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

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jeremyfiennes said:
stated my synchronization procedure in #19.

That post was completely wrong.

jeremyfiennes said:
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.
 
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PeterDonis said:
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 said:
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.
 
stevendaryl said:
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 Xb, 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.
 
jeremyfiennes said:
A clock Xb, 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:

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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.
 

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stevendaryl said:
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.
 
stevendaryl said:
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'.