B A question about relativity of simultaneity

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The discussion centers on the relativity of simultaneity, particularly in Einstein's train thought experiment, where two lightning strikes occur simultaneously from the embankment's perspective but not for an observer in the moving train. The confusion arises from the observer in the train receiving light from the front strike before the back strike, despite both light pulses traveling at the same speed. Participants clarify that this phenomenon is due to the observer moving toward the front lightning and away from the back one, leading to different rates of closure. The conversation also highlights the distinction between closing speed and velocity transformation, emphasizing that relativistic effects must be considered when analyzing events from different reference frames. Understanding these concepts is deemed essential for grasping the implications of relativity, despite the complexity of the thought experiments.
  • #31
Sagittarius A-Star said:
I don't know, what you want to Lorentz-transform.
I don't want to Lorentz-transform anything. I only want to make sure my understanding of the theory is consistent.
If what we were talking of the closing speed was true, then we could assess the linear function of the time of the moving reference frame as seen from the motionless reference frame. I did that, and got that the coefficients of x and t of t'(x, t) make a proportion ##\frac {v} {c^2-v^2}.## But, in the Lorentz transforms, they make a proportion ##\frac {v} {c^2}.## I therefore wonder if we were right speaking of the closing speed.
 
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  • #32
DmitryS said:
If what we were talking of the closing speed was true, then we could assess the linear function of the time of the moving reference frame as seen from the motionless reference frame. I did that, and got that the coefficients of x and t of t'(x, t) make a proportion ##\frac {v} {c^2-v^2}.## But, in the Lorentz transforms, they make a proportion ##\frac {v} {c^2}.##
Sorry, I don't understand your argument. What did you do in detail?

Also, you call both frames "reference frame". For a certain calculation, only one frame can be the reference. What is regarded as "moving" and what is regarded as "motionless", is then defined relative to the reference frame.
 
  • #33
The Lorentz transformation is the right tool to use to understand the relativity of simultaneity in Einstein's "train gedanken experiment" (two light signals sent from the ends of the train simultaneously wrt. the enbankment frame) or the alternative one, discussed above (a light signal sent from the middle of the train towards both ends). Just do the simple calculation, and you'll get the very same result as that given by using the "closing-speed argument" in either the train or the enbankment frame, as it must be. Also reading the kinematic part of Einstein's original 1905 paper is a good idea (not so much the part about relativistic mechanics, which is not fully worked out yet in this paper, and the part about the Maxwell equations is hard to read, because of the then very complicated notation in terms of components).
 
  • #34
vanhees71 said:
The Lorentz transformation is the right tool to use to understand the relativity of simultaneity in Einstein's "train gedanken experiment"

I think, the OP's problem is not understanding relativity of simultaneity, but understanding closing speed.
DmitryS said:
I can tell you I grasped very well the idea of the relative simultaneity which appears when you use the Lorentz transforms, but I don't understand the illustrations.
DmitryS said:
But that is the problem I'm asking about. The relative speed is greater than the speed of light. Why we can't apply the speed addition formula?
 
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  • #35
Sagittarius A-Star said:
Sorry, I don't understand your argument. What did you do in detail?

Also, you call both frames "reference frame". For a certain calculation, only one frame can be the reference. What is regarded as "moving" and what is regarded as "motionless", is then defined relative to the reference frame.
Well, that's easy. In the synchronization process, when you send a ray of light from clock to clock to be reflected back, in the proper frame of reference the travel time is the same both ways. In the relatively moving frame it is ##\frac{x'} {c-v}## forth and ##\frac{x'} {c+v}## back. The sum of these gives you ##\frac{2vx'} {c^2-v^2}## and that is twice the quantity of the 'difference-of-rate' coefficient - the partial derivative of t' with respect to x - to be more exact, it is how much this coefficient is smaller than the derivative with respect to t.
 
  • #36
DmitryS said:
In the relatively moving frame it is ##\frac{x'} {c-v}## forth and ##\frac{x'} {c+v}## back. The sum of these gives you ##\frac{2vx'} {c^2-v^2}##
No, the sum gives ##\frac{x'(c+v) +x'(c-v) }{c^2-v^2} = \frac{2cx'} {c^2-v^2}##.

I still don't understand your calculation, but I guess you want to calculate the time-difference between the two lightnings with reference to the train rest frame.

