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Homework Help: Time dilation and a rocket

  1. Nov 6, 2006 #1
    a rocket ship of length L leaves earth at a vertical speed of 4c/5. a light signal is sent vertically after it which arrives at the rocket tail at t=0 according to both rocket and earth based clock. when does the signal reach the nose of the rocket clock according to (1) the rocket clockt (2) Earth clocks




    [tex]t=\frac{L}{\frac{3}{5} c}[/tex]

    i know that this might be wrong but I'm not sure how to fix this mistake or which equation to use. any ideas?
  2. jcsd
  3. Nov 7, 2006 #2


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    I don't think you have it. This is just the light pulse travelling at c chasing the nose of the ship traveling at 4c/5. The light pulse has to go farther by one ship length, but how long is the ship according to the earth observer?
  4. Nov 7, 2006 #3
    [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}= \frac{1}{\sqrt{1-\frac{(3/5)^2 c^2}{u^2}}}= 5/4[/tex]


    so this means:

    [tex]t=\frac{\frac{4L}{5}}{{\frac{3}{5} c}}[/tex]
    [tex]t=\frac{\frac{4L}{5}}{{\frac{3}{5} c}}[/tex]

    would this be the right answer?
  5. Nov 7, 2006 #4


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    It looks to me like you used the wrong velocity to calculate gamma, but you do need the contracted length. At t = 0 the light pulse is at the tail of the rocket of length L' (the contracted length that you need to fix). During the time t the light pulse will move ct and the nose of the ship will move vt. The light pulse must move a distance L' farther than than the nose. All the relativity transformations that you need are done when you calculate L'.
  6. Nov 7, 2006 #5
    could i used the equation:

    [tex]t=\frac{t'+\frac{v x_1'}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]



    I get an answer of t= 3L/c

    does this seem correct?

    also, the time measured from a clock in the space ship would be just t'=L/c since the clock in the space ship has a relative speed of 0 wrt the spaceship right?
    Last edited: Nov 7, 2006
  7. Nov 7, 2006 #6


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    t' = L/c is correct. Inside the space ship it is just a pulse going from one end to the other with normal length L and normal time t'. Outside the ship it is

    ct = L' + vt
    (c - v)t = L'
    (1-v/c)ct = L'
    t = 3L/[5(1-v/c)c]
    t = 3L/c

    This is the same result you got by transforming the time as measured inside the ship to the time as measured outside the ship for two spatially separated events in S'. You can do it either way.
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