B A question about relativity of simultaneity

[DmitryS]

Hello. I'm new here and very much afraid of breaking rules. I would gladly post this question in the Homework section, because it's homework, but my question doesn't fit the template, it's a theory question. I hoped to find it in Relativity FAQ's, but it's not there.
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.
In Einstein's train thought experiment, we have two lightnings which simultaneously strike at the front and the back of the car. The textbook says, the observer in the center of the car will get the light of the front lightning earlier than the light from the back lightning.

I don't understand why it happens. It seems intuitive, but the velocity sum will yield the same velocity of light both from the front and the back lightning.
Why are they speaking about this thought experiments? They seem so confusing! It's so much easier to stay with Lorentz transforms.
Do we need these thought experiments at all?

 

[malawi_glenn]

The textbook says

which one?

 

[Sagittarius A-Star]

In Einstein's train thought experiment, we have two lightnings which simultaneously strike at the front and the back of the car.

The lightnings strike simultaneously with reference to the rest-frame of the embankment. They strike not simultaneously with reference to the rest-frame of the train.

Source:
https://en.wikisource.org/wiki/Rela..._I#Section_9_-_The_Relativity_of_Simultaneity

The textbook says, the observer in the center of the car will get the light of the front lightning earlier than the light from the back lightning.

I don't understand why it happens. It seems intuitive, but the velocity sum will yield the same velocity of light both from the front and the back lightning.

That is because, described with reference to the rest-frame of the embankment, the observer in the center of the car is moving towards the light pulse coming from the front. The closing speed between this light pulse and the mentioned observer is $c+v$. And he is moving away from the light pulse coming from the back side.

Why are they speaking about this thought experiments? They seem so confusing! It's so much easier to stay with Lorentz transforms.

Do we need these thought experiments at all?

I think understanding them they helps to understand and apply the LT.

 

[DmitryS]

That is because, described with reference to the rest-frame of the embankment, the observer in the center of the car is moving towards the light pulse coming from the front. The closing speed between this light pulse and the mentioned observer is $c+v$. And he is moving away from the light pulse coming from the back side.

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?

 

[jbriggs444]

But that is the problem I'm asking about. The relative speed is greater than the speed of light.

That is not a "relative speed" it is a "rate of closure".

A relative speed is the speed of one thing in the rest frame of another.

A rate of closure is the rate at which the distance between two objects changes as assessed in the rest frame of a third-party observer. In the usual case of objects on a collision course or departing from a shared event, it will be equal to the difference between the two object's velocities and, thus, will be limited to $2c$.

If I (sitting on the embankment) watch a pulse of light move from the center of a passing train rearward toward the caboose then the rate at which the distance between light pulse and caboose decreases will be a rate of closure ($v - -c = c + v$). Similarly for the rate at which a forward pulse catches up with the engine at the front of the train ($c - v$)

You can get away with just plain old velocity addition in the case of closing velocities because you are adding apples with apples. You are considering two velocities both taken from the same frame of reference. No coordinate transformations to worry about. No length contraction or time dilation as long as everything is measured against the same frame of reference. Just plain old every day positions and velocities.

It is when you are trying to deal with the velocity of object A with respect to the ground and the velocity of object B with respect to object A that you need to start worrying that the two velocities were not drawn from the same frame of reference. Then you need to transform one of the velocities to the same reference frame as the other to be able to add them correctly. That is when time dilation and length contraction (and the relativity of simultaneity) come in. The end result then is the relativistic velocity addition composition formula.

 

[Ibix]

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?

You can't add velocities like that in relativity. If you see an object doing speed $u$ and I see you doing $v$, I do not see the object doing $u+v$, but rather$$\frac{u+v}{1+uv/c^2}$$If you feed $u=c$ in then you will find that I also see the object doing $c$. Note also that if both $u$ and $v$ are much less than $c$ the denominator is nearly 1, which is why you can just add speeds in the Newtonian (low speed) approximation.

 

[Sagittarius A-Star]

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?

As others said, in case of a closing speed, the "addition formula" cannot be applied.

The "addition formula", which has a misleading name, applies only for transforming a velocity $u= \frac{dx}{dt}$ from one inertial frame to another.$$u' = \frac{dx'}{dt'} = \frac{\gamma(dx-vdt)}{\gamma(dt-vdx/c^2)}= \frac{\gamma(\frac{dx}{dt}-v)}{\gamma(\frac{dt}{dt}-\frac{dx}{dt}v/c^2)} = \frac{u-v}{1-uv/c^2}$$

If you understand the difference between closing speed and velocity transformation, then you have already learned something important from this Einstein thought experiment.

