In [thread=110633]another thread[/thread], a poster mentioned using slow transport to maintain clock synchronization.
[post=2843216]Post 42[/post]
I pointed out that
if two synchronized clocks are slowly separated, they will not stay synchronized unless they were initially at rest.
Tach disagreed, and a long argument ensued which I'm belatedly diverting into a new thread.
For reference, here are the most relevant expansions of the contested sentence:
[post=2843243]Post 47[/post]
To spell it out:
Consider two synchronized clocks at rest together in frame S.
One clock remains at rest while the other is slowly moved at velocity dv to particular distance L away.
In frame S, the resulting difference in synchronization can be arbitrarily small with an arbitrarily small separation velocity.
If you take the limit as dv approaches zero, the resulting difference in synchronization is zero.
Now consider two synchronized clocks moving at velocity v together in frame S.
One clock remains moving at v, the other clock slowly separates at velocity v+dv until it is a particular distance L away. dv is parallel to v.
Now, the resulting difference in synchronization can not be made arbitrarily small.
If you take the limit as dv approaches zero, the resulting difference in synchronization is nonzero.
Tach interpreted these scenarios as ([post=2843307]post 65[/post]:
...which of course led into a diversion about communication skills.
The next expansion was more formal:
[post=2843310]Post 66[/post]:
Two clocks, A and B, synchronized and colocated at $$t=0$$.
A moves at speed $$v + \Delta u$$.
B moves at speed $$v - \Delta v$$.
$$\Delta u$$ and $$\Delta v$$ are very small compared to $$v$$ and to c, but are not necessarily equal.
The clock velocities are parallel.
Now, at $$t = \frac{L}{\Delta u + \Delta v}$$,
- the separation between A to B is $$L$$.
- $$\tau_A$$ has elapsed on clock A
- $$\tau_B$$ has elapsed on clock B
- The difference in synchronization is $$\Delta \tau = \tau_B - \tau_A $$
The quantity in question is this:
$$\lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \left (\Delta \tau\right )$$
I maintain that the answer is a function of v and L, and equals zero if and only if either v or L is equal to zero.
And later analysed (the original post had some mistakes. Tach provided corrections, incorporated here.)
[post=2843741]Post 87[/post]
$$\begin{align}
t &= \frac{L}{\Delta u + \Delta v} \\
\tau_A &= t/\gamma(v + \Delta u) \\
\tau_B &= t/\gamma(v - \Delta v) \\
\Delta \tau &= \tau_B - \tau_A \\
&= t/\left(\gamma(v - \Delta v) + \gamma(v + \Delta u)\right) \\
&= \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right)
\end{align}$$
I don't think there's anything stopping us from immediately taking one of the limits?
$$\begin{align}
\lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \left (\Delta \tau\right ) &= \lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \ \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right) \\
&= \lim_{\Delta v \rightarrow 0} \ \frac{L}{\Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - (v/c)^2}\right) \\
&= -L \frac{d}{dv} \ \sqrt{1 - (v/c)^2} \mbox{(Definition of the derivative)}\\
&= \frac{\gamma Lv}{c^2}
\end{align}$$
