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Special Relativity- Spacetime Diagrams

  1. May 14, 2010 #1
    1. The problem statement, all variables and given/known data

    http://www.maths.ox.ac.uk/system/files/private/active/0/b07.2_c7.209.pdf [Broken]

    Q1
    (Also in Q2, what is the definition of 4 momentum for a massless particle?)


    3. The attempt at a solution

    I can do the entire question if I first derive the Lorentz transformation but I suspect that as I can't do the question directly in its intended way I don't really understand what I am doing.

    I have done ii) using Bondi's k-factor but not sure this is right as I haven't considered the path of the light from B to M2. I have just drawn a spacetime diagram with R' the inertial frame and drawn a photon leaving R' origin at time t, reaching B at t'=kt (when the light gets back to B from M2) and getting back to R' origin at k^2t=kt' and solving.

    I am just getting into a mess trying to do iii) though and absolutlely no idea how to do iv) without Lorentz transformation.

    Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 15, 2010 #2
    Given that this is a past exam, answering your questions should be within the guidelines of Physics Forums. So,

    i) What are the two postulates of the theory?
    ii) Use these postulates to derive the desired expression. Hint: Pythagorean Theorem.
    iii) The result from ii) can be used.
     
  4. May 15, 2010 #3

    vela

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    For both parts ii and iii, calculate the time it takes for the light to go from B to the mirror and back again in the moving frame and compare it to the time τ it takes in the rest frame.
     
  5. May 15, 2010 #4
    I'm really not sure how to calculate the time in the moving frame without essentially deriving Bondi's k factor. I think I must be missing something obvious.
     
  6. May 15, 2010 #5

    vela

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    It's just the distance the light travels divided by c.
     
  7. May 15, 2010 #6
    Thanks, I get ii) now by using Pythag theorem, wasn't thinking about it in the right way. Am I correct to think that the distance between B and M2 in R' is still L, just that there is time dilation?

    iii) Two equations I need are 2L=r'c and 2L'=rc then results follows using ii)

    I think the first follows as distance between B and M2 in R' is still L.
    I'm not sure how to justify second equation. As there is no triangle to consider as velocity is in the same direction as the light can we say that the light must take r to travel L'?

    I get how to do iv) now, we can just fill in the matrix entries from ii) and iii)

    Also what is the four momentum for a massless particle on Q2 and do the 3 equations still hold if the particle is massless? I can't find this defined anywhere in my book.

    Thank you so much for your help!
     
  8. May 16, 2010 #7

    vela

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    M2 is the mirror on the y-axis. Since you can assume y=y', the distance is still L.
    So you're actually interested in M1. In R', you can't assume the distance between B and M1 is still L. I don't understand your two equations since you seem to be mixing variables from the two frames, like L and r' (which is I assume the round-trip time measured in the R' frame).
    I'm not sure what exactly the problem is looking for in the definition. How does your book define the four-momentum of a massive particle?
     
  9. May 16, 2010 #8
    Well the time taken to get from B to M1 in R' is going to be L'/c and in R it takes L/c. How do I move on from here as I can't see how to relate the two equations.


    The four momentum is defined as the rest mass multiplied by the four velocity. I don't think the equations will hold for massless particle as m2 appears in the denominator when you are working out gammas.
     
  10. May 16, 2010 #9

    vela

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    You're forgetting the motion of the mirrors in R'. It'll take less time than L'/c for the light to go from B to M1 because M1 will be moving toward the light. Similarly, it'll take longer than L'/c to go from M1 back to B because B will be moving away from the light.
    OK, that definition obviously won't work for a massless particle, so perhaps they want you to specify what the components of the four-momentum are for a massless particle.
     
  11. May 16, 2010 #10
    In R', a is the time to get from the bulb to M1 and b is the time to get from M1 to the bulb.

    ca=L'-av a=L'/(c+v)
    cb=L'+bv b=L'/(c-v)


    Total time to get from bulb to m1 and back again is 2L'c/(c^2-v^2)=(2L'/c)*gamma^2 in R'.

    Answer follows as this is equal to r' as the light must return back to the bulb at the same time as that from M2.

    In iv) I can get the left column of the matrix entries ok using the co-ordinates (cr',-r'v) in R' is (cr,0) in R. The problem is in getting the two right co-ordinates. I get that (L,L) in R should be equivalent to (cL'/(c+v),cL'/(c+v)) in R' (from a above) but this doesn't yield the right answer, you need to have (cL'/(c-v),cL'/(c-v)) in R'. I can't see what I've done wrong.


    I've found a definition as E/c(1,e) where e is the unit vector of velocity.
     
    Last edited: May 16, 2010
  12. May 16, 2010 #11
    Have I got a and b the wrong way round as this would then give me the correct answer for iv)
     
  13. May 17, 2010 #12

    vela

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    It worked out for me when I used (cL'/(c+v),cL'/(c+v)) in R' and (L,L) in R.
     
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