 

[DmitryS]

If you understand the difference between closing speed and velocity transformation, then you have already learned something important from this Einstein thought experiment.

I'm afraid I didn't say everything I was thinking. If we use closing speed to assess the time difference, we get coefficients of the time transformation which are not the Lorentz coefficients.

 

[Sagittarius A-Star]

I'm afraid I didn't say everything I was thinking. If we use closing speed to assess the time difference, we get coefficients of the time transformation which are not the Lorentz coefficients.

The closing speed is described in one reference frame only, in my posting #3 with reference to the rest-frame of the embankment. There is no LT applied.

 

[PeroK]

If we use closing speed to assess the time difference, we get coefficients of the time transformation which are not the Lorentz coefficients.

So what? That's not what the Lorentz Transformation describes.

 

[DmitryS]

The closing speed is described in one reference frame only, in my posting #3 with reference to the rest-frame of the embankment. There is no LT applied.

But that's the question I am asking, don't you see? The clocks of the moving frame of reference which are located along the line of motion are not synchronous according to the observer at rest. And LT describes how they are asynchronous.
The relativity of simultaneity test like explains why this happens, and I see discrepancy between the LT and the experiment.
I'm not saying you're wrong - I believe you're right. I just can't make these things work together.

And I gave another thought to the closing speed concept, and now I'm not happy with it at all.

Let's don't talk light. Supposing there is a toy car in the middle of the traincar moving towards the back of the train at the speed w. The closing speed will be $v + w$. So, when the toy car reaches the back of the traincar, time measured by the clock of the motionless frame of reference is $$t = \frac {l} {v + w},$$ where l is a half-length of the traincar measured in the motionless frame. That means that the distance to which the train has shifted in this while is $$x = \frac {lv} {v + w}.$$ This must be the coordinate where the toy car is at the end. The toy car moved to the distance of $$l -\frac {lv} {v + w} = \frac {lw} {v + w}$$ in the stationary frame. But, at the same time, the velocity of the toy car in the stationary frame is $$u=\frac {v-w} {{1 -\frac {wv}{c^2}}$$.

If you divide the distance

$$\delta x = \frac {lw} {v + w}$$

by that, you will have different time.

I'm not sure I got you right what you said when the LT apply or don't apply, but even if you use the proper length of the traincar these don't fit together.

That's my problem, see? I think that the closing speed concept and the Einstein velocity composition cancel out.

 

[Sagittarius A-Star]

The textbook says, the observer in the center of the car will get the light of the front lightning earlier than the light from the back lightning.

I don't understand why it happens.

Description with reference to the rest-frame of the embankment:

At the time, when both lightnings happen synchronous at positions $A$ and $B$, the observer in the center $M'$ of the car is at position $M$ in the middle. Shortly afterwards, he ist on the right side of $M$.

https://en.wikisource.org/wiki/Rela..._I#Section_9_-_The_Relativity_of_Simultaneity

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 $A$.

 

[Sagittarius A-Star]

Let's don't talk light. Supposing there is a toy car in the middle of the traincar moving towards the back of the train at the speed w. The closing speed will be $v + w$. So, when the toy car reaches the back of the traincar, time measured by the clock of the motionless frame of reference is $$t = \frac {l} {v + w},$$ where l is a half-length of the traincar measured in the motionless frame. That means that the distance to which the train has shifted in this while is $$x = \frac {lv} {v + w}.$$

Here you mix up two different frames. That is a scenario not to use the closing speed, but the velocity transformation "relativistic velocity addition formula" instead.

 

[DmitryS]

Here you mix up two different frames. That is a scenario not to use the closing speed, but the velocity transformation "relativistic velocity addition formula" instead.

I do nothing of the kind! I always stay in the embankment frame of reference, as you said!
If you have the closing speed scenario which is true in the motionless frame, you have this time measured by the stationary clock. If you gave the Einstein velocity composition, you have a different time of the same event.

 

[Sagittarius A-Star]

I do nothing of the kind! I always stay in the embankment frame of reference, as you said!
If you have the closing speed scenario which is true in the motionless frame, you have this time measured by the stationary clock. If you gave the Einstein velocity composition, you have a different time of the same event.

