Quote by Doc Al I really don't know the situation you are describing any longer. I thought A and B were the light sources. What are A' and B' ? Have you lost interest in discussing Einstein's train example? I recommend getting that one straight before adding in complications.
No I am not trying to vary from the original experiment. I ask: if we replace the location of the sources with photons located at the same stationary frame location, are we changing any essential physical parameter of the experiment? If we neglect any source of photons A and B iinformation and define M when photons were at A and B are the physics changed?

I intend nothing more than to minimize experimental clutter: hence, equating the 'photon source' with 'photon' we say that "M is the midpoint of the A and B photons when the moving observer M' is located at M." We define M then, not wrt to a measured physical length of the lines AM and BM in meters, rather the distance the photon moves in time t from A to M and B to M. If you disagree then please explain. [unprimed = stationary, primed = moving]

The instantaneous locations of the essential and identifiable objects are at
1. M'(t'0,) = M(t0)
2. where M' = M'(t'1) when the B photon arrives at t'1.
3. where M' = M'(t'2) when the A photon arrives at t'2.

Hence, all previous measured events are located on the moving frames as determined by counting back from M' at t2 as: t'2 -> t'1 -> t'0 to locate the previous instantaneous locations of the photons wrt 0,0 in the moving frame.

The frame moves as before, but all positions on the frame are determined by the positions of the detected emitted photons.

Code:
A(t'0)_____A(t'1)>            MM'(t'0)            <B(t'1)____A(t'2)>____B(t'0)
Figure 1
Location scenario of the instantaneous location of MM'(t0, t'0) on the moving frame wrt A(t2).

Determine O' conclusions regarding the simultaneous emission of the A and B photons from information of the observed arrival sequence of the A and B photons in the moving frame, starting with:
1. Simultaneity equivalence with conjectured quivalence of M'(t'0) wrt M'(t'0(M'(t'2)))? then,
2. with no consideration by O' of any of her [speculated] motion wrt A and B in the stationary platfom.
3. with consideration of any of her [speculated] motion wrt A and B in the stationary frame then.
4. what is the scenario of stationary emissions such that O' would conclude the emissions of photons in the stationary frame were emitted simultaneously in the moving frame? I ask: IF not simultaneous in moving but simultaneous in stationary, then what non simultaneous sequence in stationary produces simultaneous in moving? Assuming symmetry proper?

 Quote by russ_watters I think there is a reason for adding the complications...
Besides the possibility that I am trying to trick Doc Al what is the "reason" for adding complictions? Actually, I look at it as minimizing complications.

When first learning SR did you grasp the essence and fundamentals imediately such that all SR problems and arrangments and descriptions posed were solved with ease? Isn't it a fact that at sometime in your educational process SRT stressed your logical and understanding functions?

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 Quote by geistkiesel No I am not trying to vary from the original experiment. I ask: if we replace the location of the sources with photons located at the same stationary frame location, are we changing any essential physical parameter of the experiment? If we neglect any source of photons A and B iinformation and define M when photons were at A and B are the physics changed?
Once again, I highly recommend sticking to Einstein's original simple argument. Either your more abstract one is somehow equivalent to that, or it's not. Either way, at some point I expect you to pinpoint the flaw in Einstein's argument. There is no way around that hurdle. Note that Einstein's argument requires no understanding of relativity whatsoever except the concept of the invariant speed of light.

Nonetheless... So both frames are observing photons heading towards each other. Let's see what happens...
 I intend nothing more than to minimize experimental clutter: hence, equating the 'photon source' with 'photon' we say that "M is the midpoint of the A and B photons when the moving observer M' is located at M." We define M then, not wrt to a measured physical length of the lines AM and BM in meters, rather the distance the photon moves in time t from A to M and B to M. If you disagree then please explain. [unprimed = stationary, primed = moving]
By getting rid of the light sources at A and B, which were fixed with respect to the stationary frame, you risk changing the simple scenario. M is no longer "the midpoint between the light sources A and B", which was unambiguous and agreed upon by all frames. M is also not the "midpoint of the photons A and B" since the moving frame does not agree that she is at the midpoint of the two photons when her clock reads t'=0.
 The instantaneous locations of the essential and identifiable objects are atM'(t'0,) = M(t0) where M' = M'(t'1) when the B photon arrives at t'1. where M' = M'(t'2) when the A photon arrives at t'2.
Allow me to translate:
(1) The moving observer passes by M (the midpoint of photons according to the stationary frame at t=0) at t'=0.
(2) She intercepts photon B when her clock reads t'=t'1.
(3) She intercepts photon A when her clock reads t'=t'2.
 Hence, all previous measured events are located on the moving frames as determined by counting back from M' at t2 as: t'2 -> t'1 -> t'0 to locate the previous instantaneous locations of the photons wrt 0,0 in the moving frame.
It sounds like you are assuming that the moving observer thinks the photons are at A and B when her clock says t' = 0. Not true!
 The frame moves as before, but all positions on the frame are determined by the positions of the detected emitted photons. Code: A(t'0)_____A(t'1)> MM'(t'0) ____B(t'0) Figure 1 Location scenario of the instantaneous location of MM'(t0, t'0) on the moving frame wrt A(t2).
Again, I don't understand this diagram. It looks like you are assuming that the moving observer is at the midpoint of the photons when her clock reads t' = 0. That's not true.

