Today Special Relativity dies

In summary, the conversation discusses the simultaneous emission and detection of photons in different reference frames, specifically in the case of a man standing on a movable platform bed between two photon emitters. The outcome varies depending on the reference frame and the movement of the platform. The ambiguity lies in the lack of definition of the specific reference frames in each case.
  • #281
geistkiesel said:
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).
 
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  • #282
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.
 
  • #283
geistkiesel said:
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.
 
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  • #284
ram1024 said:
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.
 
  • #285
wespe said:
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.
 
  • #286
what is "that proves that" ?

what were you trying to prove?
 
  • #287
Doc Al said:
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.


Doc al said:
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.
 
  • #288
ram1024 said:
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.
 
  • #289
Janus said:
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.
 
  • #290
Hurkyl said:
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.
 
  • #291
geistkiesel said:
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 &Delta;x and &Delta;t, 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]
 
  • #292
Doc Al said:
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.
 
  • #293
ram1024 said:
Case #7
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]

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.

Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                     <-)|[/u]
                     [u](o)                                  <-)|[/u]
                     [u](o)                               <-)|[/u]

emitter moves towards the observer.

Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                           <-)|[/u]
                     [u](o)                                              <-)|[/u]
                     [u](o)                                                 <-)|[/u]

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...

Code: 
                     [u](o)                                        <-)|[/u]
                        [u](o)                                     <-)|[/u]
                           [u](o)                                  <-)|[/u]
                              [u](o)                               <-)|[/u]

Code: 
                     [u](o)                                        <-)|[/u]
                  [u](o)                                           <-)|[/u]
               [u](o)                                              <-)|[/u]
            [u](o)                                                 <-)|[/u]

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,.
 
  • #294
ram1024 said:
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:
 
  • #295
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.
 
  • #296
Tom Mattson said:
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 traveled 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 traveled 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.
 
  • #297
geistkiesel said:
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.
 
  • #298
Hurkyl said:
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?
 
  • #299
Do you understand the concept of a distance versus time plot?

Does this look like a drawing of the experiment you describe?

[tex]
\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}
[/tex]
 
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  • #300
Tom Mattson said:
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.
 
  • #301
Who need SR in this problem?

Hurkyl said:
Do you understand the concept of a distance versus time plot?

Does this look like a drawing of the experiment you describe?

[tex]
\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}
[/tex]


Code: 
__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?
 
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  • #302
Oh wait, is (A, M, B) supposed to be the "stationary" letters?

Okay, does this diagram look like your experiment?

[tex]
\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}
[/tex]
 
  • #303
Tom Mattson said:
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.

Tom_mattson said:
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.
 
  • #304
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. ...
 
  • #305
geistkiesel said:
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?
 
  • #306
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.
 
  • #307
geistkiesel said:
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.
 
  • #308
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.
 
  • #309
ram1024 said:
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.
 
  • #310
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.
 
  • #311
Tom Mattson said:
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
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]
                     [u](o)                                        <-)|[/u]

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
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                     <-)|[/u]
                     [u](o)                                  <-)|[/u]
                     [u](o)                               <-)|[/u]

emitter moves towards the observer.


Step: 3
Code: 
                     [u](o)                                        <-)|[/u]
                     [u](o)                                           <-)|[/u]
                     [u](o)                                              <-)|[/u]
                     [u](o)                                                 <-)|[/u]

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:
Code: 
                     [u](o)                                        <-)|[/u]
                        [u](o)                                     <-)|[/u]
                           [u](o)                                  <-)|[/u]
                              [u](o)                               <-)|[/u]

Step 5:
Code: 
                     [u](o)                                        <-)|[/u]
                  [u](o)                                           <-)|[/u]
               [u](o)                                              <-)|[/u]
            [u](o)                                                 <-)|[/u]

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:
 
Last edited by a moderator:
  • #312
ram1024 said:
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.
 
  • #313
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?
 
  • #314
ram1024 said:
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:
 
  • #315
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
 

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