But if you understood your text correctly, $w$ is the toy car's speed with reference to the train's rest-frame.

 

[PeroK]

I do nothing of the kind! I always stay in the embankment frame of reference, as you said!
If you have the closing speed scenario which is true in the motionless frame, you have this time measured by the stationary clock. If you gave the Einstein velocity composition, you have a different time of the same event.

Whenever the Einstein lightning thought experiment comes up, I always say that it is particularly confusing and misleading. I always recommend that it should be ignored in favour of the much simpler examples of the RoS.

I don't know whether the experiment is confusing you or you are genuinely confused by the RoS.

The invariance of the speed of light and the invariance of simultaneity are incompatible. That's not too hard to see with a bit of basic kinematics. It doesn't need simultaneous lightning strikes or the Lorentz Transformation.

 

[DmitryS]

But if you understood your text correctly, $w$ is the toy car's speed with reference to the train's rest-frame.

I must have confused it. First, we calculate the speed of the toy car in the stationary frame, and then we can apply the closing speed scenario?

 

[PeroK]

I must have confused it. First, we calculate the speed of the toy car in the stationary frame, and then we can apply the closing speed scenario?

Even saying the "stationary" frame is misleading - the platform frame is no more a "stationary" frame than the train frame.

 

[Sagittarius A-Star]

I must have confused it. First, we calculate the speed of the toy car in the stationary frame, and then we can apply the closing speed scenario?

Yes.

 

[DmitryS]

Yes.

Thanks. What about the first part of my problem?
"The clocks of the moving frame of reference which are located along the line of motion are not synchronous according to the observer at rest. And LT describes how they are asynchronous.
The relativity of simultaneity test like explains why this happens, and I see discrepancy between the LT and the experiment.
I'm not saying you're wrong - I believe you're right. I just can't make these things work together."

 

[PeroK]

Thanks. What about the first part of my problem?
"The clocks of the moving train frame of reference which are located along the line of motion are not synchronous according to the observer at rest on the platform.

From the 1905 paper:

... suggest that the phenomena of electrodynamics as well as of mechanics possesses no properties
corresponding to the idea of absolute rest.

 

[PeroK]

But that's the question I am asking, don't you see? The clocks of the moving frame of reference which are located along the line of motion are not synchronous according to the observer at rest. And LT describes how they are asynchronous.
The relativity of simultaneity test like explains why this happens, and I see discrepancy between the LT and the experiment.
I'm not saying you're wrong - I believe you're right. I just can't make these things work together.

And I gave another thought to the closing speed concept, and now I'm not happy with it at all.

Let's don't talk light. Supposing there is a toy car in the middle of the traincar moving towards the back of the train at the speed w. The closing speed will be $v + w$. So, when the toy car reaches the back of the traincar, time measured by the clock of the motionless frame of reference is $$t = \frac {l} {v + w},$$ where l is a half-length of the traincar measured in the motionless frame. That means that the distance to which the train has shifted in this while is $$x = \frac {lv} {v + w}.$$ This must be the coordinate where the toy car is at the end. The toy car moved to the distance of $$l -\frac {lv} {v + w} = \frac {lw} {v + w}$$ in the stationary frame. But, at the same time, the velocity of the toy car in the stationary frame is $$u=\frac {v-w} {{1 -\frac {wv}{c^2}}$$.

If you divide the distance

$$\delta x = \frac {lw} {v + w}$$

by that, you will have different time.

I'm not sure I got you right what you said when the LT apply or don't apply, but even if you use the proper length of the traincar these don't fit together.

That's my problem, see? I think that the closing speed concept and the Einstein velocity composition cancel out.

I don't see, because there was a melt-down in your Latex. Can you fix that?

 

[Janus]

Animations of train experiment as seen by embankment and train observers:

Embankment observer notes that train just fits between red dots, as lightning flashes are emitted.
The train is in motion is in relative motion, so it is length contracted( if it wasn't, it wouldn't fit between the dots). The right flash meets up with the train observer when he is 4 ties past the track observer, and the left flash catches up when he is almost to the right red dot.

For the train observer, it is the tracks that are length contracted, and the train does not fit exactly between the dots. The right dot reaches the front of the train before the left one reaches the rear of the train. The train observer must agree with the track observer that the lightning struck both the ends of the train and the red dots when they were aligned with each other. Since the two ends reach their respective dots at different moments, the strikes cannot be simultaneous. Note that the train observer also agrees that the right flash reaches him when he is 4 ties from the track observer, and the left one when he is near the right red dot.

