneity corresponding to his proper time a is parallel to
< cosh a, sinh a >
because this is the unit vector orthogonal to v(a). The line is given, parametrically, by:
l(d) = r(a) + d < cosh a, sinh a>
= <cosh a, sinh a> + d <cosh a, sinh a>
= <(d+1) cosh a, (d+1) sinh a>
where d is the proper distance along the line of simultaneity. (aka the distance measured by the observer on the left edge of the train)
Setting d = 1 (corresonding to the train being constant length) and letting a vary gives us the worldline of the right edge:
r(s) = <2 cosh s, 2 sinh s>
A little calculus shows us that s is twice the proper time for the right edge.
In summary, according to the observer on the left edge of the train, the length of the train remains constant, but the clock on the right edge is running at double speed.
On a side note, the class of worldlines:
r(tau) = <k cosh (tau / k), k sinh (tau / k)>
has the property that each of them remain unchanged by any Lorentz boost that fixes the origin; in particular, these (and their translates) are the worldlines of observers that feel a constant acceleration.
if in order to make light constant you need to break other "Reality" laws such as Simultaneity, Absolute time, Absolute space, and True universal rest, then i don't need it.
Your loss.
You know, there is a group of people who refuse to consider breaking the "Reality" law of a true universal direction of up and down, They are called the "Flat Earth Society".
Actually that last one, 'True universal rest", is funny, since it hasn't been considered a "law of Reality" since the time of Galileo.
ram1024
Jun26-04, 12:23 PM
Doing all of the analysis in the stationary frame, if the train is accelerated towards the right, then no matter how the train gets acceelrated, if I consider two points that correspond to the same time in the train's frame, the right point has a greater temporal displacement and a lesser spatial displacement than the left point. Thus, if compute the proper time difference for each of the ends, the right end must have experienced more proper time than the left end.
the only reason you would think that because your measure of simultaneity is wholly derived from observation impact.
to make it simple, an acceleration from the right just creates more distance covered in comparison to the non-accelerated train. it's this distance that must be accounted for in determining this "reality" compared to the previous non-accelerated one.
the "point of simultaneity" shifts to the left, the clocks don't lose their synchronization, light speed doesn't change, just you're moving at a different speed toward the light and it is intercepted faster.
no length contraction, no "lost simultaneity", no time dialation.
just a change in relative perspective easily corrected using the proper mathematics.
SR is ridiculous :zzz:
the only reason you would think that because your measure of simultaneity is wholly derived from observation impact.
No, the reason I think that was because I was applying (Minowski) geometry.
ram1024
Jun26-04, 12:32 PM
Actually that last one, 'True universal rest", is funny, since it hasn't been considered a "law of Reality" since the time of Galileo
yet true relative rest DOES exist, we use it daily. something CAN exist completely stationary relative to something else.
the universe as a whole is comprised of every object within it. at any given instant if you averaged the locations of every atom/molecule/lepton/quark/whatever within the universe you would get the universal center. it is a concensus point where everything in the universe takes motion about that axis. anyone with omniscient information would agree with you.
Relative to that location exists universal rests. determined purely on a mathematical basis.
ram1024
Jun26-04, 12:35 PM
No, the reason I think that was because I was applying (Minowski) geometry.
but the path you took to your conclusion was towards SR. you had other options. your calculations did not retain simultaneity at the center of the train, so immediately you jumped to the conclusion that simultaneity was at fault.
Alkatran
Jun26-04, 12:39 PM
SR says, that photon B is emitted before photon A, which means the photons did not exist as a pair for a t > 0. There is no SR postuilates that will surpress the existence of the photons emitted simultaneously in the stationary by virtue of measuring the arrival times on the moving frame. Nor does SR provide for 'gost emitters' that would allow for the delayed emission of photons in the moving frame that have already been emitted. simultabneously in he stationary frame. The mere fact that SR predicts the photons were not emitted simultaneously is proof of the intrinsic error and fault and uselessness of SR.
This would be true if each observer's space wasn't distorted through time (Observer A experiences t=0 BEFORE Observer B, or vice versa!).
As for Case #7, I'll answer AGAIN:
There are 3 possible frames: Picture, Observer, and emitter.
1 - Neither is moving. From any of our 3 frames it takes the photon the same time to reach the observer.
2 - Emitter moving inward. In this case the observer and picture frames are identical. From the picture/observer frame: Light speed is independant of the speed of the emitter so it will take the same time as we observed (1). We'll cover the emitter's perspective later (it is seen in (4))
3 - Same situation as (2) except that the emitter is moving away. The photon still takes the same time as (1) from the observer and picture frames.
4 - Now the observer is moving towards the emitter. Now, this is the EXACT same thing as the emitter's frame in (2). It will take the photon less time to reach the observer because by the time it hits him he has moved closer to the emitter.
5 - Exact same thing as (4) except that this is the emitter's frame in (3).
ram1024
Jun26-04, 12:51 PM
This would be true if each observer's space wasn't distorted through time (Observer A experiences t=0 BEFORE Observer B, or vice versa!).
time is absolute. no one experiences any specific instant of time BEFORE or AFTER anyone else. if you calculate "time" using "light" then of course people receive photons from a source at different "times" depending on where they are when it hits them. light speed is not instantaneous, to measure "time" using something that propagates instead of calculating instantaneous transmission is sheer folly.
think about it this way. do we tell ourselves that light from a star 50 light years away doesn't EXIST until we see it? if this star explodes do we tell everyone "hey today this star exploded". no we do not, we tell everyone 50 years ago this star exploded, we're receiving the evidence of its explosion today.
anyone in that solar system is long dead, they did not die today.
How can two photons emitted simultaneously in one frame, now turn into photons emitted sequentially when observed in a moving frame?
Geistkiesel.. you just don't listen... ram1024 turned out to be no better than you either.. but I will still try to convince you. Consider, please:
[A_______X_______B] ->
____________________________
A,X,B is the train moving to the right according to the embankement.
X shoots two lasers in opposite directions. The lasers hit the A and B at the same time according to the train. From the embankement's perspective, the lasers hit A and B sequentially, because the train moved to the right. Now which conclusion is real?
To test it, let A and B shoot bullets when they detect the lasers. From the train's perspective, the bullets hit X at the same time, because they were fired at the same time.
From the embankement's perspective, the bullets also hit X at the same time, but they were not fired at the same time because the lasers didn't hit them at the same time, nevertheless the bullets hit X at the same time because X moved according to the embankement.
See, it's just the interpretations that change, the events don't change.
edit: If you tend to prefer the embankement's interpretation over the train's , try to put yourself in X's shoes. Don't look out of the windows. You are in a room. What you consider "at the same time" is when the bullets were fired. You see two bullets fired at the same time and they hit you at the same time.
Alkatran
Jun26-04, 01:43 PM
time is absolute. no one experiences any specific instant of time BEFORE or AFTER anyone else. if you calculate "time" using "light" then of course people receive photons from a source at different "times" depending on where they are when it hits them. light speed is not instantaneous, to measure "time" using something that propagates instead of calculating instantaneous transmission is sheer folly.
think about it this way. do we tell ourselves that light from a star 50 light years away doesn't EXIST until we see it? if this star explodes do we tell everyone "hey today this star exploded". no we do not, we tell everyone 50 years ago this star exploded, we're receiving the evidence of its explosion today.
anyone in that solar system is long dead, they did not die today.
TIME IS NOT ABSOLUTE. This is one the consequences of relativity. Of course we're going to disagree on relativity if we disagree on this point. A second for me can be two seconds for you. A second for my leg can be half a second for my arm. Everything is relative!
I refer you to that nice picture I posted earlier.
Alkatran
Jun26-04, 01:56 PM
Geistkiesel.. you just don't listen... ram1024 turned out to be no better than you either.. but I will still try to convince you. Consider, please:
[A_______X_______B] ->
____________________________
A,X,B is the train moving to the right according to the embankement.
X shoots two lasers in opposite directions. The lasers hit the A and B at the same time according to the train. From the embankement's perspective, the lasers hit A and B sequentially, because the train moved to the right. Now which conclusion is real?
To test it, let A and B shoot bullets when they detect the lasers. From the train's perspective, the bullets hit X at the same time, because they were fired at the same time.
From the embankement's perspective, the bullets also hit X at the same time, but they were not fired at the same time because the lasers didn't hit them at the same time, nevertheless the bullets hit X at the same time because X moved according to the embankement.
See, it's just the interpretations that change, the events don't change.
If you don't mind I'd just like to work that out.
(Train frame)
c=lightspeed
v = bulletspeed
w = trainspeed
d = distance(X, A), as well as distance(X, B)
time for bullet #1 to intersect: c/d + v/d
#2: c/d + v/d
(Stationary frame) (ooo oxymoron)
light goes towards right mirror
'(v+w) because the bullet appears to go faster for the observer
#1:(c-w)/d + (v+w)/d
towards left mirror
'(v-w) because now the train is "working against" the bullet.
#2:(c+w)/d + (v-w)/d
Now we check for equality
#1:
(c-w)/d + (v+w)/d = c/d+v/d
(c-w+v+w)/d = (c+v)/d
(c+v)/d = (c+v)/d
#2:
(c+w)/d + (v-w)/d = c/d+v/d
(c+w+v-w)/d = (c+v)/d
(c+v)/d = (c+v)/d
Well, that proves that.
ram1024
Jun26-04, 02:09 PM
what is "that proves that" ?
what were you trying to prove?
geistkiesel
Jun26-04, 03:49 PM
You are the one claiming to have lost a photon. What did you do with it? And where were the photons before you turned on the light? :smile:
Again? Do you even bother to read what I've written in all the threads we've been through? (I know you don't understand it, but at least make an effort to read it.) Read through them again. You'll see that I've calculated exactly where everything is according to everyone. I'm not going to do it again, until YOU give it a try.
Not much physics here is there? SR predicts that the photons will be emitted sequentially in the moving frame. THis measn that one photon cannot exist for that period of time that the B photon preeced3ed the A photon being emitted. But the stationary frame had emitted trwo photons simum,tabneously. If SR is tweaking space and time where did the lost photon go. It is a serious question Doc Al, you can't avoid it with cycnicism forever.
Not sure what you're asking here. The photons just are; they don't belong to a frame. Instead of photons, we'd be better off thinking of pulses of light (containing many photons). These pulses emanate from the light sources and can be detected in either frame. (Obviously, M' detects different photons than M. But don't get hung up on that detail. Consider them as both viewing the same pulses of light.)
You know nothing about physics, poor geistkiesel.
OK think about pulses of light. When the moving observer determines that the B photon was emitted before the A photon this has a physical meaning. It says that the A photon was suppressed from the emission process, according to the moving frame observers, but the stationary frame emitted photons simultaneously. SR is dead.
geistkiesel
Jun26-04, 03:54 PM
if in order to make light constant you need to break other "Reality" laws such as Simultaneity, Absolute time, Absolute space, and True universal rest, then i don't need it.
show me the data. you guys messed up your calculations somewhere down the line
You are 100% on Ram1024. When SR says moving observers see simultaneous emitted photons in the stationary frame turn into sequential emitted photons they are saying that for sime t > 0 thee was only one photon around. Where was the other photon that followed the first photon?
Ram1024 so much will be cleared fro you when you go through Grounded's papers. You are a foll to pass this up.
geistkiesel
Jun26-04, 04:00 PM
Your loss.
You know, there is a group of people who refuse to consider breaking the "Reality" law of a true universal direction of up and down, They are called the "Flat Earth Society".
Actually that last one, 'True universal rest", is funny, since it hasn't been considered a "law of Reality" since the time of Galileo.
So when are you going to include the observer in the motion when measuring the speed of light? If yopu don't you will always get C. You will always get an error. See the Grounded papers.
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it.
geistkiesel
Jun26-04, 04:09 PM
Yah... I started considering the situation earlier when Ram considered synching the clocks then accelerating the train. I could quickly see that it was obvious that the answer depends on how you do it, so I wondered just how you would have to do it to keep the clocks synched in their own frame.
Doing all of the analysis in the stationary frame, if the train is accelerated towards the right, then no matter how the train gets acceelrated, if I consider two points that correspond to the same time in the train's frame, the right point has a greater temporal displacement and a lesser spatial displacement than the left point. Thus, if compute the proper time difference for each of the ends, the right end must have experienced more proper time than the left end.
For my example, if we switch to units where c = 1 (for simplicity), we consider the case where the left edge of the train lies along the worldline:
r(tau) = < cosh tau, sinh tau >
(exercise: check that tau is, indeed, the proper time along the worldline)
The velocity is given by
v(tau) = < sinh tau, cosh tau >
Thus, for the observer on the left edge his line of simultaneity corresponding to his proper time a is parallel to
< cosh a, sinh a >
because this is the unit vector orthogonal to v(a). The line is given, parametrically, by:
l(d) = r(a) + d < cosh a, sinh a>
= <cosh a, sinh a> + d <cosh a, sinh a>
= <(d+1) cosh a, (d+1) sinh a>
where d is the proper distance along the line of simultaneity. (aka the distance measured by the observer on the left edge of the train)
Setting d = 1 (corresonding to the train being constant length) and letting a vary gives us the worldline of the right edge:
r(s) = <2 cosh s, 2 sinh s>
A little calculus shows us that s is twice the proper time for the right edge.
In summary, according to the observer on the left edge of the train, the length of the train remains constant, but the clock on the right edge is running at double speed.
On a side note, the class of worldlines:
r(tau) = <k cosh (tau / k), k sinh (tau / k)>
has the property that each of them remain unchanged by any Lorentz boost that fixes the origin; in particular, these (and their translates) are the worldlines of observers that feel a constant acceleration.
Hurkyl, here is a calculation with an order of magnitude of simplicity less than your own there. In the moving train experiment where the observer is at the stationary midpoint fo the photon sources when the pphotons are emitted in the stationaryframe we have a contradiciton in SR theory.
AS the observer passes from the midpoint she knows her velocity hence she is able to maintain a motorizede detector at the original midpoint. Consider this device mving in the obposite direciton from the train in the train frame and moving at velocity equals 0 wrt stationary frame. When the photons arrive first from B in the front then from A in from the back we have left out an important measurement.
Keeping a continuous watch at the midpoint the constant midpoint observer will detect the photons arriving at the stationary frame. And as the photons arrive simultaneously the moving frame observer measures the photons arriving simultaneously just like the observer would have detected the photons had she not been moving. SR is a game of contrivances.
See how simple this is. The moving observer maintains a watch at the midpioint of A and B which she can do from her moving frame. She must then see the simultanoeus arrival of the photons from her frame.
Tom Mattson
Jun26-04, 04:11 PM
So when are you going to include the observer in the motion when measuring the speed of light?
There is no need to "include the observer", because the observer is at rest in his own frame. All the observer has to do is record the time and place of the emission, and the time and place of detection. Take the ratio of Δx and Δt, and you get c.
If yopu don't you will always get C. You will always get an error.
There is no error. The rate at which a photon approaches me is the distance I measure it to cover divided by the time it took to do it.
See the Grounded papers.
Janus already knows Galilean relativity.
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it.[/QUOTE]
geistkiesel
Jun26-04, 04:12 PM
I wouldn't want to take anything away from Hurkyl's professional rep, but you must admit that you are pretty easy to confuse. :rofl:
This is becasue I make an effort to take you seriously, but you aren't in this for the ohysics, you are in this for the religious conotations.
geistkiesel
Jun26-04, 04:19 PM
Case #7
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
in this setup, we have but one emitter and one observer. keeping it simple-like. In all cases the emitter is going to emit a pulse of light on the first "frame" of the setup. assume uniform motion (no acceleration).
step1: emitter stays the same place towards the observer.
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves towards the observer.
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves away from observer.
This is simply a demonstration of what you're saying that light doesn't care what its source does, right? In all 3 cases light would reach the observer at the same time if the first "frame" were synchronized.
now we're going to do what you guys do to things...
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
we're going to take the same set ups from above and simply CHANGE the relative motion so that the emitters are stationary and the observer is the one that's moving. this shouldn't change ANYTHING as far as you guys see it right? these cases should be EXACTLY the same as the ones above, we just changed perspective....
Discuss.
Ram1024 simply add the motions of the observer to the speed of lite and the time dilations disappear,.
Tom Mattson
Jun26-04, 04:22 PM
if in order to make light constant you need to break other "Reality" laws such as Simultaneity, Absolute time, Absolute space, and True universal rest, then i don't need it.
I see you haven't learned a thing from our discourse last weekend. You still cling to your "rational" view of the world, no matter how untenable it is. You still think that what sounds reasonable to you must be right. You still think that everyday, common-sense notions must be absolute truths, despite the fact that our everyday, common-sense notions are shaped by a very limited set of circumstances (such as "normal" speeds being much, much less than that of light).
show me the data.
I will gather my resources together and post them, but there is no need for you to wait for that. Get up out of your armchair, and look for yourself. You can start by tracking down the following article:
Alvager F.J.M. Farley, J. Kjellman and I Wallin, Physics Letters 12, 260 (1964)
That group measured the speed of light from fast-moving sources, and it comes out to be 'c'. Experimental confirmation doesn't get more direct than that.
you guys messed up your calculations somewhere down the line
Do you really think that those calculations haven't been checked and re-checked? Do you really think that some amateur internet bozo has found something that full-time professional physicists have not? Give me a break. :rolleyes:
If SR is tweaking space and time where did the lost photon go.
It hasn't been created yet.
From these discussions, I think I'm beginning to understand just how big the paradigm shift from "space parametrized by time" to "space-time" was.
geistkiesel
Jun26-04, 05:24 PM
I see you haven't learned a thing from our discourse last weekend. You still cling to your "rational" view of the world, no matter how untenable it is. You still think that what sounds reasonable to you must be right. You still think that everyday, common-sense notions must be absolute truths, despite the fact that our everyday, common-sense notions are shaped by a very limited set of circumstances (such as "normal" speeds being much, much less than that of light).
I will gather my resources together and post them, but there is no need for you to wait for that. Get up out of your armchair, and look for yourself. You can start by tracking down the following article:
Alvager F.J.M. Farley, J. Kjellman and I Wallin, Physics Letters 12, 260 (1964)
That group measured the speed of light from fast-moving sources, and it comes out to be 'c'. Experimental confirmation doesn't get more direct than that.
Do you really think that those calculations haven't been checked and re-checked? Do you really think that some amateur internet bozo has found something that full-time professional physicists have not? Give me a break. :rolleyes:
OK Tom check this calculation. When the observer on the train is at the midpoint of the photon sources in the stationary frame she 0 her clock and notes her constant velocty. She has installed a motorized device that maintains the posiotion of the observer at t0 = 0 = t'0. This motorized detector maintains a constant watch on the detector set to measure thearrival of the photons from A and B. The deivice is moving at -v wrt the observer. AT t'1 she measure the photon from B and at t'2 the photons arrived simultaneously in the stationary frame at the midpoint.. AT t'1 the photon has travelled t'1c in the moving frame. At t'3 she measures the photon from A. She is physically on the train where she was when passing through the midpoint of A and B. As she analyzes later, the photons arriving at the midpoint at t'2 was unexpected as SR had predicted the photons would arrive sequentially. Of course the SR is referring to the position of the original observer who has moved..
Continuing on in this line she may assume reasonably that the photons were the photons emitted in her frame. That the photons met simultaneously at the midpoint of A and B she may assume the A photon was located at a distance -t'1v when the B photon was detected. Therefore the A photon must move a distance c(t'3 - t'1) which is equal to t'1v + t'1v + (t'3 - t'1)v. Collecting t'3 on the left we arrive at t'3 = t'1( c + v)/(c - v) . Using v= 10^-5(c) and setting c = 1, we get t'3 = t'1(1.00001)/.99999 = 1.00002t'1.
Now we substitute c(1.00002 - 1)t'1 = t'1(.00001)c = 2t'1(10^-5 ) + t'1(10^-5). Now as c = 3 10^-5, we have c10^-5 = 3x (10^-5) which is the same distance measured and travelled by the photon. The measurenment of the simultaneous arrival of the photons at the original midpoint could very well have been accomplished by a rigid series of detectors movong with the frame. These will react when the photons arrived. One detector would simultaneously detect the arrival of the photons at the original location of the midpoint of A and B in the moving frame.
Grounded's velocity addition device seems to work very niclely. What does SR predict under the present conditions? v = 10^-5 c, t'1 = 1 and c = 1.
A simple addition of the observers velocity clears up a lot of SR tweaking doesn't it?
You haven't indicated any progress in understanding Grounded's theses.
Tom Mattson
Jun26-04, 05:50 PM
Grounded's velocity addition device seems to work very niclely. What does SR predict under the present conditions? v = 10^-5 c, t'1 = 1 and c = 1.
A simple addition of the observers velocity clears up a lot of SR tweaking doesn't it?
What do you think this pointless exercise proves? You have simply declared your prediction to be right, without making any reference to the real, physical world.
You haven't indicated any progress in understanding Grounded's theses.
Grounded doesn't have "theses". The theory he is espousing is nothing other than Galilean relativity, and I understand it perfectly well. You, on the other hand, have not indicated any progress in differentiating between imaginary thought experiments and real experiments. The condition of confusing the imaginary for the real has a name. It's called "psychosis".
I am thoroughly sick of your thought experiments, and I am finished wasting my time on them. I agree with ram's comments that we aren't referring to real experiments as we should be.
geistkiesel
Jun26-04, 05:52 PM
It hasn't been created yet.
From these discussions, I think I'm beginning to understand just how big the paradigm shift from "space parametrized by time" to "space-time" was.
Hurkyl - put a motorized detector on the moving train to maintain a constant position directly next to the midpoint of the A and B detectors. Let the observer meet the photons later at t'1 and t'3. We use t'2 to be the instant the photons arrived simultaneously in the stationary platform, all witnessed by the fixed 'moving' observer at M. When analyzing the moving observer can determine the effect on the simultanous nature of emitted photons as a function of the observers motion. The motion of the observer changes, not the laws of physics, but the experimental conditions. Where M amd M' were once colocated, there can be no serious question that the photons will arrive at M' at a different time because it has moved. To call this a loss of simultaneity is foolish especially when the moving observer can easily determine the simultaneous nature of the photons without regard to the intrinsic errors in measuring the speed of light without using the velocity of the observer as a factor. Without the added velocities one is forced into the silly time dilation and mass shrinking scenario.
The thing about the stationary moving observer is that the observations are available for analysis in th emoving frame.
Hurkyl this isn't the insanity of one determined to lead a rational life where observations are what they appear to be. This is a rational analysis that just so happens to contradict SR.
Finally, Hurkyl, if the photons were emitted sequentially this means that one of the photons would have had their emission suppressed. This in light of the simultaneous emission of photons in the stationary frame leaves an insurmountable barrier doesn't ir?
Do you understand the concept of a distance versus time plot?
Does this look like a drawing of the experiment you describe?
\begin{picture}(300,240)(0,0)
\put(0,240){\line(1,-3){70}}
\put(70,20){A}
\put(100,240){\line(0,-1){210}}
\put(100,20){M'}
\put(100,240){\line(1,-3){70}}
\put(170,20){M}
\put(200,240){\line(1,-3){70}}
\put(270,20){B}
\put(0,240){\line(1,0){270}}
\put(280,240){x}
\put(0,240){\line(0,-1){210}}
\put(0,20){t}
\put(0,240){\line(1,-1){180}}
\put(200,240){\line(-1,-1){125}}
\end{picture}
geistkiesel
Jun26-04, 06:13 PM
What do you think this pointless exercise proves? You have simply declared your prediction to be right, without making any reference to the real, physical world.
Grounded doesn't have "theses". The theory he is espousing is nothing other than Galilean relativity, and I understand it perfectly well. You, on the other hand, have not indicated any progress in differentiating between imaginary thought experiments and real experiments. The condition of confusing the imaginary for the real has a name. It's called "psychosis".
