B Learning Lorentz Derivation: Solving x=vt+γx' vs. x=vt+x'/γ

[jeremyfiennes]

From a previous thread, Nugatory: "If the answers already supplied are not enough for the original poster to work out for themself why x=vt+x'/γ, and not x=vt+γx′, is the correct expression, we can have another thread devoted to only that question."
Here it is. Either Wiki is wrong or I am. In which case I need to know where my reasoning is faulty and why. My derivation:

– the frame A distance x_a' of event X from the frame B origin is x_a'=x-vt, Fig.(a)
– in frame B, everything stationary in frame A, including the distance x_a', moves at speed v, and is therefore foreshortened by γ, Fig.(b)
– whence x' = x_a'/γ = (x-vt)/γ, leading to x=vt+γx' as I maintain
– and not Wiki's x=vt+x/γ'

 

[Dale]

Can you please link to both the previous thread and the wiki?

 

[Ibix]

Your distance $x'_a$ does not transform like that. If you imagine a ruler lying at rest in O where your arrow labelled $x'_a$ is, then in O' it is moving and the ends do not meet event X and the origin of your primed coordinates simultaneously.

You have fallen for the rod-and-barn paradox, in short. The ruler plays the part of the rod, and event X and the origin of O' intersecting the left end of the rod play the parts of the barn doors.

 

[Nugatory]

In the previous thread we suggested two things that may help you find your misunderstanding:
1) Draw a Minkowski diagram showing: the event M; the worldline of the origin of frame F; and the worldline of the origin of frame F'. Draw the lines representing the distance between M and the origin of O' using both frames.
2) Try using the Lorentz transformations to convert the coordinates of the two relevant events (location of the origin of F' at the same time as M in frame F and in at the same time as M frame F' - these are different because of relativity of simultaneity!) between the two frames.

To this, I would add, "3) read @Ibix's post #3 above carefully".

However, the quickest way of stating your error so that you can correct your derivation is to repeat that length contraction is symmetrical: If the distance between O' and M is L in frame F1 then it will be $L/\gamma$ in frame F; but also if the distance is Q in frame F it will be $Q/\gamma$ in frame F'. This paradoxical-sounding fact is analogous to the way that time dilation is also symmetrical (A finds B's clock to be slow but B finds A's clock to be slow); it is a consequence of relativity of simultaneity, and the suggestions above are intended to help you see why it is this way.

 

[Nugatory]

Can you please link to both the previous thread and the wiki?

I'm not @jeremyfiennes but I can provide the backstory. The thread is https://www.physicsforums.com/threads/basic-lorentz-transformation-derivation.961038/ and the relevant wikipedia section is https://en.wikipedia.org/wiki/Deriv...rmations#Time_dilation_and_length_contraction

Deriving the Lorentz transformations from the assumptions of length contraction and time dilation is surely not the preferred approach; there's even a discussion on the article's "talk" page about whether it is legitimate at all. It is (barely, but that's a different topic).

 

[Dale]

the frame A distance xa' of event X from the frame B origin is xa'=x-vt, Fig.(a)

This is horrendously confusing. In Fig. (a) there is no frame A or B.

And I am completely bewildered by the confusing, unexplained, and completely unnecessary $x’_a$. There should be only x, x’, t, t’, v, and $\gamma$. Anything else is superfluous.

From the picture it seems like maybe it is just an algebraic shorthand for x-vt, but then it is confusing that an algebraic quantity defined exclusively from quantities in the unprimed frame would be given a primed symbol. So maybe it is supposed to have some other significance, but it is not clear at all.

frame B, everything stationary in frame A, including the distance xa',

Assuming frame B is frame F’ and further assuming $x’_a=x-vt$ then clearly $x’_a$ is not stationary in A (F). The vt term clearly makes one side move at speed v.

Anyway, with all of this confusion it is not particularly surprising that you made a mistake. Notational clarity is important for something like this.

 

[jeremyfiennes]

Can you please link to both the previous thread and the wiki?

Sorry, I was out of the loop for a while. I see Nugatory has already answered this. But since I had already looked the the refs up and pasted them in: the original thread is <https://www.physicsforums.com/threads/basic-lorentz-transformation-derivation.961038/>. My original statement of the problem was in post #8.
The Wiki article is <https://en.wikipedia.org/wiki/Deriv...rmations#Time_dilation_and_length_contraction>. The equation I disagree with is the second in the section "From physical principles: time dilation and length contraction".