 

[Sagittarius A-Star]

"The clocks of the moving frame of reference which are located along the line of motion are not synchronous according to the observer at rest. And LT describes how they are asynchronous.

The same applies to Einstein's train/embankment thought experiment. It describes with reference to the rest-frame of the embankment, why the observer in the center of the car receives the two light-pulses at different times.

The conclusion for the description with reference to the rest-frame of the train:

The train is at rest and the embankment is moving backwards.

The observer is at rest at $M'$. This location is in the middle between the locations $A'$ and $B'$ in the train, where the lightnings happend. Both light-pulses moved over equal distances with equal speeds $c$ towards this observer at rest at $M'$. Because the observer receives the two light-pulses at different times and first the light-pulse from $B$, the lightning at location $B'$ must have happened earlier than the lightning at location $A'$.

The clocks of the moving embankment-frame, which are located along the line of motion, are not synchronous according to the observer at rest at $M'$ in the train. (And LT describes how they are asynchronous.)

 

[vanhees71]

It's MUCH simpler to just do the Lorentz transformations of the trajectories of the wave fronts and the points $M$ and $M'$. In the embarkment frame those are
$$x_{A/B}=(ct,\mp L/2 \pm c t), \quad x_{M}=(c t,0), \quad x_{M'}=(ct,v t).$$
Now just do the Lorentz transformation and, check when the light signals reach $M$ and $M'$ from the point of view of the observer in the train. As a byproduct you'll also find the length contraction, i.e., the length of the train as seen from the embarkment is shorter than the length in the train's rest frame.

 

[PeroK]

I think we need to put a valid calculation on the table, as it were. We have a train wagon of length $L$, as measured in a frame in which it is traveling with speed $v$ to the right. A source in the centre of the wagon emits a pulse of light towards both ends. When the light hits each end of the wagon, a clock at each end of the wagon is set to $0$.

Under the assumption that the light is at the centre of the wagon in the wagon's rest frame (exercise to justify this), in the wagon frame the clocks at either end are set to $0$ simultaneously.

Now, let's look at the scenario in the frame where the wagon is moving.

The light hits the rear of the train at time $t_1 = \frac{L/2}{c + v}$ and the front of the train at time $t_2 = \frac {L/2}{c-v}$. That's the relativity of simultaneity right there, in a nutshell. No need for miraculously simultaneous lightning strikes!

And, we can calculate $$\Delta t = t_2 - t_1 = \frac{Lv}{c^2 - v^2} = \frac{Lv}{c^2(1 - v^2/c^2)} = \frac{Lv\gamma^2}{c^2}$$Now, if we know about time dilation, then the rear clock will have advanced only$$\Delta t' = \frac{\Delta t}{\gamma} = \frac{Lv\gamma}{c^2}$$And, we can see that when the front clock reads $0$, the rear clocks reads $\Delta t' = \frac{Lv\gamma}{c^2}$. Note that $L$ is not the rest length of the wagon here.

Now, let's check the Lorentz transformation. We have two events in the frame in which the wagon is moving: $(0, 0)$ and $(\Delta t, L+v\Delta t)$. We get this by putting the origin at the rear of the train.

These transform to $(0,0)$ and $$(\gamma(\Delta t - \frac{v(L+v\Delta t)}{c^2}), \gamma(L + v\Delta t - v\Delta t)) = (0, \gamma L)$$ in the wagon frame. As expected. Note that we also get the rest length of the wagon as $\gamma L$ from this. Finally, let's just check that calculation of the time coordinate in the last step:
$$\Delta t - \frac{v(L+v\Delta t)}{c^2} = \Delta t(1 - \frac{v^2}{c^2}) - \frac{vL}{c^2} = \frac{\Delta t}{\gamma^2} - \frac{vL}{c^2} = \frac{vL}{c^2} - \frac{vL}{c^2} = 0$$And everything is hunky dory.

 

[DmitryS]

The same applies to Einstein's train/embankment thought experiment. It describes with reference to the rest-frame of the embankment, why the observer in the center of the car receives the two light-pulses at different times.

The conclusion for the description with reference to the rest-frame of the train:

The train is at rest and the embankment is moving backwards.

