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Two clocks are positioned at the ends of a train of length $L$ (as measured in its own frame). They are synchronized in the train frame.

The train travels past you at speed $v$. It turns out that if you observe the clocks at simultaneous times in your frame, you will see the rear clock showing a higher reading than the front clock. By how much?

picture of clocks in train

The solution to this exercise is given in the book, and it denotes that we put "a light source on the train, but let’s now position it so that the light hits the clocks at the ends of the train at the same time in our frame." As in this figure:

picture of photon in train

I don't understand:

  1. Why is there a difference in the clocks in the first place?
  2. Why are the fractions are multiplied by half? i.e. what is the origin of the $2$ in $$\frac{L(c+v)}{\textbf{2}c}\text{?}$$

Note: This one is taken from David Morin textbook for mechanics.

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In our frame, the light beam approaches the rear end of the train at speed $c+v$, because both are moving and in opposite directions. Note that this does not mean anything is traveling faster than light. Similarly, the light beam approaches the front of the train at speed $c-v$. We want to choose the position of emission so that they arrive at the same time. If $df$ is the distance to the front of the train from the emission point in our frame and $dr$ is the distance from the rear of the train to the emission point in our frame, we have $$df+dr=L\\ \frac {df}{c-v}=\frac {dr}{c+v}\\df+df\frac{c+v}{c-v}=L \\2cdf=L(c-v)\\df=\frac{L(c-v)}{2c}$$ as required. The $2$ comes from solving the equations. Note that if $v=0$ this simplifies to $df=L/2$ because the light source is at the center of the train. This is the same $2$.

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