I am thoroughly sick of your thought experiments, and I am finished wasting my time on them. I agree with ram's comments that we aren't referring to real experiments as we should be.
Are you telling me, or trying to tell me that SR isn't an imaginary discipline? And where do you get Psychosis? I know we refer to each other as crazy, but I would like to know where you arrive at psychosis? Just because I or others disagree with you? You read below and tell me thios is the ranting of a mad man as witnessed by his lack of reasoning in his analytic posture.
Where have you heard Grounded's theses? Who in your career has said we must add the velocities of the observer into the calculations? And who contiued along in the line taken by grounded in calculating frequency abnd wave length? Who said this and where was she dismemebred?
If you understand Grounded very well then why do you have to resort to some vague reference to the Galilean nature of his theses? Your mathematical description a hwile back was impressive, as I noted. Clear, straightforward, careful use of proper parameters in the equatons, all of that which is expected of one who knows his theory, which you obviously do. Tom how many times have you pondered the limitations of mathematics to adequately describe physical activity? Just because the math says you can do it doesn't mean it is physically proper. A train station is stationaary and a train is mobile, moving. To say you can mathematically swap reference frames while being a physical impossibility, then why even broach the matter in analysis? Let me guess because you get the right answer doing it that way?
You mention real experiments and I look them up and see they are flawed with the same erronoeus reasoning that you manifest. It seems you are getting tired Tom,. It must be a chore being the smartest one in the room all the time, with no reasonbale relief in sight.
It would do you well tom to get yourself into Grounded's shoes. Also, you might tell me where the photon that follows the first emitted photon in the moving frame is hiding when the first is emitted,. You do rememebr that the photons were emitted simultaneously in the stationary frame don't you?
The words 'galilean reference frame' is your latest mantra Tom.
geistkiesel
Jun26-04, 06:26 PM
Do you understand the concept of a distance versus time plot?
Does this look like a drawing of the experiment you describe?
\begin{picture}(300,240)(0,0)
\put(0,240){\line(1,-3){70}}
\put(70,20){A}
\put(100,240){\line(0,-1){210}}
\put(100,20){M'}
\put(100,240){\line(1,-3){70}}
\put(170,20){M}
\put(200,240){\line(1,-3){70}}
\put(270,20){B}
\put(0,240){\line(1,0){270}}
\put(280,240){x}
\put(0,240){\line(0,-1){210}}
\put(0,20){t}
\put(0,240){\line(1,-1){180}}
\put(200,240){\line(-1,-1){125}}
\end{picture}
__A________________M______|________|_______B__
| | |
M'(t'0) M'(t'1) M'(t'3)
M'(t'2) B A
Right A M and B are stationary points. M the midpoint of photon sources A and B The terms below the line are in the moving frame. The M' represent the actial location of the observer at the indicated times. As you see thee isn't any reason to define loss of simultaneity as it was done.
M' represents the position of the moving observer at various times. We keep a moving observer at M, the midpoint of the photon sources at A and B. Hw ei is motorinzed and moving at -v with respect to the Observer in the mopving frame. We can consider the M'(t'2) the 0 point of the coordinate system in ythe moviong frame. M'(t'0) is when the observer was at the midpoint in the stationary frame. M'(t'1) is the measurement of the B photon. M'(t'2) the measurement of the simultaneous arrival at M'(t'2) of the emitted photons from A and B at the stationary midpoint, all observed by the M'(t'2) observer in the moving frame. M'(t'3) the A photon is detected by the observer.
Who needs SR to look at this problem?
And where did was missing photon?. The one that was left behind when the B photon was emitted first?
Oh wait, is (A, M, B) supposed to be the "stationary" letters?
Okay, does this diagram look like your experiment?
\begin{picture}(240,240)(0,0)
\put(0,240){\line(0,-1){210}}
\put(0,20){A,t}
\put(100,240){\line(0,-1){210}}
\put(100,20){M}
\put(200,240){\line(0,-1){210}}
\put(200,20){B}
\put(100,240){\line(1,-3){70}}
\put(170,20){M'}
\put(0,240){\line(1,-1){180}}
\put(200,240){\line(-1,-1){125}}
\put(0,240){\line(1,0){210}}
\put(220,230){x}
\end{picture}
geistkiesel
Jun26-04, 06:35 PM
There is no need to "include the observer", because the observer is at rest in his own frame. All the observer has to do is record the time and place of the emission, and the time and place of detection. Take the ratio of Δx and Δt, and you get c.
There is no error. The rate at which a photon approaches me is the distance I measure it to cover divided by the time it took to do it.
Remember the grounded example using his automobiles in motion wrt each other? When adding the observer's velocity the proper length of the automobiles was determined from measurment. Using only the velocity of one of the automobile resultes in a shortening of the car's length. For the same reason you measure a shorter wave length when counting the full wave length increments passing through your eye. Make some calculations.
Janus already knows Galilean relativity.
geistkiesel spouts
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it. Get yourself in a higher state of energy and you measure round, from orbit.
ram1024
Jun26-04, 06:43 PM
http://homepage.sunrise.ch/homepage/schatzer/space-time.html
wow. i found a page that pretty much sums up all of our arguments thus far, and explains my position for "galilean relativity" pretty precisely. (even though i had no idea that's what it was)
it's kind of a long read though, i'll pull some highlight parts that are noteworthy from my side of the fence.
... Briefly, whereas Galilean space-time allows the realization of faster-than-light speeds, at least in principle, Minkowski space-time does not. What is the reason for this difference? In the next sections it is exposed that the key point is the conception of global time, ie. the physical significance of the term simultaneity...
... Because of this absolute time the global notion of past, present and future is the same in all reference frames. If two events are simultaneous in one particular reference frame, this means that they are also simultaneous in all reference frames...
... Although the one-way speed of light is not constant in general (ie. when expressed in an arbitrary reference frame), the mean-speed c of a round-trip is again constant [2], what is in accordance with all experiments (like Michelson-Morley a.s.o.). It should be emphasized again that there has been no experiment which determined the one-way speed of light [3], since this would require the possibility of synchronizing physical clocks by some other means than finite-speed signals...
... Moreover, further arguments can be found which might motivate the reintroduction of (Galilean) absolute time to physics:
First, if there exists a physical absolute time, then the number of fundamental constants is reduced by one, since the (one-way) speed of light is not a constant any longer. This leads to a simplification and a new interpretation of the physical quantities and constants...
...If our universe has a Newtonian background, ie. if there is an absolute time underlying the space-time continuum, then there is no threat on causality by superluminal processes, because time travel and its paradoxes are excluded a priori. And thus, within this framework, faster-than-light travel is possible, at least in principle. ...
Tom Mattson
Jun26-04, 06:46 PM
Remember the grounded example using his automobiles in motion wrt each other? When adding the observer's velocity the proper length of the automobiles was determined from measurment. Using only the velocity of one of the automobile resultes in a shortening of the car's length.
You are still confused on the same point that befuddled Grounded. It's not that we have to add the observer's velocity, it's that we have to take the correct velocity of the car with respect to the observer. Using the ground speed to make a length calculation is incorrect in either Galilean relativity or SR. Everyone who knows physics already knows this, and I have already addressed it.
Length contraction is predicted by SR even when you take the correct relative velocity. Furthermore, the length contraction is not equal to the "missing distance" that Grounded calculated. Even he realizes that now.
For the same reason you measure a shorter wave length when counting the full wave length increments passing through your eye.
No, it's not "for the same reason". The wavelength of light is also predicted to change in SR when the correct relative velocity of the source is used.
Make some calculations.
Make some calculations yourself. I performed these exercises a long time ago, and I have moved on to other things. It's time for you to do the same.
geistkiesel spouts
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it. Get yourself in a higher state of energy and you measure round, from orbit.
WTF???
It should be emphasized again that there has been no experiment which determined the one-way speed of light [3], since this would require the possibility of synchronizing physical clocks by some other means than finite-speed signals...
I learned something today. Assuming http://www.phys.virginia.edu/classes/109N/lectures/spedlite.html is right, the very first measurement of the speed of light was one-way.
First, if there exists a physical absolute time, then the number of fundamental constants is reduced by one, since the (one-way) speed of light is not a constant any longer.
I don't see how this follows.
If our universe has a Newtonian background, ie. if there is an absolute time underlying the space-time continuum, then there is no threat on causality by superluminal processes, because time travel and its paradoxes are excluded a priori. And thus, within this framework, faster-than-light travel is possible, at least in principle.
There's a word for believing something because you want it to be true; gullible.
Tom Mattson
Jun26-04, 07:32 PM
Are you telling me, or trying to tell me that SR isn't an imaginary discipline?
There's not a single line of what I wrote that says anything about that, one way or the other. But since you ask, SR is a theoretical discipline. It just so happens that this particular theoretical discipline makes predictions that are agreed with by experiments that aren't imaginary.
And where do you get Psychosis? I know we refer to each other as crazy, but I would like to know where you arrive at psychosis? Just because I or others disagree with you? You read below and tell me thios is the ranting of a mad man as witnessed by his lack of reasoning in his analytic posture.
I already explained why I used the term "psychosis". It is a result not of disagreeing with me, but of confusing the imaginary for the real.
Where have you heard Grounded's theses? Who in your career has said we must add the velocities of the observer into the calculations? And who contiued along in the line taken by grounded in calculating frequency abnd wave length? Who said this and where was she dismemebred?
Grounded is espousing Galilean relativity. Ask me again, and I'll tell you the same.
If you understand Grounded very well then why do you have to resort to some vague reference to the Galilean nature of his theses?
Your mathematical description a hwile back was impressive, as I noted. Clear, straightforward, careful use of proper parameters in the equatons, all of that which is expected of one who knows his theory, which you obviously do.
You are contradicting yourself right here. If you absorbed my mathematical presentation so well, then why do you think my references to Galilean relativity is vague? I spelled Galilean relativity out very clearly, and when I refer to it, I am referring back to that post.
Tom how many times have you pondered the limitations of mathematics to adequately describe physical activity? Just because the math says you can do it doesn't mean it is physically proper.
No kidding. That's why we do experiments. And guess what? The experiments say that Grounded's ideas are not "physically proper".
A train station is stationaary and a train is mobile, moving. To say you can mathematically swap reference frames while being a physical impossibility, then why even broach the matter in analysis? Let me guess because you get the right answer doing it that way?
You are wrong. The observer on the train is perfectly free to say that he is stationary, and that the station is moving away from him.
You mention real experiments and I look them up and see they are flawed with the same erronoeus reasoning that you manifest.
LOL, name a single experiment published in any reputable journal that you've looked up, and why you think it is flawed. Then, I'll explain to you why you are wrong.
It would do you well tom to get yourself into Grounded's shoes.
Again: I've already done this, in this very thread. I have shown many pages ago why the velocity transformation he uses cannot be right.
Also, you might tell me where the photon that follows the first emitted photon in the moving frame is hiding when the first is emitted,. You do rememebr that the photons were emitted simultaneously in the stationary frame don't you?
You seem to be referring to a discussion in which I have not taken part.
The words 'galilean reference frame' is your latest mantra Tom.
Only because I keep getting the same stupid questions about it.
ram1024
Jun26-04, 07:38 PM
You are wrong. The observer on the train is perfectly free to say that he is stationary, and that the station is moving away from him.
not when light is involved, i have demonstrated this in a past exercise.
because of light's "independent of source" nature, when you change reference frames it creates a fundamental dissimilariy for the situation if perceived as equivalent to the "real" reference frame.
Tom Mattson
Jun26-04, 07:40 PM
not when light is involved, i have demonstrated this in a past exercise.
No. Any inertial observer is free to consider himself at rest. If you "demonstrated" otherwise, then you made a mistake.
ram1024
Jun26-04, 07:41 PM
There's a word for believing something because you want it to be true; gullible.
could say the same thing about your side of the fence with your crazy length contractions, time dialations, and reference based simultaneities.
ram1024
Jun26-04, 07:43 PM
No. Any inertial observer is free to consider himself at rest. If you "demonstrated" otherwise, then you made a mistake.
well since you HAD to bring it up, here's exactly why
Case #7
Step: 1
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
in this setup, we have but one emitter and one observer. keeping it simple-like. In all cases the emitter is going to emit a pulse of light on the first "frame" of the setup. assume uniform motion (no acceleration).
emitter stays the same place towards the observer.
Step: 2
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves towards the observer.
Step: 3
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
emitter moves away from observer.
This is simply a demonstration of what you're saying that light doesn't care what its source does, right? In all 3 cases light would reach the observer at the same time if the first "frame" were synchronized.
now we're going to do what you guys do to things...
Step 4:
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Step 5:
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
we're going to take the same set ups from above and simply CHANGE the relative motion so that the emitters are stationary and the observer is the one that's moving. this shouldn't change ANYTHING as far as you guys see it right? these cases should be EXACTLY the same as the ones above, we just changed perspective....
http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif
Your steps 1, 2, and 3 seem to take a view from A's frame. But steps 4 and 5 take a view from a frame in which A is moving. Of course that frame will measure different times. It should be no surprize to you by now that time measurements are frame dependent.
that is exactly why picture 2 is NOT the same situation as picture 1. :surprise:
Tom Mattson
Jun26-04, 07:43 PM
could say the same thing about your side of the fence with your crazy length contractions, time dialations, and reference based simultaneities.
No, you couldn't say the same thing. We don't believe those things because we want them to be true, we believe them because they are consistent with experimental results. The "crazy" one is the one who refuses to accept reality for what it is.
could say the same thing about your side of the fence with your crazy length contractions, time dialations, and reference based simultaneities.
Maybe I'm forgetting; could you point out where I've stated anything about what I (or anyone on my side of the fence) wish was true? Let alone suggesting that this wish is a good reason to believe SR?
Tom Mattson
Jun26-04, 07:45 PM
that is exactly why picture 2 is NOT the same situation as picture 1. :surprise:
So what? This doesn't prove your assertion about absolute motion or absolute rest. It just proves that the world looks different from the two different perspectives.
To that, I can only say: No kidding! :surprise:
ram1024
Jun26-04, 07:46 PM
now what IS reality is changing the perspective but using the "center of emission" as a relative reference for frame swapping.
any observer motion towards THAT is equal to motion of THAT towards the observer in a true equivalent reference frame swap
Tom Mattson
Jun26-04, 07:48 PM
now what IS reality is changing the perspective but using the "center of emission" as a relative reference for frame swapping.
any observer motion towards THAT is equal to motion of THAT towards the observer in a true equivalent reference frame swap
One more time, in English? :confused:
Once again, the "center of emission according to picture #1" and the "center of emission according to picture #2" are different things. There should be no surprise that they don't act as if they were both the same thing.
ram1024
Jun26-04, 07:53 PM
So what? This doesn't prove your assertion about absolute motion or absolute rest. It just proves that the world looks different from the two different perspectives.
To that, I can only say: No kidding! :surprise:
it doesn't "look different" the physics ARE different.
the first three steps are the observer stationary to the object emitting the light. in each of these steps the amount of time it would take for the photon to reach the observer would be the same.
in the last two "steps" the transition to "observer moving" from "emitter moving" has huge consequences. the time of a photon to reach the destination is very very different based upon the movement.
the reason is because the transition did NOT take into account the relative motion of the observer to "the geometric calculated source location" of the emission. in case 1, 2 and 3, there is no relative motion towards it or away from it, in cases 4 and 5 there is.
they're completely different situations BECAUSE light is NOT tied to its source in REALITY. 4 and 5 would not be "the observer's perspective of the same situation" but instead completely different situations.
Ram, I have already pointed out your mistake. You can't combine steps 4 and 5 as your pictures suggest. My guess is the last frame in step 4 corresponds to first frame in step 5. The distance light has to travel is the same for all steps.
ram1024
Jun26-04, 07:58 PM
http://www.imagedump.com/index.cgi?pick=get&tp=87816
this picture displays the 'centers of emissions' for you, Tom
Ram, I have already pointed out your mistake. You can't combine steps 4 and 5 as your pictures suggest. My guess is the last frame in step 4 corresponds to first frame in step 5. The distance light has to travel is the same for all steps.
i'm not combining 4 and 5, i'm contrasting 4 and 5 to 2 and 3, as you guys have said that simply changing the roles of observer and emitter is "proper" for relative frames,
you can see quite clearly that changing JUST the observer to BE moving changes the distance that light will travel hence change the intercept times.
sorry if i didn't make it quite clear enough that THAT was what i was accomplishing.
ram1024: do you agree that this diagram, a plot of time vs distance, accurately represents picture #1?
(specifically, time increases as you go down, and left-right is spatial position)
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}
\end{picture}
The red lines are the emitters, the yellow and blue lines are the observers, and the green lines are the photons. (Ignore the black line for now)
Tom Mattson
Jun26-04, 08:01 PM
it doesn't "look different" the physics ARE different.
If you reached that conclusion then you made a mistake. One of the postulates of relativity is that the physics is the same for all inertial observers, and this is perfectly consistent with the idea that any inertial observer can regard himself at rest. Just look at the derivation of the Lorentz transforms from the postulates, as applied to Maxwell's equations.
I'm not particulary interested in looking at your trains anymore. If you want to show how the denial of absolute rest leads to the denial of the invariance of physical laws, then show me the math.
ram1024
Jun26-04, 08:04 PM
ram1024: do you agree that this diagram, a plot of time vs distance, accurately represents picture #1?
(specifically, time increases as you go down, and left-right is spatial position)
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(200,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}
\end{picture}
The red lines are the emitters, the yellow and blue lines are the observers, and the green lines are the photons. (Ignore the black line for now)
not really, the yellow and blue line have to start halfway between the first emitter and the midpoint and then the yellow line would angle to intercept the midpoint crossing of the two green lines in the center <if i'm reading this correctly>
ram1024
Jun26-04, 08:05 PM
sorry i mean the blue line would.
well whichever... one of them would ;D
I drew the wrong picture. :frown: it's been fixed; is it right now?
ram1024
Jun26-04, 08:08 PM
I drew the wrong picture. :frown: it's been fixed; is it right now?
lemme try
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(166.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}\end{picture}
Grr, I had the train going the wrong way too, I think... Bah, lemme look at picture #1 again to be sure!
ram1024
Jun26-04, 08:12 PM
\begin{picture}(400,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(110,210){\textcolor{red}{\line(0,-1){210}}}
\put(120,210){\textcolor{red}{\line(0,-1){210}}}
\put(130,210){\textcolor{red}{\line(0,-1){210}}}
\put(140,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){75}}}
\put(136.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(146.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(156.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(176.7,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){210}}
\end{picture}
HARRRR Lines!!!
ram1024
Jun26-04, 08:14 PM
sorry... i'm too easily amused :O
Ok, I think I have the image right.
http://www.physicsforums.com/latex_images/243709-0.png
That's the link to it, just to save on server space. :smile:
We have the two red sources are stationary, as is the yellow observer at their midpoint. The two sources emit green photons simultaneously. The blue train observer happens to arrive at the yellow observer just as green photons from both directions do as well.
P.S. good thing you haven't learned "multiput" yet. :devil:
Ok, now to start getting to the point of drawing this diagram.
Rather than thinking of the universe as being some spatial "thing" that only exists "now", we can think of the universe as consisting not only of now, but the past (and the future?).
The thing I'm getting at is that instead of thinking of the animation in picture #1 which shows the universe at a given instant of time, and show how the universe changes as time changes, we can think in terms of the diagram I draw which shows the whole of space-time all at once.
In this representation of the universe, we can recover our classical notion of "now"; a point in time corresponds to a horizontal line on this diagram, and everything occupying that line is what we, classically, think of the universe looking like at that time.
Agree?
ram1024
Jun26-04, 08:41 PM
if the moving picture stops 1/2 way through that diagram where the green lines cross then yeh i'd say that looks pretty accurate. :smile:
ram1024
Jun26-04, 08:42 PM
horizontal cut = now
vertical cut = where
yesh... proceed
[url]
this picture displays the 'centers of emissions' for you, Tom
i'm not combining 4 and 5, i'm contrasting 4 and 5 to 2 and 3, as you guys have said that simply changing the roles of observer and emitter is "proper" for relative frames,
you can see quite clearly that changing JUST the observer to BE moving changes the distance that light will travel hence change the intercept times.
sorry if i didn't make it quite clear enough that THAT was what i was accomplishing.
OK, according to observer (o), there are two different emitters (a) and (b)
(o)----------(a)->
(o)--------<-(b)
You draw step 4 and 5 like this
(o)-----------(a)
(o)------------(a)
(o)-------------(a)
(o)-----------(b)
(o)----------(b)
(o)---------(b)
It gives the illusion that you can combine them and light will have to travel different distances:
But you can't combine them. You should draw from a's perspective like this:
(o)-----------(a)
(o)------------(a)
(o)-------------(a)
(o)-----------(b)
(o)----------(b)
(o)---------(b)
and from b's perspective like this:
(o)-----------(a)
(o)------------(a)
(o)-------------(a)
(o)-----------(b)
(o)----------(b)
(o)---------(b)
So the distance light travels is the same for a and b.
ram1024
Jun26-04, 08:53 PM
no idea what you're getting at since i never combined those two :biggrin:
no idea what you're getting at since i never combined those two :biggrin:
then why do you say "changing JUST the observer to BE moving changes the distance that light will travel hence change the intercept times." ?
ram1024
Jun26-04, 08:59 PM
Case #7
Step: 1
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
======================================
Step: 2
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Step 4:
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
===============================
Step: 3
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
Step 5:
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
i move them around so you can do the comparison easier.
ram1024
Jun26-04, 09:03 PM
for each step, pulse a light from the emitter at line1 and assume it hits the observer on line 4.
it's pretty easy to see that the simple changing of "who's moving" doesn't make for an equivalent picture. in 2 and 3, no relative distance is made <progress> of the observer towards the location of the light emission source (not the emitter, but where the emitter WAS in line1 when it pulsed)
in 4 and 5 relative motion IS noted between the light emission source and the observer(this time since the emitter source is stationary, the light emission source occupies the same location)
Case #7
i move them around so you can do the comparison easier.
Yes, the distance is measured by observer and emitters differently, and so is the elapsed time, due to length contraction and time dilation. So what? I will provide you numerical examples to make you understand what value everyone will measure. Wait..
\begin{picture}(500,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){180}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){360}}
\put(325, 135){\circle{4}}
\end{picture}
Ok, I've extended the diagram a little so I can talk about some additional things.
Once we permit ourselves to think about the universe in terms of a space-time, we can ask a "what if" question...
We know from geometry that all lines are the same. Sometimes we pick two perpendicular lines to be axes, but we can always rotate and pick different axes. So...
What if we allow lines that aren't horizontal to be "when" and lines that aren't vertical to be "where"?
Nonvertical lines reprsenting "where" have been around a long time; the blue line represents a "where"... the train. Other lines parallel to the blue line also represent "where"; places at a fixed position relative to the train.
The thing that makes SR different than Gallilean relativity is that SR also permits diagonal lines to represent "when". The black line I've drawn on the diagram is a line that, according to SR, represents "when" in the train's frame.
Now, before you jump on this and call it stupid, consider this: even if you think it's stupid to let a diagonal line represent "when", we can identify this black line in reality. We could define a collection of lines parallel to the black line, and define something by these lines.
So, whether or not you consider this a measurement of time, it is certainly a measurement of something physically meaningful in reality.
In this diagram, I hope you can see how SR can say that in the train's frame that the emissions aren't simultaneous. If we're using these diagonal lines to measure "time", instead of horizontal lines like we're "supposed to", then it's clear that this different way of measuring "time" will not measure the two emission events as occuring at the same time. In particular, if we're using these diagonal lines to measure time, then the emitter on the right will fire before the emitter on the left!
