A question about relativity of simultaneity

In summary, the textbook says that the observer in the center of the car will get the light of the front lightning earlier than the light from the back lightning. It is confusing, and it is easier to stay with Lorentz transforms. The thought experiments are helpful in understanding the LT, but they are not necessary.
  • #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.
 
  • #61
Sagittarius A-Star said:
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
I think this is a bit misstated.

The units of the constant ##\alpha## in the derivation you refer to are inverse velocity squared. So assuming that ##\alpha > 0## is equivalent to assuming that there is a finite invariant speed. It is true that this, by itself, does not tell you that that finite invariant speed is the speed of light; but since there can be only one finite invariant speed, anything that has an invariant speed must have that invariant speed, including light. With that taken into account, assuming ##\alpha > 0## is really equivalent to assuming invariance of the speed of light.

The interesting part of the derivation is showing that assuming ##\alpha > 0## implies ##t' \neq t##. But you only know that by already having the transformation equations in terms of ##\alpha##. So I don't think this is quite the same as deriving the LT from relativity of simultaneity without using the invariance of the speed of light; that would imply that relativity of simultaneity by itself, without using the invariance of the speed of light, or something logically equivalent to it (like ##\alpha > 0##), could tell you the form of the transformation equations. But the derivation of the transformation equations in terms of ##\alpha## doesn't use relativity of simultaneity at all.
 
  • #62
PeterDonis said:
So assuming that ##\alpha > 0## is equivalent to assuming that there is a finite invariant speed. It is true that this, by itself, does not tell you that that finite invariant speed is the speed of light
I don't assume that ##\alpha > 0##. This is the only remaining possibility after excluding the cases ##\alpha < 0## (from assumed invariance of causality, as I stated) and excluding ##\alpha = 0## (from experiment disproving absolute time, for example showing relativity of simultaneity = excluding Galileo transformation).

It is true, that this derivation does not also derive, that the invariant speed, called ##c##, is equal to the physical speed of light in vacuum. That would need an additional experiment.

PeterDonis said:
The interesting part of the derivation is showing that assuming ##\alpha > 0## implies ##t' \neq t##.
It is the opposite direction of reasoning, see above.

PeterDonis said:
But you only know that by already having the transformation equations in terms of ##\alpha##.
I get this transformation equations in terms of ##\alpha## from assuming that the velocity composition law is commutative, as stated. This follows from transformation group laws, as stated in the paper "Nothing but Relativity", which I linked at the top of the derivation.
 
  • #63
Sagittarius A-Star said:
I don't assume that ##\alpha > 0##.
"Assume" is just a form of words. You can say "exclude all of the other possibilities" if you like that better.

Sagittarius A-Star said:
This is the only remaining possibility after excluding the cases ##\alpha < 0## (from assumed invariance of causality, as I stated) and excluding ##\alpha = 0## (from experiment disproving absolute time, for example showing relativity of simultaneity = excluding Galileo transformation).
You exclude ##\alpha = 0## to exclude the Galilean case, but the only reason you can equate that with excluding "absolute time" (which you actually can't really--see below) is by looking at the specific form of the transformation equation for ##t'##. However, you can equate ##\alpha > 0## with a finite invariant speed just by looking at the units required for ##\alpha##, without making use of any specific form of the transformation equations.

Also, you can't exclude absolute time, or more precisely absolute simultaneity, by experiments, because simultaneity is a convention. You can, however, establish that there is a finite invariant speed by experiments, if you can find a physical phenomenon that moves at that speed, such as light.

Sagittarius A-Star said:
It is the opposite direction of reasoning
I disagree. You first derive the transformation equations in terms of a parameter ##\alpha##. Then you observe from the transformation equation for ##t'## that ##\alpha > 0## implies ##t' = t##. There is no reasoning that let's you go in reverse, because just assuming ##t' = t## doesn't tell you what the form of the transformation equations is in terms of ##\alpha##. In fact, even if you take the transformation equations as given, by itself ##t' = t## doesn't require ##\alpha > 0##, it just requires ##\alpha \neq 0##. Nor did you have to assume ##t' = t## in order to introduce ##\alpha## as a parameter in the equations or derive their form.
 
  • #64
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?
What is unclear with my calculation in #42? You have to use the Lorentz transformation to get the record straight, because it follows from the usual Einsteinian synchronization convention in an inertial frame of reference.

What's missing is, when the light signals reach the observer at the platform. It's easy to calculate that the light signal sent from A reach the observer at ##M## at the time ##t_{MA}'=(1+\beta) \gamma L/c## and the signal from ##B## at ##t_{MB}'=(1-\beta) \gamma L/(2c)##.
 
  • #65
PeterDonis said:
(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.
I don't agree that he assumes what you say, or rather, that he only assumes what you say, but let that be for a moment.
Basically, as I was saying, the starting point of his derivation is the same setup as that of the simultaneity test. I already posted the screenshot. And that setup leads to some transforms, which are not LT:
1664965889726.png

https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

These are very much not the LT. I would be glad if you can explain to me how you can get from those to the LT. I tried, and I failed.
That's, as I am saying, the result of using the simultaneity test as the starting point of derivation. That's why I think the simultaneity test contradicts the LT.
 
