# Horizontal light clock

1. Sep 26, 2014

### center o bass

Consider a Horizontal light clock of length $L_0$ lying at rest in a frame K. There are two important events: (A) a photon gets emitted from the left mirror and (B) it gets reflected at the right mirror.
Another frame K' is moving by at velocity v and the frames are in standard configuration such that event (A) is assigned coordinates $(t_A,x_A) = (t'_A,x'_A) = (0,0)$. Clearly, $t_B = L_0$ (c=1), but now the question is, what is $t_B'$?

My intuitive reasoning would be as follows: in K' the length of the light clock is contracted to a length of $L_0/\gamma$ and photon has the same velocity c=1. But now the right mirror is moving towards the photon at velocity v, and hence it takes a time
$$t_B' = \frac{L_0}{\gamma (1-v)}$$
for the photon to reach the right mirror.

However, according to
exercise 4 at p. 24 in http://www.uio.no/studier/emner/matnat/astro/AST1100/h14/undervisningsmateriale/lecture7.pdf
$$t_B' = L_0 \gamma(1-v).$$