You have fallen for the rod-and-barn paradox,

What is this? Maybe I have.

 

[jeremyfiennes]

This is horrendously confusing. In Fig. (a) there is no frame A or B. And I am completely bewildered by the confusing, unexplained, and completely unnecessary x′axa′x’_a.

Sorry, there was a duplication of nomenclature. Frame A = frame F = unprimed. Frame B = frame F' = primed. The 'O's and 'M' were from the Wiki article. . I have cleaned up the diagram:

x_(a)' is the length x-vt in frame A. x' is the same length in frame B, where it moves at speed v and is therefore foreshortened by γ, giving x'=(v-t)γ. This, however, according to Wiki is incorrect. The question is not whether some other approach gives the 'correct' answer. But rather: why my reasoning, which as far as I can see conforms strictly to SR principles and is completely valid, should lead to a non-standard result.

 

[Nugatory]

The question is...: why my reasoning, which as far as I can see conforms strictly to SR principles and is completely valid, should lead to a non-standard result.

Because your reasoning contains an error that has been pointed out to you a half-dozen times already by multiple posters, most recently in post #4 above. You have not responded to (or even shown any sign that you have bothered to read) these posts. This is wasting everyone's time including yours, and the thread will be closed if you don't show at least some sign that you are listening.

 

[Nugatory]

You have fallen for the rod-and-barn paradox, in short.

What is this? Maybe I have.

It's also called the "ladder-barn paradox", so if you google for either "pole barn relativity" or "ladder barn relativity" you will find plenty of hits, and we also have some threads here. It's a common and standard thought experiment that shows how length contraction works (not the way you're using it in your attempted derivation!) and why relativity of simultaneity is so important.

 

[stevendaryl]

Sorry, there was a duplication of nomenclature. Frame A = frame F = unprimed. Frame B = frame F' = primed. The 'O's and 'M' were from the Wiki article. . I have cleaned up the diagram:
View attachment 235065
x_(a)' is the length x-vt in frame A. x' is the same length in frame B, where it moves at speed v and is therefore foreshortened by γ, giving x'=(v-t)γ. This, however, according to Wiki is incorrect. The question is not whether some other approach gives the 'correct' answer. But rather: why my reasoning, which as far as I can see conforms strictly to SR principles and is completely valid, should lead to a non-standard result.

I already tried to explain this, but let me try again. To make it concrete, let's suppose that observer A has a stick of constant length $x$ (as measured in his own frame) that he holds out straight in front of him. At the end of the stick, there is a clock, which is synchronized according to A's frame. Then let's consider the event where that clock shows time $t$. In A's coordinate system, the coordinates are $(x,t)$.

Now, let's switch to frame $B$. In his coordinate system, the location of the clock at the end of the stick is only a distance of $x/\gamma$ away from observer $A$. So the location of the clock is equal to the location of A + $x/\gamma$. What is the location of A? In B's coordinate system, A starts off at $x'=0$ when $t'=0$ and is traveling in the negative-x direction at speed $v$. So at a time $t' > 0$, A is at location$- v t'$. The clock is a distance $x/\gamma$ ahead, so it is at location$-vt' + x/\gamma$.

Therefore, the point with coordinates $(x,t)$ in frame A corresponds to the point $(x', t')$ in frame B where $x' = -vt' + x/\gamma$.

$x'$ is NOT the length of the stick, it's the location of the clock on the end of the stick. The length of the stick is $x/\gamma$, and the location is $-vt' + x/\gamma$.

So the relation between $x'$ and $x$ is $x' = -vt' + x/\gamma$, which can be rearranged to give: $x = \gamma (x' + vt')$.

A similar argument shows that the relationship between $x'$ and $x$ can also be written as: $x = x'/ \gamma + vt$, which can be rearranged to: $x' = \gamma(x - vt)$

So the two equations are:

1. $x' = \gamma (x - vt)$
2. $x = \gamma(x' + vt')$

You can rewrite 2. using 1. to get:
$t' = (x/\gamma - x')/v = (x/\gamma - \gamma x + \gamma vt)/v = \gamma (x(1/\gamma^2 - 1)/v + t) = \gamma (-vx/c^2 + t)$

 

[jeremyfiennes]

If you google pole barn-relativity"

I did. Thanks.

 

[jeremyfiennes]

A similar argument shows that ..

Thanks. Due to time dilation t'=t/γ, so your x=γ(x'+vt') is equivalent to my x=γx'+vt. On this we agree. But what "similar argument"?