I extended the picture so I could also talk about the "center of emission". Again, if I'm using these diagonal lines to define "when", then the points where the right photons intersect the black line correspond to the same "time". The little dot I drew is the midpoint between these two intersections. If I'm using the black line to define "when", then that dot should be the "center of emission", because it is the point midway between the photons.
Again, this diagram shows how we can be referring to the same reality, but disagree on the "center of emission".
Each frame of animation #2 can be generated by drawing one of these diagonal lines. The black line I drew corresponds to the frame in animation #2 when the left emitter fires. Notice that, along this black line, the right photon is much closer to the yellow observer than the left photon is, the train is located at the midpoint of the photons, and the right center of emission is located a little to the right of the right source. If you look at animation #2, the frame where the left emitter fires will conform precisely to this description. (including your green fuzzy dot!)
ram1024
Jun26-04, 09:16 PM
no, there is no distance measured differently... light CLEARLY travels farther if you switch the frames like this.
in the emitter reference we have the emitter moving away from the "light emission source" which has NO impact on when the observer will receive the photon.
in the observer reference we have the observer moving away from the "light emission source" which has TOTAL impact on when the observer will receive the photon.
not "time gets skewed" because the observer is moving but "it's not even the same situation"
ram1024
Jun26-04, 09:21 PM
\begin{picture}(500,240)(0,0)
\put(100,210){\textcolor{red}{\line(0,-1){210}}}
\put(300,210){\textcolor{red}{\line(0,-1){210}}}
\put(100,210){\textcolor{green}{\line(-1,-1){75}}}
\put(100,210){\textcolor{green}{\line(1,-1){180}}}
\put(300,210){\textcolor{green}{\line(-1,-1){180}}}
\put(300,210){\textcolor{green}{\line(1,-1){180}}}
\put(200,210){\textcolor{yellow}{\line(0,-1){210}}}
\put(166.7,210){\textcolor{blue}{\line(1,-3){70}}}
\put(100,210){\line(3,-1){360}}
\put(325, 135){\circle{4}}
\end{picture}
i don't see how the black line can be the train's "where" because it never gets hit by the light from emitter(L)...
explain the black line to me once more :(
ram1024
Jun26-04, 09:25 PM
ah okay nevermind you said it's a "when" line
ram1024
Jun26-04, 10:07 PM
it was informative, but it doesn't tell me much except that non-simultaneity for that frame is a direct consequence of light speed measured constant for the observer.
allow for a moment light to be subjected to normal physics in the sense that the emitters impart whatever velocity they have to the light when it is emitted, as would be if light were a normal mass attribute object.
here is my version of events allowing inertial velocity to be added to the light sources (making the light travel with the emitters) in order to preserve a true reference frame according to the stationary train as an observer.
http://www.imagedump.com/index.cgi?pick=get&tp=89150
<changed the background, white hurts my eyes>
[edit] explaining that a little bit for you, the white line is the skewed "when" line for the now "stationary" train. i shifted the train back over to the original location, since it "doesn't move". i allowed the emitters to carry their inertia from movement over to the photons so that they're moving along with added velocity of the source (if you check my earlier case #7 you'll see why this is necessary to ensure constancy). the red lines are the moving emitters. the "dotted" red lines are theoretical emitter locations if they were not moving.
Informing was my goal. I wasn't trying to convince anyone of anything; I was mainly trying to clear up how, according to SR, the two different frames could be corresponding to the same "reality" despite the disagreement over things like simultaneity and centers of emission.
I just recall that, when I was learning this stuff, I never really "got" SR until I tried understanding it from these diagrams instead of from wordy descriptions and algebra. Thought it might be helpful while we wait for Tom to dig up the meaty stuff. :smile:
kurious
Jun27-04, 04:31 AM
Today special relativity dies.....
Kurious:
It will die in space-time...
ram1024
Jun27-04, 06:04 AM
my diagram makes sense right? i've never messed with this space-time stuff, it was good of jscd<sp> to point out how useful it could be for explaining stuff. :smile:
in any case, i hope you can see that on my diagram <allowing for relative motion of the observer towards the "light emission centers"> his space/time is skewed but he still calculates simultaneous "when" emissions on his skewed line
i think grounded has the mathematics for such a transform already in place, he's just missing the "light emission centers" bit in there which according to how we'd plotted it would be a simple geometric rotation of the observer's spacetime "worldline"
ram1024
Jun28-04, 04:23 AM
woops my bad, jdavel suggested space-time diagrams
no, there is no distance measured differently... light CLEARLY travels farther if you switch the frames like this.
in the emitter reference we have the emitter moving away from the "light emission source" which has NO impact on when the observer will receive the photon.
in the observer reference we have the observer moving away from the "light emission source" which has TOTAL impact on when the observer will receive the photon.
not "time gets skewed" because the observer is moving but "it's not even the same situation"
Ram, I now think "time gets skewed" is the resolution to your apparent paradox.
I found that other link again. Please take a look:
http://casa.colorado.edu/~ajsh/sr/paradox.html
Note that vermilion and cerulean do not agree on where the centre of emission is. So the centre of emission is relative too.
In your example, no one will agree on times and distances due to SR effects, that's normal. But, what confused me was: time dilation and length contraction don't depend on direction of speed. So they don't explain the difference between 2/4 and 3/5. But, relative simultaneity does depend on direction, and that is why there is a skew in time, not a shift. So, I think the resolution to your paradox is the skew. I hope you can work it out with the spacetime diagrams.
ram1024
Jun28-04, 07:10 PM
In your example, no one will agree on times and distances due to SR effects, that's normal. But, what confused me was: time dilation and length contraction don't depend on direction of speed. So they don't explain the difference between 2/4 and 3/5. But, relative simultaneity does depend on direction, and that is why there is a skew in time, not a shift. So, I think the resolution to your paradox is the skew. I hope you can work it out with the spacetime diagrams.
see the main thing i see as "the problem" is that light itself does NOT obey physical laws of the universe, so when you put it in a position where the frames get switch you cannot expect light to behave the same.
case in point the picture i submitted most recently which is a spacetime diagram of the train stationary, but the light sources moving AND the photon light emission centers being tied to the emitters.
in this diagram it is clearly obvious <by taking "whens" on that diagonal <the train's view> you can see that even according to the train, emission IS simultaneous.
but BECAUSE light is not tied to its source <not the case with other physical mass bearing things in the universe> if you shift the frames and don't compensate for this, it's not the same situation.
i had the same problem when i was prooving .999~ doesn't equal 1. people were functioning under the limit that infinity was an absolute limit, then got all confused when numbers at the limit didn't behave the same way as numbers anywhere else on the line. remove the limit and everything still works the same.
same thing here, compensate for your limitations and everything still works fine. no time dialation, no length contraction, no simultaneity nonsense.
no relative to the viewer light speed, but i do believe that's solely an error based on how they're calculating their measurments. we'll see when the data arrives.
Alkatran
Jun28-04, 08:35 PM
But .999~ does equal 1?
see the main thing i see as "the problem" is that light itself does NOT obey physical laws of the universe, so when you put it in a position where the frames get switch you cannot expect light to behave the same.
I wouldn't say light does not obey physical laws of the universe. It's a special case because of its zero rest mass. Then again you might say light is not physical but it does interact with mass.
case in point the picture i submitted most recently ........
I don't have the time to decipher your diagram so I can't comment..
same thing here, compensate for your limitations and everything still works fine. no time dialation, no length contraction, no simultaneity nonsense.
But everything doesn't work fine with Galilean relativity. GPS for example. MMX null result. And of course experiments in particle accelerators. What about muon decay? Muons live longer and travel longer when they are moving fast. You can't ignore all these.
no relative to the viewer light speed, but i do believe that's solely an error based on how they're calculating their measurments. we'll see when the data arrives.
Are we expecting new data? What are the errors? Do you have a consistent theory to correct the errors?
ram1024
Jun28-04, 10:08 PM
But everything doesn't work fine with Galilean relativity. GPS for example. MMX null result. And of course experiments in particle accelerators. What about muon decay? Muons live longer and travel longer when they are moving fast. You can't ignore all these.
without data i can only offer conjecture on these topics.
my thoughts are as follows
GPS still works because of triangulation. it doesn't work PERFECTLY, but then again it's all consistent so if they're running off the same system and all give the same error who's to know it's actually an error? being peak technology for that kind of thing the only way you could check it would be to use cruder technology
MMX null result is of no consequence because i don't believe in "aether" anyways.
does a muon decay at the same rate in space as it does on earth, subjected to atmospheric pressure, gravity, and element interaction? not enough data to conclude.
so without that kind of data, i can't give you much
Alkatran
Jun28-04, 10:30 PM
GPS still works because of triangulation. it doesn't work PERFECTLY, but then again it's all consistent so if they're running off the same system and all give the same error who's to know it's actually an error? being peak technology for that kind of thing the only way you could check it would be to use cruder technology
I think they would notice if the GPS satellites made an error. Even if they all made the same error people would notice when it said they were 2 km from where they actually were.
ram1024
Jun28-04, 10:32 PM
well... the only way they'd be able to detect it is if you had other GPS satellites going the OTHER direction (orbit) and i don't think they do
ahrkron
Jun28-04, 10:51 PM
In a sense, they already do. The GPS satellites orbit Earth in a nice symmetric pattern. Here's a representation:
http://www.gisillinois.org/gps/GPSDEF/sat.htm
If there was a systematic shift, how much it would affect you would depend on where and when you took the reading. Sometimes it would point one direction, and sometimes the opposite, depending of where the visible satellites were with respect to you at the time of the evaluation.
It would be quite chaotic, which it isn't.
ram1024
Jun28-04, 11:31 PM
i went and looked up alot of references on GPS and relativity, and apparantly there's still alot of controversy as to what is going on up there.
apparantly regardless of how they correct the system they still get errors, so the whole thing has to be continually recalibrated.
they're saying that how things operate up there is MORE consistant with Lorentz Relativity than Special Relativity.
but General relativity seems to be holding for gravity effects <shrug>
now i gotta go figure out what the difference is between Lorentz Relativity and Galilean/Einstein relativities :O
Ram, I think your skepticism is fine. But one thing seems certain: there is no going back to Galilean relativity. Current data fits into Einstein's relativity much better.
they're saying that how things operate up there is MORE consistant with Lorentz Relativity than Special Relativity.
If I'm not mistaken, SR includes LET as a subset but LET includes an aether which is just another frame in SR, if LET is what you mean by Lorentz relativity.
ram1024
Jun29-04, 09:28 AM
still trying to figure out what it means myself... :D
russ_watters
Jun29-04, 09:34 AM
i had the same problem when i was prooving .999~ doesn't equal 1. Oy - not only is every physicist wrong, but every mathematician is wrong too? Ram, you have a problem that we can't help you with here.apparantly regardless of how they correct the system they still get errors, so the whole thing has to be continually recalibrated. Every system has errors. GPS clocks are quite good (easily good enough that they notice time dilation), but they still need to be synchronized periodically.
One little tidbit: GPS satellite clocks are calibrated according to the predictions of GR/SR and as such, when on the ground, they don't keep accurate time. Once in orbit, they do keep accurate time. What does that tell you?
In any case, it is clear from your statements that you don't know how GPS works, nor do you know its implications for SR/GR. You also don't understand the significance of the Lorentz transformaions an their relationship to SR. You should choose to learn these things instead of just assuming that you are right, choosing to believe that despite your admitted ignorance you know something that thousands of scientists don't, and choosing not to look at the mountain of evidence that says otherwise. Maybe thats not a choice you are capable of making. I don't know.
ram1024
Jun29-04, 10:02 AM
i told you i was investigating into it
i told you i was waiting on more conclusive data to make real conclusions
maybe you need to calm down :D
Oy - not only is every physicist wrong, but every mathematician is wrong too? Ram, you have a problem that we can't help you with here.
an expected response from someone indoctrined to believe...
ram1024
Jun29-04, 10:57 AM
so i opened up 3dsMax and started messing with spheres to try and figure out geometrically how GPS would work.
with one sphere and the earth, you can't really tell very much, basically the only information gleaned is that you're on the part of the earth within direct line of sight with the satellite.
with two spheres, the intersection is a circle, more importantly where it intersects the earth would be a line.
with 3 spheres you get a line, more importantly where that line intersects the earth would be a point.
i'm assuming the fourth satellite <sphere> is for discerning altitude (but not really as it is a point on the line and the line might be coming in non vertical)
in each case time would be determined for each satellite, probably encoded into the pulse data. they run the time data from the satellites to generate 4 expanding spheres, the point where they all intersect gets the "you are here"
i think what that guy said about the fourth satellite being used for SR correction is false. just by messing with the sphere myself it doesn't seem to be the case.
ram1024
Jun29-04, 11:02 AM
just so you don't mistake me i'm not saying they DON'T use SR correction, just that it's most likely all on the clock/time side
ram1024
Jun29-04, 06:39 PM
meh... i've scared Tom, Hurkyl, and Doc Al away.... wespe you're all i have left!
hold me...
meh... i've scared Tom, Hurkyl, and Doc Al away.... wespe you're all i have left!
hold me...
We love you too Ram
But, is there an argument left to discuss? (Not case 7. The center of emission is relative so it's not a paradox)
Your last posts indicate you are trying to figure out how GPS works. That's not an argument and there are no questions, so you shouldn't be expecting a reply.
russ_watters
Jun29-04, 11:21 PM
so i opened up 3dsMax and started messing with spheres to try and figure out geometrically how GPS would work.....
i think what that guy said about the fourth satellite being used for SR correction is false. just by messing with the sphere myself it doesn't seem to be the case. You're operating based on a preconcieved notion of how you think it might work, which is (unsurprisingly) wrong.
When I said you should "choose to learn" these things, maybe I wasn't specific enough. What I meant was that you should choose to learn how these things actually work, not how you think they should work in your nonexistent universe.
Read up on how GPS calculates a position. Read up on how the clocks are synchronized. Read up on what the implications are for Relativity.
Or you could simply admit that you will refuse to learn real science and end this. That is likely the reason why the others have abandoned these discussions.
ram1024
Jun30-04, 02:28 AM
you're talking as if you're absolutely positively sure that i have no idea what i'm talking about.
do YOU know exactly how GPS works? If NOT then you can't be sure i'm not correct. If SO, it would have been easier to post where i went wrong rather than post a useless paragraph condemning my learning habits.
be part of the solution not part of the problem, MmmKay?
Your last posts indicate you are trying to figure out how GPS works. That's not an argument and there are no questions, so you shouldn't be expecting a reply.
yeh i was waiting for Hurkyl to interpret my spacetime diagram, and Tom to come back with data.
i think they both went on vacation though, i haven't seen them post in other threads either.
As for Doc Al, i think i wore him out :(
the GPS is a good distraction while i wait for their return
Alkatran
Jun30-04, 06:11 AM
Does the fact that you have been arguing your case, and we have been arguing ours as well, for 10 days, and yet no progress has been made? It looks to me like one of us is either refusing to accept a notion, or that ... well actually there's just the one option.
I thought I didn't really have much to say on this thread.
Anyways, http://math.ucr.edu/home/baez/RelWWW/wrong.html is a great page, and it has a section on GPS vs relativity.
ram1024
Jun30-04, 11:47 AM
wonderfully informative page.
even figured out where i went wrong with my spheres. without knowledge of the sphere's locations relative to each other (satellites) you would need 4 to determine the locations of intersections, but they do a precalculation and send it down as part of the signal. i thought the signal would be simple and the calculations would be done on the receiver end. but they're asserting that the signal carries 9 layers of information and the receivers actually do very little of their own computation. in this sense they only need 3 for triangulation, and the 4th supplies the receiver with the time of the earth based GPS clock.
good stuff
ram1024
Jun30-04, 12:19 PM
i need you to look over my spacetime diagram Hurkyl :D
yeh i was waiting for Hurkyl to interpret my spacetime diagram, and Tom to come back with data.
i think they both went on vacation though, i haven't seen them post in other threads either.
As for Doc Al, i think i wore him out :(
the GPS is a good distraction while i wait for their return
oh then that was the expected data.. I didn't understand what you meant.
but russ_watters is right IMO, you could learn more efficiently if you did more search and reading, or else you will wear people out
thanks for the link Hurkyl
ram1024
Jun30-04, 03:16 PM
heh, you guys are under the false impression that i do no reading and investigating on my own :D
i do quite alot, but as far as stuff online it's quite hard to judge facts from fiction most of the time, which is why before i go into researching i lay out my own "how i think it would work" scenario as a guidestone, and then work the research data into my theory as i progress, using each new piece to compare and contrast to find out how stuff actually works and where i went wrong <if anywhere>
a lot of times i will get conflicting data from sites so generally i tend to go with the ones that DO take the commonly accepted theories "point of view" as being the truth.
if i got bogged down with contradictions i'd never make it anywhere.
and i am truly sorry if i wear you guys out. :(
russ_watters
Jun30-04, 03:51 PM
you're talking as if you're absolutely positively sure that i have no idea what i'm talking about. It would appear (with a gentle nudge from Hurkyl) that you answered your own question. Why couldn't you have just read up on it like I (we) suggested instead of guessing and getting combative when you guessed wrong? Its simpler, faster, and less painful for all of us....I lay out my own "how i think it would work" scenario as a guidestone, and then work the research data into my theory as i progress, using each new piece to compare and contrast to find out how stuff actually works and where i went wrong <if anywhere>. And this is why you are having such a hard time learning these things. Its human nature to not want to be wrong and so I understand your frustration. You are trying to learn by a process that ensures that for you to learn something you must first make a guess and be wrong (and accept that you were wrong). On this forum it manifests itself as you making guesses, us trying to help you, and you responding with the natural human instinct of getting defensive/combative. Why put yourself through that? Just learn the right way the first time. Don't guess.
Yes, can be tough to find credible info on the web. We can help. But if you don't trust us, its not hard to figure out on your own what is credible and what isn't.
ram1024
Jun30-04, 04:07 PM
the process of me guessing helps me to understand the ramifications of why things are the way they are more intuitively.
this is the way that i learn best. <shrug>
take case in point the GPS system. they're doing it with 4 satellites in view, using 3 for triangulation and the 4th to match GPS local time for the earth based station.
my system would use 4 satellites but we wouldn't need to know the positions of the satellites or local earth time, as 4 expanding spheres with calibrated time only intersect in 2 places, one would be your position on earth and the other would be somewhere way off in space.
i don't know if that would make computations any easier or harder, but it is a system that would work, fashioned entirely from my brain in a way that i personally can understand it.
relaying that information in a way that you or anyone else could understand it is indeed a bit more difficult which is why it takes 25 pages for me to get my point across to you guys.
in the end if you guys are simply trying to "teach me" and not also "learn what the heck i'm talking about" then there's a failure on your end as well. because from my standpoint i have taken everything "taught" to me thus far and incorporated it into my "self theory" as retained knowledge. you have simply clung to what you believe and not made an attempt to "learn" what i might have to offer.
...
relaying that information in a way that you or anyone else could understand it is indeed a bit more difficult which is why it takes 25 pages for me to get my point across to you guys.
in the end if you guys are simply trying to "teach me" and not also "learn what the heck i'm talking about" then there's a failure on your end as well. because from my standpoint i have taken everything "taught" to me thus far and incorporated it into my "self theory" as retained knowledge. you have simply clung to what you believe and not made an attempt to "learn" what i might have to offer.
Ram, I did learn something from your case#7 (maybe others already knew, but I didn't): that the center of emission is not a space buoy. So, thank you for that. But.. the other cases were a waste of time, due to a lack of knowledge on your side. That's the part that could be eliminated. I didn't say you read nothing, I just said more reading would be better. Sorry if that offended you.
ram1024
Jun30-04, 05:38 PM
the center of emission HAS to be a space buoy. if it's moving then what's its speed and direction?
the whole "light is not connected to its source" means it either A> has to be stationary absolutely or B> has to move, but relative to what in your guys SR case, in Galilean it would move with the imparted velocity from the source. which is where a lot of confusion comes into play i'm sure. still need the data to verify my postulates.
i will say again, if light is independant of its source then the space buoy HAS to work.
the center of emission HAS to be a space buoy.
If the center of emission is a space buoy, then speed of light can't be the same for both emitter and receiver.
if it's moving then what's its speed and direction?
it's relative to frames and it's not a real object.
ram1024
Jun30-04, 07:35 PM
If the center of emission is a space buoy, then speed of light can't be the same for both emitter and receiver.
it's relative to frames and it's not a real object.
so in the moving observer frame, the center of emission is stationary
in the stationary observer frame the center of emission is moving.
that would be consistant right?
but it's not so according to the way you guys are calculating it. you have it stationary in the moving frame and stationary in the stationary frame.
you say it's not a real thing, but it has to be real enough in such that light emitted from that point will spread "at light speed" in all directions. that is a given.
let's analyze that further.
(o) <-)|
(o) <-)|
we're putting two observers next to each other. a flash of light is emitted at the same time from emitters same distance from them. <this can be the same emitter so don't worry about synchronicity>
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
one of the observers moves towards the emitters. it's guaranteed he gets hit before the other observer due to less distance light has to travel. correct?
now we swap frames
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
or
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
or
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
these are the three relative frames i managed to come up with.
the first defining stationary as the point between O1 and O2, having each other move away at same speed. the emitters have to close some of the distance.
the second being O2 stationary, so O1 must move away and the emitters must close the distances
the third being O1 stationary and O2 moving. this is exactly the same case as outlined in the first example.
let's define a length that light travels at a certain time to be light's speed in this scenario.
here is that length
(o) length <-)|
(o) <-)|
apply that length to the scenarios outlined above as being a constant speed.
you'll see that this length segment coresponds to a space-time such that in the original case allows both observers to be hit by the photon.
in case 2 and 3 the first observer can NOT be hit by his photon during this allotted space-time.
there is only ONE allowable case at constant speed of light that makes it so that length of space time encompasses both observers and that is case 4, which is identical to case 1. so the only answer from an SR standpoint can be that case 2 and 4 are incorrect as reference frames. case 1 is what actually happens.
russ_watters
Jun30-04, 08:51 PM
my system would use 4 satellites but we wouldn't need to know the positions of the satellites or local earth time, as 4 expanding spheres with calibrated time only intersect in 2 places, one would be your position on earth and the other would be somewhere way off in space.
i don't know if that would make computations any easier or harder, but it is a system that would work, fashioned entirely from my brain in a way that i personally can understand it. If your system doesn't take Relativity into consideration, it might work in your universe, but it won't work in ours. in the end if you guys are simply trying to "teach me" and not also "learn what the heck i'm talking about" then there's a failure on your end as well. because from my standpoint i have taken everything "taught" to me thus far and incorporated it into my "self theory" as retained knowledge. you have simply clung to what you believe and not made an attempt to "learn" what i might have to offer.This process certainly does help us to learn (thats part of the reason I'm here), but its learning by finding your errors. You get us to think about these subjects, pick apart your ideas, and find the flaws.
Alkatran
Jun30-04, 08:54 PM
Alright, I'll solve these with my ultimate solvination program: paint.
Just a minute.
Alkatran
Jun30-04, 09:15 PM
time increases upwards
distance between obs and emit = 1m
listspeed = 1m/s for simplicity (if you want I can just make it more than 1 meter, or say it's the meter #2)
I can't believe the imit is 400x400 I had to crunch up my pic and mess everything up!
Anyways this is what would happen.
*edit* it was 402*398 ... fixed now
The "rel.jpg" is just the proff that your buoy won't work
ram1024
Jun30-04, 09:15 PM
omg i'm painting TOO!!!
we're of like mind, Alka :D
ram1024
Jun30-04, 09:16 PM
attachments never get approved. better host it somewhere....
OK, here's what I could come up with (someone correct me if I'm mistaken)
Suppose the second emitter was comoving with the second observer.
The emitters emit photons when they meet.