Just for the record, let's find out where the moving observer thinks those photons are when her clock reads t'=0.

I will assume that in the "stationary" frame (O) at time t=0, the photons are at A (x = -L; t = 0) and at B (x = L; t = 0). (I assume that A and B are a distance L from the midpoint M, according to the stationary frame.)

So where are these photons at t'=0 according to the moving frame?
Photon A (at t'=0) is at:
$$x\'\; = -\gamma L -\gamma \frac {vL}{c}$$

Photon B (at t'=0) is at:
$$x\'\; = \gamma L -\gamma \frac {vL}{c}$$

Where, as usual, $\gamma = 1/\sqrt{(1 - \frac{v^2}{c^2})}$.

Now, is your scenario really less complicated than Einstein's?

 Quote by Doc Al Once again, I highly recommend sticking to Einstein's original simple argument. Either your more abstract one is somehow equivalent to that, or it's not. Either way, at some point I expect you to pinpoint the flaw in Einstein's argument. There is no way around that hurdle. Note that Einstein's argument requires no understanding of relativity whatsoever except the concept of the invariant speed of light. Nonetheless... So both frames are observing photons heading towards each other. Let's see what happens... By getting rid of the light sources at A and B, which were fixed with respect to the stationary frame, you risk changing the simple scenario. M is no longer "the midpoint between the light sources A and B", which was unambiguous and agreed upon by all frames. M is also not the "midpoint of the photons A and B" since the moving frame does not agree that she is at the midpoint of the two photons when her clock reads t'=0. Allow me to translate: (1) The moving observer passes by M (the midpoint of photons according to the stationary frame at t=0) at t'=0. (2) She intercepts photon B when her clock reads t'=t'1. (3) She intercepts photon A when her clock reads t'=t'2. It sounds like you are assuming that the moving observer thinks the photons are at A and B when her clock says t' = 0. Not true! Again, I don't understand this diagram. It looks like you are assuming that the moving observer is at the midpoint of the photons when her clock reads t' = 0. That's not true. Just for the record, let's find out where the moving observer thinks those photons are when her clock reads t'=0. I will assume that in the "stationary" frame (O) at time t=0, the photons are at A (x = -L; t = 0) and at B (x = L; t = 0). (I assume that A and B are a distance L from the midpoint M, according to the stationary frame.) So where are these photons at t'=0 according to the moving frame? Photon A (at t'=0) is at: $$x\'\; = -\gamma L -\gamma \frac {vL}{c}$$ Photon B (at t'=0) is at: $$x\'\; = \gamma L -\gamma \frac {vL}{c}$$ Where, as usual, $\gamma = 1/\sqrt{(1 - \frac{v^2}{c^2})}$. Now, is your scenario really less complicated than Einstein's?

One correction only: The given from Einstein is that the moving oberver O' is at M' on th etrain which is at M in the stationary frame when the photons were emitted from A and B in the stationary frame. Whatever the moving observer concludes later is her business, but let us stick to the original experiment.
This inserted as an afterthought.

Let us return to the given. When the observer O' is at M' in the train, M' is moving through M the very instant the photons were emitted from A and B in the stationary frame. This you and I know.

Take two cases:
1. The moving observer knows the photons were emitted in the stationary frame when M'(t'0) was colocated with M in the stationary frame and that M was the midpoint of A and B in the stationary frame.
2. The moving observer knows nothing about the emitted photons, only that she moves through M when M'(t'0) was colocated with M.

You have answered these questions and we both agree SR predicts, or states that the A and B emitted photons were not perceived as emitted simultaneously in her moving frame in both cases above, correct?

I wanted to remove the sources from the problem to remove ambiguity. I see you have an objection.

Let me form the final question using both cases above, which differ only by the information the moving observer has regarding the emitted photons being simultaneous in the stationary frame. The time now is one week after O' has finished her calculations and we are all sitting aorund a table discussing th experimental results. Some idiot proposes the following scenario for consideration:

At the instant the B photon was detected, t'1 in the moving frame, the A photon is assumed to have been emitted at some time before t'1 and is located some place along the A'B' line, correct? (where the A'B' line is the moving frame reference line). We know the values t'0, t'1, t'2, and v', where v' is measured wrt the stationary frame (meters'/second' are moving frame values). We know that the photons from A and B both move at the same velocity in the moving frame, or v(A') = v(B') correct? We are not saying v(A') = v(A), only that the moving observer will measure v(A') = v(B'), correct?

