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??

  • $\begingroup$ "I need to derive the fact that the train in the rear appears faster by.." . Do you mean clock in the rear of the train? $\endgroup$
    – cms
    Commented Feb 8, 2018 at 17:05
  • $\begingroup$ @cms yes, typo there $\endgroup$
    – JohnDD
    Commented Feb 8, 2018 at 17:06
  • $\begingroup$ "the clock in the rear appears faster by...". Can you write out exactly what the expected answer is in that context? Do you mean $t_2 = t_1 + l_0/c - vl_0/c^2$ in the rest frame? $\endgroup$
    – cms
    Commented Feb 9, 2018 at 0:12
  • $\begingroup$ I'm not allowed to ask questions any more. Einstein said that simultaneity is relative. So event A may seem to occur before event B in one frame of reference. In the next frame it may seems as though event A and B occurred at the same time. In the next it may seem as though event B occurred before event A. Now Einstein had two wives, one after another. But, from some other frame of reference did it seem as though Einstein had two wives at the same time. LOL. $\endgroup$
    – Sean
    Commented Feb 13, 2018 at 8:47
  • $\begingroup$ @Sean Nope. Time-like separated events preserve their order of occurrence as seen from all inertial frames. You could have understood this from reading almost any page in Wikipedia on SR like this. $\endgroup$ Commented Feb 15, 2018 at 15:11

1 Answer 1


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): minkowski diagrams 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: proper minkowski diagram 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.