This is o1 frame:
(o1)----------------------------------------<-)|
This is o2 frame:
(o2)----------------------------------------<-)|
But o1 sees the o2 frame length contracted:
(o2)----------------------------------<-)|
Also o2 sees the o1 frame length contracted:
(o1)----------------------------------<-)|
According to o1 this is what happened:
observers meet first:
(o1)----------------------------------------<-)|
(o2)----------------------------------<-)|
(o1)----------------------------------------<-)|
(o2)----------------------------------<-)|
(o1)----------------------------------------<-)|
(o2)----------------------------------<-)|
then emitters meet:
(o1)----------------------------------------<-)|
(o2)----------------------------------<-)|
later o2 detects photon. later o1 detects photon
(o1)----------------------------------------<-)|
(o2)----------------------------------<-)|
According to o2 this is what happened:
emitters meet first:
(o1)----------------------------------<-)|
(o2)----------------------------------------<-)|
(o1)----------------------------------<-)|
(o2)----------------------------------------<-)|
(o1)----------------------------------<-)|
(o2)----------------------------------------<-)|
then observers meet:
(o1)----------------------------------<-)|
(o2)----------------------------------------<-)|
later o2 detects photon. later o1 detects photon
------------------------------<-)|
(o2)----------------------------------<-)|
So, o2 detects photon before o1 in both frames, but they won't agree on times and distances and simultaneity.
does this make sense?
edit: Your mistake is assuming the photons were emitted at the same time for both observers, which is the time the two observers were at the same location. That would violate relative simultaneity/time. For the moving observer, the photons were emitted some time ago before the observers were at the same location (because of length contraction as I showed above).
Alkatran
Jun30-04, 09:36 PM
Alright! I fixed it.
The green and blue lines are the two different observers.
Blue is the stationary (relative to emitter) observer and green is the moving
The top frame is blue's frame and the bottom is green's
The thick lines are the opposite observer's frame
(you see what they mean by skewed?)
ram1024
Jun30-04, 10:14 PM
i said already, don't use attachment, they don't get approved. we'll never be able to see them
host it on imagedump or something :|
Alkatran
Jun30-04, 10:43 PM
You're kidding me? You can't see any attachments until they're approved? Even pictures?! (Can you even put damaging stuff in jpeg's? I mean that will be run)
Anyways, I modified the picture a bit, pretty much the same except the thick lines are boxes now.
Your case as seen from obs1 (above) and obs 2 (below) (http://www.vbforums.com/attachment.php?s=&postid=1726678)
Red Dot: The buoy
Blue Dot: photon heading towards buoy from emitter
Green Dot: photon heading towards buoy from emitter
Why your space buoy won't work (http://www.vbforums.com/attachment.php?s=&postid=1726680)
I'd also like to point out that this picture is "impossible" because it's drawn as if the buoy were moving relative to it and yet light doesn't move normally (for no speed). It's as if we are observing a frame from a a different frame.
ram1024
Jun30-04, 11:44 PM
have NO idea what i'm looking at in your first diagram.
in the second one it just seems to prove that it SHOULD work. in order for the lights to simultaneously meet in the center the light from blue travels at 1m/s towards red which travels at 1m/s towards it (1+1=2) and light from green travels at 3m/s but red travels away from it at 1m/s so (3-1=2) light is travelling at 2m/s <lightspeed> relative to the observer <red>
but wait a minute. that means light was really <to no observer> travelling at 3 in one instance and 1 in another instance... what gives?
just to reiterate, the space-buoy does NOT work, but that is not why it does NOT work :D
Alkatran
Jun30-04, 11:54 PM
Do you notice how the light intersects at the buoy (this is a spacetime diagram) and is always equi-distant from it on both sides? That's why ti doesn't work.
The first is also a space time diagram. First ignore the large square (dark blue/dark green) and look. The black line (going straight up in the first, since it is unmoving) is the emitter, red is light (same in both), green is one observer and blue is another.
The dark squares are the other frames. (rotate your head to look at them like that and they look similar to the other diagram, but it is not an accurate representation)
the space buoy picture as you have drawn it is not how SR sees the events.
if it was then it would prove the inconsistancy of light speed, 3 m/s from the left and 1m/s from the right.
SR asserts that the left light would be emitted first, then the right light emitted later, so that they both meet when red dot reaches 1. but even doing so, red has to be moving away from green at SOME speed so light from the direction of green will always be travelling at "greater than light speed" in ANY case where motion of the apparatus is assumed.
SR elites have yet to attack my space-buoy because they see the paradox that occurs when the theory is applied to it.
gotta goto work today :( laters!
Ram, your pictures are still based on absolute time and space. They give the illusion that everything is happening in parallel in both frames. If you could examine #385, I have shown this is not so.
Specifically, your bad assumption is:
we're putting two observers next to each other. a flash of light is emitted at the same time from emitters same distance from them. <this can be the same emitter so don't worry about synchronicity>
Even if there is only one emitter, you still have to worry about length contraction. The two observers don't measure the distance the same. Therefore light cannot be emitted at the same time for both observers.
Alkatran
Jul1-04, 07:33 AM
Did you even read my post explaining that we're watching one frame from another? If we were watching the non-moving frame, THAT'S when there woul dbe loss of simultanity because it IS simultaneous in the moving frame. geez
In fact, for any speed of the buoy you get the same thing in the buoy frame: The light is emitted at the same time and received at the same time. All you'll prove is the buoy is at rest relative to itself (duh).
In fact, for any speed of the buoy you get the same thing in the buoy frame: The light is emitted at the same time and received at the same time. All you'll prove is the buoy is at rest relative to itself (duh).
how so? in your very own picture you have light approaching from the rear at 3m/s <measured> and from the front at 1m/s <measured>
put a stationary observer at the final interception point. light from the left covers a REAL distance of 3 to get to him from the left and a REAL distance of 1 to get to him from the right. there is no length contraction craziness, these would be real measurements to the stationary viewer. both of these observers would be hit by the same photons at the same time <occupying the same space> so there can be no discrepancies as far as where and when the lights were emitted.
Even if there is only one emitter, you still have to worry about length contraction. The two observers don't measure the distance the same. Therefore light cannot be emitted at the same time for both observers.
length contraction is bunk. working on a picture :yuck:
Alkatran
Jul1-04, 10:26 PM
Ugh, did you even comprehend what I wrote? Here, picture this: The same picture, except the buoy doesn't move (since we are in its frame!) and light approaches from both sides at the same speed, intercepts at the same speed.
There, that's how the buoy sees it, it won't work.
length contraction is bunk.
This is not an argument.
working on a picture :yuck:
Have fun.
Ugh, did you even comprehend what I wrote? Here, picture this: The same picture, except the buoy doesn't move (since we are in its frame!) and light approaches from both sides at the same speed, intercepts at the same speed.
There, that's how the buoy sees it, it won't work.
are you saying that's how the buoy sees it. no matter what?
because picture 2 buoys. one stationary and one moving away relative to the other.supposeing we use the same measurements as the other instance 2m/s lightspeed and the second buoy is moving away at 1m/s. they buoys start at the same place and fire at the same time.
determine the the reception times for photons of the second buoy to hit buoy 1 and vice versa.
geistkiesel
Jul2-04, 06:08 AM
Ram, your pictures are still based on absolute time and space. They give the illusion that everything is happening in parallel in both frames. If you could examine #385, I have shown this is not so.
Specifically, your bad assumption is:
Even if there is only one emitter, you still have to worry about length contraction. The two observers don't measure the distance the same. Therefore light cannot be emitted at the same time for both observers.
Length contraction comes directly from simultaneity concepts that form the basis of relativity and light propagation. In this sense simultaneity precedes Sr. You have no loss of simultaneity, you have no SR, In the case where two photons are emitted simultaneously from A and B and meet at the midpoint M (of A and B) which has deflecting mirrors (\/) that immediately deflect both photons into the passing moving frame there is no loss of simultaneity. Einsteins definition the "events simultaneous with respect to the stationary frame are not simultaneous with respect to the moving frame." must be discarded as contrary to physical law and experimental results..
M
A -->\/<--B
||
||
########## Detectors in the moving frame --> motion
A'M'B'
We assume the photons move only a few wave lengths before being detected in the moving frame. Einstein's example which says: The observers ". . .must, therfore, come to the conclusion that . . ." the photons were not emitted simultaneouisly in the moving frame. This is because the B photon was detected before th A photon caught up from the rear of the frame in AE's original gedunken, which is obviously flawed. It is a difficult point to get accrioss, but one ought not take dogma so seriously. This definition was described by AE as a "natural definition" that operated to discard absolute space and all the rest.
How are the photons not going to be observed emitted simultaneously in the moving frame? Except by operation of the definition?
Alkatran
Jul2-04, 08:17 AM
are you saying that's how the buoy sees it. no matter what?
because picture 2 buoys. one stationary and one moving away relative to the other.supposeing we use the same measurements as the other instance 2m/s lightspeed and the second buoy is moving away at 1m/s. they buoys start at the same place and fire at the same time.
determine the the reception times for photons of the second buoy to hit buoy 1 and vice versa.
What's the difference?? You're still only going to be finding the speed of Buoy #2 relative to #1 or vice versa! :mad:
After all these hundreds of posts, it seems like very little progress has been made. :smile:
so in the moving observer frame, the center of emission is stationary
in the stationary observer frame the center of emission is moving.
that would be consistant right?
Wrong. The "center of emission" is stationary in every frame. This is a direct consequence of the invariant speed of light.
but it's not so according to the way you guys are calculating it. you have it stationary in the moving frame and stationary in the stationary frame.
That's correct.
you say it's not a real thing, but it has to be real enough in such that light emitted from that point will spread "at light speed" in all directions. that is a given.
The center of emission is not a "thing". It is just the place where the source was (with respect to an observer) when the light was flashed. The source may move on (depending upon what frame we are describing) but the center of emission never does: it is always fixed in the observer's frame.
let's analyze that further.
(o) <-)|
(o) <-)|
we're putting two observers next to each other. a flash of light is emitted at the same time from emitters same distance from them. <this can be the same emitter so don't worry about synchronicity>
I presume the source is in the "stationary" frame (the O1 frame). Again, saying that the flash is emitted "at the same time" is ambiguous. You certainly realize by now that only the O1 frame would observe the flash being emitted at the same time as the two observers pass each other. O2 would disagree. Simultaneity is key to understanding how to relate the observations made in different frames (along with length contraction and time dilation).
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
(o) <-)|
one of the observers moves towards the emitters. it's guaranteed he gets hit before the other observer due to less distance light has to travel. correct?
Both observers agree that the light hits O2 before it hits O1.
now we swap frames
OK, now I presume those diagrams are your attempt to illustrate the same events from three different frames. But your diagrams ignore length contraction. And since they don't show the center of emission, or where the pulse is at the moment of observation, or the extent of length contraction, it is difficult to draw any meaningful conclusions.
You pick three frames: O1, O2, and a third in which observers O1 and O2 are moving apart (at the same speed I presume?). Let's call that third observer O3. OK, so what?
these are the three relative frames i managed to come up with.
the first defining stationary as the point between O1 and O2, having each other move away at same speed. the emitters have to close some of the distance.
OK, this is the one I call O3. If the emitter moves with O1, then of course it will move in the O3 frame. So what?
the second being O2 stationary, so O1 must move away and the emitters must close the distances
Yeah, yeah. What do you mean "emitter must close the distance"? It's moving! More interesting would be to show (1) the location of the center of emission (2) the location of the light pulse at this moment (3) the length contraction of the O1 - emitter distance that a moving observer would find.
the third being O1 stationary and O2 moving. this is exactly the same case as outlined in the first example.
OK. And the point is?
let's define a length that light travels at a certain time to be light's speed in this scenario.
Huh? I hope you realize by now that light travels at the speed c in every frame!
Your ensuing discussion of length (or something) I cannot decipher, but you manage to conclude that some frames don't work with SR. Well, do it over! All inertial frames are equally valid in SR.
correct me if i'm wrong in saying the following.
SR believes that aether travels the speed and direction of the observer.
SR believes that light travels through Aether at approximately 300,000 km/s (true value defined by Maxwell Equation)
What's the difference?? You're still only going to be finding the speed of Buoy #2 relative to #1 or vice versa! :mad:
|(< (o) >)|
|(< (o) >)|
|(< (o) >)|
|(< (o) >)|
|(< (o) >)|
|(< (o) >)|
|(< (o) >)|
|(< (o) >)|
you're afraid to do it, i don't blame you. it makes you confused doesn't it?
if photons are emitted simultaneously for the first frame and hit both observers at the same time <according to what you've told me> then a photon has travelled both one distance and another completely different distance AT THE SAME TIME.
do photons have a probabilistic existance?
Alkatran
Jul2-04, 10:11 PM
if photons are emitted simultaneously for the first frame and hit both observers at the same time <according to what you've told me> then a photon has travelled both one distance and another completely different distance AT THE SAME TIME.
Listen VERY CAREFULLY: IF the photons are emited simultaneously for the first frame they are NOT for the other!
Your problems don't confuse me at all, they are all the same thing drawn slightly differently. Stop refusing to learn the basics and pick up a book.
Listen VERY CAREFULLY: IF the photons are emited simultaneously for the first frame they are NOT for the other!
Your problems don't confuse me at all, they are all the same thing drawn slightly differently. Stop refusing to learn the basics and pick up a book.
what are you saying, that makes no sense.
Step 1: photons emitted simultaneously
|(< (o) >)|
|(< (o) >)|
Step 2: motion
|(< (o) >)|
|(< (o) >)|
Step 3: more motion
|(< (o) >)|
|(< (o) >)|
Step 4: final step <interception>
|(< (o) >)|
|(< (o) >)|
Light from Observer 2's LEFT Emitter would have had to have travelled MORE distance in the SAME amount of time to meet observer 2 in the center of HIS moving frame this is compared to light from Observer1's LEFT emitter which would have travelled to the center of observer 1's STATIONARY frame.
yet in both cases light was EMITTED from the same spot at the SAME TIME.
light travelling 2 different distances at the same interval = NOT constant.
Alkatran
Jul2-04, 11:20 PM
How is this any different than any other of your cases? Here, I'll draw ANOTHER picture... sigh...
Lemme make it easier for you. instead of 4 emitters we have 2 explosions, one on each side.
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
light is independant of source right? it doesn't matter if i don't move the explosions with the second observer.
this should yield exactly the same results right? observer 1 and observer 2 both witness the photons simultaneously, right?
Alkatran
Jul2-04, 11:49 PM
Look at this:
Now, HERE'S WHAT's HAPPENING:
Top Graph: The observer who isn't moving in the picture.
Bottom graph: Second observer, duh
Now, it looks like the graphs contradict each other, until you remember that, gasp*, they don't have the same time, or distance.
http://www.vbforums.com/attachment.php?s=&postid=1728412
Michael F. Dmitriyev
Jul3-04, 12:08 AM
Your discussion follows beyond relativity, guys.
The light is excluded from this theory. Its speed is absolute initially.
The theory of relativity examines movement of objects concerning light but not on the contrary.
Relativity can be applied only for objects having mass, because the mass is a source of this phenomenon.
Michael
Look at this:
Now, HERE'S WHAT's HAPPENING:
Top Graph: The observer who isn't moving in the picture.
Bottom graph: Second observer, duh
Now, it looks like the graphs contradict each other, until you remember that, gasp*, they don't have the same time, or distance.
http://www.vbforums.com/attachment.php?s=&postid=1728412
now put a stationary observer at the point where 2 intercepts light simultaneously from HIS 2 sources.
Observer 3 is co-located at that point where Observer 2 gets hit simultaneously.
Observer 3 MUST get hit by the photons simultaneously as well.
Observer 3 is NOT centered between the 2 "emitters" during the time of emission, yet BECAUSE of observer 2 he DOES get hit simultaneously by light from 2 differing distances?
Ooooh it's like MAGIC. just by having someone LOOK at the light and running BACKWARDS really FAST we can make it intercept objects between us and the light SOONER!
i think you guys are smoking and not sharing...
Alkatran
Jul3-04, 01:23 AM
now put a stationary observer at the point where 2 intercepts light simultaneously from HIS 2 sources.
Observer 3 is co-located at that point where Observer 2 gets hit simultaneously.
Observer 3 MUST get hit by the photons simultaneously as well.
Observer 3 is NOT centered between the 2 "emitters" during the time of emission, yet BECAUSE of observer 2 he DOES get hit simultaneously by light from 2 differing distances?
Ooooh it's like MAGIC. just by having someone LOOK at the light and running BACKWARDS really FAST we can make it intercept objects between us and the light SOONER!
i think you guys are smoking and not sharing...
You know what? I'm done in this thread. It's obvious you don't want to learn.
As for someone running backwards and having light intercept faster: That's BECAUSE HE REACHES THAT POINT IN SPACETIME SOONER. There, happy? Oh wait, you aren't going to read and understand a thing I wrote.
you're the one who isn't understanding the consequences of your own theories.
you refuse to admit that light travelling the same relative speed for any observer causes light to travel different distances for all observers, (Even stationary ones!).
1. Observer 1 stationary in the center getting hit simultaneously
2. Observer 2 moving and STILL getting hit simultaneously
3. Observer 3 stationary and co-located to the position that observer 2 is hit simultaneously, getting hit simultaneously.
this leads to the obvious and clear conclusion, that it doesn't matter where you are between two light sources, they will hit you at the same time. all you need is someone from the middle to travel to your location?
You know what? I'm done in this thread. It's obvious you don't want to learn.
You don't even understand your own theory, how do you expect to teach anyone?
Lemme make it easier for you. instead of 4 emitters we have 2 explosions, one on each side.
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
¤¤¤¤¤ (o) ¤¤¤¤¤
Your 4 emitters are all fixed with respect to O1. (So you really only need 2; the extra 2 add nothing.) Do those explosions occur simultaneously? According to who? O1 or O2? They don't agree.
light is independant of source right? it doesn't matter if i don't move the explosions with the second observer.
Yes, light speed is independent of the source. But what matters is when the explosions occur. If they are simultaneous in the O1 frame, they cannot be simultaneous in the O2 frame.
this should yield exactly the same results right? observer 1 and observer 2 both witness the photons simultaneously, right?
Wrong. Obviously O1 sees O2 traveling towards the oncoming light: so O1 knows that O2 will detect the oncoming light before he does. Both O1 and O2 agree that O2 detects the light before O1.
Alkatran
Jul3-04, 11:27 AM
Yes, I'll have to apologize for my faulty picture. I was confused for a minute that the emition was simultaneous in both frames, which it can't be. :redface:
Maybe this will get the point accross:
http://www.geocities.com/janus58.geo/simultaneous.html
Michael F. Dmitriyev
Jul3-04, 02:27 PM
Your dispute about not existing thing.
The relativity of speed of light is absurdity.
HallsofIvy
Jul3-04, 02:35 PM
"relativity of speed of light "??? The whole point of relativity is that the speed of light is the one thing that is not relative!
Michael F. Dmitriyev
Jul3-04, 03:51 PM
"relativity of speed of light "??? The whole point of relativity is that the speed of light is the one thing that is not relative!
I have said more definitely:
- It is absurdity.
So?
Alkatran
Jul3-04, 04:24 PM
I like Michael's argument style:
It's right 'cause I said so!
Your 4 emitters are all fixed with respect to O1. (So you really only need 2; the extra 2 add nothing.) Do those explosions occur simultaneously? According to who? O1 or O2? They don't agree.
fine, O1 then. but doesn't O2 AGREE that O1 receives photons simultaneously?
Wrong. Obviously O1 sees O2 traveling towards the oncoming light: so O1 knows that O2 will detect the oncoming light before he does. Both O1 and O2 agree that O2 detects the light before O1.
Good. so i set my clocks when i'm "dead in space" so that emitted photons will hit me simultaneously. then whatever relative motion i make, light will NOT hit me simultaneously because i would be moving to intercept it, correct?
Alkatran
Jul3-04, 11:31 PM
fine, O1 then. but doesn't O2 AGREE that O1 receives photons simultaneously?
Good. so i set my clocks when i'm "dead in space" so that emitted photons will hit me simultaneously. then whatever relative motion i make, light will NOT hit me simultaneously because i would be moving to intercept it, correct?
Yes, he will agree he received them simultaneously. But only because he perceived the photons as being emitted non-simultaneously. All events at the same position and time are simultaneous to all observers. If there's a difference in position or time they won't be (for all)
light is independant of source right? it doesn't matter if i don't move the explosions with the second observer.
this should yield exactly the same results right? observer 1 and observer 2 both witness the photons simultaneously, right?
This reveals you still haven't understood a single thing. Otherwise you wouldn't make this wrong guess.
If the explosions occured simultaneously in the "moving" frame, observer2 would see them at the same time.
Observer2 didn't see them at the same time, therefore the explosions didn't occur simultaneously in that frame.
Logically, if a then b, if not b then not a.
What problem is there? Is it that the explosions did occur simultaneously in the "stationary" frame? But Observer1 did see them at the same time. You just have to understand that simultaneity is relative, instead of coming up with different examples for the same thing.
geistkiesel
Jul4-04, 05:15 AM
This reveals you still haven't understood a single thing. Otherwise you wouldn't make this wrong guess.
If the explosions occured simultaneously in the "moving" frame, observer2 would see them at the same time.
Observer2 didn't see them at the same time, therefore the explosions didn't occur simultaneously in that frame.
Logically, if a then b, if not b then not a.
What problem is there? Is it that the explosions did occur simultaneously in the "stationary" frame? But Observer1 did see them at the same time. You just have to understand that simultaneity is relative, instead of coming up with different examples for the same thing.
Simultaneity is just a word. How do you prove simultaneity without just repeating the mantra that it comes from special relativity. How do you prove simultaneity? Surely it is more than sheer belief, is it?
geistkiesel
Jul4-04, 05:28 AM
Your discussion follows beyond relativity, guys.
The light is excluded from this theory. Its speed is absolute initially.
The theory of relativity examines movement of objects concerning light but not on the contrary.
Relativity can be applied only for objects having mass, because the mass is a source of this phenomenon.
Michael
Michael, How do you prove simultaneity, briefly? This is what they are discussing isn't it?
Does Einstein's stationary - platform/moving - train prove it? or is this just a definition?
Thanx, Geistkiesel.
geistkiesel
Jul4-04, 05:33 AM
You know what? I'm done in this thread. It's obvious you don't want to learn.
As for someone running backwards and having light intercept faster: That's BECAUSE HE REACHES THAT POINT IN SPACETIME SOONER. There, happy? Oh wait, you aren't going to read and understand a thing I wrote.
It is all to do with simultaneity right? Other than merely state that it is SR, how do you prove simultaneity? Briefly. Is Einstein's definition enough?
I am asking around about this question,
thanx, geistkiesel.
PS how do you create and post your attachments?
Thanx again.
Simultaneity is just a word. How do you prove simultaneity without just repeating the mantra that it comes from special relativity. How do you prove simultaneity? Surely it is more than sheer belief, is it?
Einstein's method to determine simultaneity is the most logical one possible. Consider two events separated by a distance. How else can you prove that they happened at the same time? Be realistic; you can't see or know everything instantly (like in the thought experiments). Your best option is to assign a midpoint observer to be the judge, and use light signals. If this will make more sense, think of the midpoint observer actually watching both events at the same time (using mirrors). If he sees the events occur at the same time, the events are simultaneous.
I know, of course, you will now say that if the midpoint observer is moving, he can't be the judge. You know, of course, I will say that he considers himself stationary and also the emission points of light stationary. Therefore the only requirement is that the observer was at the midpoint when the events occured and his relative speed to something else doesn't matter. Then, of course, you won't agree, just like the hundred times you were told this before. OK, but please discuss properly, don't start with your unintelligible essays again please. Quote what you don't agree with and provide your reasoning.
geistkiesel
Jul4-04, 07:42 AM
There is no need to "include the observer", because the observer is at rest in his own frame. All the observer has to do is record the time and place of the emission, and the time and place of detection. Take the ratio of Δx and Δt, and you get c.