 

[Dale]

x' is the same length in frame B,

Here is your logical mistake then. Because $x-vt$ is changing and because of the relativity of simultaneity these two quantities are not the same.

 

[jeremyfiennes]

Here is your logical mistake then.

I'm not sure where you got my "x' is the same length in frame B" from. I was probably saying that the length x-vt in frame A corresponds to x' in frame B. Applying length contraction gives x'=(x-vt)/γ. Leading to x=vt+γx', agreeing with stevendaryl's eq.2, and contradicting Wiki's x=vt+x'/γ

 

[stevendaryl]

Thanks. Due to time dilation t'=t/γ

No, it's NOT! That's not correct. $t'$ is not equal to $t/\gamma$.

$t'$ is not the time on B's clock. It's a coordinate. The time $t'$ associate with the event $(x,t)$ is the time that would show on a clock that: (1) Is synchronized with B's clock, and (2) Is at rest relative to B. Of course, there doesn't actually have to be a clock at that event; B can compute what such a clock would show by the following calculation:

• Let $t'_{received}$ be the time that the light from the event reached B,.
• Let $x'$ be the location (in B's coordinate system) that the event took place.
• Then the calculated time $t'$ is given by: $t' = t'_{received} - x'/c$

That is definitely not the same thing as $t' = t/\gamma$

 

[stevendaryl]

Thanks. Due to time dilation t'=t/γ, so your x=γ(x'+vt') is equivalent to my x=γx'+vt. On this we agree. But what "similar argument"?

If A has a stick of length $x$, and there's a clock on the end showing time $t$, then the location of the clock as computed by $B$ is given by:

$x' = x/\gamma - v t'$

If B has a stick of length $x'$, and there's a clock on the end showing time $t'$, then the location of the clock as computed by $A$ is given by:

$x = x'/\gamma + v t$

Those two equations imply the Lorentz transformations:

$x' = \gamma (x-vt)$
$t' = \gamma (t - \frac{vx}{c^2})$

 

[Dale]

I'm not sure where you got my "x' is the same length in frame B" from.

From here (it is word for word a direct quote):

x_(a)' is the length x-vt in frame A. x' is the same length in frame B.

I am not objecting to x’, but your quantity $x’_a$ needs to be removed. It is completely superfluous and your use of it is contributing to your mistake.

I was probably saying that the length x-vt in frame A corresponds to x' in frame B. Applying length contraction gives x'=(x-vt)/γ.

Here the error is clear. You cannot apply length contraction to x-vt; there is no frame where both ends are at rest.

 

[jeremyfiennes]

No, it's NOT! That's not correct.

Omaigodd! Then I'm all mixed up again. On the original setup I considered there were clocks at 1) frame A space origin, 3) frame B space origin, 3) the event spatial location, which is x in frame A. All three clocks were synchronised at t=0 (the coincidence of the frames A and B spatial origins) via a light signal sent from a midpoint x/2 in frame A. From then on clock B travels at speed v relative to frame A, and is dilated by γ relative tp clock A. Giving at subsequent instants t'=t/γ, where t is frame A time and t' is frame B time. Where is that wrong?

"The time t′ is the time that would show on a clock that: (1) Is synchronized with B's clock, and (2) Is at rest relative to B."
That is exactly the definition of the t'=t/γ I am considering.

• "Let t′_received be the time that the light from the event reached B." I was informed in the previous thread that photon travel times have no place the Lorentz transformation. Times t are those on any synchronized clock stationary in the frame.

 

[jeremyfiennes]

I am not objecting to x’, but your quantity x_(a)' a needs to be removed.

I agree it is not essential. But rather than refer each time to "the frame A correlate of the frame B space coordinate x' ", I found it more convenient to give it a name. This also enables it to be identified on the diagram.

 

[stevendaryl]

"The time t′ is the time that would show on a clock that: (1) Is synchronized with B's clock, and (2) Is at rest relative to B."
That is exactly the definition of the t'=t/γ I am considering.

Well, that's wrong. I just went through the proof that it's wrong. The correct relationship between $t'$ and $t$ depends on $x$.

I was informed in the previous thread that photon travel times have no place the Lorentz transformation. Times t are those on any synchronized clock stationary in the frame.

What you were informed of is exactly what I'm informing you of now, which is that the time $t'$ that B assigns to an event is NOT equal to the time $t'_{received}$ that he sees the light from the event. They are related, though:

$t'_{received} = t' + x'/c$

where $x'$ is the location of the event in B's coordinate system.

