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

[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]

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