There is no error. The rate at which a photon approaches me is the distance I measure it to cover divided by the time it took to do it.
Excuse the late rehash, but I've been thinking about this.
Yes but it approaching you means you areapproaching it. If from the instant the Obsever was at M until measuring B at t1v, the photon has travelled ct1. You are eliminating the relative velocity of the observer, naturally you will calculate C. See below for a corected vesion of your statement. :smile:
When the obsever on the moving frame (MF) detects the oncming B photon ahe notes the time in her frame. T1 is measured from zero when she crossed hrough the midpoint of A and B in the stationary Frame (SF). Simlarly T2 measues he time the A photon arrives from behind. Using the fact that the photon from A at T1 is equidistant to M and must cover c(T2 -T1) space during the train moving from T1v to T2v or v(T2 -T1), which is T1v + T1v + (T2 - T1)v we arive at T2 = T1(C + v)/(C - v). :smile:
You must mean that the distance the train moved to the B detection spot was T1v, during which time the light moved from B to T1C.
The place of emission of the B photon to the right of M is L - T1v = T1C and then L = t1(c + V), right? And from the rear, from symmetry considerations, the distance the photon travel from T 1 to T2, when detected, is, C(T2 - T1) = L + t1v + v(T2 - T1), or L = C(T2 - T1) -vT1 - v(T2 - T1), or T2 = T1(C + v)/(c - v).
The velocity of C then in terms of v, T1 and T2 is simply
c= v(t1 + T2)/( T2 - T1) :smile:
Another way to say all this:
Maintain a series of photon detectors along the train and measure when the A and B photons arrive simultaneously at M. What ever that time happens to be it is a singularly measured time. If the photon detector in the SF and the MF are separated by one wavelength then the photons are measured (after reflecting thrug a distance of one wavelength) as simultaneously being emitted from the SF into the MF, correct? We find then that the relative velocity C is just = v(t1 + T2)?t2 - T1).
Janus already knows Galilean relativity.
Maybe Janus should be nstructed in the art of including the osbervers velocity, speed.
quoted by geistkiesel in a snit of impatience..
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it.
Tom,
Take a look at my post to "Has anyone seen the lost simultaneity?" #55. It is much more complete.
geistkiesel
Jul4-04, 08:24 AM
Oh wait, is (A, M, B) supposed to be the "stationary" letters?
Okay, does this diagram look like your experiment?
\begin{picture}(240,240)(0,0)
\put(0,240){\line(0,-1){210}}
\put(0,20){A,t}
\put(100,240){\line(0,-1){210}}
\put(100,20){M}
\put(200,240){\line(0,-1){210}}
\put(200,20){B}
\put(100,240){\line(1,-3){70}}
\put(170,20){M'}
\put(0,240){\line(1,-1){180}}
\put(200,240){\line(-1,-1){125}}
\put(0,240){\line(1,0){210}}
\put(220,230){x}
\end{picture}
Hurkyl, I may have overlooked answering you. Yes the drawing here is correct. First the B photon is detected in the moving Frame then the A detected This appears to be your drawing. As this condition is really a test of simultaneity we needn't introduce any SR at this point correct? It will, must, show up in the analysis. (See Has anyone seen the lost simultaneity post # 55).
During he time the photon travelled to its detection point it travelled cT1 while the train travelled vt1, During the time the A photon travelled to the detection point is v(t2 - t1). From synmmetry the A photon was at -t1v when the B photon was detected. Shortening the algebra we get that, for t1 = 1, C = 1, and v = .1,
t2 = t1(C + V)/(C -V) = 19, or that C = V(t2 + t1)/(t2 - t1) = 1.
Who needs SR to get the right answer?
Don't tell me you are going to bootsrap a reply using time dilation etc. You should be able, you must be able, to get it from this analysis. The biggest problem in SR is the very long reach of assuming that because the different imes that the observer measured the arrival of the A and B photons, you jump to a loss of simultabneity, discarding absolute time, insering time dilation and mass shrinking.
let us put an observer that measures the simultaneous arrival of the A and B photons at M (arriving into reflecting mirrors reflecting to M and M' at the same time). Just put your slanted line for the world events located at M and M = M'. The spatial difference in detectors is one wavelength of the photons. Are you still going to insist that the moving frame observer will detect the photon arriving on the moving frame at different times?
Or said another way,
Are events simultaneous in the stationary frame not simultaneous in the moving frame? Where the events, of course, is the simultaneous emission of photons in the stationary frme at A and B. :smile:
geistkiesel
Jul4-04, 08:29 AM
It would appear (with a gentle nudge from Hurkyl) that you answered your own question. Why couldn't you have just read up on it like I (we) suggested instead of guessing and getting combative when you guessed wrong? Its simpler, faster, and less painful for all of us. And this is why you are having such a hard time learning these things. Its human nature to not want to be wrong and so I understand your frustration. You are trying to learn by a process that ensures that for you to learn something you must first make a guess and be wrong (and accept that you were wrong). On this forum it manifests itself as you making guesses, us trying to help you, and you responding with the natural human instinct of getting defensive/combative. Why put yourself through that? Just learn the right way the first time. Don't guess.
Yes, can be tough to find credible info on the web. We can help. But if you don't trust us, its not hard to figure out on your own what is credible and what isn't.
Russ_watters, you have to be one obnoxious snob, talking like ou do in yopur trust you? Is this your argument or are you the reincarnation of Ptolemy, are you? If so, please excuse the 'snob' reference, I can understand your personal position.
geistkiesel
Jul4-04, 08:51 AM
There is no need to "include the observer", because the observer is at rest in his own frame. All the observer has to do is record the time and place of the emission, and the time and place of detection. Take the ratio of Δx and Δt, and you get c.
There is no error. The rate at which a photon approaches me is the distance I measure it to cover divided by the time it took to do it.
Excuse the late rehash, but I've been thinking about this.
Yes but it approaching you means you areapproaching it. If from the instant the Obsever was at M until measuring B at t1v, the photon has travelled ct1. You are eliminating the relative velocity of the observer, naturally you will calculate C. See below for a corected vesion of your statement. :smile:
When the obsever on the moving frame (MF) detects the oncming B photon ahe notes the time in her frame. T1 is measured from zero when she crossed hrough the midpoint of A and B in the stationary Frame (SF). Simlarly T2 measues he time the A photon arrives from behind. Using the fact that the photon from A at T1 is equidistant to M and must cover c(T2 -T1) space during the train moving from T1v to T2v or v(T2 -T1), which is T1v + T1v + (T2 - T1)v we arive at T2 = T1(C + v)/(C - v). :smile:
You must mean that the distance the train moved to the B detection spot was T1v, during which time the light moved from B to T1C.
The place of emission of the B photon to the right of M is L - T1v = T1C and then L = t1(c + V), right? And from the rear, from symmetry considerations, the distance the photon travel from T 1 to T2, when detected, is, C(T2 - T1) = L + t1v + v(T2 - T1), or L = C(T2 - T1) -vT1 - v(T2 - T1), or T2 = T1(C + v)/(c - v).
The velocity of C then in terms of v, T1 and T2 is simply
c= v(t1 + T2)/( T2 - T1) :smile:
Another way to say all this:
Maintain a series of photon detectors along the train and measure when the A and B photons arrive simultaneously at M. What ever that time happens to be it is a singularly measured time. If the photon detector in the SF and the MF are separated by one wavelength then the photons are measured (after reflecting thrug a distance of one wavelength) as simultaneously being emitted from the SF into the MF, correct? We find then that the relative velocity C is just = v(t1 + T2)?t2 - T1).
Janus already knows Galilean relativity.
Maybe Janus should be nstructed in the art of including the osbervers velocity, speed.
quoted by geistkiesel in a snit of impatience..
OK smart guy, measure the flat of the planet on the surface,. You will never be able to measure round. The ships coming to port over the horizon are measurements at a distance. Within limits the planet is measured flat man. Get used to it.
Tom,
Take a look at my post to "Has anyone seen the lost simultaneity?" #55. It is much more complete.
geistkiesel
Jul4-04, 09:04 AM
Einstein's method to determine simultaneity is the most logical one possible. Consider two events separated by a distance. How else can you prove that they happened at the same time? Be realistic; you can't see or know everything instantly (like in the thought experiments). Your best option is to assign a midpoint observer to be the judge, and use light signals. If this will make more sense, think of the midpoint observer actually watching both events at the same time (using mirrors). If he sees the events occur at the same time, the events are simultaneous.
I know, of course, you will now say that if the midpoint observer is moving, he can't be the judge. You know, of course, I will say that he considers himself stationary and also the emission points of light stationary. Therefore the only requirement is that the observer was at the midpoint when the events occured and his relative speed to something else doesn't matter. Then, of course, you won't agree, just like the hundred times you were told this before. OK, but please discuss properly, don't start with your unintelligible essays again please. Quote what you don't agree with and provide your reasoning.
You took the words right out of my typing fingers. See the post #55 in "Has anyonme seen the lost simultaneity?'
I put mirrors at M, the midpoint, in the stationary frame, reflecting into the moving train just as the detecters in the moving frame pass the midpoint M when the photons arrive simultaneously from A and B sources. The photons are immediately directed into the moving frame and immediately detected after a distance travelled of one wavelength of the photon light.
To say it gently: "The photons emitted simultaneously in the stationary frame are simultaneously emitted in the moving frame."
I demand that the midpoint observer be the judge, absolutely.
For a detailed solution to your conundrum wespe check out my #55. and try to keep your mind open. Unintelligble? posts? only to those unable to reason past their robotic limits.
I always recognized that the B photon is detected before the A photon, I have just been arguing that the naive conclusions of Einstein, and yourself for instance, are jsut that, naive.
Einstein's method to determine simultaneity is the most logical one possible. Consider two events separated by a distance. How else can you prove that they happened at the same time? Be realistic; you can't see or know everything instantly (like in the thought experiments). Your best option is to assign a midpoint observer to be the judge, and use light signals. If this will make more sense, think of the midpoint observer actually watching both events at the same time (using mirrors). If he sees the events occur at the same time, the events are simultaneous
but he's not a midpoint observer if he moves relative to the sources. you can't possibly tell me that an observer who moves to intercept photons AT the midpoint can call himself stationary and believe he was always at the midpoint the whole time. measure the distances, it's that simple
:confused:
geistkiesel
Jul4-04, 01:54 PM
but he's not a midpoint observer if he moves relative to the sources. you can't possibly tell me that an observer who moves to intercept photons AT the midpoint can call himself stationary and believe he was always at the midpoint the whole time. measure the distances, it's that simple
:confused:
Ram. In einstein's train the single observer O' is the focus of the argument, only!!.
What about the passengers that are sitting adjacent to the midpoint just as the photons from A and B arrive there, after the intercepion of the B photon by O' but before O' detects the A photon? For these passengers the photons are detected at the very instant the photons arrive at the midpoint. Hence, these passengers do observe the simultaneous arrival of the photons in the movng frame.
But according to SR all the other passengers "must. therefore" conclude something different. Something for everybody I suppose.
I wonder, did the passenegrs on Einstein's train argue this out among themselves?
but he's not a midpoint observer if he moves relative to the sources. you can't possibly tell me that an observer who moves to intercept photons AT the midpoint
He doesn't move to intercept the photons at the midpoint of the stationary frame. He remains at the midpoint in his frame. He also doesn't see the events occuring at the same time, therefore concludes they did not occur simultaneously in his frame.
can call himself stationary and believe he was always at the midpoint the whole time. measure the distances, it's that simple :confused:
In the moving frame, he is stationary and always at the midpoint of the two events. Explained below.
First, don't think of the moving frame consisting only the observer. Extend it so that it covers the locations the events occured. It's a train with an observer in it, sufficiently long to include both events.
Don't think of the events belonging to the stationary frame. The events also occur in the train frame. That's why lightenings were used in the gedanken. They occur in an instant and they can leave burning marks in both frames.
The midpoint observers in both frames are always the same distance from the marks in their frame. That's how we can say the observer in the moving frame is and always was at the midpoint of the two events. That's why both can be the judge in determining simultaneity in their frame.
Of course, the burning marks in the two frames don't remain co-located after the lightenings strike (and both were never co-located at the same time in both frames, but that will be the conclusion). Since the lightenings don't belong to either frame, and speed of light is independent of its source, the relative motion between the frames is irrelevant to the events.
that's ridiculous and wrong.
the observer did NOT start out in the middle of the two events, he moved to BE at the midpoint WHEN the light hit.
<talking about the train gedunken experiment portrayed in the animated gifs>
even in HIS frame he was not in the midpoint of the two emitters. you're basically telling me that via length contraction NOTHING in the universe moves EXCEPT light
<talking about the train gedunken experiment portrayed in the animated gifs>
Post #425 which you replied in #431 was not talking about animated gif. Do you even care to read anything others write?
pallidin
Jul4-04, 05:05 PM
This is a great discussion, but I have a problem with it.
"Thought experiments" are of great potential value to be sure, but often entail the concept of "perfect" circumstances.
More often than not, these "perfect" circumstances exclude or inappropriately minimize important influences associated with reality; often rendering experimental validation impossible, so what are we left with?
In all fairness, thought experiments should be considered a valued commodity in the scientific community, as most foundations of development began with this.
if i was replying to that post i would have quoted it. i am replying to your most recent post, of course
:yuck:
if i was replying to that post i would have quoted it. i am replying to your most recent post, of course
:yuck:
Assuming this was a reply to post #435 (why don't you press the quote button?)
Here's the history:
#422 by Geistkiesel (asks "how do you prove simultaneity?")
#425 by Wespe (talks about Einstein's gedanken to prove simultaneity)
#431 by Ram1024 (asks something about an animated gif without mentioning it)
#433 by Wespe (replies thinking of Einstein's gedanken)
#434 by Ram1024 (says it's wrong according to the animated gif)
#435 by Wespe (says he wasn't talking about animated gif)
#437 by Ram1024 (quoted above)
doesn't matter if you were talking about the animated gif or NOT
you made a statement that it fallacious applied to previously accepted SR arguments "In the moving frame, he is stationary and always at the midpoint of the two events." in the case of:
http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif
Alkatran
Jul4-04, 07:25 PM
doesn't matter if you were talking about the animated gif or NOT
you made a statement that it fallacious applied to previously accepted SR arguments "In the moving frame, he is stationary and always at the midpoint of the two events." in the case of:
http://home.teleport.com/~parvey/train1.gif
http://home.teleport.com/~parvey/train2.gif
"OH MY GOD"
I know I said I would stay away from this thread but this is unimaginable!
The midpoint of the two events is the midpoint of the two expanding spheres/circles of LIGHT! Light moves at c for all observers! The mid point is different depending on frame FOR THE LIGHT!
He was talking about the midpoint between the EMITTERS.
but in this case he is NOT always in the midpoint between the two emitters.
OR the light sources for that matter.
but he still receives simultaneous light. if he assumes that the light he receives simultaneously came at him from the same distances THEN light speed would be constant for him as a stationary observer. simultaneous reception for him is NOT simultaneous emission only because you guys have no idea how to measure things when you switch frames.
Alkatran
Jul4-04, 07:49 PM
but in this case he is NOT always in the midpoint between the two emitters.
OR the light sources for that matter.
but he still receives simultaneous light. if he assumes that the light he receives simultaneously came at him from the same distances THEN light speed would be constant for him as a stationary observer. simultaneous reception for him is NOT simultaneous emission only because you guys have no idea how to measure things when you switch frames.
Thanks for the insult, I'm sure everyone here "can't measure things".
Look, just face it, you have no idea what you're talking about. All your arguments are the same thing.
Thanks for the insult, I'm sure everyone here "can't measure things".
Look, just face it, you have no idea what you're talking about. All your arguments are the same thing.
the imperfection is yours.
you guys bring forth an argument and when i prove it doesn't and CAN'T work that way, you say i'm wrong without saying why i'm wrong.
why am i wrong then?
Alkatran
Jul4-04, 08:03 PM
First of all, how can we say you're wrong when you won't even tell us your theory?
Secondly, relativity applies (remarkably accurately) to reality.
K i've figured out exactly where you went wrong with this "light speed constant to the observer" stuff
i don't even need the data anymore.
in every case you're taking the data with the OBSERVER as stationary and it doesn't matter where the light source is moving because you're "ASSUMING" the sources are stationary with the observer BECAUSE you believe light to NOT be dependant on source.
Essentially you're saying that MOVING, the SOURCE will always MOVE with the observer. <reverse frame>
well of course you're always going to measure constant light speed if you use that definition to measure it. it's like saying a car can only move at 5 miles an hour towards me, but i cannot make any motion towards it.
Light speed is only constant BECAUSE you're assuming a stationary observer and applying your measurements to that frame <with a stationary source>
i have already proven WHY this does not accurately reflect reality when something is indeed moving.
Alkatran
Jul4-04, 09:06 PM
K i've figured out exactly where you went wrong with this "light speed constant to the observer" stuff
i don't even need the data anymore.
in every case you're taking the data with the OBSERVER as stationary and it doesn't matter where the light source is moving because you're "ASSUMING" the sources are stationary with the observer BECAUSE you believe light to NOT be dependant on source.
Essentially you're saying that MOVING, the SOURCE will always MOVE with the observer. <reverse frame>
well of course you're always going to measure constant light speed if you use that definition to measure it. it's like saying a car can only move at 5 miles an hour towards me, but i cannot make any motion towards it.
Light speed is only constant BECAUSE you're assuming a stationary observer and applying your measurements to that frame <with a stationary source>
i have already proven WHY this does not accurately reflect reality when something is indeed moving.
Congradulations! You've just realized that light is at c relative to yourself, meaning that, technicly, everyone is stationary relative to light travelling at c. :wink:
Don't think too hard, wouldn't want to go down the right path.
It's like saying that if you start moving towards the car (moving away from you) it speeds up.
Congradulations! You've just realized that light is at c relative to yourself, meaning that, technicly, everyone is stationary relative to light travelling at c. :wink:
Don't think too hard, wouldn't want to go down the right path.
It's like saying that if you start moving towards the car (moving away from you) it speeds up.
exactly... it DOES speed up. but not relative to you, because YOU ARE STATIONARY.
it's an exercise in futility. like catching those motes in your eye... you track them and they move away,,,
of course i have already proven why this reasoning is false, it helps now that i understand why you think this way, so i can better explain it to others who may have the same misconceptions.
please go back and read why this is NOT reality.
Alkatran
Jul4-04, 09:21 PM
exactly... it DOES speed up. but not relative to you, because YOU ARE STATIONARY.
it's an exercise in futility. like catching those motes in your eye... you track them and they move away,,,
of course i have already proven why this reasoning is false, it helps now that i understand why you think this way, so i can better explain it to others who may have the same misconceptions.
please go back and read why this is NOT reality.
I've read through this entire thread and I can tell you that it's amazing you've somehow managed to avoid learning the entire time. You seemed so close with the space-time graphs... but alas..
Acceleration has to do with general relativity, and this thread is about special relativity, so let's just ignore that, alright?
You see, the main concept you need to grasp is this: If observer A is moving relative to observer B, events seperated by distance that are simultaenous for A will not be for B, and vice versa.
That's it. That's all you need to accept, no, CONSIDER, before the learning really begins.
wrong.
events that APPEAR to be simultaneous for a stationary observer will not APPEAR to be simultaneous for a moving observer.
but that's a GIVEN in ANY relativity. i've never disputed THAT.
but you guys are given to the notion that if something APPEARS to be simultaneous then it IS simultaneous, which is quite illogical considering the multitudes of situations where things contradict that logic.
Light can move at a constant calculated speed and still not screw up anything that you guys hold precious.
the whole relative light speed is a miscalculation. that is all
Alkatran
Jul4-04, 09:43 PM
wrong.
events that APPEAR to be simultaneous for a stationary observer will not APPEAR to be simultaneous for a moving observer.
but that's a GIVEN in ANY relativity. i've never disputed THAT.
but you guys are given to the notion that if something APPEARS to be simultaneous then it IS simultaneous, which is quite illogical considering the multitudes of situations where things contradict that logic.
Light can move at a constant calculated speed and still not screw up anything that you guys hold precious.
the whole relative light speed is a miscalculation. that is all
What in the world are you talking about?
How can something not be simultaneous but "APPEAR" to be simultaneous? If you include a few calculations based on relativity, things that are simultaneous for you are "simultaneous" for me. But simultanity is frame-dependant (because it has to be at the same time FOR YOU).
Please give an example where something "appears" to be simultaneous but isn't.
Train2
just look at the gif
Train2
just look at the gif
you look at the modified gif
http://www.imagedump.com/index.cgi?pick=get&tp=87816
the green dots you added are the point of events which are stationary.
This gif is not the same as Einstein's gedanken.
http://www.bartleby.com/173/9.html
In this gif, the observer detects the photons at the same time, but doesn't conclude the events were simultaneous in his frame, because he is not at the midpoint (of the green dots). In Einstein's gedanken, observer starts at the midpoint (and remains there in his frame), doesn't detect photons at the same time, therefore concludes events were not simultaneous.
The two train gifs just demonstrate how things look different in different frames.
no, read your Einstein Gedanken again.
he was at the center of point of emissions WHEN the lights flashed <and the embankment is used as the "stationary" point of simultaneity. BUT the train is moving with velocity V -> and "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A"
that is standard movement, i'm guessing Newtonian.
The gif does NOT describe Einstein's Gedanken, as you said. but the Gif does display a pretty convincing argument as to why simultaneity is not at fault, just faulty calculations
no,
No to what? Quote please.
No to what? Quote please.
No
just kidding :D
In Einstein's gedanken, observer starts at the midpoint (and remains there in his frame)
it was in reference to that. in the Einstein Gedanken the observer moves towards an emitter <lightning> and away from another, starting at the midpoint.
the "remains there in his frame" is wrong. As per Case #7.
i can elaborate with gifs if need be...
in the Einstein Gedanken the observer moves towards an emitter <lightning> and away from another, starting at the midpoint.
the "remains there in his frame" is wrong. As per Case #7.
Let me clarify.
The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement.
Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant. But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.
About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.
Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant. But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.
if he was at the midpoint of the "events" and light speed is constant relative to him, it stands to reason that HE views the light as striking simultaneously as well, does it not? he would not lose simultaneity in this case...
About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.
but the law cannot be frame inviolate as i have outlined, the times for light to reach an observer from "an event" is different if the observer is moving than if the emitter is moving. the emitter does not impart relative change in motion of an "event" but the observer DOES.
in other words, the light is NOT tied to it's SOURCE in an event, it IS however tied to an Observer.
Make it so Observers are both Observers AND Emitters. Make one of them move and the other one stationary. ping the emitters simultaneously in a "midpoint" frame. One of them <the moving one> will be hit with the opposing emitter's photon first <by virtue of closing a distance towards the emitter>
now switch "frames" make the moving one stationary. all of a sudden the OTHER emitter (now the moving one) gets hit by the opposing photon first.
Holy Rusted Metal, Batman
Paradox
still working on the gif for this one... but i figured the explanation should be good enough to get the point across for the time being
stupid animated gif programs not being free :O
if he was at the midpoint of the "events" and light speed is constant relative to him, it stands to reason that HE views the light as striking simultaneously as well, does it not? he would not lose simultaneity in this case...
Yes "he was at the midpoint of the events" and yes, "light speed is constant relative to him". Therefore, yes, IF he saw lights striking simultaneously, he would rightfully conclude the events occured simultaneously. But he does NOT see them simultaneously, therefore concludes they were not simultaneous events in his own frame. You may think "why not change the observation rather than the conclusion. How are we sure he does not see them at the same time?". Because that's what the stationary frame sees: that the moving observer meets the lights at different points so he can't have possibly seen them at the same time. I could think of a paradox if this was not the case.
but the law cannot be frame inviolate as i have outlined,
I have stated my objection to that above so let's not continue until that's clear.
