Derivation of Relativity of Simultaneity 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??
 A: 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:


*

*When two events occurs at the same location but different times in the moving frame $\bar S$, then the rest frame sees a direct relationship between those times: $\Delta t = \gamma \Delta \bar t$. This is the traditional time dialation. 

*When two events occur at the same time but different locations in the moving frame $\bar S$, then the rest frame sees a direct relationship between those locations: $\Delta x = \Delta \bar x / \gamma$. This is the traditional length contraction. 

*When two events occur at different times and different locations in the moving frame $\bar S$, then the rest frame sees a mixture of time and location relationships; you must use the full-blown lorentz transformation.


The mistake the naive approach makes is a violation of the last bullet.
