1. Jun 13, 2004

### geistkiesel

In the following the observer at the M' locations does not know if she is moving or stationary. All data is available for analysis one week after it is acquired.

1).
Code (Text):

A__________________M_____M'_____________B

Light sources at A and B emit photons simultaneously both directed to M, the midpoint of the sources. A detector in the stationary frame is located at M' detects the B photon at t = M'B/c then the A photon at t = M'A/c.

Does the observer at M' determine the lights were emitted simultaneously at A and B?

Explain.

2).
Code (Text):

A___________________M________M'B________M'A_____A

Same as above only with two detectors each rigged such that the M'B detects only from the B direction, the M'A detects only from the A direction. M'B is located to the right of M.

Does the observor conclude the photons were emitted simultaneously at A and B?

Explain.

3)

Same as 2), except M' is moving in the direction of B and the light from A was pulsed on before the light from B such that each photon reached M'B and M'A simultaneously.

Does the observer conclude the photons were emitted simultaneously.

4). Same as 3), except the photons were emitted simultaneously when M' was at M and the M'B detected photons before the M'A detected photons..

Does the observer at M' conclude the photons were emitted simultaneously?

Explain.

2. Jun 14, 2004

### Staff: Mentor

Not sure what you mean by that, since you state in each example whether she is moving or not.
Note that in all of these examples the only thing directly measured by the observer is the time of arrival of the pulses according to her clocks. Any other conclusions drawn (for example, about whether the pulses were emitted simultaneously or not) must be based upon additional information that she is presumed to possess used in conjunction with the laws of physics.
Yes. Based the observer's knowledge of where the lights are and their distance from her, she will conclude that they were emitted simultaneously at t = 0 according to her clock.
Yes. No difference. (I assume that M'B and M'A are collocated, otherwise two observers are involved.)

No she does not. What she concludes depends on where she is when she detects the lights and on her relative velocity. For example, say she intercepts the light just as she reaches B. Obviously she must conclude that the light from A was emitted before the light from B.
When you say "emitted simultaneously" I assume you mean according to the stationary observers.

The moving observer will conclude that the lights were not turned on simultaneously according to her clocks. If they were turned on simultaneously, just as she passes their midpoint, then she would receive the pulses simultaneously--which she does not.

3. Jun 14, 2004

### geistkiesel

No this is the same as the first, with two locations as described. The data is available one week after the photons ae detected. We will exclude this quesion.

I agree. The light from behind is received before the light from in front, just as we rigged the experiment. Obvious, too obvious to be a trick question.

yes simultaneously in the stationary frame as M' was at M, the stationary midpoint..

I have been with you for the first three hypos, which weren't all that instructive. Here, though you should check your response. The question had the photons emitted simultaneously as M' passed M. How could she receive the photons simultaneously in the moving frame if the photons were emitted simultaneously when M' passed through M? Only the stationary M midpoint will receive the photons simultaneously. I don't think you intended the answer you gave. It sounds like you meant to say, if she "received the lights at the midpoint".
Correct me if I am wrong.

4. Jun 14, 2004

### Staff: Mentor

5. Jun 14, 2004

### geistkiesel

This is what you repeated alright. Isn't this the Einstein train experiment that you and I, among others, were arguing to death? If i misread my own post I concede, but I reread it and it is as you understand.

here is the 4th.
"4). Same as 3), except the photons were emitted simultaneously when M' was at M and the M'B detected photons before the M'A detected photons..

Does the observer at M' conclude the photons were emitted simultaneously?"

I am not arguing with you (yet) but if she ispassing through M as the photons were emittd simultaneously in the stationary frame, would not she detect the photons sequentially later as she approached B and moved away from A?

Yet the first part of the statement said she would not conclude the lights were turned on simultaneously. Do you see where I might be a tad confused by your statement? When you conclude she would conclude simultaneity for the same condition? That is the photons were emitted simultaneously as she passed through the midpoint M in the stationary frame?