This train/embankment thought experiment contains an easier clock synchronization procedure than the original one from Einstein. Two clocks are regarded as synchronous in the reference frame, in which they are both at rest, if an observer in the middle of them receives simultaneously their light pulses, that were sent of at their same clock-times.

Based on this, I continue my "closing speed" calculation from posting #12:
Sagittarius A-Star said:
The light-pulse from the front needs the following time-interval to reach the observer in the center of the car:
##\overline{BM}/(c+v)##

The light-pulse from the back needs the following time-interval to reach the observer in the center of the car:
##\overline{AM}/(c-v)##

Because the distances ##\overline{AM}## and ##\overline{BM}## are equal, the observer in the center of the car sees the light-pulse from ##B## earlier than that from ##A##.
This calculation was done with reference to the embankment frame. I continue with this reference frame:

The time-difference for the observer in the center of the car is:
##\overline{AM}/(c-v) - \overline{BM}/(c+v) = \frac{L}{2} \frac{2v}{c^2-v^2} = \gamma^2 v\frac{L}{c^2}##
To convert this time-difference to the train frame, this needs to be devided by ##\gamma## because of time-dilation. Then compare to ##t'_A- t'_B## in my posting #29:

Sagittarius A-Star said:
##x'_A = \gamma (x_A - vt_A) = -\gamma \frac{L}{2}##, ##\ \ \ \ \ \ t'_A= \gamma (t_A - vx_A/c^2) =\gamma \frac{L}{2}v/c^2##.

Transform event B coordinates:
##x'_B = \gamma (x_B - vt_B) = +\gamma \frac{L}{2}##, ##\ \ \ \ \ \ t'_B= \gamma (t_B - vx_B/c^2) =-\gamma \frac{L}{2}v/c^2##.
 
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  • #37
Sagittarius A-Star said:
The time-difference for the observer in the center of the car is:
##\overline{AM}/(c-v) - \overline{BM}/(c+v) = \frac{L}{2} \frac{2v}{c^2-v^2} = \gamma^2 v\frac{L}{c^2}##
To convert this time-difference to the train frame, this needs to be devided by ##\gamma## because of time-dilation.
Ah, I see the problem now. I want the experiment to comply with the Lorentz transforms. You get ##\gamma^2## and then divide it by another ##\gamma##. What I want, however, is that we did not take LT for granted, but saw them confirmed by the experiment.
 
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  • #38
No, no, no, wait a bit. You said lightnings. I was thinking about a different experiment.
Let's say, it's two flashes on board the train. They will get to the center simultaneously in the train's frame, and at different times to that center in the embankment frame.
Then all those ##c+v## and ##c-v## will refer to the embankment observer, and they will get that ##\gamma^2## instead of ##\gamma##. And they will have no reason to divide it by ##\gamma## because all those calculations that you use for the lightning case are now true in the embankment frame.
 
  • #39
So what? It's of course a different situation, and you get different results. It's well worth to consider these example from different points of view. The most simple approach is to parametrize the various worldlines involved and transform from one frame of reference to another with the Lorentz transformation. The other way is to think about the situation in each reference frame. Of course you should get the same result as with the Lorentz transformation. Last but not least, you can also draw a Minkowski diagram with the two reference frames.
 
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  • #40
DmitryS said:
I was thinking about a different experiment.
Let's say, it's two flashes on board the train. They will get to the center simultaneously in the train's frame, and at different times to that center in the embankment frame.
For symmetry reasons, this would describe exactly the same experiment when renaming the embankment into "train", renaming the train into "embankment" and reversing the directions of the x- and x'-axes. The math would be completely the same.
 
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  • #41
DmitryS said:
Let's say, it's two flashes on board the train. They will get to the center simultaneously in the train's frame, and at different times to that center in the embankment frame.
That's impossible. If both flashes reach the centre of the train at the same time in one frame, they must reach the centre of the train in all frames. This is because they are described by the same coordinates (time and space).
 
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  • #42
Ok, let's do this version of the train gedanken experiment:

In the restframe of the train there are two light signals sent simultaneously to an observer in the center of the train. The train is moving with velocity ##v## relative to the platform.

In the restframe of the train:
$$x_A=\begin{pmatrix} c t \\ -L/2+c t \end{pmatrix}, \quad x_B=\begin{pmatrix} c t \\ L/2-c t \end{pmatrix}, \quad x_M={c t,0}.$$
Here ##x_A## and ##x_B## are the world lines of the light signals and ##x_M## is the world line of the center of the train.