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  • #66
DmitryS said:
That's, as I am saying, the result of using the simultaneity test as the starting point of derivation. That's why I think the simultaneity test contradicts the LT.
Those are intermediate variables. He ends up with the LT further on in the paper.

Are you trying to learn SR or trying to prove its's wrong, if you don't mind me asking?
 
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  • #67
DmitryS said:
These are very much not the LT. I would be glad if you can explain to me how you can get from those to the LT. I tried, and I failed.
They're just intercept calculations between the light ray and a mirror, aren't they?
 
  • #68
PeroK said:
Those are intermediate variables. He ends up with the LT further on in the paper.

Are you trying to learn SR or trying to prove its's wrong, if you don't mind me asking?
I don't mind you asking anything as long as we remain within the protocol of mutual respect.
I think I made my point abundantly clear from the start. I find the Lorentz mathematics very much consistent, but I don't like the idea of the physical evidence of relative simultaneity. It seems to me a flop. If I remember correctly, you agreed with me about this point.
Now, to give it a rigorous proof, I refer to Einstein's 1905 paper. Mind that I'm only voicing an opinion without claiming it right. If you think it is wrong, feel free to prove it.
I will put down my points and supporting references in order, so you could tell me where I am wrong.
1) The experiment with the ray going from 0 to x' and its accompanying mathematics is simply an extension of simultaneity test.
2) With that mathematics, Einstein ends up with the transforms
1664987136431.png


which are not LT. They are pretty much the logical finale of Einstein's setup.
3) You say: "Those are intermediate variables. He ends up with the LT further on in the paper". That's exactly the point that in my opinion needs a proof on your part. What I see from the text - and I provided the link to the online PDF - is that Einstein simply writes down the LT after these transforms. I won't make a snapshot, you can easily find it yourself, here's the link:
https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf
To me, that looks like he assumed the LT to be true, whatever his own reasoning might have led to. If you can derive the LT from the relations above, show me how - and I will confess I was wrong.
 
  • #69
DmitryS said:
I find the Lorentz mathematics very much consistent, but I don't like the idea of the physical evidence of relative simultaneity. It seems to me a flop. If I remember correctly, you agreed with me about this point.
Not at all. I just don't like the Einstein simultaneous lightning strikes thought experiment. It's over-complicated and potentially misleading.

I gave a much cleaner example of the incompatibility of universal simultaneity with the invariance of the speed of light in a previous post.
DmitryS said:
Now, to give it a rigorous proof, I refer to Einstein's 1905 paper. Mind that I'm only voicing an opinion without claiming it right. If you think it is wrong, feel free to prove it.
I will put down my points and supporting references in order, so you could tell me where I am wrong.
1) The experiment with the ray going from 0 to x' and its accompanying mathematics is simply an extension of simultaneity test.
2) With that mathematics, Einstein ends up with the transforms
View attachment 315069

which are not LT. They are pretty much the logical finale of Einstein's setup.
3) You say: "Those are intermediate variables. He ends up with the LT further on in the paper". That's exactly the point that in my opinion needs a proof on your part.
I'm not sure what's to prove.
DmitryS said:
What I see from the text - and I provided the link to the online PDF - is that Einstein simply writes down the LT after these transforms. I won't make a snapshot, you can easily find it yourself, here's the link:
https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf
To me, that looks like he assumed the LT to be true, whatever his own reasoning might have led to. If you can derive the LT from the relations above, show me how - and I will confess I was wrong.
There's a derivation of the LT here:

http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

There is an issue if you take Einstein's 1905 paper as some sort of Gospel. It was the first paper on SR, but the theory has been developed and refined for over 100 years since then. I love the 1905 paper, but it's not an ideal source from which to learn SR as a student. Your confusions bear this out to some extent.

I don't want to get into a game of having to justify everything Einstein wrote in 1905. We all see flaws in that paper, when it comes to it.

But, an incompatibility between the LT and RoS? That's just nonsensical. The LT encapsulates the RoS along with time dilation and length contraction. That I can prove, but not by dissecting the 1905 paper.

PS both @vanhees and I have proved the compatibility of the LT and RoS in various posts in this thread.
 
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  • #70
PeterDonis said:
Also, you can't exclude absolute time, or more precisely absolute simultaneity, by experiments, because simultaneity is a convention. You can, however, establish that there is a finite invariant speed by experiments, if you can find a physical phenomenon that moves at that speed, such as light.
You are right, I made an error in posting #59. This thought experiment works under the assumption, that the LT is correct. Instead, for example, a twin paradox experiment with muons in a particle storage ring could disprove GT / Newtonian physics.

PeterDonis said:
You first derive the transformation equations in terms of a parameter ##\alpha##. Then you observe from the transformation equation for ##t'## that ##\alpha > 0## implies ##t' = t##.

No, I wrote the opposite:
Sagittarius A-Star said:
  1. ##\alpha < 0##
  2. ##\alpha = 0##
  3. ##\alpha > 0##
...
Case 2, the GT, can be excluded when assuming ##t' \neq t##, which is the opposite of Newton's assumption of an "absolute time" (see equation 9).
 

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