Yes "he was at the midpoint of the events" and yes, "light speed is constant relative to him". Therefore, yes, IF he saw lights striking simultaneously, he would rightfully conclude the events occured simultaneously. But he does NOT see them simultaneously
:confused: Einstein's Gedanken right? observer moving FROM the center of the light points AT THE INSTANT that lights are fired simultaneously in the stationary frame. Make that observer stationary. Lights are STILL fired in the first instant of the experiment when the distance between observers and is equal on both sides, light does not move with sources (in this case lightning doesn't leave a "source" to move with anyways). observer does not move either. midpoint DOES move.
conclusion? observer receives light simultaneously.
I have stated my objection to that above so let's not continue until that's clear.
and my continuation should prove beyond a doubt that my reasoning in case #7 holds true. Postulate: "Motion of light is independant of the motion of the source, but DEPENDANT on the motion of the observer"
Alkatran
Jul5-04, 07:07 AM
if he was at the midpoint of the "events" and light speed is constant relative to him, it stands to reason that HE views the light as striking simultaneously as well, does it not? he would not lose simultaneity in this case...
but the law cannot be frame inviolate as i have outlined, the times for light to reach an observer from "an event" is different if the observer is moving than if the emitter is moving. the emitter does not impart relative change in motion of an "event" but the observer DOES.
in other words, the light is NOT tied to it's SOURCE in an event, it IS however tied to an Observer.
Make it so Observers are both Observers AND Emitters. Make one of them move and the other one stationary. ping the emitters simultaneously in a "midpoint" frame. One of them <the moving one> will be hit with the opposing emitter's photon first <by virtue of closing a distance towards the emitter>
now switch "frames" make the moving one stationary. all of a sudden the OTHER emitter (now the moving one) gets hit by the opposing photon first.
Holy Rusted Metal, Batman
Paradox
Oh, I get what you mean now. No, that's not right. They actually do go the same distance... here allow me to explain:
Observer 1 is stationary relative to the emitter, observer 2 is moving towards it.
The Emitter releases a photon at 0.
Observer 1 receives this photon at t (in M), after Observer 2. Observer 2 receives it at t' (in M'), before Observer 1.
Now, what distances have these photons travelled? Well, they've travelled the same distance in each frame (t in M and t' in M') because you don't measure the distance between the emitter and the observer when he receives the light, because the speed of the emitter doesn't affect the photon's speed!
This means you need to use the emitter's original location >IN THAT FRAME<.
In M' (the moving frame) the emitter's original location is behind the emitter's location in M. So the distances are equal.
i have no idea what you mean by that. But i just woke up so i'm slow starting :D
according to SR, both frames are equally valid. we KNOW from experimentation <i'm going to assume SOMEONE did Einstein's Gedanken before> that an observer moving towards a light receives photons before a person stationary at the same start location.
but it doesn't work in this case. if you switch frames so that the moving observer is "the stationary one" then the stationary observer in reality (now the moving one) gets hit by photon first. How can you determine who gets hit first in this case?
No if you switch frames ram the observer who was stationary in the frame in your first paragraphy is now moving away from the light source in the frame in your second parargraph, consequently in this frame also it is this obsrever who receives the light last.
nah we've moved on to Case #8 now, jcsd
lemme diagram it for you
Case #8
(o) | (o)
(o) | (o)
(o) | (o)
(o) | (o)
The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers. ObserverB then moves towards ObserverA. Predict who receives a photon first.
(o) | (o)
(o) | (o)
(o) | (o)
(o) | (o)
now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one. in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment. How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?
Alkatran
Jul5-04, 02:53 PM
nah we've moved on to Case #8 now, jcsd
lemme diagram it for you
Case #8
(o) | (o)
(o) | (o)
(o) | (o)
(o) | (o)
The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers. ObserverB then moves towards ObserverA. Predict who receives a photon first.
(o) | (o)
(o) | (o)
(o) | (o)
(o) | (o)
now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one. in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment. How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?
They don't agree on the order because the two events are seperated by distance. :rolleyes:
I told you you weren't reading my posts.
:confused: Einstein's Gedanken right? observer moving FROM the center of the light points AT THE INSTANT that lights are fired simultaneously in the stationary frame. Make that observer stationary. Lights are STILL fired in the first instant of the experiment when the distance between observers and is equal on both sides, light does not move with sources (in this case lightning doesn't leave a "source" to move with anyways). observer does not move either. midpoint DOES move.
conclusion? observer receives light simultaneously.
Einstein's gedanken corresponds to your case#3. The moving observer does not see the lightenings at the same time (since they occured simultaneously in the stationary frame, they can't have occured simultaneously in the train frame). I am 100% sure this is SR's prediction. Plus, if he saw them at the same time, there would be a paradox, because the non-moving midpoint observer sees them at the same time at another point in space.
But if you make the moving observer stationary, of course he sees the lightenings at the same time, because he would be in the stationary frame now. That would correspond to case#1.
I don't understand why you say "midpoint DOES move". The midpoint of events never move within any frame, because the location of events never move within a frame. Location of an event in a frame may look like moving from another frame (like burning marks on the train as seen from the embankement), but they would move at the same speed as the train.
Postulate: "Motion of light is independant of the motion of the source, but DEPENDANT on the motion of the observer"
For the hundreth time, there is no motion of the observer in the observer's frame, observer is at rest wrt itself. Of course separation speed changes when looked from another frame, but that's irrelevant. What changes in a frame is the simultaneity of events, not how light moves. That's the whole point you are missing.
Yes it does matter, if he is at the midpoint he will detect both simultaneously, if he's not he won't.
Example: the two following gifs show what happens according to the frame the emitters are in and the frame the man is in for the same situation. (the actual animation deals with a railway car moving along a track, but we will just assume that the track represents the platform. In this situation we have to men, one that stays stationary to the platform and one that moves with it.
We further stipulate that the moving man is next to the stationary man when the light is first detected by either.
The first animation shows things from the perspective of the emitters and the stationary man:
http://home.teleport.com/~parvey/train1.gif
The photons expand at c in two spherical fronts that reach the mid point at the same time by the observation of both men (both men see the photons arrive at the same time.)
The second animation shows what happens according to the man movig relative to the platform/track.
Since the speed of light is invarient for all observers, he also must see photons exapand outward from the point of emmission as a sphere in his frame. But from his perspective, the emitters do not stay at the the point of emmission. Therefore, in order for him to detect the emissions from both emitters at the same time, and at the same time as the man stationary to the platform, in his frame, the emiiters do not emit simultaneously, but one emits after the other.
http://home.teleport.com/~parvey/train2.gif
The reason things have to be this way is that the two men, being at the same point at the same time according to both of them, and both seeing the flash from the emitters arrive simultaneously is a spacetime event that is invarient and must be agreed upon by everyone.
I guess you dropped the football alot in high school huh? . When a quarterback throws a ball to a "stationary" receiver and the football is intercepted by a running safety as they collide, you must think the ball was thrown at a different time for the receiver than the safety, that's why one guy misses it right?
They don't agree on the order because the two events are seperated by distance. :rolleyes:
I told you you weren't reading my posts.
read the case. simultaneity is verified by a midpoint observer. in one "view" the midpoint observer DOES see the events simultaneously. in the other "view" he doesn't. but it's the SAME CASE. we're not doing the experiment twice.
simultaneity at a point is INVIOLATE.
you're obviously not reading MY posts. I'll let you ruminate on that subject for a while.
For the hundreth time, there is no motion of the observer in the observer's frame, observer is at rest wrt itself. Of course separation speed changes when looked from another frame, but that's irrelevant. What changes in a frame is the simultaneity of events, not how light moves. That's the whole point you are missing.
you seem to be dodging case #8, wespe. please give it a try
Alkatran
Jul6-04, 12:20 AM
read the case. simultaneity is verified by a midpoint observer. in one "view" the midpoint observer DOES see the events simultaneously. in the other "view" he doesn't. but it's the SAME CASE. we're not doing the experiment twice.
simultaneity at a point is INVIOLATE.
you're obviously not reading MY posts. I'll let you ruminate on that subject for a while.
you seem to be dodging case #8, wespe. please give it a try
Oh, sorry, I forgot to tell you AGAIN that the photons aren't emitted simultaneously in the moving frame, making the middle observer receive them at the same time. :rofl:
And this is because the two events (the emitions) are SEPERATED BY DISTANCE.
you seem to be dodging case #8, wespe. please give it a try
You have dodged cases #1 to #7 and some 400 replies. Please go back and read all the thread again.
OK, case #8, involves acceleration. ObserverB first shines light and then starts moving, therefore switching to a different frame. So he can't declare himself stationary all the time (or a pseudo gravity field must be added). Similarly to the Twins Paradox, see:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
So, explaining case #8 as is requires invoking General relativity, which I can't. To avoid this, assume ObserverB never moves but an inertial ObserverC was passing near ObserverB as the light was shined. Then, midpoint always receives the lights at the same time. Only for ObserverC they were not emitted simultaneously, just as with the previous cases.
You have dodged cases #1 to #7 and some 400 replies. Please go back and read all the thread again.
i'm the one making the cases, i don't have to answer them. i'm finding out what SR thinks of the situations. i already KNOW what I think of the situations...
So, explaining case #8 as is requires invoking General relativity, which I can't. To avoid this, assume ObserverB never moves but an inertial ObserverC was passing near ObserverB as the light was shined. Then, midpoint always receives the lights at the same time. Only for ObserverC they were not emitted simultaneously, just as with the previous cases.
fine make ObserverB an inertial viewer. don't need to add more observers into the mess.
ObserverB moving towards ObserverA shines a light in such a fashion that the light he shines and the light ObserverA shines hits the midpoint viewer simultaneously.
switch view frames so ObserverB is NOT a moving observer.
Midpoint Observer and ObserverA have to move in this frame. Midpoint viewer AND ObserverA both do not receive photons in the correct time frame in this "view"
please resolve this paradox.
what was once simultaneous interception for the midpoint viewer is no longer such. just by changing "views"
the obvious conclusion is changing views does NOT work when dealing with light BECAUSE light is observer motion dependant
QED
Einstein's gedanken corresponds to your case#3. The moving observer does not see the lightenings at the same time (since they occured simultaneously in the stationary frame, they can't have occured simultaneously in the train frame). I am 100% sure this is SR's prediction. Plus, if he saw them at the same time, there would be a paradox, because the non-moving midpoint observer sees them at the same time at another point in space.
In Janus's animated presentation, she argues (and you agreed) that the moving observer observes the lightning flashes simultaneously (at M')because they were not emitted simultaneously in the moving frame. If I read you correctly, you are saying here that the moving observer does not receive the lightning flashes simultaneously (at M') because they were not emitted simultaneously in the moving frame. Sounds like a contradiction here- care to explain?
In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occured simultaneously in the stationary frame, they can't have occured simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B travelled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events. And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.
Eyesaw: they define simultaneity at a distance different from normal people. For them something at a distance only happens "at the same time" if it can be witnessed by an observer in the center of "the events" "at the same time"
thus the source of 30 pages of confusion :D
finally have the whole thing nailed down as a simple postulate error involving frame changes. pretty sure all "relative light speed to the observer" calculations were done as observer stationary frame, which is wrong.
Eyesaw: they define simultaneity at a distance different from normal people. For them something at a distance only happens "at the same time" if it can be witnessed by an observer in the center of "the events" "at the same time"
thus the source of 30 pages of confusion :D
finally have the whole thing nailed down as a simple postulate error involving frame changes. pretty sure all "relative light speed to the observer" calculations were done as observer stationary frame, which is wrong.
I think you are right.
well thank you, it's nice to have a favorable vote once in a while :D
i'm the one making the cases, i don't have to answer them. i'm finding out what SR thinks of the situations. i already KNOW what I think of the situations...
Well you aren't just asking questions. You are presenting arguments and drawing conclusions. When you are answered with a counter argument and you don't have anything left to say, you simply change the case while you should be accepting that your initial argument was incorrect. Therefore you aren't learning anything. That's what I meant by dodging. Please do read http://www.ephilosopher.com/Sections-article36-page1.html
fine make ObserverB an inertial viewer. don't need to add more observers into the mess.
ObserverB moving towards ObserverA shines a light in such a fashion that the light he shines and the light ObserverA shines hits the midpoint viewer simultaneously.
switch view frames so ObserverB is NOT a moving observer.
Midpoint Observer and ObserverA have to move in this frame.
It's fine up to this point.
Midpoint viewer AND ObserverA both do not receive photons in the correct time frame in this "view"
You draw this conclusion without showing me the reasoning. It is a wrong conclusion. The setup is that midpoint receives both photons at the same time. Switching the view won't change this fact. Why did you think so? Please explain.
please resolve this paradox.
No paradox. From observerB's perspective, ObserverA doesn't shine light simultaneously with ObserverB. ObserverB shines light at a certain time such that the midpoint receives the photons at the same time. I could show you how this is so (requires adding another observer in ObserverB frame), but first tell me what made you think:
what was once simultaneous interception for the midpoint viewer is no longer such. just by changing "views"
My guess is that you didn't think; you just used your intuition.
In Janus's animated presentation, she argues (and you agreed) that the moving observer observes the lightning flashes simultaneously (at M')because they were not emitted simultaneously in the moving frame. If I read you correctly, you are saying here that the moving observer does not receive the lightning flashes simultaneously (at M') because they were not emitted simultaneously in the moving frame. Sounds like a contradiction here- care to explain?
Well, there were different cases discussed in this thread. One of them is an animated gif which shows the moving observer seeing the flashes at the same time. In that gif, the observer is not at the midpoint. The one you quoted is different; the observer is at the midpoint and does not see the flashes at the same time.
In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occured simultaneously in the stationary frame, they can't have occured simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B travelled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events.
So your suggestion is :the train observer must prefer the burned wood on the tracks over the burn marks on the train, as the location of events. Why? What is special about the tracks?
And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.
Speed of light is constant relative to all observers (do you disagree at this point?). Then, if light from both events travel the same distance (that is, the observer is at the midpoint), the order of detection of light signals can be used to conclude about the sequence of events. Why do you think not?
Fine, you want my observations this time, you can have them
1. before the frame shift, ObserverA is stationary and the Midpoint Observer is stationary. ObserverB is moving towards ObserverA <and the Midpoint Observer>
2. In this "Frame" Lights are emitted and progress is made by ObserverB Towards the opposing source of light. no progress is made by ObserverA towards HIS opposing light source. Light for ObserverB will be intercepted before ObserverA. the midpoint observer being stationary and at the midpoint of the emission sources will receive the emissions simultaneously.
3. Frames are switched. ObserverB is now stationary. ObserverA and Midpoint Observer move to preserve "reality" and "relative movements". in this frame, ObserverB makes no motion towards the light sources. ObserverA and Midpoint Observer DO make progress. The Midpoint Observer now progressing in a direction towards a light source will no longer receive photons simultaneously in this case, as he is no longer stationary at the midpoint of the two light sources at the time of emission and reception.
No paradox. From observerB's perspective, ObserverA doesn't shine light simultaneously with ObserverB. ObserverB shines light at a certain time such that the midpoint receives the photons at the same time. I could show you how this is so (requires adding another observer in ObserverB frame), but first tell me what made you think:
please do show me how the midpoint observer can still receive photons simultaneously in this case. i can almost guarantee you'll come to the same conclusion / revelation that i've been holding out of your reach for 30 pages.
oh and remember to preserve the speed of light relative to the midpoint observer ;D
Fine, you want my observations this time, you can have them
1. before the frame shift, ObserverA is stationary and the Midpoint Observer is stationary. ObserverB is moving towards ObserverA <and the Midpoint Observer>
2. In this "Frame" Lights are emitted and progress is made by ObserverB Towards the opposing source of light. no progress is made by ObserverA towards HIS opposing light source. Light for ObserverB will be intercepted before ObserverA. the midpoint observer being stationary and at the midpoint of the emission sources will receive the emissions simultaneously.
3. Frames are switched. ObserverB is now stationary. ObserverA and Midpoint Observer move to preserve "reality" and "relative movements". in this frame, ObserverB makes no motion towards the light sources. ObserverA and Midpoint Observer DO make progress.The Midpoint Observer now progressing in a direction towards a light source
fine up to this point
will no longer receive photons simultaneously in this case,
this is your wrong conclusion. your reasoning is:
as he is no longer stationary at the midpoint of the two light sources at the time of emission and reception.
true, no longer at the midpoint, but you also assume that in ObserverB frame, ObserverA emitted light simultaneously with ObserverB. Not true. According to ObserverB, ObserverA emitted light before ObserverB, such that the moved mid-observer received the lights at the same time.
please do show me how the midpoint observer can still receive photons simultaneously in this case. i can almost guarantee you'll come to the same conclusion / revelation that i've been holding out of your reach for 30 pages.
I explained above how midpoint observer still receives photons simultaneously in ObserverB frame. Now I am expecting you to present your counter-argument. No more cases, enough is enough.
oh and remember to preserve the speed of light relative to the midpoint observer ;D
From ObserverB's perspective, the SEPARATION speed between two light signals and mid-observer are of course different than c. But from mid-observer's perspective, relative speed of light signals are still c.
Originally Posted by Eyesaw
In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occured simultaneously in the stationary frame, they can't have occured simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B travelled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events.
So your suggestion is :the train observer must prefer the burned wood on the tracks over the burn marks on the train, as the location of events. Why? What is special about the tracks?
Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.
Quote:
Originally Posted by Eyesaw
And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.
Speed of light is constant relative to all observers (do you disagree at this point?). Then, if light from both events travel the same distance (that is, the observer is at the midpoint), the order of detection of light signals can be used to conclude about the sequence of events. Why do you think not?
Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.
Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.
I don't understand your point. What kind of difference? Elaborate please..
Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.
Yes, both frog and duck are fried at the same time in the stationary frame. And, both frog and duck will be fried in the train frame, despite at different times. But their frying at different times doesn't mean you can prevent the frog or duck from frying after the other one is fried. That would require sending a message faster than light, which means going back in time (does bring up paradoxes, but FTL is not possible according to SR).
You still seem to think that all you need to do to switch a diagram from one frame to another is just slide things over so that the pictures line up with the other observer. If things were that easy then everyone would understand relativity! :smile:
You keep ignoring simultaneity, length contraction, and other relativistic effects.
nah we've moved on to Case #8 now, jcsd
lemme diagram it for you
Case #8
(o) | (o)
(o) | (o)
(o) | (o)
(o) | (o)
The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers.
So the midpoint is in the frame of Observer A. And the lights are switched on simultaneously according to Observer A, but not Observer B. The diagram shows the view from the Observer A frame.
ObserverB then moves towards ObserverA. Predict who receives a photon first.
According to the A frame clocks, Observer B detects a photon first. But so what? The two events--(1) Observer B detects a photon and (2) Observer A detects a photon--are not causally connected and happen at different locations. Their time order is frame dependent.
(o) | (o)
(o) | (o)
(o) | (o)
(o) | (o)
now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one.
You think you are showing the view from B's view, but you're not drawing it accurately. So the diagram is useless for drawing any meaningul conclusions. An accurate diagram would show length contraction, time dilation, and most importantly the lack of simultaneity of the two light flashes.
in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment.
All this demonstrates is that you don't know how to draw the diagram from B's viewpoint. It's not easy.
How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?
I'm glad you realize that that would make no sense! But the problem lies with your analysis, not with relativity. Both frames agree that the photons arrive at M simultaneously.
Alkatran
Jul6-04, 08:31 AM
Well, RAM. You've had a few different people say exactly the same thing for an entire page. That's how special relativity sees it. It seems like you have your "own" special relativity that you're trying to disprove to prove your own theory.
You can't change an existing theory to disprove it.
Been wracking my brain for two days trying to find SOME way to get you guys to understand me.
i may have stumbled upon the answer, however, so try and hear me out once more.
How does one measure distance? it is a measurement based upon a constant "movement" over an allotted "time"
speed is the direct proponent of this calculation including both distance AND time.
now when you say the speed of light is constant relative to the observer, that's all well and good, but only if the observer is stationary. stationary in respect to WHAT you may ask. stationary in respect to THE SPEED OF LIGHT says SR. The measurement "method" of SR is to assume that the observer is ALWAYS stationary, and since any "motion" of a light source has no effect, light will always be stationary as well.
think of how anything else moves through the universe however. If something IS moving, then it covers a distance TOWARDS something else.
Hence if i have 3 positions on a line A, B and C. and a person at point C throws a ball to point B while A moves to point B to reach that point the same time the ball reaches there.
the relative velocity of the ball, to ParticipantA is the velocity of the distance C-B over the time it took the ball to get from C-B PLUS the velocity of the distance A-B divided by the time it took ParticipantA to get from A-B. basically A-B/t + B-C/t = A-C/2t
Even taking ParticipantA as stationary, now ParticipantB has a velocity of B-A and the ball has a velocity C-B making the ball's velocity towards A, A-C/2t
now the problem with relativity is you have to abandon this "stationary observer" method of measurement, or account for the problem caused by light being unbound to its source. The problem you run into is as follows.
A is making progress to a point B, reaching there at the same time as a photon from C. The relative velocity of light for A in THIS case is of course easily measured as above. A covers a distance A-B/t and the photon covers a distance C-B/t. the relative velocity of the photon is of course A-C (total distance covered)/2t (total time elapsed)
case1
http://www.imagedump.com/index.cgi?pick=get&tp=93222
now move to make A a stationary observer.
I'll present to you two pictures and let you guess which is right. which is wrong. which is SR prediction.
case2
http://www.imagedump.com/index.cgi?pick=get&tp=93223
case3
http://www.imagedump.com/index.cgi?pick=get&tp=93224
Now looking at these two pictures you can note a few things.
Case2 asserts that A is stationary, it does NOT move, which means B must move relatively to its position and light from C intercepts them both at that location.
Case3 asserts THE SAME THING, except it asserts A is already IN B's position. well then where is B? obviously B has to move relatively to get to A so we move it over to maintain consistancy. C, which is not an object, but light from a source does NOT move, so lo and behold the measurement of light from C to A is the same as it was from C to B in the previous case.
Case3 ignores a very real distance, that is a distance covered from A-B in the time it took light to cover the distance C-B. Case 3 is an incorrect measurement of the speed of light.
Alkatran
Jul7-04, 06:50 PM
Judging entirely by sight, the first one can't be true because light is exceding c according to the observer. (Assuming the yellow line is a photon)
So, by elimination, the second one is true. But I fail to see your point.
Are you going to give another argument about the observers being in two places at once? Because I've already told you it's because of their "when" line being skewed.
by "second one" do you mean case3 ?
Are you going to give another argument about the observers being in two places at once? Because I've already told you it's because of their "when" line being skewed.
no, not at all. in case3 the setup applied was that there has to exist relative gains between Participants A and B so if A is already in position B <to keep lightspeed constant> then logic dictates B is closer to C than the original picture. but we know this is NOT the case, so the picture is wrong.
Alkatran
Jul7-04, 07:13 PM
What in the world are you talking about? In ALL the pictures B is ALWAYS closer to C than A.
BC in case1 is a different distance than BC in case3
simply
Alkatran
Jul7-04, 07:25 PM
That would be because of *gasp* length contraction. Due to the "when" line being skewed WHICH I ALREADY SAID!
no DUH it's because of length contraction
but the length contraction is wholly based on the wrongness of the picture. Length contraction, Time Dialation, constant light speed are COMPLETELY based off of using the wrong picture to measure distances and times.
look at case2 which is the RIGHT way to measure things. that is the way we measure EVERYTHING else in the universe. why measure light different?
because you're all a bunch of lunatics :D
Alkatran
Jul7-04, 07:39 PM
Because all measurements of light have shown that that is the only way to measure it. Duh.