 

[Dale]

But rather than refer each time to "the frame A correlate of the frame B space coordinate x' ",

You should not refer to it at all. It is messing you up and you are using it inappropriately. You should refer only to x, x’, t, t’, v, and $\gamma$. It would also be OK to refer to the frames and axes themselves, and if you feel compelled to use unnatural units then you can include c also.

 

[jeremyfiennes]

Well, that's wrong. I just went through the proof that it's wrong. The correct relationship between t′ and t depends on xxx

The question seems to have reduced to this: Einstein with his station-and-truck-observer thought experiment showed that a clock in a primed frame B, moving at speed v relative to an unprimed frame A, runs slow by a factor γ, irrespective of its space location x. Giving t'=t/γ. Whereas Lorentz gave the relation as t'=γ(t–vx/c²). The two disagree. But since Einstein gave a logical geometric derivation for his expression, whereas Lorentz apparently provided none, on this basis one would prefer Einstein's version as being more scientifically correct.

 

[Ibix]

The question seems to have reduced to this: Einstein with his station-and-truck-observer thought experiment showed that a clock in a primed frame B, moving at speed v relative to an unprimed frame A, runs slow by a factor γ, irrespective of its space location x. Giving t'=t/γ. Whereas Lorentz gave the relation as t'=γ(t–vx/c²). The two disagree. But since Einstein gave a logical geometric derivation for his expression, whereas Lorentz apparently provided none, on this basis one would prefer Einstein's version as being more scientifically correct.

Um... no. Einstein derived the Lorentz transforms exactly as Lorentz discovered them. There's no difference, except Einstein provided a good reason instead of "well they fix Maxwell's equations".

 

[Nugatory]

Einstein with his station-and-truck-observer thought experiment showed that a clock in a primed frame B, moving at speed v relative to an unprimed frame A, runs slow by a factor γ, irrespective of its space location x. Giving t'=t/γ.

No. Einstein derived exactly the same Lorentz transformations as Lorentz did.

You are confusing the time dilation formula with the Lorentz transformation equation for the $t$ coordinate.

It is correct that the moving clock runs slow, but writing that as $t'=t/\gamma$ is sloppy notation (people do it all the time because they understand and expect their audience to understand that the $t$ and $t'$ in that formula do not mean the same thing as the $t$ and $t'$ in the Lorentz transformations - it's still sloppy). Things are more clear we write the time dilation formula without sloppiness: $\Delta{T}'=\Delta{T}/\gamma$. $\Delta{T}$ is by definition $t_1-t_0$, $\Delta{T}'$ is by definition $t_1'-t_0'$ and:
- $t_0$ and $t_1$ are two successive times at which an observer at rest in the unprimed frame checks his clock. Clearly the time interval between these readings in the unprimed frame will be $\Delta{T}$.
- $t_0'$ is what a clock at rest in the primed frame reads at the same time (using the unprimed frame's definition of "at the same time") that the unprimed clock reads $t_0$. It is related to $t_0$ by the Lorentz transformation, including the $vx/c^2$ term.
$t_1'$ is what a clock at rest in the primed frame reads at the same time (using the unprimed frame's definition of "at the same time") that the unprimed clock reads $t_1$. It is related to $t_1$ by the Lorentz transformation, including the $vx/c^2$ term.

It should be clear that $\Delta{T}'$ is the time that passes between the two clock readings in the primed frame so $\Delta{T}'=\Delta{T}/\gamma$ properly captures the time dilation relationship.

A common source of confusion, almost certainly your difficulty here, is that when we derive time dilation using a light clock or the station/truck thought experiment, we generally set things up so that the first clock reading is done when the two clocks are colocated and have just been set to 0 - that is $x_0=t_0=x_0'=t_0'$. That causes a bunch of terms to disappear: $\Delta{T}$ becomes equal to $t_1$, and because the second clock reading has the coordinates (using the unprimed frame) $(0,t_1)$ it's very easy to confuse $\Delta{T}$, the interval between two clock readings, with $t_1$, the time coordinate of the second clock reading when made under these special conditions. That's how you ended up writing $t'=t/\gamma$ and thought you had a relationship between time coordinates.

 

[stevendaryl]

The question seems to have reduced to this: Einstein with his station-and-truck-observer thought experiment showed that a clock in a primed frame B, moving at speed v relative to an unprimed frame A, runs slow by a factor γ, irrespective of its space location x.