Using the information given above determine the location of the A' photon wrt M'(t'0) when the B' photon was detected at t'1. Wherever A'(t'1) was located it had to travel a distance (t'2 - t'1)c to reach the point referenced by the t'2 point, correct? The train moved a distance v'(t'2- t'1) during this same time span, correct? We assume that M' in the train remains constant wrt the train. This is where all train times are measured.

So where was A'(t'1) wrt M'(t'0) and B'(t'1) 'in terms of the primed values?

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 Quote by geistkiesel One correction only: The given from Einstein is that the moving oberver O' is at M' on th etrain which is at M in the stationary frame when the photons were emitted from A and B in the stationary frame. Whatever the moving observer concludes later is her business, but let us stick to the original experiment.[/b] This inserted as an afterthought.
Ah...so we're back to discussing Einstein's original setup? Good. (But you still haven't addressed Einstein's argument yet.) So by "train" I assume you mean the moving observer O'. And, once again, M is the midpoint between the two light sources at A and B. O' passes M at the moment that the photons are emitted according to the stationary frame. That moment is at time t' = 0, according to O'. The stationary frame thinks that the lights were turned on at t = 0.
 Let us return to the given. When the observer O' is at M' in the train, M' is moving through M the very instant the photons were emitted from A and B in the stationary frame. This you and I know.
Once again, precision is essential: O' passes M at the moment that the clock at M says t = 0 (and the O' clock says t' = 0). The stationary frame claims that the lights at A and B were emitted simultaneously according to the stationary frame at t = 0.
 Take two cases:The moving observer knows the photons were emitted in the stationary frame when M'(t'0) was colocated with M in the stationary frame and that M was the midpoint of A and B in the stationary frame. The moving observer knows nothing about the emitted photons, only that she moves through M when M'(t'0) was colocated with M.
Again, the state of knowledge of O' is irrelevant.
 You have answered these questions and we both agree SR predicts, or states that the A and B emitted photons were not perceived as emitted simultaneously in her moving frame in both cases above, correct?
If the stationary frame observes the lights emit simultaneously, then the moving frame will disagree.
 I wanted to remove the sources from the problem to remove ambiguity. I see you have an objection.
 Let me form the final question using both cases above, which differ only by the information the moving observer has regarding the emitted photons being simultaneous in the stationary frame. The time now is one week after O' has finished her calculations and we are all sitting aorund a table discussing th experimental results. Some idiot proposes the following scenario for consideration: At the instant the B photon was detected, t'1 in the moving frame, the A photon is assumed to have been emitted at some time before t'1 and is located some place along the A'B' line, correct? (where the A'B' line is the moving frame reference line).
OK. We will assume that the photon from A has been emitted before the photon at B is detected at time t' = t'1 according to the moving frame O'. (Note that this implies that v < c/2.)
 We know the values t'0, t'1, t'2, and v', where v' is measured wrt the stationary frame (meters'/second' are moving frame values). We know that the photons from A and B both move at the same velocity in the moving frame, or v(A') = v(B') correct? We are not saying v(A') = v(A), only that the moving observer will measure v(A') = v(B'), correct?
I don't know what you mean by "v". If you are talking about the speed of the photons, then ALL observers will measure the speed of the photons with respect to their own frames as having speed c.
 Using the information given above determine the location of the A' photon wrt M'(t'0) when the B' photon was detected at t'1. Wherever A'(t'1) was located it had to travel a distance (t'2 - t'1)c to reach the point referenced by the t'2 point, correct?
Wherever the photon from A is when the photon from B is detected by O' (at t' = t'1) then that photon must travel a distance (t'2 - t'1)c according to the moving frame before it reaches O'.
 The train moved a distance v'(t'2- t'1) during this same time span, correct?
No. The moving frame claims that the stationary frame has moved a distance v(t'2 - t'1) during that time. (v is the relative speed of the two frames.) The stationary frame will disagree.
 We assume that M' in the train remains constant wrt the train. This is where all train times are measured.
I assume you mean that O' remains at x' = 0 in her moving frame? Right.
 So where was A'(t'1) wrt M'(t'0) and B'(t'1) 'in terms of the primed values?
Not sure what you are asking here. O' is at location x' = 0 in her frame. When she detects the photon from B (I assume that's what you mean by B'(t'1)) then that photon is at x' = 0. (And when she detects the photon from A, that photon will be at x' = 0.) If you are asking: Where is the photon from A at the moment that O' detects the photon from B according to the moving frame? Here you go:
$$x' = -2\gamma \frac {vL}{c}$$
Did you have a point to make here?

