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Horizontal light clock

  1. Sep 26, 2014 #1
    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
    the answer should be
    $$t_B' = L_0 \gamma(1-v).$$
    So, where am I going wrong in my argument?
     
  2. jcsd
  3. Sep 26, 2014 #2

    ghwellsjr

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    Gold Member

    The text says the light clock is moving at v but you said frame K' is moving at v which makes the light clock move at -v. If you change the sign of v in your equation, you will get the same answer.
     
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