Suppose you have a train moving forward relative to an inertial observer at velocity $v$. Suppose you have a clock 1 at the front of the train and a clock 2 at the back, and, in the frame of the train, you synchronize the clocks via
$$t_2 = t_1 + \frac{l_0}{c}$$
where $t_1$ is the omission time. Using only length contraction, the fact that light moves at a speed $c$ in all reference frames, and time dilation, I need to derive the fact that the clock in the rear appears faster by $$\frac{vl_0}{c^2}.$$
A similar procedure found here: http://galileo.phys.virginia.edu/classes/252/synchronizing.html
which makes use of relative velocity $c+v$ to find the relativity of simultaneity constant, but the procedure is from the middle of the train.
I'm a bit confused why this "relative velocity" use is acceptable.
When I try to apply this to the version I have, I get that the difference between the front clock and the rear clock is
$$\frac{l_0}{\gamma (c+v)} – \frac{l_0}{\gamma c}$$ which is not correct.
My reasoning is:
denote $s_1$ by the time the light is emitted from the front of the train as seen by the track observer. Then $ s_1 = \gamma t_1$.
Upon receiving the light, the back of the train clock is set to
$$s_B = s_1 + \frac{l_0}{\gamma(c+v)}$$
since the time to reach the back in the ground frame is given by
$$ct + vt = \frac{l_0}{\gamma}$$
and by this time, the front clock reads
$$s_F = s_1 + \frac{l_0}{\gamma c},$$ since the track observer the length of the train altered by $\gamma^{-1}$.
What is my issue here and why does the given link do this "additive velocity" business that relativity is supposed to avoid??
Best Answer
I'm not sure why your question didn't get more attention, since it seems to be a good question. There was hope someone else would respond but it seems that isn't going to happen; I will give my best interpretation of the mistake, but take it with a bit of skepticism. To answer your question, allow me to first show the simplest way to get the correct solution:
The lorentz transformation must be used in this problem. It takes events $(t,x)$ in the rest frame (denoted $S$) and converts them to the coordinate system of the moving frame $(\bar t, \bar x)$ denoted $\bar S$:
$$ \begin{array}{c|c} \text{Lorentz Transformation} & \text{Inverse Transformation} \\ \bar t = \gamma (t - vx/c^2) & t = \gamma (\bar t + v\bar x / c^2) \\ \bar x = \gamma (x - vt) & x = \gamma(\bar x + v\bar t) \end{array}$$
We will be using the inverse transformation. Let event $A$ be the emission of the light from the clock at the front of the train and $B$ the reception of light from the clock at the rear of the train. Then The coordinates of $A$ and $B$ in $\bar S$ are: $$ \bar A = (\bar t_1, \bar x_2) = (\bar t_1, \bar \ell)$$ $$ \bar B = (\bar t_2, \bar x_2) = (\bar t_2, 0)$$
Where I have used the 'true' length of the train to be denoted as $\bar \ell$ instead of $\ell_0$ (a random '$0$' subscript is dangerous here). Running these events through the inverse transformation yields: $$\begin{align} A &= (t_1, x_1) = ( \gamma(\bar t_1 + v \bar \ell / c^2), \gamma(\bar \ell + v \bar t_1) ) \\ B &= (t_2, x_2) = ( \gamma \bar t_2, \gamma v \bar t_2) \end{align} $$ To find the difference in time, we subtract $B-A$: $$\begin{align} t_2 - t_1 &= \gamma(\bar t_2 - \bar t_1 - v\bar \ell /c^2) \\ x_2 - x_1 &= \gamma(v(\bar t_2 - \bar t_1) - \bar \ell) \end{align}$$ Using the fact that $\bar t_2 - \bar t_1 = \bar \ell / c$, then the answer materializes out of the first coordinate: $$ t_2 - t_1 = \gamma( \frac{\bar \ell}{c} - \frac{v \bar \ell}{c^2}) $$
why the naive approach doesn't work:
Let's draw minkowski diagrams for the moving frame $\bar S$ and the rest frame $S$ (which has an X on it because it is wrong):
The rear of the train is placed at the origin of $\bar S$. Note that the vertical axis is the time axis, and the positions of the front/rear of the train are unchanging in $\bar S$. In $S$ they move with speed $v$ so they have slopes $1/v$. The light ray travels at a $45^\circ$ from $A$ to $B$ (a true statement in both $\bar S$ and $S$).
The naive approach (that I first tried too) is to look at the $S$ diagram and say the light ray travels to the back of the train from $x_1 = \ell = \bar \ell / \gamma$ to $x_2 = v (t_2 - t_1)$. It does this at speed $c$, so $ c(t_2 - t_1) = x_2 - x_1 = v(t_2 - t_1) - \bar \ell / \gamma$. However, this is wrong, as can be seen by the true values of $x_1$ and $x_2$ in the lorentz transformation.
The correct way to draw the minkowski diagram for this problem is with both $\bar S$ and $S$ overlaid:
Where the axis of $\bar S$ have been properly transformed by tilting them inward at an angle $\tanh(\alpha) = v/c$. You can see from this picture the event $A$ doesn't occur immediately (at the $t=0$ axis) from the perspective of $S$; the train actually moves forward a bit to position $x_1$ before the front clock emits a light pulse directed at the rear clock.
the take-away from this problem
What I took away from this problem can be summed up fairly concisely as follows:
The mistake the naive approach makes is a violation of the last bullet.