6. Jun 15, 2004

### Staff: Mentor

Yep. You keep bringing it up, although we've discussed this many times before.
Not sure what you mean by this statement.
Of course she would detect the photons sequentially. I explicitly state that she does not detect the pulses simultaneously.
Yes, that is the conclusion she would draw based upon her understanding of physics and the invariance of light speed. Paralleling Einstein's own argument, the moving observer can reason as follows: "IF the lights were turned simultaneously (according to my clocks) as I passed the midpoint, THEN I would detect them simultaneously (at a later time). But since I do not detect them simultaneously, they must not have been turned on simultaneously (according to my clocks)."
I'm not clear where your confusion lies or what question you are asking. Are you asking, Why doesn't she conclude that the lights were turned on simultaneously? If so, see my reasoning above. If not, rephrase your question.

7. Jun 16, 2004

### geistkiesel

Aha, I see the point of non-contention. When I was asking you whether she would see the lights 'simultaneusly' I was referring not to whether she would detect the lights simultaneously later, I was referring to whether she would conclude from the manner in which she did detect the lights whether the lights were emitted simultaneously in a). The statiionary frame and/or b). her moving frame.

I wouldn't want to hold you to responding to this as I just (re)wrote it having been framed differently than in the original post. I am, however, truly interested in determining the 'official rational' for the conclusions of SRT as it applies in this case. She passes M, the midpoint of the source of lights at A and B in the stationary frame, just as the photons are emitted from A and B. She notes this time in her frame, and later detects the photons arriving first from B, then from A. She knows her velocity wrt the stationary frame.

Does she perceive, or detect that the photons were emitted simultaneously in her frame?
Using SR what must her conclusion be?

Take the two cases where
1. She knows the photons were emitted simultaneously in the stationary frame from A and B when she was at M and,
2. She was ignorant of the time, or times, the photons were emitted from A and B in the stationary frame when she was at M.

8. Jun 17, 2004

### Staff: Mentor

As I've said repeatedly: The moving observer will conclude that the lights were not turned on simultaneously according to her clocks.
She reaches the same conclusion in either case. The fact that she detects the pulses from A and B at different times allows her to deduce that they were not simultaneously emitted at the moment she passed the midpoint according to her clock.

9. Jun 17, 2004

### geistkiesel

Einstein's train hypothetical revisited.

I am only asking the questions in this way so I can reduce any ambuguous understanding I have regarding simultaneity. For what it is worth I "stipulate" that SR will make the predictions you have stated here. I have my concerns regarding SR, but I do not challenge what SR will do in this case.

First question. Does the moving observer need to make any SR calculations in concluding that the photons were not emitted simultaneously in her moving frame or is the fact of the staggered arrival of the photons from A and B sufficinent for her concluding nonsimultaneous emission of the photons in her frame?

Second: Does her knowledge that the photons were emitted simultaneously in the stationary frame when she was at the midpoint of the A and B sources when the photons were emitted in the stationary frame have any bearing on her concluding that the photons were not emitted simultaneously in her moving frame? I would assume no, and that she is quite content to accept the "naive" contradiction based on the staggered arrival times of the photons in her frame.

Third: You mention that she knew the time the photons were emitted in the moving frame from her clocks, which she obviously looked at at when she was located at M, the midpoint of A and B in the stationary frame. Were her clocks used in concluding the photons were not emitted simultaneously in her frame in calculations using SR Theory. Or would the calculations be redundant, or merely confirmation of the nonsimultaneous emission of the photons in her frame which she concluded from her observing the staggered arrival times?

Fourth: Would accurately determining that the photons were emitted simultaneously in her frame be unconditionally based on the simultaneous arrival of the photons in her frame?. In other words would the simultaneous emission of photons in her frame be predicated on her observing the simultaneous arrival of the photons in her frame also?

Five: If she knew that when the B photon arrived that M, as determined by her clocks, was at that instant the midpoint of A and B in her frame, what would she conclude about the simultaneous emission of photons in her moving frame?
Code (Text):
A ___________>            M             M'B'<_____________   B
We show M'B' the arrival point of the B photon in her frame. These conditions may be different from the previous hypotheticals above, but M is now the midpoint of A and B in her frame. Of course she would not have this information available at the instant the condition was as described by the figure until later, say the next week. However, she obviously still detects the staggered arrival of the photons.