You've asked quite a few questions, but you still haven't answered mine. Where is the flaw in Einstein's train gedanken experiment?

 Quote by Doc Al Ah...so we're back to discussing Einstein's original setup? Good. (But you still haven't addressed Einstein's argument yet.) So by "train" I assume you mean the moving observer O'. And, once again, M is the midpoint between the two light sources at A and B. O' passes M at the moment that the photons are emitted according to the stationary frame. That moment is at time t' = 0, according to O'. The stationary frame thinks that the lights were turned on at t = 0. Once again, precision is essential: O' passes M at the moment that the clock at M says t = 0 (and the O' clock says t' = 0). The stationary frame claims that the lights at A and B were emitted simultaneously according to the stationary frame at t = 0. Again, the state of knowledge of O' is irrelevant. If the stationary frame observes the lights emit simultaneously, then the moving frame will disagree. That just added complexity. OK. We will assume that the photon from A has been emitted before the photon at B is detected at time t' = t'1 according to the moving frame O'. (Note that this implies that v < c/2.) I don't know what you mean by "v". If you are talking about the speed of the photons, then ALL observers will measure the speed of the photons with respect to their own frames as having speed c. Wherever the photon from A is when the photon from B is detected by O' (at t' = t'1) then that photon must travel a distance (t'2 - t'1)c according to the moving frame before it reaches O'. No. The moving frame claims that the stationary frame has moved a distance v(t'2 - t'1) during that time. (v is the relative speed of the two frames.) The stationary frame will disagree. I assume you mean that O' remains at x' = 0 in her moving frame? Right. Not sure what you are asking here. O' is at location x' = 0 in her frame. When she detects the photon from B (I assume that's what you mean by B'(t'1)) then that photon is at x' = 0. (And when she detects the photon from A, that photon will be at x' = 0.) If you are asking: Where is the photon from A at the moment that O' detects the photon from B according to the moving frame? Here you go:$$x' = -2\gamma \frac {vL}{c}$$Did you have a point to make here? You've asked quite a few questions, but you still haven't answered mine. Where is the flaw in Einstein's train gedanken experiment?
V is the velocity of the moving frame wrt the stationary frame.

Why do you say that the moving frame sees the platform moving? Isn't this just an arbitrary move on her part? She determined her velocity as v wrt the stationary frame. She detected the B photon at t1, where was the A photon at this instant in terms of t0,t1,t2 and v?

 Quote by Doc Al No. The moving frame claims that the stationary frame has moved a distance v(t'2 - t'1) during that time. (v is the relative speed of the two frames.) The stationary frame will disagree. I assume you mean that O' remains at x' = 0 in her moving frame? Right.
Doc Al, use the word "stationary" to mean what the word says, v = 0.

The observer is always at M' where all the measurements are taking place.
v is the velocity of the moving frame wrt the stationary frame. I have never used the terms x' = 0. I have consistently said that O' the observer is located at M' where she makes all measurements.

Here is one objection: When we say stationary frame this means "stationary" . Now for her to say : It is moving not me, is changing a fact that she cannot do so easily. Stationary means v = 0. She can't switch. Even if she does, why avoid the simple task of defining A in terms of all moving frame experimentally collected values? v is in terms of m/s, all of which are dilated, I assume. Her clocks are working. It should be a straight forward calculation to find A wrt the known measured parameters. Certainly there are enough to solve the problem as requested. S do it.

Again, it is not absolutely necessary that she now says that the platform is moving and she is stationary. [Sounds like a smoke screen just went up.] The A photon is sitting out there heading to O' the observer at M', where M' was colocated at M when the moving frame passed through M and when the photons were emitted from A and B in the stationary frame. I grant your SR says the photons were not emitted simultabneously. Therefore the calculation you were requested to make should confirm SR, right?

It seems so simple to just look at the problem from the moving frame, she sees the B photon before the A photon, which has nothing to do with the stationary frame, now not so "stationary". The photon from B was detected at t1. The photon from A was detected at t2. She determined the photons were not simultaneously emitted in her frame because she detected the sequential arrival of the photons in her moving frame. She is not being consistent with her calculations and is trying to hide something. Maybe she doesn't want to be the one who pointed out the contradiction in SR? What do you think?

If, as you say, the platform is now the moving entity, why is the A photon detected after the B photon in the direction past where B was detected in the direction A was moving?

If the moving frame is now stationary, when B is detected, then the A photon will be detected where? Cannot you carry the calculations to its logical end? Are you saying that she cannot determine the location of the A photon in terms of t0,t1,t2 and v, when the B photon was detected? Does SR prohibit this? Is this just a convenience for you to avoid discussing the inability to determine where A was when B was detected? as requested? I give you more credit than that Doc Al, much more. Show how cionsistent the calculations are.

