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Varying the action of a point particle in curved spacetime

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    What I have to do is vary the action
    [tex] S =-mc \int ds = -mc \int \sqrt{-g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt}} \ \ dt [/tex]
    to find the equation of motion for [tex]x^\mu(t)[/tex]

    3. The attempt at a solution

    Now, to begin with, I have to admit that I am having quite some trouble with this course. I have done some exercises varying the action before, but they were a lot easier. What I 'learned' from that was that I'll probably have to linearize using a taylor expansion, and in the end I'll have to use partial integration.

    But my main problem is, how do I start? I want to take S(x+dx) - S(x), but how do I do that here?

    I apologize if this is not what one would call attempting to solve the question. I'm just pretty much lost when it comes to this. Pointing me towards a textbook in which this is explained is just as awesome (or even more) as giving me a helping hand as to start solving this problem.

    (note: the actual question is much longer than this, but if I can't solve this part, I'll get nowhere)
  2. jcsd
  3. May 14, 2013 #2


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    What you have to use is the Euler-Lagrange equations. And what you are doing here is deriving the geodesic equation from the action S. That should be in any number of books.
  4. May 14, 2013 #3
    Well, I checked with the teacher actually, and he really wants us to actually vary the action, were not allowed to use the Euler-Lagrange equations. But I will try and look up a source that does this for the geodesic equation, thank you :)
  5. May 15, 2013 #4
    Hm, one thing I forgot to add is that t is an arbitrary parameter here, so after the variation I can replace it with tau, proper time, and clear most of the things out, just to be left with a c. But sadly, I still don't see how to get there. (Does this make sense?)
  6. May 15, 2013 #5


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    It's hard to say until you show more of your work.
  7. May 16, 2013 #6
    I understand, I just don't know where to begin really.
    Looking at my syllabus, it seems as if the variation is effectively just a derivative. So would it be right to say that
    [tex]\delta S[/tex] is just 1 divided by two times the square root, multiplied by three terms (from the chain rule)? I apologize for such rough work, I am really not comfortable with these calculations as of now.
  8. May 16, 2013 #7


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    Roughly, sure. You are going to have derivatives with respect to ##x^\mu## and ##\dot x^\mu## and yes, variation is effectively a derivative. It's explicitly spelled out in the Euler-Lagrange equation. They are the same thing as a 'variation'. Look up some examples of using that.
  9. May 16, 2013 #8
    Well, the exercise is really deriving those equations, I guess.
    What I've done so far can be written as
    [tex] \delta S = -mc \int \frac{-(\frac{g_{\mu\nu}(x(t))}{dx^ \lambda }\delta x^ \lambda \frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt})-(g_{\mu\nu}(x(t))\frac{d\delta x^\mu(t)}{dt} \frac{dx^\nu(t)}{dt})-(g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{d \delta x^\nu(t)}{dt})}{2\sqrt{-g_{\mu\nu}(x(t))\frac{dx^\mu(t)}{dt} \frac{dx^\nu(t)}{dt}}} \ \ dt [/tex]

    which is a bit neater with xdot notation and such, but I'm not that good at latex yet, so I apologize.

    But from here on, I don't know how to continue. I can combine two of the terms up top if the metric is symmetric (which I think it is?), but then..

    (I did this literally applying the chain rule, I hope that's ok)
    Last edited: May 16, 2013
  10. May 19, 2013 #9
    Ah, I got it. I can use a trick to make the square root term into a c, replacing t by proper time. Then up top, I just need to combine the two terms at the end, partially integrate and relabel a bit from there. I do indeed end up with the equations of motion for curved space. Awesome :) Thanks for taking the time to look at this, next time I'll try and post more work. (Where I got stuck was that you could just take [tex]\delta S[/tex] to be a derivative of some sorts, instead of having to do the taylor expansions)
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