Obviously the signals reach the center of the train for ##x_A^1=0##, i.e., at time ##t_{AM}=L/(2c)## and for ##x_B^1=0## at ##t_{BM}=L/2c##. The light signals reach the center of the train simultaneously, which shouldn't be a big surprise.

The world lines seen from the platform rest frame are given by the Lorentz transformation
$$\hat{\Lambda}_{-v}=\begin{pmatrix} \gamma & \gamma \beta \\ \gamma \beta & \gamma \end{pmatrix}.$$
The world lines thus read
$$x_A'=\hat{\Lambda} x_A=\gamma \begin{pmatrix} \beta L/2 +(c-v) t, L/2-(c-v) t\end{pmatrix}, \quad x_B'=\hat{\Lambda} x_B=\gamma \begin{pmatrix} -\beta L/2 +(c+v) t, -L/2+(c+v) t\end{pmatrix}, \quad x_M'=\hat{\Lambda} x_M=\gamma \begin{pmatrix}c t,v t \end{pmatrix}.$$
The emission times of the light signals are given by ##t=0##, i.e.,
$$t_A'=L/(2 c), \quad t_B'=-L/(2c).$$
and they reach the center of the train for ##t=L/(2c)##, i.e. also at the same time in the rest frame of the platform:
$$t_{AM}'=t_{BM}'=\gamma L/(2 c).$$
Here the point is that the emission times are not the same from the point of view of an observer on the platform but they reach the center of the train also at the same time ##t_{AM}'=t_{BM}'## with the time-dilation Lorentz factor ##\gamma## as compared to the time they reach the center of the train as observed by and observer on the train.
 
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  • #43
PeroK said:
That's impossible. If both flashes reach the centre of the train at the same time in one frame, they must reach the centre of the train in all frames. This is because they are described by the same coordinates (time and space).
Thank you very much for bringing this up! it's exactly what's worrying me.
It seems that even as we stage the thought experiments, we already know that the relative simultaneity is there. The setup of the experiment already has it.
So, this is either the circular argument, or all those thought experiments are just illustrations and not thought experiments. This brings us back to the question I asked at the beginning: Why do we need them at all?
And, rewording your statement... The light of the two lightnings reaches the observer on the embankment at the same time. This means that this point of simultaneity exists even inside the traincar. It's not the center, but it is there. So, there is at least one observer who belongs to the traincar frame and who thinks that the lightnings are simultaneous. Granting that the clocks of the traincar frame are all synchronous for that frame, we have the time when the lightnings are simultaneous in the traincar frame?
 
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  • #44
Sagittarius A-Star said:
For symmetry reasons, this would describe exactly the same experiment when renaming the embankment into "train", renaming the train into "embankment" and reversing the directions of the x- and x'-axes. The math would be completely the same.
All I am asking is, can you actually derive LT from relative simultaneity test without assumption that the LT must be the correct transformation?
 
  • #45
DmitryS said:
Thank you very much for bringing this up! it's exactly what's worrying me.
It seems that even as we stage the thought experiments, we already know that the relative simultaneity is there. The setup of the experiment already has it.
So, this is either the circular argument, or all those thought experiments are just illustrations and not thought experiments. This brings us back to the question I asked at the beginning: Why do we need them at all?
And, rewording your statement... The light of the two lightnings reaches the observer on the embankment at the same time. This means that this point of simultaneity exists even inside the traincar. It's not the center, but it is there. So, there is at least one observer who belongs to the traincar frame and who thinks that the lightnings are simultaneous. Granting that the clocks of the traincar frame are all synchronous for that frame, we have the time when the lightnings are simultaneous in the traincar frame?
I've no idea what any of that means. SR is just Minkowski spacetime in the end. A relatively simple geometric model.

You're the one who insisted on studying this Einstein thought experiment. I advised you to ignore it!
 
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  • #46
DmitryS said:
can you actually derive LT from relative simultaneity test without assumption that the LT must be the correct transformation?
The relativity of simultaneity in your scenario is a consequence of assuming that the speed of light is the same in all reference frames and in all directions. The LT can be derived from that assumption. So both relativity of simultaneity and the LT can be viewed as consequences of the speed of light being invariant.
 
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  • #47
PeroK said:
I've no idea what any of that means. SR is just Minkowski spacetime in the end. A relatively simple geometric model.