You recognize that you swapped reference frames while the photons were onflight, don't you?

This is the first time in this thread that you have swapped reference frames. I see no absolute requirement that you do this.There isn't am absolute requirment. This isn't absolutely necessary is it Doc Al? She sees her velocity wrt to the stationary frame and she can calculate it both ways if you are correct, so why not calculate if both ways? Because there is a contrradicition isn't there?

A is located at -t1 measured wrt M'(t0) when the B photon was detected at t1. This is how the calculation comes out, right?

A then must move (t2 - t1)c to reach t2. After the algebra A is located at -t1 when B was detected, which places M'(t0) at the midpoint of the A and B photons at the instant B was detected. If the photons are equidistant from M'(t0) at this instant, then A and B photons have always been equiidstant from M'(t0) because the A and B photons are moving at the same velocity and have covered the same distance in equal time segements.

OK you wanted a contradiction, there you have it. The photons were emitted simultaneously in both frames by one calculation, but not by both..

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 Quote by geistkiesel V is the velocity of the moving frame wrt the stationary frame.
And vice-versa. V is the relative speed of the two frames--it works both ways.
 Why do you say that the moving frame sees the platform moving? Isn't this just an arbitrary move on her part? She determined her velocity as v wrt the stationary frame. She detected the B photon at t1, where was the A photon at this instant in terms of t0,t1,t2 and v?
You're joking right? You surely realize that each observer is entitled to view herself as being at rest? This has nothing to do with SR--Galileo understood this.

As far as where photon A is at the moment that O' detects photon B (according to the moving frame): read my last post where I spell it all out for you.
 Doc Al, use the word "stationary" to mean what the word says, v = 0.
You do realize that "stationary" and "moving" are just terms of convenience: the stationary frame is not really stationary.
 The observer is always at M' where all the measurements are taking place. v is the velocity of the moving frame wrt the stationary frame. I have never used the terms x' = 0. I have consistently said that O' the observer is located at M' where she makes all measurements.
For convenience and simplicity of calculation, I have chosen x' = 0 as the location of the moving observer in the moving frame.
 Here is one objection: When we say stationary frame this means "stationary" . Now for her to say : It is moving not me, is changing a fact that she cannot do so easily. Stationary means v = 0. She can't switch.
That's silly. The moving observer never "switches": to her she is always at rest.
 Even if she does, why avoid the simple task of defining A in terms of all moving frame experimentally collected values? v is in terms of m/s, all of which are dilated, I assume. Her clocks are working. It should be a straight forward calculation to find A wrt the known measured parameters. Certainly there are enough to solve the problem as requested. S do it.
I have done the calculation. Read my last post again.
 Again, it is not absolutely necessary that she now says that the platform is moving and she is stationary. [Sounds like a smoke screen just went up.] The A photon is sitting out there heading to O' the observer at M', where M' was colocated at M when the moving frame passed through M and when the photons were emitted from A and B in the stationary frame. I grant your SR says the photons were not emitted simultabneously. Therefore the calculation you were requested to make should confirm SR, right?
My calculations most certainly "confirm" SR. Or don't you understand what we've been talking about?
 It seems so simple to just look at the problem from the moving frame, she sees the B photon before the A photon, which has nothing to do with the stationary frame, now not so "stationary". The photon from B was detected at t1. The photon from A was detected at t2. She determined the photons were not simultaneously emitted in her frame because she detected the sequential arrival of the photons in her moving frame. She is not being consistent with her calculations and is trying to hide something. Maybe she doesn't want to be the one who pointed out the contradiction in SR? What do you think?
What do I think? I think you have no understanding whatsoever of SR and what it says or doesn't say. I have provided all the calculations you have requested.
 If, as you say, the platform is now the moving entity, why is the A photon detected after the B photon in the direction past where B was detected in the direction A was moving?
Sorry, I can't decode that sentence. Have you forgotten that the moving observer (O') does not agree that the photons were emitted simultaneously?
 If the moving frame is now stationary, when B is detected, then the A photon will be detected where? Cannot you carry the calculations to its logical end? Are you saying that she cannot determine the location of the A photon in terms of t0,t1,t2 and v, when the B photon was detected? Does SR prohibit this? Is this just a convenience for you to avoid discussing the inability to determine where A was when B was detected? as requested? I give you more credit than that Doc Al, much more. Show how cionsistent the calculations are.
Why do you keep saying that O' cannot determine the location of the photon from A at any time she chooses? I've provided that calculation. (Done correctly, mind you!)
 You recognize that you swapped reference frames while the photons were onflight, don't you?
Now what are you talking about? O' makes all her own measurements--where does she "switch" frames?
 This is the first time in this thread that you have swapped reference frames. I see no absolute requirement that you do this.There isn't am absolute requirment. This isn't absolutely necessary is it Doc Al? She sees her velocity wrt to the stationary frame and she can calculate it both ways if you are correct, so why not calculate if both ways? Because there is a contrradicition isn't there?
Show me where I've "swapped reference frames". In fact, please outline exactly how I did my calculations so you can pinpoint my error. I'm sure folks would find that most instructive.
 A is located at -t1 measured wrt M'(t0) when the B photon was detected at t1. This is how the calculation comes out, right?
What do you mean "located at -t1"? t1 is a time measurement, not a location. The location of the photon from A is provided in my previous post.
 A then must move (t2 - t1)c to reach t2.
Interpreted properly, this could make sense: According to O', the photon from A must move a distance (t2 - t1)c between the time that O' detects photon B and then photon A. Right. So what?
 After the algebra A is located at -t1 when B was detected, which places M'(t0) at the midpoint of the A and B photons at the instant B was detected. If the photons are equidistant from M'(t0) at this instant, then A and B photons have always been equiidstant from M'(t0) because the A and B photons are moving at the same velocity and have covered the same distance in equal time segements.
Not sure what you are saying here. Obviously, when photon B is detected it is collocated with observer O' (what you call M', but I call x' = 0). So how could they be equidistant from M'??? Gibberish.