You see, it's all math until you apply it to REALITY, that's when it becomes Physics.
no, you could measure it THE CORRECT WAY
craziness.
i'm going to have to make machines to do it properly aren't i. dammitall
Alkatran
Jul7-04, 07:49 PM
:rofl:
Oh, you're so willing to learn, aren't you? The mere idea that time can be so lined with space drives you mad.
You go and make your machines to measure c, but there's no point, people much more intelligent than you have tried it before. Guess what? They all found out the same thing (except maybe the ones who made mistakes): Eistein was right.
no einstein assumed that the motion of the observer makes no difference which is a ludicrous proposition AT BEST.
all calculation done for speed of light were taken WITH a stationary observer completely discarding that motion/distance
time/space is fine, but not with light as absolute speed defining it.
Quote:
Originally Posted by Eyesaw
Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.
I don't understand your point. What kind of difference? Elaborate please..
Quote:
Originally Posted by Eyesaw
Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.
Yes, both frog and duck are fried at the same time in the stationary frame. And, both frog and duck will be fried in the train frame, despite at different times. But their frying at different times doesn't mean you can prevent the frog or duck from frying after the other one is fried. That would require sending a message faster than light, which means going back in time (does bring up paradoxes, but FTL is not possible according to SR).
Reread Einstein’s gendanken in his Special Relativity. The experimental set up is such that the rear and front of the train are located at A&B when the lightning flashes occur, i.e. those events happen simultaneously in the absolute sense for both the platform frame and the train frame. What is not simultaneous from one frame to the other are the reception of the signals from the events to the observers in the different frames. This however is hardly surprising since the train is moving towards one signal and away from the other signal when compared with the observer of the signals on the platform.
What is special about the platform frame is that since the observer is supposed to be at the midpoint of the flashes, she can only receive the signals simultaneously if she was really absolutely at rest- otherwise she would have to conclude that the speed of light measured from two different directions are not the same, contradicting the postulates of SR. So since an absolute rest frame was found it definitely makes life easier to measure all speeds relative to it. In reality the train experiment is of course flawed since the Earth is actually moving through space so that even the platform observer cannot receive the signals simultaneously.
But for amusement purposes, let’s assume that Einstein wasn’t talking about an absolute rest frame in his gedanken. It’s still absurd to use the reception time of signals as a representation of the sequence of how actual events occurred. If we see a star explode today from our telescope, who would be gullible enough to think that it happened today? So why do you make the case for relativity of simultaneity based on the reception time of the signals if in the absence of an absolute rest frame, it would be nigh impossible to determine the actual place in space when the light was produced nor our velocity through space, which surely must affect our determination of the speed of light since if the speed of light was truly constant in vacuo, if we have any motion through space, we will misjudge its true velocity. I don’t think this is a good reason to rid of absolute time do you? Really, which of you SR die hards can claim to have experienced any event that wasn’t absolutely simultaneous for everyone? If events were not absolutely simultaneous, your hand should have fell off from your wrist and your head from your body long before you finished typing your response.
... (post#485)
Ram,
I hate to tell you this but (A-B)/t + (B-C)/t = (A-C)/2t is wrong, it equals to (A-C)/t. I hope you won't discuss this or we have serious problems here.
Apart from that, I looked at your first picture. I'm not very familiar with spacetime diagrams. But, the picture looks upside down; conventionally, time axis should increase in the up direction.The angle between red and yellow lines should be 45 degrees (time and space units are chosen so that speed of light is 1). You should label things to avoid any confusion. See http://www.cord.edu/dept/physics/credo/spacetime.html
Also specify the values of speed, time, distance (chose convenient numbers).
Do that and ask what the diagram would look like from A's perspective according to SR. Obviously you don't know the answer. Just ask, don't draw conclusions. I'm not saying I can draw the correct skewed diagram, but I know the angle between the observer and light should always be 45 degrees. Perhaps someone here will help you if you show willingness to learn.
Reread Einstein’s gendanken in his Special Relativity. The experimental set up is such that the rear and front of the train are located at A&B when the lightning flashes occur, i.e. those events happen simultaneously in the absolute sense for both
sorry, we can't communicate if what you comprehend is that from reading http://www.bartleby.com/173/9.html
http://www.imagedump.com/index.cgi?pick=get&tp=93385
happy now?
sorry, we can't communicate if what you comprehend is that from reading http://www.bartleby.com/173/9.html
well that IS what it says.
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment. But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A —> B on the travelling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M’ in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
The conclusion is based off of not KNOWING he has moved any distance. Basically because of the distance unknown he has to assume his distances <measured on the train> are correct and that simultaneity is at fault, not any of his measurements.
well that IS what it says.
I'm sorry Ram, I can't communicate with you either.
It does NOT say "experimental set up is such that .. those events happen simultaneously in the absolute sense for both frames" as Eyesaw claims. It says the opposite.
Nice that you now corrected your diagram. Hope someone helps you with that. If not, you could try another forum.
i'll break it down into parts with my translations, tell me where i go wrong.
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment.
Observers taking the Embankment as a reference frame <stationary> will see that light from A and B hit M mid-point simultaneously
But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A —> B on the travelling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train
Assigning a point M' for the train reference frame such that at the precise instant any first flash of lightning hits "anywhere" M' coincides with M perfectly.
If an observer sitting in the position M’ in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated
If he didn't make any relative motion in relation to the embankment he would receive simultaneous light just as the embankment viewers.
Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.
Reality is defined as REAL motion being made TOWARDS target light B and AWAY from target light A. Naturally this follows that he will see light B before light A.
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
BUT if he takes his own reference frame as stationary, to him light would have to cover the SAME DISTANCES to reach him, but yet he receives light in staggered intervals. since light speed is constant, and the distances are the same <because he measures the distances in HIS frame (not including the distances he travelled in the embankment frame) he MUST conclude that God didn't turn the lights on at the same time.
<end interpretation>
geistkiesel
Jul8-04, 08:15 PM
i'll break it down into parts with my translations, tell me where i go wrong.
.
Observers taking the Embankment as a reference frame <stationary> will see that light from A and B hit M mid-point simultaneously
Assigning a point M' for the train reference frame such that at the precise instant any first flash of lightning hits "anywhere" M' coincides with M perfectly.
If he didn't make any relative motion in relation to the embankment he would receive simultaneous light just as the embankment viewers.
Reality is defined as REAL motion being made TOWARDS target light B and AWAY from target light A. Naturally this follows that he will see light B before light A.
BUT if he takes his own reference frame as stationary, to him light would have to cover the SAME DISTANCES to reach him, but yet he receives light in staggered intervals. since light speed is constant, and the distances are the same <because he measures the distances in HIS frame (not including the distances he travelled in the embankment frame) he MUST conclude that God didn't turn the lights on at the same time.
<end interpretation>
Yes, in a sense. Here Einstein concludes that the mere fact that the moving observer measures the photon arrival at staggered intervals that the passengers on the train "must, therefore come to the conclusion" the photons were not emitted simultaneously in the moving frame. If this is true, your God solution fits the glove like a hand.
And are you familiar with the study that showed the differences in praying practices performed by persons praying in chrches versus those praying in gambling casinos?
The gamblers really mean it!
geistkiesel
Jul8-04, 08:40 PM
Einstein's gedanken corresponds to your case#3. The moving observer does not see the lightenings at the same time (since they occured simultaneously in the stationary frame, they can't have occured simultaneously in the train frame). I am 100% sure this is SR's prediction. Plus, if he saw them at the same time, there would be a paradox, because the non-moving midpoint observer sees them at the same time at another point in space.
Your 100% assurance is not correct.
AE did not invoke SR to come to his conclusion. You are bootstrapping, it is called circuitous reasoning. AE came to the conclusion that because the B photon was detected before the A photon was detected that this alone is why the passengers "must therefore come to the conclusion" that the photons were not emitted simultaneously in the moving frame. Any SR imperatives, even if applicable and they aren't, would be swamped by the time difference between the time the B and A photon were detected. Ene Doc Al agrees with this.
But if you make the moving observer stationary, of course he sees the lightenings at the same time, because he would be in the stationary frame now. That would correspond to case#1.
I don't understand why you say "midpoint DOES move". The midpoint of events never move within any frame, because the location of events never move within a frame. Location of an event in a frame may look like moving from another frame (like burning marks on the train as seen from the embankement), but they would move at the same speed as the train.
For the hundreth time, there is no motion of the observer in the observer's frame, observer is at rest wrt itself. Of course separation speed changes when looked from another frame, but that's irrelevant. What changes in a frame is the simultaneity of events, not how light moves. That's the whole point you are missing.
Looking at these probkem by stopping one frame and moving another is grossly and physically impossible. Where do you get justification for this? Do you kinow? It is from hand me down mathematics isn't it?
This is not a photon problem. As stated above AE came tot he conclusion he diod based purely on the fact the B photon was detected earlier than the A photon. But AE did not discuss passengers at a and b located at A and B just as the photons were emitted and ercorded the train time of the events. The a and b clock are synchribnized wrt the moving frame. This inoformation senet immediately tio O' will arrive just as the A photon arrives. Also, observers loated at M just as the photons A and B arrived at M also record the time of arrival and send thir tiems tio O'. O' now has the arival time of A, the recorded times of the A and B emissions from a and b, and the recorded time the photons arrived simultaneously at M in the stationary frame. The clock information is convincing that the photons arrived in the moving frame simultaneously as measured by twp sets of passengers. Likewise, O' can calculate the simultaneity himself. After a little algebra,
t3 = t1(C + v)/(C - v).
Draw your own time and distance map, this is what you will get, if you keep SR out of the gthe picture, or even if you insist SR will be swamped. Try it out.
t1 is the time of arrival of the B photons measured from when the O' observer was at the midpoint M when the photons were emitted and v the known velocity of the train. t3 is also a measured quantity the time the A photon was detected by O'. calculation can determine the accuracy of the measured vs. calculated time t3.
Again SR affects, if any, will be swamped by the staggered times of the B and A photon arrival.
Your 100% assurance is not correct.
AE did not invoke SR to come to his conclusion. You are bootstrapping, it is called circuitous reasoning. AE came to the conclusion that because the B photon was detected before the A photon was detected that this alone is why the passengers "must therefore come to the conclusion" that the photons were not emitted simultaneously in the moving frame. Any SR imperatives, even if applicable and they aren't, would be swamped by the time difference between the time the B and A photon were detected. Ene Doc Al agrees with this.
Nonsense. The reasoning that Einstein uses in the train gedanken is SR in action, my friend. What Einstein invokes is the invariant speed of light: a key premise of SR. And what he concludes--The relativity of simultaneity--is a key result of SR.
And what are you babbling about SR "imperatives" being swamped by time differences? (Do you seriously think I agree with this gibberish?)
You, geistkiesel, are the true master of circular reasoning.
geistkiesel
Jul8-04, 08:57 PM
Let me clarify.
The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement. Correct so far.
Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant.
Wrong, even from SR theory.
What is being measured is "events simultaneous in the staionary frame are not simultaneous in the moving frame". When M' was at M, the midpoint when the photons were emitted simultaneously does not allow the moving observer to say the moving iobserver was not at M when the photons were emitted simultaneously.This is a given, remember, you can't change the givesn. You can add observers, measuring equipment and the like, but changiong the giovens is a no no. This is one problem that SR invokes when "considering a moving frame stationary, a physical impossibility.
The moving observer concludes the photons were not emitted simultaneously, not from SR theory, but from the simple staggered measurement of the photons, first B then A, nothing else. SR follows this, it does not precede this experiment. You are close to being there.
But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.
About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.[/QUOTE]
geistkiesel
Jul8-04, 09:05 PM
Let me clarify.
The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement. Correct so far.
Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant.
Wrong, even from SR theory. M' concludes the lighning doesn't strike simultaneously only from his staggered measurement of the B and A photon, nothing else. Stick to the experiment. read it again befopre yopu start showing off your Sr theory knowledge, which has a nother day or two to survive and that is just becasue ram1024 is extending a proferssional courtesy.
What is being measured is "events simultaneous in the staionary frame are not simultaneous in the moving frame". When M' was at M, the midpoint when the photons were emitted simultaneously does not allow the moving observer to say the moving was not at M when the photons were emitted simultaneously.This is a given, remember, you can't change the given. You can add observers, measuring equipment and the like, but changing the givens is a no no. This is one problem that SR invokes when "considering" a moving frame stationary, a physical impossibility.
The moving observer concludes the photons were not emitted simultaneously, not from SR theory, but from the simple staggered measurement of the photons, first B then A, nothing else. SR follows this, it does not precede this experiment. You are close to being there.
But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.
Maybe so, but the moving observer will still detect the B photon before the A photon. The experiment doesn't change.The light apeed deos not depend on the velocity of the source.
About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.
I suppose so.
geistkiesel
Jul8-04, 09:10 PM
Let me clarify.
The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement.
Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant. But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.
About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.
I brought the point up in response to another post of yours. The observer only determines the photons were not emitted simultaneously from the staggered detection of the B and A photons, nothing more or less. You cannot stop moving frame and consider the other moving. This is SR stuff. This problem is not SR. The staggered detection of the B and A photon will swamp any SFR aspects of the experiment, which there aren't any anyway.
OK, one more try for Geistkiesel:
Suppose we have two space-trains passing by each other. They are moving inertially, but we don't know which one had accelerated in the past. No experiment can determine which train is "really" moving. We only know that they have a relative speed wrt each other. Now, suppose two explosions is space occur near the trains (but not tied to any of the trains, just instant explosions, at any two times, we don't know yet). The explosions create burn marks on both trains.
SR's claim is: if there were two observers at the mid point of those burn marks (measured in each train separately), they could determine the simultaneity of the explosions according to their train, by looking at the order they see the explosions, because speed of light is constant and distances are the same.
Now, if one of the observers see the explosions at the same time, the other will not. Because, the burn marks and the midpoints on two trains will have a relative speed wrt each other, and therefore the light from the explosions cannot hit both observers at the same time.
So tell me, if one observer sees them at the same time, and both events were simultaneous in both frames, and speed of light is independent of its source, why won't the other see them at the same time? You can't use the excuse "the trains were moving" or "one of them was stationary", because then you assume aether, and then you can't determine simultaneity in any frame with this method without knowing speed wrt aether. Are you saying that simultaneity is affected by speed wrt aether? So what method can you tell me to determine if two events occured simultaneously, if we can't determine speed of a frame wrt aether? (of course in fact there is no sign of aether whatsoever)
geistkiesel
Jul8-04, 10:02 PM
Nonsense. The reasoning that Einstein uses in the train gedanken is SR in action, my friend. What Einstein invokes is the invariant speed of light: a key premise of SR. And what he concludes--The relativity of simultaneity--is a key result of SR.
No SR is the key result of simultaneity, read the history sir.
BS. NASCARS are running invariant in speed in my example and you ignored that example. Eintein does not use any SR postulate in arriving at his conclusion. He uses only the the staggrered arrival of theh B photon and A photon as detected by the O' observer. You have said on many occasions that this is what all agree.
quote Doc Al "The fact (agreed by all) the light from B hits O' before the light from A leads the O' observer to conclude the lights could not flash simultameously."
Which leg standeth thee on oh master mentor?
And what are you babbling about SR "imperatives" being swamped by time differences? (Do you seriously think I agree with this gibberish?)
You, geistkiesel, are the true master of circular reasoning.Quote:
Originally Posted by geistkiesel
I just came up with another proof that the moving observer must conclude the photons were emitted in the moving frame simultaneously with the emitted photons in the stationary frame.
Yeah, "another" one.
Let's cut through the nonsense, once again. Let's say a and b are the observers on the train located right next to the flashing lights at A and B when they flash. When the see the lights flash, they check the time. No need for any "relaying" of clock times anywhere. Assuming, like you did, that all clocks on the train are synchronized then--like it or not--observers a and b will record different times for the two photon emissions.
How do you know this? They must check the times they are on opposite ends of the train, but each immediately relay their findings to O' who receives the data along with the simultaneous observation of the A and B photons arriving at M simultaneously.
You are invoking SR time dilation and mass shrinking right? What else could you be invoking to make your statement that the observers at A and B will not record the same time of the emitted photons. The photons know nothing of SR theory or even that there are two observers dutifully recording the times the photons are emitted. the a and b observers record the same time.
Remember, what you said back in the Lost simultaneoity thread. let me remind you:
quote by Doc Al
"If you understood the Eisntein simple argument you would know that nowhere does "time dilation" or " mass shrinking" enter into it"
geistkiesel
Jul8-04, 10:05 PM
Quote:
Originally Posted by Eyesaw
Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.
Reread Einstein’s gendanken in his Special Relativity. The experimental set up is such that the rear and front of the train are located at A&B when the lightning flashes occur, i.e. those events happen simultaneously in the absolute sense for both the platform frame and the train frame. What is not simultaneous from one frame to the other are the reception of the signals from the events to the observers in the different frames. This however is hardly surprising since the train is moving towards one signal and away from the other signal when compared with the observer of the signals on the platform.
What is special about the platform frame is that since the observer is supposed to be at the midpoint of the flashes, she can only receive the signals simultaneously if she was really absolutely at rest- otherwise she would have to conclude that the speed of light measured from two different directions are not the same, contradicting the postulates of SR. So since an absolute rest frame was found it definitely makes life easier to measure all speeds relative to it. In reality the train experiment is of course flawed since the Earth is actually moving through space so that even the platform observer cannot receive the signals simultaneously.
But for amusement purposes, let’s assume that Einstein wasn’t talking about an absolute rest frame in his gedanken. It’s still absurd to use the reception time of signals as a representation of the sequence of how actual events occurred. If we see a star explode today from our telescope, who would be gullible enough to think that it happened today? So why do you make the case for relativity of simultaneity based on the reception time of the signals if in the absence of an absolute rest frame, it would be nigh impossible to determine the actual place in space when the light was produced nor our velocity through space, which surely must affect our determination of the speed of light since if the speed of light was truly constant in vacuo, if we have any motion through space, we will misjudge its true velocity. I don’t think this is a good reason to rid of absolute time do you? Really, which of you SR die hards can claim to have experienced any event that wasn’t absolutely simultaneous for everyone? If events were not absolutely simultaneous, your hand should have fell off from your wrist and your head from your body long before you finished typing your response.
Where in the hell have you been hiding?
OK, one more try for Geistkiesel:
Suppose we have two space-trains passing by each other. They are moving inertially, but we don't know which one had accelerated in the past. No experiment can determine which train is "really" moving. We only know that they have a relative speed wrt each other. Now, suppose two explosions is space occur near the trains (but not tied to any of the trains, just instant explosions, at any two times, we don't know yet). The explosions create burn marks on both trains.
SR's claim is: if there were two observers at the mid point of those burn marks (measured in each train separately), they could determine the simultaneity of the explosions according to their train, by looking at the order they see the explosions, because speed of light is constant and distances are the same.
Now, if one of the observers see the explosions at the same time, the other will not. Because, the burn marks and the midpoints on two trains will have a relative speed wrt each other, and therefore the light from the explosions cannot hit both observers at the same time.
So tell me, if one observer sees them at the same time, and both events were simultaneous in both frames, and speed of light is independent of its source, why won't the other see them at the same time? You can't use the excuse "the trains were moving" or "one of them was stationary", because then you assume aether, and then you can't determine simultaneity in any frame with this method without knowing speed wrt aether. Are you saying that simultaneity is affected by speed wrt aether? So what method can you tell me to determine if two events occured simultaneously, if we can't determine speed of a frame wrt aether? (of course in fact there is no sign of aether whatsoever)
Two trains have a relative speed. They both have burn marks on the front and rear and both observers are at the midpoint of their train. So if the speed of light is constant in vacuo how can either, or any observer for that matter, detect the explosion flashes simultaneously inside their own frame unless they were at absolute rest ? I'd like to hear your response on this point before I comment on the rest. I'm sure your response will involve a discussion about length contraction/time dilation so I'm curious how this will work. Seems to me that one half of the train should be contracted while the other half expanded for this to work, and how much contraction/expansion dependent on the frame observing the rest frame. Well, sorry but this sounds ridiculous.
Another point I am confused with is this: if any inertial frame can consider themselves to be at the center of arbitrary emission events, doesn't that equate with the emitters of the light being at rest with respect to each inertial frame? If so, how can any inertial frame detect doppler shifting of light from two continuous emission sources? That is, if there is never any relative velocity between emission sources and the detection frame, how could light be doppler shifted in different inertial frames? Or if Doppler shifting of light was allowed, surely the observer that detects the flashes as non-simultaneous can deduce that his frame was in motion relative to the light emitters and hence the actual time of occurrence of the events?
Where in the hell have you been hiding?
Well you and ram are tough acts to follow so I've basically been reading.
heh, geist is scary. he'll be all quiet and calm for a week then BAM he'll jump in with a sledgehammer and hit you with pages worth of arguments.
Geist writes very well, wish I can do that. Those defending SR here are also brilliant but they have a much more difficult task imo because they are trying to make something that is pretty nonsensical sound rational.
geistkiesel
Jul9-04, 12:10 AM
Two trains have a relative speed. They both have burn marks on the front and rear and both observers are at the midpoint of their train. So if the speed of light is constant in vacuo how can either, or any observer for that matter, detect the explosion flashes simultaneously inside their own frame unless they were at absolute rest ? I'd like to hear your response on this point before I comment on the rest. I'm sure your response will involve a discussion about length contraction/time dilation so I'm curious how this will work. Seems to me that one half of the train should be contracted while the other half expanded for this to work. Well, sorry but this sounds ridiculous.Excuse me for butting in like this but here is one possible soluiton.If I understand the problem we could have a situation like the following.
o ____________m1____________0 ->v1
\ / \ /
x1 x2
/ \ / \
<-v2 o___________m2_____________o
As I have drawn the situation the explosions at x1 and x2 are simultaneous and arrive at the midpoints of each moving frame, by sheer luck, but it is consistent.
The diagram is a set of equilateral triangles here each moving frame is located such that the photons from the explosions arrive simultaneously at the midpoints and end points of each train simultaneously. However the photons do not necessarily arrive on each train at the same time. The arrive on the trains simultaneously. Well maybe with a slight discussion.
The bottom train where the triangles are true equilaterals the photons arrive at the end points and the center point simultaneously. In the other train which may be say twice as far from the explosions as the first, the photons can arrive at the midpoint before the photons arrive at the end points. However, each conclusion will be the same the photons were emitted simultaneoulsy as this is what was observed. It would be a mistake to link the two frames which are completely independent of each other to say that if one sees the explosions simultaneously the other cannot see the explosions simultaneously,
It might help to have clocks that stop when the explosions reach the trains, which would end the speculation.
As I contrived the experiment it is possible for the frames, or an infinite number of frames, to see the explosions as simultaneous. A contrived situation, but possible.
which has a nother day or two to survive and that is just becasue ram1024 is extending a proferssional courtesy
i'm wondering how the heck he knew i was holding something back :O
trying to be unpredictable and he's reading my game
geistkiesel
Jul9-04, 12:22 AM
Another point I am confused with is this: if any inertial frame can consider themselves to be at the center of arbitrary emission events, doesn't that equate with the emitters of the light being at rest with respect to each inertial frame? If so, how can any inertial frame detect doppler shifting of light from two continuous emission sources? That is, if there is never any relative velocity between emission sources and the detection frame, how could light be doppler shifted in different inertial frames? Or if Doppler shifting of light was allowed, surely the observer that detects the flashes as non-simultaneous can deduce that his frame was in motion relative to the light emitters and hence the actual time of occurrence of the events?
I am not sure i have you exactly as you mean, but here is a try.
