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But they're not and that's whole point; you can't use the assumptions of Galilean relativity to disprove special relatvity.
Today Special Relativity dies
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ram1024
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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...
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...
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ram1024 said:
No, it is a postulate. A postulate that has been experimentally confirmed in numerous ways.
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ram1024 said:
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!
The point that you aren't getting is that you can't simply declare[/color] 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[/color] be simultaneous in any other frame.
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
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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.
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.
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a vbiew from one rest frame:
A view from another:
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 [itex]x^{\mu}_1[/itex] and [itex]x^{\mu}_2[/itex] (where [itex]x^0 = t[/itex])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:
[tex]x^{\nu}_1 = \left(\begin{array}{c}0\\{\frac{l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x^{\nu}_2 = \left(\begin{array}{c}0\\{\frac{-l}{2}}\\0\\0\end{array}\right)[/tex]
Let [itex]x'^{\mu}_1[/itex] and [itex]x'^{\mu}_2[/itex] 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:
[tex]x'^{\mu}_1 = {\Lambda^{\mu}}_{\nu}x^{\nu}_1[/tex]
[tex]x'^{\mu}_2 = {\Lambda^{\mu}}_{\nu}x^{\nu}_2[/tex]
therefore:
[tex]x'^{\mu}_1 = \left(\begin{array}{c}{\frac{-\gamma l\beta}{2c}}\\{\frac{\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x'^{\mu}_2 = \left(\begin{array}{c}{\frac{\gamma l\beta}{2c}}\\{\frac{-\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
So in the first frame the photons are emitted both at t = 0
but in the second frame the're emitted at [itex]\frac{-\gamma l\beta}{2c}[/itex] and [itex]\frac{\gamma l\beta}{2c}[/itex] respectively which are only equal if u = 0.
They're not the same clock as they're not local.
Code:
|> o <|
Code:
|-------->o<--------|A view from another:
Code:
|> 0 |
Code:
----|-------->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 [itex]x^{\mu}_1[/itex] and [itex]x^{\mu}_2[/itex] (where [itex]x^0 = t[/itex])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:
[tex]x^{\nu}_1 = \left(\begin{array}{c}0\\{\frac{l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x^{\nu}_2 = \left(\begin{array}{c}0\\{\frac{-l}{2}}\\0\\0\end{array}\right)[/tex]
Let [itex]x'^{\mu}_1[/itex] and [itex]x'^{\mu}_2[/itex] 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:
[tex]x'^{\mu}_1 = {\Lambda^{\mu}}_{\nu}x^{\nu}_1[/tex]
[tex]x'^{\mu}_2 = {\Lambda^{\mu}}_{\nu}x^{\nu}_2[/tex]
therefore:
[tex]x'^{\mu}_1 = \left(\begin{array}{c}{\frac{-\gamma l\beta}{2c}}\\{\frac{\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x'^{\mu}_2 = \left(\begin{array}{c}{\frac{\gamma l\beta}{2c}}\\{\frac{-\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
So in the first frame the photons are emitted both at t = 0
but in the second frame the're emitted at [itex]\frac{-\gamma l\beta}{2c}[/itex] and [itex]\frac{\gamma l\beta}{2c}[/itex] respectively which are only equal if u = 0.
They're not the same clock as they're not local.
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ram1024 said:
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[/color] 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?
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).
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.
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The problem is that you are trying to get people to accept something that they know isn't acceptable.ram1024 said:
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
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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
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
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ram1024 said:
No, I meant what I said: It doesn't work according the real universe.
No, you aren't disproving it. You are simply denying it. Try to understand the difference.
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[/color] those premises.
And the answer to both sets of questions is the same: Because that's the way it is.[/color]
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ram1024 said:
They don't BECOME unscyhronised, they ARE unsychronised in different inertial reference frames.
ram1024
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sure i can win any argument with "because i said so" too :|
that tells us nothing
that tells us nothing
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ram1024 said:
!
That's exactly 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.
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[/color]". It says so when we ask it, via experimientation.
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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
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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.
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.
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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.
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ram1024 said:
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[/color] 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[/color] discern that the clocks don't tick at the same rate for all observers.
ram1024
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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:
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
let's move onto stage 2. i think i have enough of an idea of where we're at to continue now.
Case #4:
Code:
observer1
[u]|(-> (o) <-)|[/u]
[u]|(-> (o) <-)|[/u]
[u]|(-> (o) <-)|[/u]
[u]|(-> (o) <-)|[/u]
Code:
observer2
[u]|(-> (o) <-)|[/u]
[u]|(-> (o) <-)|[/u]
[u]|(-> (o) <-)|[/u]
[u]|(-> (o) <-)|[/u]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
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ram1024
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stupid "code" screwing up my pretty trains
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ram1024 said:
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.
It is ludicrous to the moving observer. You have absolutely no basis for telling him that what he measures isn't real.
In which frame?
In which frame?
In which frame?
ram1024
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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?
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?
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ram1024 said:
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.
In which frame?
ram1024
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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
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
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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?
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
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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 ?
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
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ram1024 said:
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
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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
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
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ram1024 said:
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.
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ram1024 said:
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
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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