Or perhaps you mean: At t' = t1, photon A and photon B (which is at x' = 0) are equidistant from M (the midpoint in the stationary frame) as measured in the moving frame? If so, still wrong. I'll leave it to you to figure out where that midpoint M is according to the moving frame. (I already told you where photon A would be.)
 OK you wanted a contradiction, there you have it. The photons were emitted simultaneously in both frames by one calculation, but not by both..
Not by a longshot! SR remains unscathed by your pre-galilean arguments.

 Quote by Doc Al Ah...so we're back to discussing Einstein's original setup? Good. (But you still haven't addressed Einstein's argument yet.) So by "train" I assume you mean the moving observer O'. And, once again, M is the midpoint between the two light sources at A and B. O' passes M at the moment that the photons are emitted according to the stationary frame. That moment is at time t' = 0, according to O'. The stationary frame thinks that the lights were turned on at t = 0.
the stationary frame thinks nothing. It is given that A and B enit photons when M' passes through M, at that instant. We have always been discussing the same gedunken.

 Quote by Doc Al Once again, precision is essential: O' passes M at the moment that the clock at M says t = 0 (and the O' clock says t' = 0). The stationary frame claims that the lights at A and B were emitted simultaneously according to the stationary frame at t = 0. Again, the state of knowledge of O' is irrelevant.
when A and B emit photons, M hasn't received the photons until they arrive.

 Quote by Doc Al If the stationary frame observes the lights emit simultaneously, then the moving frame will disagree.
The stationary frame actually observes the photons after the moving frame detects the B photons coming from the front of the train. All the moving frame has is the time of arrival of the photons B and A, the time M'=M which is where the moving frame notes the time on her clock and I assume the moving frame determined her velocity v (which could be at any time, right?) . The moving frame effectively measures the time the B photon arrives at M'(t1) or where O the observer was located at t1, which just happens to be the negative of the place (negative or -t'1) where the A photon is after the B photon was measured at t1.

 Quote by Doc Al You've asked quite a few questions, but you still haven't answered mine. Where is the flaw in Einstein's train gedanken experiment?

 Quote by Doc Al No. The moving frame claims that the stationary frame has moved a distance v(t'2 - t'1) during that time. (v is the relative speed of the two frames.) The stationary frame will disagree. I assume you mean that O' remains at x' = 0 in her moving frame? Right.
Not exactly. Using a simpleminded law of physics and asking the question can the moving frame legitimately consider it self as stationary when the erstwhile staionary platform was the planet earth, and the moving frame the minisicule train? SR theory and practice are being subverted, especially when everyone knows, including O' that she was at one time v = 0, i.e. stationary and that she accelerated while the stationary platform remained stationary.

 Quote by Doc Al Not sure what you are asking here. O' is at location x' = 0 in her frame. When she detects the photon from B (I assume that's what you mean by B'(t'1)) then that photon is at x' = 0. (And when she detects the photon from A, that photon will be at x' = 0.) If you are asking: Where is the photon from A at the moment that O' detects the photon from B according to the moving frame?
No the B photon is at vt'1 in the moving frame offset from M'(t'0) when detected in the moving frame.
yes that is the question, however it specifically asked for the position of A in terms of t'0, t'1, t'2, and v. Can you provide this?

 Quote by Doc Al Did you have a point to make here?
Yes read on and of course there is the previos post that I wriote in response to this post of yours here -- A sort of double barreled enquiry.

