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Special Relativity - time dilation and length contraction

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    The highest energy protons have gamma factors around ##1.0*10^{12}##.
    (a) Our galaxy has a disk diameter of 30 kpc, which is ##9.3*10^{20}m##. If a photon and one of these high energy protons start traversing the galaxy at the same time, by how long will the arrival of the photon on the other side precede the arrival of the proton, in the rest frame of the galaxy? Give your answer in seconds.
    (b) To an observer in the reference frame of the travelling proton, what is the measured diameter of the galaxy in kpc?
    (c) To an observer in the reference frame of the travelling proton, how long does the journey across the galaxy take? Give your answer in seconds.

    2. Relevant equations
    ##t'=\gamma (t-\frac{xv}{c^2})##
    ##x'=\gamma (x-vt)##
    3. The attempt at a solution
    (a) I have managed to get an answer of ##1.55*10^{-12} s## for this question using a binomial expansion.
    (b) I'm not sure as to how to obtain the diameter of the galaxy in the reference frame of the travelling proton.
    I'm assuming it should be shorter in the reference frame of the proton due to length contraction
    Do I have to use the formula ##x'=\gamma(x-vt)## where x is ##9.3*10^{20}m## and ##t=\frac{d}{v} =\frac {9.3*10^{20}}{(3*10^8 - 1.55*10^{-12})}## ? When I do I get an answer which isn't contracted, but is dilated.

    (c) ... Once I've understood how to do (b) I'm sure I can work out (c)!

    Thanks in advance for any help !
     
  2. jcsd
  3. Dec 3, 2015 #2

    vela

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    Your calculation should give you x'=0, which should make sense. Because the proton is at rest in its rest frame, it remains at x'=0.

    On a spacetime diagram, you should be looking for the length along the x' axis. You're looking for the distance between the opposite ends of the galaxy at the same instant in time in the proton's frame.

     
  4. Dec 4, 2015 #3
    Is it simply ##x'=\frac{x}{\gamma}=\frac{9.3*10^{20}}{1*10^{12}}=9.3*10^{8}m = 3*10^{-8} pc## then?

    In which case (c) would also simply be ##t=\frac{t'}{\gamma} = \frac{(3.1*10^{12}+1.55*10^{-12})}{1*10^{12}} ≈ 3.1 s## ?
    Doesn't seem right to me..

    EDIT :: ##t'=\frac{d}{c}+1.55*10^{-12}=\frac{9.3*10^{20}}{3*10^8}+1.55+10^{-12}## which I found in part (a), I forgot to explain where my t' came from, sorry.
     
    Last edited: Dec 4, 2015
  5. Dec 4, 2015 #4

    vela

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    Yes, that's right. Note if you divide x' by c, you also get 3.1 s, which matches the answer you got for (c). This is the calculation an observer at rest relative to the proton would make based on measurements in that reference frame.
     
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