I have used the dopller shift argument to explain away the claim that two frames cannot determine which is moving with respect to whom. In the standard Einstein train station moving train experiment the flashes can occur in either frame. The frame not holding the source of the emitted photons would naturally see a doppler shift.
Another way to tell how each frame is in motion with respect to the other is to look at the acceleration data of each frame. If three frames were at one time at rest wrt the earth, for instance and each acceleratss, one for 1 g for 5 minutes, another 1 g for 7 minutes and the thirid 1 g for 10 minutes and knowing the mass of each, (easy if standardized frames), relative velocity could be defined by such information.
Two trains have a relative speed. They both have burn marks on the front and rear and both observers are at the midpoint of their train. So if the speed of light is constant in vacuo how can either, or any observer for that matter, detect the explosion flashes simultaneously inside their own frame unless they were at absolute rest ? I'd like to hear your response on this point before I comment on the rest.
From that, I guess you are imagining that speed of light is constant with respect to "aether". No, speed of light is measured the same in all frames (counter intuitive, but experimental fact). Also, I guess you are imagining that the explosions occured simultaneously according to "aether". I didn't say that. They just occur at two times, that's not given. We are investigating wheter they occured simultaneously according to the two trains, by using the two midpoint observers. And they won't agree about simultaneity, that's the main point.
Does that clarify?
geistkiesel
Jul9-04, 12:51 AM
i'm wondering how the heck he knew I was holding something back :O
trying to be unpredictable and he's reading my game
I have a mofdified ram1024 buoy. It might be the same so let me know. It is a six sided cube, for three directions of motion to cover. Assume each set of parallel sides has its own photon source at the midpoint of the sources. If the buoy is moving in one direction, say along the x axis the return reflections from the two sides is detected at the midpoint and a relative velocity wrt the light speed is detected.
The surfaces of the mirrors have photon detectors that determine the time the photons arrived and hence determine the speed and direction direction (hey that's velocity) and by applying a little braking force the x direction velocity can be minimized to what, zero, of course. (I am using a modified Einstein gedanken; The photons moving in the diretion of motion will take longer to reach its intended surface than the light directed at the surface closing in on the oncoming photon. Sounds like more of an engineering problem than physics.)The other two directions can be manipulated the same. In fact the calculations of the velocity in the three directions can be calculated more or less at the same time, ergo, the buoy can be brought to zero velocity, wrt to the speed light., fairly quickly. This would give a very accurate zero velocity - 1 part in 3x10 ^8m/sec for +- 1 meter, better if using centimeter or millilmeter, or what evere the creative genius of the current inertial frame engineers determines.
The rest of the moving stellar objects in the universe can move as they will, but one buoy can determine a zero velocity wrt c.
Is this in line with the ram1024 version? Each frame can radiate a number giving its relative absolute velocity or error correction information. Why would any SRist want to complain about at least trying something like the ram1024 buoy? Hell we've got all the money in the universe, effectively, we being the us of a.
I would only caution about trying to correct all three directions at the same time, after all Mother Nature's last act before launching her creation as a universe was to add time so everything wouldn't happen all at once.
heh it is quite similar to the URF Space-Buoy, but that was a creation i made knowing it would fail, not according to SR precepts per se, but due to my own.
the problems that arose with it were the need to calibrate it accordingly. i had used clocks synchronized at zero-point and then moved into positions using and intricate and precise technique, but SR people refused to let that be the true deciding factor for the device.
they did allow me to synchronize using the midpoint as simultaneous light reception. what this would do would allow the device not to be able to find the Universal Reference Frame (Absolute Stationary), but instead allow the device to calibrate to one specific inertial frame, and be able to detect any deviation from this frame precisely.
in this sense we could define frames throughout the universe simply by knowing the values of deviation from our defined "earth frame". A buoy with known "Earth Frame" values could be set up in another system light years away to measure "Earth Frame" in respect to "That Other System's Frame" just by letting the buoy jet until it reaches "Earth Frame Equilibrium"
hmm didn't mean to write a novel :D but yeah, technically according to SR's constant light speed thing the device you described would have to work... but only if they allowed zero-point synchronization
Another point I am confused with is this: if any inertial frame can consider themselves to be at the center of arbitrary emission events, doesn't that equate with the emitters of the light being at rest with respect to each inertial frame? If so, how can any inertial frame detect doppler shifting of light from two continuous emission sources?
An observer can consider himself to be at rest with the emission point of a single photon, regardless of the relative speed of the photon source. Also same with successively emitted photons, but the emission point for those successive photons will be getting closer/farther as the source moves, so the final effect is doppler shift.
That is, if there is never any relative velocity between emission sources and the detection frame, how could light be doppler shifted in different inertial frames?
I'm a bit lost here. If there is no relative velocity between emitter and detector, ther is no doppler shift. Are you imagining that light itself gets modified? Detected frequency/wavelength is a relation (can't find a better word for this right now) between emitter and detector, light itself remains same.
Or if Doppler shifting of light was allowed, surely the observer that detects the flashes as non-simultaneous can deduce that his frame was in motion relative to the light emitters and hence the actual time of occurrence of the events?
If I don't misunderstand you, no, doppler shift can't be used to detect absolute motion, because it doesn't matter if the emitter or detector is "moving", only relative speed matters.
heh, geist is scary. he'll be all quiet and calm for a week then BAM he'll jump in with a sledgehammer and hit you with pages worth of arguments.
I'd say more like frustrating. At some point you will wear everyone out. What you guys are doing isn't anything like physics. You are just rejecting SR because it seems to defy common sense, which we all understand (no we aren't crazy). But you can't ignore experiment results which are in accordance with SR, not the intuitive Gallilean relativity (I'm not sure Geistkiesel understands even that). Other people aren't stupid, not to mention 100 years of brain power combined. You can see I'm frustrated with you Ram and Geistkiesel too.
geistkiesel
Jul9-04, 06:08 AM
I'd say more like frustrating. At some point you will wear everyone out. What you guys are doing isn't anything like physics. You are just rejecting SR because it seems to defy common sense, which we all understand (no we aren't crazy). But you can't ignore experiment results which are in accordance with SR, not the intuitive Gallilean relativity (I'm not sure Geistkiesel understands even that). Other people aren't stupid, not to mention 100 years of brain power combined. You can see I'm frustrated with you Ram and Geistkiesel too.
No I don't reject SR just because it defies common sense, even though it surely does that, at least you do see that. If you have been following my threads on the subject I use a hell of a lot more than "violations of common sense" if you are paying attention like you claim you are. You might not agree with me and my techniqoues and you might be frustrated, so what? Do you see that ram1024, myself, grounded and eyesaw, to name a few, even yourself once, remember wespe, are all approaching from a different point of view, but guess what there is a focus and we aren't as far apart as we were last month last week or yesterday, right? I suspect each has a secet desire fror recognition at the very least. There is a purpose, personal I mean, in ewhat anybody does, right? And for physics forums, theory development it sure isn't the money, is it? Are you getting paid? I wopuld be soely poissed if Wspe was getting some quid pro quo and I was only getting my regular monthly DSL bills. :cry: :surprise: I can assure you that my pockets are as bare as they were when I started this.
You were a lot more interesting and energetic, to yourself as I observed, even when you realized how far off the nornmal path you actually were. But wasn't it energetic, stimulating, got the old gray cells moving in ways you didn't know they could move?
:surprise:
Wespe no one is saying experimental results aren't what they are.
A stupid trout example. Let us say you have a watch that runs fast. The watch is expensive, the best in town. The only thing you use the watch for is its designed purpose: measuring the speed that trout swim in cold water. Let us say the watch has trout moving 1.2 times what was measured using ordinary watches. Since your watch was so expensive and calibrated, wrongly it seems, it was hard to reject such expensive technology. Trout around the world are now believed to swim in cold water 1.2 faster than heretofore thought since All trout swimmer measurers got a wespe watch that has the same built in error. So when a trout is found not to swim 1.2 faster than wespe's watch expects, it is probably due to using a watch other than wespe's so someone comes up with the excuse that wespe's watch which is used under water etc. is the only accurate measure, therefore the trout found out of sink, (pun intended) get explained away by some inane and inexplicable nonsnese, like dark matter, or ptolemy's circles within circles, restricted fin action, you know hre drill. But every time wespe's watch is used the trout swim at 1.2v as expected. Therefore, only use wespe's watch and all the associated techniques of using the watch will get ya back on the proper physical tack if ya wanna measure trout swimming speed according to physical law: go wespe's watch!!. After a while truth takes on a different turn that is coaxed along with each finding of a conflict. Dissidents are exiled to Theory Development forums, oh the horror, the horror..... :yuck:
I said the example was stupid, but the metaphorcal inanity is in synch with built in "always observed in experiental results" rhetoric.
One thing you have never done is publically or privately, made a decent and honest effort, your own standards, in taking soemone's appraoch and see if there is anything reasonable or appropriate. Check out some of Doc Al responses to my posts. Sometimes he is on, other times he is just screaming in my face and he claims it is physics I am getting. You know all about that don't ya?
I consider tom_mattson a very bright guy, but he doesn't always approach physics as I feel he should, he would agree at least partially on this (not openly, after all he is a mentor and like America's greatest thepian of all times, yes I speak of John Wayne, was fond of saying, "Don't apologize it is a sign of weakness." He might have used the word "pilgrim' in this context here I am not sure.) sometimes just like a normal person he goes screaming at someone repeating SR mantras, do ya dig it? If it weren't for dissidents of fundamental concepts of science and our beloved physics all worhtwhile activity of human kind would stagnate, wouldn't it, hasn't it?
I can tell from this post it is way past my bedtime.
russ_watters
Jul9-04, 08:22 AM
Wespe no one is saying experimental results aren't what they are. What you and the others don't understand or simply won't accept is that what the experimental results say directly contradicts you. When this is pointed out and we ask that you give an example of one experiment that contradicts Relativity, you guys turn conspiracy theorist, claiming the results are faked, experimenters lie, and there is a vast conspiracy among scientists to hide the flaws in Relativity. This is why I (and finally, just about everyone else) have stopped playing this little game of obfuscate-the-thought-experiment with you guys. Are you dishonest? Are you blind? At this point, I don't really care.
No SR is the key result of simultaneity, read the history sir.
Gibberish.
BS. NASCARS are running invariant in speed in my example and you ignored that example.
Unless your NASCARS are running at the speed of light, their speeds are not invariant.
Eintein does not use any SR postulate in arriving at his conclusion.
Of course he does: the invariant speed of light.
He uses only the the staggrered arrival of theh B photon and A photon as detected by the O' observer. You have said on many occasions that this is what all agree.
That "staggered arrival" by itself means nothing. But when coupled with Einstein's postulate of the invariant speed of light, then you can conclude that simultaneity is relative.
quote Doc Al "The fact (agreed by all) the light from B hits O' before the light from A leads the O' observer to conclude the lights could not flash simultameously."
Right--assuming that the speed of light is frame invariant.
How do you know this? They must check the times they are on opposite ends of the train, but each immediately relay their findings to O' who receives the data along with the simultaneous observation of the A and B photons arriving at M simultaneously.
I know it because I know relativity. Realize that your sleight of hand of "relaying their findings" doesn't change their findings.
You are invoking SR time dilation and mass shrinking right?
(What the hell is mass shrinking?) What else could you be invoking to make your statement that the observers at A and B will not record the same time of the emitted photons.
You are babbling again: Observers at A, B, a, and b all record the time that the photons were emitted. A and B record the same times; a and b record different times. The photons know nothing of SR theory or even that there are two observers dutifully recording the times the photons are emitted.
The photons don't have to know SR; but if you wish to make an accurate prediction of how they behave, you'd better know SR. You don't.
the a and b observers record the same time.
On what basis do you make that wild and incorrect claim?
Remember, what you said back in the Lost simultaneoity thread. let me remind you:
quote by Doc Al
"If you understood the Eisntein simple argument you would know that nowhere does "time dilation" or " mass shrinking" enter into it"
Right! Let's be perfectly clear:
Does Einstein need to invoke "time dilation" or "length contraction" to prove that simultaneity is frame dependent in his Train Gedanken: No! All he needs to invoke is the invariant speed of light.
Does that mean that "time dilation" and "length contraction" are irrelevant to a complete analysis of the Train Gedanken? No! To go beyond Einstein's simple argument, and completely describe all events from each frame you must include the effects of time dilation, length contraction, and the relativity of simultaneity. For example: Where is observer a on the train? observer b? What times do their clocks read when they detect those photons being emitted from A and B? Where is the train when these photons are detected? All these questions are from the train frame: to get the answers, you need to understand SR.
geistkiesel
Jul9-04, 10:57 AM
Einstein tells us that the mere staggered arrival time of the photons and that c is constant as everybody knows, is all that the passenger have that they "'must, therfore come to the conclusion" the photons were not emitted simultaneously. The same would hold for the NASCAR hypothetical.
That "staggered arrival" by itself means nothing. But when coupled with Einstein's postulate of the invariant speed of light, then you can conclude that simultaneity is relative.
AE said the staggered arrival times was everything. So you alter AE's conclusions do you? To win an argument, or to keep it going from your naive prerspective and to hell with physics. AE did not inlcude the invariant speed og light as you are trying to stuff it in some crack here in the conversation where it doesn't belong and where only a classic DOc Al smog bank is created. Your light invariant argument doesn't get off the grouind. It was redlined Doc. That kite doesn't fly.
Einstein says: "Hence, the observer will see the the beam of light emitted from B earlier than he wills see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash at B took place earlier than the lightning flash A. We thus arrive at the important result: Events which are simultaneous with reference to the embankment are not simultaneous wrt the train and vice versa."
Also if O' weren't moving O' would see the lights emitted simultaneously.
I read, motion, staggered arrival and the speed of light is c.
So what I infer from this is that the observers that see the lights arrive simultaneously at M', here frame wise no different than where O' had just measured the arrival of B, there would be no loss of simultaneity as the photons would have arrived simultaneously. All your postulates and arguments are included. I also conclude that to the O' observer the photons were not emitted simultaneously in the moving frame, but only by definition, and that for the a|b observers at the midpoint M when the photons A and B arrive simultaneously, the emitted photons were simulaneously emitted in the stationary and moving frame.
Therefore the Sr theory predicts that on the same frame there are a multitude of observers seeing the simultaneous arrival of the photons as well as a multitude seeing no simultaneity.
The observers at A and B, a and b, are also recorded the times the photons were emitted and must also conlcude the photons were emitted simultaneously. Assuming the observes a and b know they are measuring photons that are simultaneously emitted in the stationary frame, they must conclude they are measuring the same as their partner on the olther end of the frame. The a and b clocks are moving frame synchronized, therefore each clock always reads the same. Therefore the a and b clocks a must read the same as the clocks used by the observers at M viewing the arrival of the A and B photons arriving simultaneously there. and of course, O'.
Another problem you have. Assume the stationary frame emits ionly one photon instead of two. Will the photon anticipate which photons it was that had photons been emitted simultaneously from both sources?
If the B photon is exclusively emitted will it be measured the same instant of emittance as in the case the A photon were emitted also? How do the photons know they are a pair and that one must precede the other? Nonlocal entanglement?
How do you come to the conclusions you do? If the a and b observers and the a|b observers all relay their times to O' the signals will arrive simultaneously with the A photon. Hence O' will have two unambiguous data sources indicating simultaneously emitted photons. But you know different, why?
I know it because I know relativity. Realize that your sleight of hand of "relaying their findings" doesn't change their findings.
I never said it did change their findings. I was merely giving you some alternative to observations of passengers on Einstein's train that conclude the photons were emitted simultaneously in the moving frame, obviously.
This hypo of AE is no different tha the conclusion spectators at the NASCAR race I described will also conclude the NASCARS emitted form A and B emitted sinmultaneously as determined by the moving Ford Futura, the analog of O', as well as 100,000 spectators viewing the event.
(What the hell is mass shrinking?)
You are babbling again: Observers at A, B, a, and b all record the time that the photons were emitted. A and B record the same times; a and b record different times.
The photons don't have to know SR; but if you wish to make an accurate prediction of how they behave, you'd better know SR. You don't.
Prove the a and b and the a|b passengers will not measure the same as the A and B observers. "I know it because I know relativity." This is your physics, "I know", is your physics?. The forum is there for you to put your knowledge on the line and to prove what you know.
What do we do in the alternative, just start calling each other names and making the wise cracks? By the way, if you were looking for the job of smart ***, that job is already filled, and by a real professional. Need I say more?
On what basis do you make that wild and incorrect claim?
Wild guess, maybe or finally, but it turned out OK, like some of the other wild guesses that turn out ok. Incorrect though, you must show those with more than "I know".
From the reason and logic learned in the study of physics. Einstein didn't use a gamma in his gedanken, nor did he infer the existence of a gamma regarding the conclusions derivable form the gedanken, but Doc Al can hardly wait. However,I anticipate no such physics will be forthcoming from Doc Al.
Right! Let's be perfectly clear:
Does Einstein need to invoke "time dilation" or "length contraction" to prove that simultaneity is frame dependent in his Train Gedanken: No! All he needs to invoke is the invariant speed of light.
That is what I thought, there is no definitive SR postulate or imperative that proves the a and b observers did not observe the photons emitted simultaneously, and the same ogres for th a|b [passengers agt the midpoint when A and B pohoton s arrive there simultaneously.
Does that mean that "time dilation" and "length contraction" are irrelevant to a complete analysis of the Train Gedanken? No! To go beyond Einstein's simple argument, and completely describe all events from each frame you must include the effects of time dilation, length contraction, and the relativity of simultaneity.
I knew it, you flim flammed us, or are trying to.
Try looking at this again:
"Hence, the observer will see the the beam of light emitted from B earlier than he wills see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash at B took place earlier than the lightning flash A. We thus arrive at the important result: Events which are simultaneous with reference to the embankment are not simultaneous wrt the train and vice versa."
You mean that if we just use different observers than O', then you have to use time dilation and length contraction, AKA mass shrinking.? Sounds like a fast swimming trout story to me.
Bogus Doc, you can do much better. You seem to think nobody remembers what you just posted above, in this post. AE didn't need it but Doc Al needs it in order to smog up the discussion and keep some semblance of SR alive. Chill DOc. ram1024 has given everybody another couple of days, Relax, prepare yourself for the inevitable. Don't embarrass yourself.
For example: Where is observer a on the train? observer b?
a and b are at A and B when the photons were emitted into the moving frames, which is recorded by their clocks. a nad b obseved the phtons being emitted. a and b recorded the time of the emitted photons when emitted simulataneoslh at A and B.
What times do their clocks read when they detect those photons being emitted from A and B?
t' = 0. The same as the O' observer who zeroed her clock when she arrived at the midpoint just as the photons were emitted simultaneously from A and B. Long before the O' detected the staggered arrival of the B and A photons. Come on we have done this scene many times before.
Where is the train when these photons are detected?
The train? Come on Doc, please. Doc the passengers are riding on the train. a is at A, b is at B and O' is at M, when the clocks were set to zero, remember?
All these questions are from the train frame: to get the answers, you need to understand SR.
To get the answers? I do know SR. I even understand it. It's just isn't true that's all. The a and b observer record the A and B photon simultaneously when emitted at A and B, as do the a|b observers when the A and B photons arrive at M simultaneously.
You think you are going to blind side the maths ignorant geistkiesel do you? Try me Doc. I dare you to try me.
Ese el mentor!!: la li'nea habe dibujado en el desierto con lamis espada. Aqui! ahora!
Question Doc Al: How do the a and b observers know there are two photons emitted from the stationary frame?
or said another way, how do the photons know they must change their emittance protocol because the photons are about to be emitted into the moving frame? How do the photons know which one must emit first, and at what what time? I think you are digging yourself a perception of the observers hole instead of building a phyisics arguement.
.
Einstein tells us that the mere staggered arrival time of the photons and that c is constant as everybody knows, is all that the passenger have that they "'must, therfore come to the conclusion" the photons were not emitted simultaneously.
Yes, and your point is what? "c is constant" means "the speed of light is invariant".
AE said the staggered arrival times was everything. So you alter AE's conclusions do you? To win an argument, or to keep it going from your naive prerspective and to hell with physics. AE did not inlcude the invariant speed og light as you are trying to stuff it in some crack here in the conversation where it doesn't belong and where only a classic DOc Al smog bank is created. Your light invariant argument doesn't get off the grouind. It was redlined Doc. That kite doesn't fly.
It's time to up the dosage on your medication, geistkiesel. What do you think "constant speed of light" means?
The observers at A and B, a and b, are also recorded the times the photons were emitted and must also conlcude the photons were emitted simultaneously.
Wrong.
Assuming the observes a and b know they are measuring photons that are simultaneously emitted in the stationary frame, they must conclude they are measuring the same as their partner on the olther end of the frame.
Wrong. All a and b do is observe the times of emission and record it. They observe different times.
The a and b clocks are moving frame synchronized, therefore each clock always reads the same. Therefore the a and b clocks a must read the same as the clocks used by the observers at M viewing the arrival of the A and B photons arriving simultaneously there. and of course, O'.
After all these posts you still haven't the foggiest notion of what's going on. Clocks a and b are syncronized in the O' frame; so of course they read "the same time" in the O' frame. But emissions at A and B do not happen at the same time in the O' frame.
Another problem you have. Assume the stationary frame emits ionly one photon instead of two. Will the photon anticipate which photons it was that had photons been emitted simultaneously from both sources?
Huh? Totally incoherent. What are you talking about?
If the B photon is exclusively emitted will it be measured the same instant of emittance as in the case the A photon were emitted also? How do the photons know they are a pair and that one must precede the other? Nonlocal entanglement?
Huh? Totally incoherent. What are you talking about?
How do you come to the conclusions you do? If the a and b observers and the a|b observers all relay their times to O' the signals will arrive simultaneously with the A photon. Hence O' will have two unambiguous data sources indicating simultaneously emitted photons. But you know different, why?
You keep claiming that a and b measure the same time? Where do you get this??? Read Einstein's train gedanken one more time.
Try looking at this again:
"Hence, the observer will see the the beam of light emitted from B earlier than he wills see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash at B took place earlier than the lightning flash A. We thus arrive at the important result: Events which are simultaneous with reference to the embankment are not simultaneous wrt the train and vice versa."
I think you are the one who should look at this again. It obviously hasn't sunk in yet.
a and b are at A and B when the photons were emitted into the moving frames, which is recorded by their clocks. a nad b obseved the phtons being emitted. a and b recorded the time of the emitted photons when emitted simulataneoslh at A and B.
a and b record the times that A and B emitted the photons. OK, so what time is that, geistkiesel? I want the actual time seen on the O' clocks in terms of L, v, and c. (Hint: look back over the many posts I've made: I spell it all out for you.) A and B only flash simultaneously in the O frame. (Or have you already forgotten the Einstein quote you just entertained us with?)
t' = 0. The same as the O' observer who zeroed her clock when she arrived at the midpoint just as the photons were emitted simultaneously from A and B. Long before the O' detected the staggered arrival of the B and A photons. Come on we have done this scene many times before.
Yes we have. And yet you still get it wrong!
Clocks at A and B time the flashes to occur at t = 0. All of the clocks in O' (including a and b) are synchronized in the O' frame. We agree that when M' passes M their two clocks both read zero. Everything else must be deduced.
Question Doc Al: How do the a and b observers know there are two photons emitted from the stationary frame?
or said another way, how do the photons know they must change their emittance protocol because the photons are about to be emitted into the moving frame? How do the photons know which one must emit first, and at what what time? I think you are digging yourself a perception of the observers hole instead of building a phyisics arguement.
What are you babbling about now? What's this nonsense about photons changing their "emittance protocol"? And photons conspiring together? :rofl:
Einstein's train gedanken is as simple a set up as you are going to get (but obviously not simple enough!). Two lights flashing. No conspiracy. No "emittance protocol". Just the invariant speed of light.
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