The moving frame is constructed as follows:

The moving frame is a flat plate circle with M’ at the center of the circle. Around the circumference are photosensitive detectors with a bandwidth of f/1000, where f is the frequency of the photons of light at A and B. These photosensitive devices are also distributed along the radius to a value ~ 2xAB. Therefore, when the moving frame M’ location, the circle center, passes through M, the midpoint of the photon sources at A and B, there will be photosensitive devices collocated with the emitted photons in the moving frame.

For a direction of motion 45 degrees with the AB line direction, the moving frame will first detect the emitted photon from B. Then the observer at M’ will read the time the photons were emitted wrt the moving frame. Finally the M’ will detect the photon from A. Straight forward geometry.

From the simple geometry, the moving frame detects photons emitted simultaneously in the stationary frame, regardless of time dilation and frame shrinking..

Take the case where M’ is moving 90 degrees to the direction of motion of the photons. This moving observer will detect the photons from A and B simultaneously, unambiguously, as the wave fronts from A and B meet along the path of the moving frame. Symmetry demands the conclusion that the A and B photons emitted into the moving frame simultaneously with the photons in the stationary frame.

Now then, some will say that the slightest variation from a pure 90 degrees orthogonal direction of frame and photons brings the specter of SR into the real physics world. Who are these that doth speaketh such blasphemy? And what do they say about simultaneity here?

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You seem a little confused with your notation. I hope you realize that what you call M' is the location in the moving frame where observer O' sits--with respect to the moving frame, it doesn't change. I call that location x' = 0... why not? So, in the moving frame, where is M' at t' = 0? at x' = 0, at t' = t1? at x' = 0, at t' = t2? at x' = 0. Get it? O' doesn't move in her own frame.

 Quote by geistkiesel The stationary frame actually observes the photons after the moving frame detects the B photons coming from the front of the train. All the moving frame has is the time of arrival of the photons B and A, the time M'=M which is where the moving frame notes the time on her clock and I assume the moving frame determined her velocity v (which could be at any time, right?).
We both agree that the relative speed of the frames is v.
 The moving frame effectively measures the time the B photon arrives at M'(t1)
Yes, moving frame measures the time that the B photon arrives at M' (x' = 0); that time we call t' = t1.
 or where O the observer was located at t1,
Now you're starting to get sloppy. Yes, the position of M as measured by O' has moved from the position it had when t' = 0 (that position was at x' = 0, of course). How much did it move? According to O', M has moved a distance v*t1 to the left, which means it is now at position x' = - v*t1.
 which just happens to be the negative of the place (negative or -t'1) where the A photon is after the B photon was measured at t1.
Huh? What are you talking about? According to O', where is that A photon when the B photon is detected? (Hint: I already told you!)
 Not exactly. Using a simpleminded law of physics and asking the question can the moving frame legitimately consider it self as stationary when the erstwhile staionary platform was the planet earth, and the moving frame the minisicule train? SR theory and practice are being subverted, especially when everyone knows, including O' that she was at one time v = 0, i.e. stationary and that she accelerated while the stationary platform remained stationary.
Nope. There is no acceleration whatsoever in Einstein's train gedanken.
 No the B photon is at vt'1 in the moving frame offset from M'(t'0) when detected in the moving frame.
Huh? When the B photon is detected in the moving frame, it is obviously at M' (otherwise known as x' = 0). What are you talking about?
 yes that is the question, however it specifically asked for the position of A in terms of t'0, t'1, t'2, and v. Can you provide this?
Sure. But why be so mysterious? Let's assume (like we did many posts ago) that the light sources A and B are each a distance L from the midpoint M as measured in the "stationary" frame. Now there's no need for any nonsense, since v is given. You should be able to answer these questions without any handwaving:
(1) When does O' detect photon B? Don't just say t' = t1, give me the real answer (in terms of L and v).
(2) When does O' detect photon A? Again, don't just say t' = t2, give me the real answer (in terms of L and v).
(3) Where does O' say that photon A is when she detects photon B? Give me the actual coordinates according to O' in terms of L and v. (Call the position of O' (M') as x' = 0.)
(4) Where does O' say that midpoint M is when she detects photon B? Give me the actual coordinates according to O' in terms of L and v. (Call the position of O' (M') as x' = 0.)
Please provide the answers both ways. First, give your answers. Then provide the answers according to SR. Since I assume you believe your answers do not agree with those of SR, I'm sure you must be able to produce the SR answers at will. (Otherwise, how can you claim that SR is wrong?) Kid stuff, right? Go for it! (Careful you don't "swap" frames! )

 Quote by Doc Al You seem a little confused with your notation. I hope you realize that what you call M' is the location in the moving frame where observer O' sits--with respect to the moving frame, it doesn't change. I call that location x' = 0... why not? So, in the moving frame, where is M' at t' = 0? at x' = 0, at t' = t1? at x' = 0, at t' = t2? at x' = 0. Get it? O' doesn't move in her own frame.
No O moves with M'. She is always there at lest as t is increasing. In our math we don't have to include her, but I wanted her here where he action was.

