Question

A train with proper length L has clocks at the front and back. A photon is fired from the front to the back. Working in the train frame, we can easily say that if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c. Now consider this setup in the ground frame, where the train travels by at speed v. Rederive the above frame-independent result (namely, if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c) by working only in the ground frame.

Answer 1

Explanation:

In train's rest frame, the speed of photon is $c$ and the proper length of the train is $L$. The time taken by the photon to cross the train is $t=\frac{L}{c}$

In ground frame, the speed of the photon is given as follows:

$v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}$

$=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c$

The speed of light or photon remains same in every frame of reference.

Now, the speed of train is very less as compared to the speed of photon so that $v$ So that, $\frac{v}{c} \ll 1$

The length contraction in the ground frame is given as follows:

$L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}$

$=L$

Time taken by the photon to travel the length of the train in ground frame is .