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General Relativity - Circular Orbit around Earth

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Find the proper time in the rest frame of particle
    (b) Find the proper time in the laboratory frame
    (c) Find the proper time in a photon that travels from A to B in time P


    2vte9g6.png

    2. Relevant equations


    3. The attempt at a solution

    Part(a)

    The metric is given by:

    [tex] ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2 [/tex]

    Circular orbit implies that ##dr^2 = 0##, so
    [tex]ds^2 = c^2 d\tau^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 [/tex]
    [tex] \left( \frac{d\tau}{dt} \right)^2 = \left( 1 - \frac{2GM}{c^2R} \right) [/tex]
    [tex] \frac{d\tau}{dt} = \sqrt { \left( 1 - \frac{2GM}{c^2R} \right) } [/tex]

    Since the time between event A and B is ##dt = P##, the time experienced in the rest frame must be ## d\tau = \sqrt { \left( 1 - \frac{2GM}{c^2R} \right) } P ##?

    I'm not sure how to approach parts (b) and (c)..
     
  2. jcsd
  3. Feb 1, 2015 #2

    TSny

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    Does ##dx^2 + dy^2 +dz^2 = 0## for two neighboring points on the orbit?
     
  4. Feb 1, 2015 #3
    They are at the same radius, so yes.
     
  5. Feb 1, 2015 #4

    TSny

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    For general motion in the xy plane (z = 0), how would you express ##dx^2+dy^2## in polar coordinates ##(r, \theta)##?
     
  6. Feb 1, 2015 #5
    [tex] r^2sin^2 \theta d\phi [/tex]
     
  7. Feb 1, 2015 #6

    TSny

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  8. Feb 1, 2015 #7
    Oh I must have confused the r's. It should be:

    [tex] ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2 [/tex]

    Since for ##z=0## we let ##\theta = \frac{\pi}{2}##.
     
  9. Feb 1, 2015 #8

    TSny

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    On the circular orbit of radius R, what is the value of little r?
     
  10. Feb 1, 2015 #9
    r = R
     
  11. Feb 1, 2015 #10

    TSny

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    Right.
     
  12. Feb 1, 2015 #11
    [tex]ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2 [/tex]

    Using ## d\phi = \omega dt = \frac{2\pi}{P} dt ##

    [tex] ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2[/tex]
     
  13. Feb 1, 2015 #12

    TSny

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    Not quite. Does ##R^2P^2dt^2## have the correct dimensions of distance squared?
     
  14. Feb 1, 2015 #13
    Using ## d\phi = \omega dt = \frac{2\pi}{P} dt ##,

    [tex] ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2 [/tex]
    [tex] -c^2 d\tau^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2 [/tex]
    [tex] \left( \frac{d\tau}{dt} \right)^2 = \left( 1 - \frac{2GM}{c^2R} \right) - \left( 1 + \frac{2GM}{c^2 R} \right) R^2 \left(\frac{2\pi}{Pc}\right)^2 [/tex]
     
  15. Feb 1, 2015 #14

    TSny

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    Looks good.
     
  16. Feb 1, 2015 #15
    I am stuck on part (b) though..
     
  17. Feb 1, 2015 #16

    TSny

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    What is ##dx^2 + dy^2 + dz^2## for a fixed point on the orbit?
     
  18. Feb 1, 2015 #17
    For a fixed point, it is ##0##, so ##\frac{d\tau}{dt} = 1 - \frac{2GM}{c^2R}##.
     
  19. Feb 1, 2015 #18

    TSny

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    ...and ##ds^2 = ##?
     
  20. Feb 1, 2015 #19
    Solving, we get ## \frac{d\tau}{dt} = 1 - \frac{2GM}{c^2R} ##.
     
  21. Feb 1, 2015 #20
    For part (c), since beam of light travels at speed ##c## in all frames, is ##dr^2 = c^2 dt^2 = c^2 P^2##?
     
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