 Quote by Doc Al We both agree that the relative speed of the frames is v. Yes, moving frame measures the time that the B photon arrives at M' (x' = 0); that time we call t' = t1. Now you're starting to get sloppy. Yes, the position of M as measured by O' has moved from the position it had when t' = 0 (that position was at x' = 0, of course). How much did it move? According to O', M has moved a distance v*t1 to the left, which means it is now at position x' = - v*t1.
No, the frame is moving to the right from M'(t0) to M'(t1) . Come on Doc Al, you're swapping directions and I suspect you know it. She is moving with the train frame, the stationary platform is stilla stationary platform.

 Quote by Doc Al Huh? What are you talking about? According to O', where is that A photon when the B photon is detected? (Hint: I already told you!)
Tell me again in terms of t0,t1, t2, and v is what you were requested to do.

 Quote by Doc AL Nope. There is no acceleration whatsoever in Einstein's train gedanken.
You mean the train has been moving at v forever? It never comes to rest wrt to the stationary platform? Get reaL.

 Quote by Doc Al Huh? When the B photon is detected in the moving frame, it is obviously at M' (otherwise known as x' = 0). What are you talking about?
That would be at M'(t1) right? Doc Al I don't change your notation and try to get you confused and perhaps the readers also, so why do you do it?

 Quote by Doc Al Sure. But why be so mysterious? Let's assume (like we did many posts ago) that the light sources A and B are each a distance L from the midpoint M as measured in the "stationary" frame. Now there's no need for any nonsense, since v is given. You should be able to answer these questions without any handwaving:(1) When does O' detect photon B? Don't just say t' = t1, give me the real answer (in terms of L and v).
M + vt1 to the right of M.
t1 is the clock in the moving frame, remember? This isn't good enough for you? Einstein didn't use L and you are switching frames, aren't you?

 Quote by Doc Al (2) When does O' detect photon A? Again, don't just say t' = t2, give me the real answer (in terms of L and v).
t2 is a real time from a real clock in themoving rframe. No need to jump outsie and swap frames is there? M + vt1 + vt2. to the right of M.
 Quote by Doc Al (3) Where does O' say that photon A is when she detects photon B? Give me the actual coordinates according to O' in terms of L and v. (Call the position of O' (M') as x' = 0.)
She doesn't know until she gets the time t2 and at this time she has all the information she needs to calulate where A was at t = t1.
 Quote by Doc Al (4) Where does O' say that midpoint M is when she detects photon B? Give me the actual coordinates according to O' in terms of L and v. (Call the position of O' (M') as x' = 0.)
In terms of the moving frame is M'(t1) is simply vt1 in front of the B photon.
 Quote by Doc Al Please provide the answers both ways. First, give your answers. Then provide the answers according to SR. Since I assume you believe your answers do not agree with those of SR, I'm sure you must be able to produce the SR answers at will. (Otherwise, how can you claim that SR is wrong?) Kid stuff, right? Go for it! (Careful you don't "swap" frames! )
Doc Al I amj going to use your expertise on this as you have already stated that the photons were not emitted simultaneously in The moving frame, yet that is exacly what MY calculations show here. You remember the math you gave up earlier don't you? I even used only the moving frame measurements, including those of the speed of light. You are trying to shame me? After I have been pronouncing myself as a dissident anti-SRist? Knowing enough just to get a little hole started in the dike, you are aware of this aren't you?

How can I claim SR is wrong? because Doc Al just told me so. Can you suggest a better source? I don't think so.

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 Quote by geistkiesel How can I claim SR is wrong? because Doc Al just told me so. Can you suggest a better source? I don't think so.

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 Quote by geistkiesel No O moves with M'. She is always there at lest as t is increasing. In our math we don't have to include her, but I wanted her here where he action was.
I don't have the time to sort out every one of your errors. But, since you seem to want to use the stationary frame for your measurements, here are a few extra credit problems for you, geistkiesel. (Believe me, you need the extra credit--you are flunking my course!) These should be easy.
(5) According to the stationary frame, what time does the moving observer detect photon B? (Let's call this time t = t1*.) Find t1* in terms of L and v.
(6) According to the stationary frame, what time does the moving observer detect photon A? (Let's call this time t = t2*.) Find t2* in terms of L and v.
(7) According to the stationary frame, where is observer O' at t1*? Answer in terms of L and v.
(8) According to the stationary frame, where is observer O' at t2*? Answer in terms of L and v.
Just to be clear, let's call our previous answers (according to the moving frame) for when O' detects photon B t' = t1' and photon A t' = t2'. Primed time means measured by O'; starred time means measured by O. Got it?

Pop quiz: Does t1' = t1*? Does t2' = t2*?