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About Schwarzschild geodesics

  1. Nov 8, 2008 #1
    I was looking into the geodesic equations for the Schwarzschild metric and I noticed a discrepancy between two sources: According to http://www.mathpages.com/rr/s5-05/5-05.htm (near the bottom) the second derivative of the azimuth angle is


    \frac {d^{2} \phi}{d \lambda ^2}=\frac {-2}{r} \frac {dr}{d \lambda} \frac {d \phi}{d \lambda}+ \frac {2}{tan (\theta )}\frac {d \theta}{d \lambda} \frac {d \phi}{d \lambda}


    where [tex]\lambda[/tex] is something proportional to proper time [tex]\tau[/tex]. In Relativity Demystified (David McMahon, 2006 McGraw Hill, page 218) however, the corresponding equation has the second term negative instead of positive. (The other three equations match exactly.) I also have a copy of Lillian R. Lieber's The Einstein Theory of Relativity (Paul Dry Books, 2008) but in her version (p. 269) the second term is missing altogether.

    I intend to try to derive these equations myself but I would like to know in advance what the correct versions are so when I'm done I can see if I made any mistakes. Does anyone happen to know which source is right?
  2. jcsd
  3. Nov 8, 2008 #2

    George Jones

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    I only have two books home with me that treat Schwarzschild geodesics, and neither uses Christoffel symbols to find the geodesics, so their equations are a little further on in the process.
    As is the case for Newtonian orbits, motion takes place "in a plane", and, WLOG, it's convenient to take [itex]\theta = \pi / 2[/itex]. This wipes out the last term in the equation.
  4. Nov 8, 2008 #3


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    Sean Carroll gives his version in equations (7.29), (7.33) and (7.37).

    Carroll, Lecture Notes on General Relativity, http://arxiv.org/abs/gr-qc/9712019
  5. Nov 9, 2008 #4
    Thanks; I'll take a look at the Carroll notes as soon as I can get to a computer with Adobe. Unfortunately my operating system is so old that Adobe is no longer available for it.
  6. Nov 9, 2008 #5


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  7. Nov 9, 2008 #6
    Hey thanks again, atyy. It looks like Carroll agrees with mathpages.com instead of McMahon. I'll have to do the computations myself to be sure, but in the meantime I suspect that he's right since frankly that McMahon text doesn't seem very reliable. It's written in a very easy-to-read style but is definitely not error-free.
  8. Nov 16, 2008 #7
    I'm one of the editors of the new edition of Lillian and Hugh Lieber's The Einstein Theory of Relativity. I don't know if the reader above (snoopies622) checked out the additional notes Bob Jantzen and I added to the book, but they may help him or her work through the Schwarzschild derivation.

    My thanks to George Jones who pointed out that \theta could be set equal to \pi/2 without loss of generality. Just as in the standard Newtonian Binet equation the motion takes place in a plane, so too in the relativistic version; the Schwarzschild orbits are two dimensional problems.
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