Last edited: Jun 17, 2004
10. Jun 17, 2004

### Staff: Mentor

Her observation of the staggered arrival--coupled with her knowledge of the invariant speed of light--is all she needs to correctly conclude that (according to her frame) the photons could not have been emitted simultaneously at the moment she passed the midpoint. No calculations needed.
Such knowledge is irrelevant. And, since she understands physics, she sees no contradiction whatsoever between observations made in different frames.
If you mean: "For her to conclude that the photons were simultaneously emitted (in her frame) at the very moment she crossed the midpoint, must she detect the photons arriving simultaneously?" -- Yes.
I don't understand this question. Both frames agree that M is the midpoint between A and B.

11. Jun 17, 2004

### geistkiesel

A slight difference. From the drawing M is the midpoint of the two photons when when the B photon is detected at M'B'. This may be d9ifferent from all the previous situation.
Code:
A ___________> M M'B'<_____________ B

Here M is the midpoint of the photons in the moving frame at the instant the B photon was detected at M'B'.
1. No other conditions other than the ones stated are imposed. In other words, the sources of A and B could be lightyears removed from the locality here and M need not necessarily be the midpoint of the sources of the photons, but M is the midpoint of the photons at the instance shown.
2. As in the first situation with the added condition That M is also the midpoint of the sources of A and B in The stationary frame.
3. The A photon arrives at where the M'B' location is at the present instant (M' has moved forward during the time the A photon was detected at the ol M'B'

12. Jun 18, 2004

### Staff: Mentor

fifth question

1. Is this what you are saying: A and B, at rest with respect to each other and some arbitrary distance apart, emit pulses of light at some arbitrary times (not necessarily simultaneously). A moving observer intercepts the light from B at some point and time; at that very instant (according to her moving frame) the light from A happens to be at another point. Location M happens to be midway between those two points according to the moving frame. What can she conclude?

Beats me! It's not obvious that she can conclude anything significant.

Was there a point you were trying to make with this example?

13. Jun 18, 2004

### geistkiesel

doesn't she see the photons arriving at different tines? If so then she would preceive the photonds were not emitted in the moving frame simultaneously with the staiobary photons, I guess?

14. Jun 18, 2004

### Staff: Mentor

Just because she intercepts the photons at different times is not sufficient for her to conclude anything about whether they were emitted simultaneously in her frame. After all, they could have been emitted simultaneously at different distances from her.

15. Jun 18, 2004

### geistkiesel

Could the locations of the photons ever have been colocated at A and B in the stationary frame, or even A' and B' in the moving frame without regard to the origin of the photon's source?
This would be equivalent to the photon 'sources' location be at A and B, correct?

16. Jun 18, 2004

### Staff: Mentor

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.

17. Jun 18, 2004

### Staff: Mentor

I think there is a reason for adding the complications...

18. Jun 19, 2004

### 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?

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 (Text):
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:
• Simultaneity equivalence with conjectured quivalence of M'(t'0) wrt M'(t'0(M'(t'2)))? then,
• with no consideration by O' of any of her [speculated] motion wrt A and B in the stationary platfom.
• with consideration of any of her [speculated] motion wrt A and B in the stationary frame then.
• 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?

19. Jun 19, 2004

### geistkiesel

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?

20. Jun 19, 2004

### Staff: Mentor

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?

21. Jun 20, 2004

### 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.
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?

22. Jun 21, 2004

### Staff: Mentor

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.
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?

23. Jun 21, 2004

### geistkiesel

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?

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..

24. Jun 21, 2004

### Staff: Mentor

And vice-versa. V is the relative speed of the two frames--it works both ways.
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.
You do realize that "stationary" and "moving" are just terms of convenience: the stationary frame is not really stationary.
For convenience and simplicity of calculation, I have chosen x' = 0 as the location of the moving observer in the moving frame.
That's silly. The moving observer never "switches": to her she is always at rest.
I have done the calculation. Read my last post again.
My calculations most certainly "confirm" SR. Or don't you understand what we've been talking about?
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.
Sorry, I can't decode that sentence. Have you forgotten that the moving observer (O') does not agree that the photons were emitted simultaneously?
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!)
Now what are you talking about? O' makes all her own measurements--where does she "switch" frames?
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.
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.
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?
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.)
Not by a longshot! SR remains unscathed by your pre-galilean arguments.

25. Jun 21, 2004

### geistkiesel

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.

when A and B emit photons, M hasn't received the photons until they arrive.

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.

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.

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?

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?