Yes, but $t'$ is NOT the time on B's clock. It's the time coordinate. To go from time on a clock to a time coordinate, you need a synchronization convention for distant clocks.

To relate A's time coordinate, $t$ to B's time coordinate, $t'$, let's pick a clock that it is a constant distance $x'$ away from B, as measured in B's frame. What is the time showing on that clock when the coordinate time, according to $A$, is $t$?

It's this: $t' = t/\gamma + \Delta t'$

The meaning of this equation is that the difference between $t'$ and $t$ depends on two factors: (1) the rate at which the clock advances, (which we're assuming is $1/\gamma$), and (2) an offset, $\Delta t'$. Think about comparing a clock in France to a clock in New York. Even if they are running at the same rate, they won't show the same time, because of time zones. France is -- I don't know -- 6 hours ahead.

So we need to figure out what $\Delta t'$ is for a clock that is $x'$ away from $B$. How do we do that? Well, B "thinks" that he is the one at rest. So from his point of view, he can synchronize the clock at location $x'$ through the following procedure:

1. At time $t' = 0$, he sends a light signal toward the clock at location $x'$.
2. He calculates that it will arrive at time $x'/c$. So when it arrives, that clock should be set to $x'/c$.

But let's look at things from the point of view of A. From his point of view, two things are different:

1. The distance between B and the clock is $x'/\gamma$, not $x'$.
2. The clock is moving away from the light signal, so instead of taking time $x'/(\gamma c)$ to arrive, it takes time $x'/(\gamma (c-v))$.

So the light signal takes time $x'/(\gamma c)$ to arrive at the clock, and in the meantime, B's clock advances less than that, by a factor of $\gamma$ (because it's running slow). So when the light signal arrives, B's clock shows time $t' = (x'/(\gamma (c-v)))/\gamma = \frac{x'}{c-v} \frac{1}{\gamma^2}$

The clock is then set to $x'/c$, so this clock is set to a DIFFERENT time than B's clock, by an amound $\Delta t' = x' (\frac{1}{c} - \frac{1}{c-v} \frac{1}{\gamma^2})$. Using $\frac{1}{\gamma^2} = 1 - \frac{v^2}{c^2}$, we have: $\Delta t' = - \frac{v x'}{c^2}$
So the relationship between $t$ and $t'$ is not $t' = t/\gamma$, but $t' = t/\gamma - \frac{vx'}{c^2}$

From A's point of view, there are two things "wrong" with B's clocks: (1) they are running slow, and (2) they are not synchronized correctly. You need both of these in order to have consistent transformation equations that respect the principle of relativity.

 

[stevendaryl]

So the relationship between $t$ and $t'$ is not $t' = t/\gamma$, but $t' = t/\gamma - \frac{vx'}{c^2}$

If we plug in: $x' = \gamma (x - vt)$, then we get the Lorentz expression:

$t' = t/\gamma - \gamma \frac{v (x - vt)}{c^2} = \gamma (t (\frac{1}{\gamma^2} + \frac{v^2}{c^2}) - \frac{vx}{c^2} = \gamma (t - \frac{vx}{c^2})$

 

[jeremyfiennes]

To go from time on a clock to a time coordinate, you need a synchronization convention for distant clocks.

I stated my synchronization procedure in #19.

Yes, but t' is NOT the time on B's clock. It's the time coordinate.

The time coordinate of a frame is the time on any clock stationary in that frame. Einstein's truck clock is stationary in frame B. And therefore shows the time coordinate for that frame.

 

[PeterDonis]

On the original setup I considered there were clocks at 1) frame A space origin, 3) frame B space origin, 3) the event spatial location, which is x in frame A.

Is clock #3 at rest relative to clock #1 (at rest in frame A) or clock #2 (at rest in frame B)?

 

[Dale]

The time coordinate of a frame is the time on any clock stationary in that frame.

No, it is still different. The proper time on a clock is only defined on the worldline of that clock. The coordinate time is defined throughout spacetime. Where both are defined they do agree, but they remain distinct because they are defined in different regions of spacetime.

 

[stevendaryl]

stated my synchronization procedure in #19.

That post was completely wrong.

The time coordinate of a frame is the time on any clock stationary in that frame.

No, it's not. Do you understand that a time coordinate means an assignment of times to every event. Different events have different time coordinates. B's clock only assigns times to events that take place at B's location. A's clock only assigns times to events that take place at A's location. For an event that takes place anywhere else, you have to either do a calculation, or you have to use synchronized clocks.