The observer is at rest at $M'$. This location is in the middle between the locations $A'$ and $B'$ in the train, where the lightnings happend. Both light-pulses moved over equal distances with equal speeds $c$ towards this observer at rest at $M'$. Because the observer receives the two light-pulses at different times and first the light-pulse from $B$, the lightning at location $B'$ must have happened earlier than the lightning at location $A'$.

The clocks of the moving embankment-frame, which are located along the line of motion, are not synchronous according to the observer at rest at $M'$ in the train. (And LT describes how they are asynchronous.)

That's right. And if you assess the effect with (c+v) and (c-v), you get the coefficients which are not LT. That's the problem I'm talking.

 

[PeroK]

That's right. And if you assess the effect with (c+v) and (c-v), you get the coefficients which are not LT. That's the problem I'm talking.

What don't you understand about post #26? The Lorentz Transformation encapsulates time dilation, length contraction and the RoS. Where's the problem you are talking about?

 

[Sagittarius A-Star]

That's right. And if you assess the effect with (c+v) and (c-v), you get the coefficients which are not LT. That's the problem I'm talking.

I don't know, what you want to Lorentz-transform.

You could for example Lorentz-transform the coordinates of the two lightning events from the (unprimed) embankment rest frame into the (primed) train rest frame.

Event A: $\ \ \ \ \ \ x_A=-\frac{L}{2}$, $\ \ \ \ \ \ t_A=0$.
Event B: $\ \ \ \ \ \ x_B=+\frac{L}{2}$, $\ \ \ \ \ \ t_B=0$.

Transform event A coordinates:
$x'_A = \gamma (x_A - vt_A) = -\gamma \frac{L}{2}$, $\ \ \ \ \ \ t'_A= \gamma (t_A - vx_A/c^2) =\gamma \frac{L}{2}v/c^2$.

Transform event B coordinates:
$x'_B = \gamma (x_B - vt_B) = +\gamma \frac{L}{2}$, $\ \ \ \ \ \ t'_B= \gamma (t_B - vx_B/c^2) =-\gamma \frac{L}{2}v/c^2$.

From $t'_B < t'_A$ you see, that events B happens earlier than event A, with reference to the train rest frame.

 

[Ibix]

That's right. And if you assess the effect with (c+v) and (c-v), you get the coefficients which are not LT. That's the problem I'm talking.

As noted upthread, $(c+ v)$ and $(c-v)$ are the closing rates of the light pulses and the observer on the train as measured in the embankment frame. The closing rate is not the same thing as the relative speed. The closing rate is the change in the distance I measure in my frame between you and some object. The relative speed is the change in distance you measure from you to an object in your frame (note that there are only two things in this description, while there were three in the description of closing rate). These quantities are distinct. In Newtonian physics they are degenerate and have the same value. In relativity they are not the same.

So you seem to be correctly calculating the closing rate and worrying that it is not the same as the relative velocity. Don't. They are not the same.

You might ask why they are not the same. Fundamentally it is because we are choosing to divide spacetime into space and time in different ways (which is what "using different frames" means). The direct consequence of that is that I see your rulers as length contracted, and your clocks as time dilated and incorrectly synchronized. Thus I am not surprised that you do not measure $c\pm v$ for the speed. You would say the same of my clocks and rulers, so should not be surprised that I don't measure $c$ for the closing rate.

 

[DmitryS]

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.

 

[Sagittarius A-Star]

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.

 

[vanhees71]

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).

 

[Sagittarius A-Star]

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.

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.

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?

 

[DmitryS]

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.

 

[Sagittarius A-Star]

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:

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:

$x'_A = \gamma (x_A - vt_A) = -\gamma \frac{L}{2}$, $\ \ \ \ \ \ t'_A= \gamma (t_A - vx_A/c^2) =\gamma \frac{L}{2}v/c^2$.

Transform event B coordinates:
$x'_B = \gamma (x_B - vt_B) = +\gamma \frac{L}{2}$, $\ \ \ \ \ \ t'_B= \gamma (t_B - vx_B/c^2) =-\gamma \frac{L}{2}v/c^2$.

 

[DmitryS]

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.

 

[DmitryS]

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.

 

[vanhees71]

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.

 

[Sagittarius A-Star]

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.

 

[PeroK]

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).

 

[vanhees71]

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.

 

[DmitryS]

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?

 

[DmitryS]

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?

 

[PeroK]

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!

 

[PeterDonis]

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.

 

[DmitryS]

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.

 

[PeroK]

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.

 

[PeterDonis]

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.

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.

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.

 

[DmitryS]

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.

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