You're the one who insisted on studying this Einstein thought experiment. I advised you to ignore it
I beg your pardon. I didn't quite follow what you were saying - I was in the one-to-one conversation mostly, and thought you were tracking it.
Yes, that was my point from the start! I'm not happy with these thought experiments at all. So, you agree that they are an unwanted burden?
But others would say then there's nothing 'physical' about the spacetime.
 
  • #48
DmitryS said:
Yes, that was my point from the start! I'm not happy with these thought experiments at all. So, you agree that they are an unwanted burden?
The Einstein lightning thought experiment is a train crash, IMHO. The first part of post #26 was intended to show how simple it is that the relativity of simultaneity is a conseqence of the invariance of the speed of light.

You need thought experiments in SR, because macroscopic objects cannot be accelerated to near light speed relative to each other.
 
  • #49
DmitryS said:
I'm not happy with these thought experiments at all.
Then, as @PeroK has already said, you should not be bothering yourself with them. You should focus on things that you find useful.

DmitryS said:
So, you agree that they are an unwanted burden?
The fact that you don't find these thought experiments useful does not mean nobody else does.

Einstein's original purpose with these thought experiments was not to derive the Lorentz Transformation from relativity of simultaneity (which, as I have already remarked, cannot be done). It was to show that the simple experimental fact (as shown by the Michelson-Morley experiment) of the speed of light being the same in all inertial frames, implies relativity of simultaneity. And that means that Newtonian mechanics, which has absolute simultaneity, cannot be exactly right.

You might not need to be convinced that simultaneity is relative or that Newtonian mechanics is not exactly right. If so, these thought experiments indeed will not be useful to you, since you are already convinced of the things that these thought experiments were intended to convince people of. But there are plenty of people who are not convinced of those things and for whom thought experiments like these might be useful.

DmitryS said:
But others would say then there's nothing 'physical' about the spacetime.
Minkowski spacetime is perfectly "physical" in the sense that any sufficiently small patch of any spacetime looks like a small patch of Minkowski spacetime. This fact is made use of all over the place in General Relativity.
 
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  • #50
PeterDonis said:
Einstein's original purpose with these thought experiments was not to derive the Lorentz Transformation from relativity of simultaneity (which, as I have already remarked, cannot be done). It was to show that the simple experimental fact (as shown by the Michelson-Morley experiment) of the speed of light being the same in all inertial frames, implies relativity of simultaneity. And that means that Newtonian mechanics, which has absolute simultaneity, cannot be exactly right.
And here, I am afraid, you contradict the facts.
1664913402276.png

And so on, and so forth. It is with basically the same idea as lies behind the simultaneity test Einstein is deriving the Lorentz transforms.
It's from his 1905 article.

https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf
 
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  • #51
DmitryS said:
And here, I am afraid, you contradict the facts.
I have no idea what you are talking about. Everything I said is perfectly consistent with Einstein's 1905 paper.
 
  • #52
DmitryS said:
Thank you very much for bringing this up! it's exactly what's worrying me.
I think you misunderstood the reply of @PeroK to your posting #38 "to that center in the embankment frame".

Your statement in #38 may be interpreted by a reader in such a way, that you mix-up Einstein's ##M## and ##M'##. They are different objects, that move relative to each other. With reference to the embankment rest frame, ##M## and ##M'## meet only at that instance of time, when the lightnings at ##A## and ##B## occur.

DmitryS said:
This means that this point of simultaneity exists even inside the traincar. It's not the center, but it is there. So, there is at least one observer who belongs to the traincar frame and who thinks that the lightnings are simultaneous.
No! That the lightnings are simultaneous, could only be concluded by an observer at rest in the train, who receives both light-pulses simultaneous, if he is located in the center of the train. But in the center of the train the light-pulses are received not simultaneously.
 
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  • #53
DmitryS said:
It's from his 1905 article
This article, being a published physics paper, does not discuss underlying motivations much; it focuses on the physics. (That said, as I noted in post #51, the physics discussed is perfectly consistent with my post #49. In Section 2 of the paper, Einstein derives relativity of simultaneity, along with length contraction and time dilation, from the constancy of the speed of light. In Section 3 of the paper, he derives the Lorentz Transformation equations from the constancy of the speed of light. He never derives the LT from relativity of simultaneity. Exactly as I said.)