You seem to have missed a critical fact, which is that clocks that are synchronized in A's frame are NOT synchronized in B's frame. You seem to think that you can use a single clock to synchronize clocks in both frames. That is not possible.

 

[jeremyfiennes]

Is clock #3 at rest relative to clock #1 (at rest in frame A) or clock #2 (at rest in frame B)?

In frame A, clocks A(#1) and X(#3) are stationary, and B(#2) is moving at v.

 

[stevendaryl]

In frame A, clocks A(#1) and X(#3) are stationary, and B(#2) is moving at v.

So you don't have multiple clocks at rest relative to B. Then you don't have a coordinate system set up for B.

 

[jeremyfiennes]

That post was completely wrong.

It seems I am also confused about synchronization. I defined the time origin t=0 as the instant at which frames A and B space origins are coincident, with clock X at frame A x. A synchronizing signal from the frame A midpoint x/2 reaches all three clocks simultaneously. Since this procedure is independent of x, one can have a frame with as many synchronized clocks as one likes. I understood from the previous thread that the time coordinate of a frame is that on anyone of these clocks.

 

[jeremyfiennes]

So you don't have multiple clocks at rest relative to B. Then you don't have a coordinate system set up for B.

A clock X_b, stationary in B, could be synchronized in the same way. Synchronization being an instantaneous procedure, velocities are irrelevant. The present question is however the B coordinates of an event stationary in A.

 

[stevendaryl]

A clock X_b, stationary in B, could be synchronized in the same way. Synchronization being an instantaneous procedure, velocities are irrelevant.

No, it is not. If the third is stationary in A's frame, then it CAN'T be synchronized with B's clock.

The present question is however the B coordinates of an event stationary in A.

An "event" is a single point in space and time. It's not stationary in any frame, because it's not an object.

Anyway, I've tried to explain to you how to compute $t'$ in terms of $t$. Here's a picture of the situation:

The event with coordinates $(x,t)$ in frame A is the event that would take place at the end of a stick of length $x$ held out by A, and on the end of that stick is a clock, synchronized with A's clock, showing time $t$. The same event has coordinates $(x',t')$ in frame B, which would be an event that would take place on a stick of length $x'$ held out by B, on the end of which is a clock showing time $t'$. We arrange it so that when B's clock shows time $t'$, A's clock shows time $t$, and they are right beside each other. So $(x,t)$ and $(x',t')$ are two different coordinates describing the same event.

Shown from the point of view of A's frame, we can reason as follows:

1. The distance between B and his remote clock is $x'$ in his own frame.
2. The distance between A and his remote clock is $x-vt$.
3. Because B's stick is length-contracted, we have: $x-vt = x'/\gamma$
4. In A's frame, his remote clock is synchronized with his personal clock, right next to him. So they both show time $t$
5. In A's frame, B's personal clock shows time $t_1' = t/\gamma$, because B's clock is running slower than A's by a factor of $\gamma$
6. In A's frame, B's remote clock shows time $t' = t_1' + \Delta t'$, where $\Delta t'$ is the offset of B's remote clock, relative to B's personal clock.

So we have the following relationship between A's coordinates $(x,t)$ and B's coordinates $(x', t')$:

$x' = \gamma (x-vt)$
$t' = t_1' + \Delta t' = t/\gamma + \Delta t'$

So we have our transformation, except that we need to compute $\Delta t'$, the offset, as measured in A's frame, between B's remote clock and B's personal clock.

I already went through how to compute $\Delta t'$. There is only one choice that is consistent with light having speed $c$ in B's frame:

$\Delta t' = - \frac{v x'}{c^2} = -\gamma \frac{v (x - vt)}{c^2}$

With that choice, B will measure time $x'/c$ for light to travel from him to the remote clock, and time $x'/c$ for it to travel back.

 

[jeremyfiennes]

No, it is not. If the third is stationary in A's frame, then it CAN'T be synchronized with B's clock.

Not with you. A synchronizing signal sent from frame A (x/2,0) synchronises all three clocks - A and B at their here coincident origins, and X at frame A x.

 

[jeremyfiennes]

An "event" is a single point in space and time. It's not stationary in any frame, because it's not an object.

Agreed. But it is here defined in terms of its frame A location x, rather than the corresponding frame B x'.

 

[Nugatory]

Closed for moderation.
It will reopen later today