If you want a better source for Einstein's underlying motivations for the "train and lightning flashes" thought experiment, along the lines I described in post #49, try his popular book Relativity: A Clear Explanation That Anyone Can Understand.
 
  • #54
PeterDonis said:
This article, being a published physics paper, does not discuss underlying motivations much; it focuses on the physics. (That said, as I noted in post #51, the physics discussed is perfectly consistent with my post #49. In Section 2 of the paper, Einstein derives relativity of simultaneity, along with length contraction and time dilation, from the constancy of the speed of light. In Section 3 of the paper, he derives the Lorentz Transformation equations from the constancy of the speed of light. He never derives the LT from relativity of simultaneity. Exactly as I said.)
I don't have this impression from reading Einstein. To make sure we are talking about the same thing -- what's the blueprint of Einstein's derivation of LT in your opinion?
 
  • #55
Sagittarius A-Star said:
No! That the lightnings are simultaneous, could only be concluded by an observer at rest in the train, who receives both light-pulses simultaneous, if he is located in the center of the train. But in the center of the train the light-pulses are received not simultaneously.
PeroK said that if something happens in one point of spacetime, that's an event that must exist in any frame of reference. I agree to this, because this is in line with the LT - and that means that inside the train frame of reference there must be a point where the light from both lightnings come simultaneously.
It is not the center, we agree about that. It must be closer to the back of the train, and it is easy to calculate where this point should be based on the concept of closing speed.
 
  • #56
DmitryS said:
PeroK said that if something happens in one point of spacetime, that's an event that must exist in any frame of reference. I agree to this, because this is in line with the LT - and that means that inside the train frame of reference there must be a point where the light from both lightnings come simultaneously.
It is not the center, we agree about that. It must be closer to the back of the train, and it is easy to calculate where this point should be based on the concept of closing speed.
It is the centre of train. In the platform frame the light emission events are not simultaneous.

That a single event in one frame is a single event in all frames is a requirement for physical consistency of the theory.
 
  • #57
DmitryS said:
and that means that inside the train frame of reference there must be a point where the light from both lightnings come simultaneously.
It is not the center, we agree about that. It must be closer to the back of the train, and it is easy to calculate where this point should be based on the concept of closing speed.
The condition, that both light pulses are received by the observer simultaneously, is not sufficient to conclude, that the lightnings happened also simultaneously. Also the distances, the light-pulses travel with ##c##, must be equal. In the rest frame of the train, the locations ##A'## and ##B'## of the lightning events are permanently at both ends of the train.
 
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  • #58
PeroK said:
It is the centre of train. In the platform frame the light emission events are not simultaneous.

That a single event in one frame is a single event in all frames is a requirement for physical consistency of the theory.
It isn't the center. I'm talking about lightnings.
If the light of both lightnings meets at the same time at the place where the observer on the embankment is, then it is a single event. If it is a single event, it must be a single event in the train frame. I agree it's not the centre of the traincar. It must be closer to the tail.
 
  • #59
DmitryS said:
All I am asking is, can you actually derive LT from relative simultaneity test without assumption that the LT must be the correct transformation?
Yes, under a few conditions. A real relative simultaneity test can prove, that the opposite of Newton's assumption of an absolute time is true, that means in general ##t' \ne t##. In addition you need SR postulate 1 (principle of relativity) and assuming isotropy of the space, homogeneity of space and time, invariance of causality and that the velocity composition law is commutative.

Under these conditions, SR postulate 2 (invariance of the speed of light in vacuum) is not needed to derive the LT.

You can find such a derivation of the LT here:
https://www.physicsforums.com/threa...rom-commutative-velocity-composition.1017275/

Edit: I must correct my answer: This thought experiment works under the assumption, that the LT is correct. So the answer must be "no".
 
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  • #60
DmitryS said:
what's the blueprint of Einstein's derivation of LT in your opinion?
If you want a brief summary of Section 3 of the paper, sure:

(1) He assumes we have two inertial coordinate systems, with one (the "moving" system) moving in the positive ##x## direction with speed ##v## with respect to the other (the "stationary" system).

(2) He assumes that the transformation equations between the two systems must be linear.

(3) He assumes a light ray that goes from the spatial origin of the moving system out to some positive ##x'##, is reflected there, and returns to the spatial origin of the moving system.

(4) He analyzes the motion of this light ray and uses the fact that the speed of the light ray must be the same in both frames to derive the transformation equations.
 

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