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Choice of action to geodesic equations

  1. Jan 7, 2015 #1
    Hello,

    I am quite new to GR and I have a question regarding the construction of the action to find the geodesic equation.
    In pretty much every book, you'll find:

    ##S=-m ∫ dS##

    using: ## dS=dS\frac{d\tau}{d\tau}=\sqrt{g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau## with ##dS=\sqrt{g_{\mu \nu}dx^{\mu}dx^{\nu}}##

    one gets: ##dS= -m∫\sqrt{g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}d\tau##

    Minimisation of this quantity gives rise to the geodesic equations.
    I understand that contruction: given a certain metric, the particule moves in a "straight" line, minimizing proper time.

    Here's my problem, when dealing with the Schwarzchild metric, my prof. uses:
    ##S= -m∫g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}d\tau##
    Which makes the computation much easier, I've also been able to re-find the geodesic equation starting from this expression.


    My questions would be:
    What is the diffence between ##d\tau## and ##dS## and what is the difference between these to actions ? I feel like I'm confusing proper time, interval ##dS## and the action differential...

    It would be great if someone could explain me the meaning of this!
    Thank you very much
     
  2. jcsd
  3. Jan 7, 2015 #2

    Orodruin

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    The second choice is equivalent to the first if you reparameterise your curve such that the curve parameter is an affine parameter (i.e., essentially you are taking the curve length as the parameter). This is of course a choice you can always make and it will not change the curve that results.
     
  4. Jan 7, 2015 #3

    stevendaryl

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    The two approaches are not exactly the same. If you minimize the first action integral, you obtain the equations of motion:

    [itex]\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} + \frac{dx^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}}) = 0[/itex]

    where [itex]\Gamma^\mu_{\nu \lambda}[/itex] is a combination of derivatives of [itex]g_{\mu \nu}[/itex].

    The second action integral leads instead to:

    [itex]\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} = 0[/itex]

    They are only the same if
    [itex]\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0[/itex]

    which means

    [itex]\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}} = K[/itex]

    for some constant [itex]K[/itex], which in turn implies:

    [itex]d\tau = \frac{1}{K} \sqrt{g_{\nu \lambda} dx^\nu dx^\lambda}[/itex]

    which means that [itex]\tau[/itex] is linearly related to proper time.

    The first integral works for any parameter [itex]\tau[/itex], while the second only works when [itex]\tau[/itex] is proper time.
     
  5. Jan 7, 2015 #4

    Orodruin

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    Just to nitpick, I suspect you intended to write:
    $$
    \frac d{d\tau}\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0
    $$
     
  6. Jan 7, 2015 #5
    First thank you both for you answers.

    It is very strange, I've done the computations yesterday and both gave me the exact same geodesic equations. I've must have made the assumption ##d\tau=dS## during the second one without knowing it. I'll take a look at it again.

    So what is the difference between proper time ##d\tau## and space-time interval ##dS## ? I can't figure it out.
    Also, does that mean that I can take any action ##dS^{z}##?
     
  7. Jan 7, 2015 #6

    stevendaryl

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    Yes.
     
  8. Jan 7, 2015 #7

    Orodruin

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    The relation between proper time and the space-time interval is the same as the relation between curve length and the space interval in Riemann geometry. In Riemann geometry, you could always pick the length along the curve as your curve parameter. In the same way, you can here pick the proper time along the world line as your curve parameter.
     
  9. Jan 7, 2015 #8
    Does that mean I can always pick ##dS=d\tau## ?

    That would mean ##g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\mu}}{d\tau}=1## right ?
     
  10. Jan 7, 2015 #9

    stevendaryl

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    It's the first integral where you have to make the assumption that [itex]d\tau = dS[/itex] in order to get rid of the term

    [itex]\frac{d x^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}})[/itex]

    Using the Lagrangian equations of motion, with [itex]L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}[/itex], you find

    [itex]\frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{g_{\mu \nu} \frac{dx^\nu}{d\tau}}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}}[/itex]

    Now, take [itex]\frac{d}{d\tau}[/itex] and you get two terms:
    [itex]\frac{d}{d\tau} \frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{\frac{d}{d\tau} (g_{\mu \nu} \frac{dx^\nu}{d\tau})}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}} + g_{\mu \nu} \frac{dx^\nu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}})[/itex]
    The first term leads to the usual geodesic equation (once you multply through by [itex]g^{\mu \mu'} \sqrt{g_{\mu'' \nu}\frac{dx^{\mu''}}{d\tau} \frac{dx^\nu}{d\tau}}[/itex]). The second term is an additional term not found in the geodesic equation that you derive from the second action integral.

    In the first integral, [itex]\tau[/itex] is not necessarily proper time. It is any parameter whatsoever that increases continuously along the path. The first integral is invariant under parameter changes.
     
  11. Jan 7, 2015 #10

    stevendaryl

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    Except that in the first integral, it isn't necessary to pick [itex]\tau[/itex] to be proper time.
     
  12. Jan 7, 2015 #11

    stevendaryl

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    Yes, and that allows you to throw away any derivatives of that expression (since they'll be zero).
     
  13. Jan 7, 2015 #12

    Orodruin

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    Which is why I used "can" and not "must". :)
    It is the same in Riemann geometry, you can, but must not, use the curve length as the curve parameter. For example, I could parameterise the line ##x = y## as ##x = t^3## and ##y = t^3##.
     
  14. Jan 7, 2015 #13

    pervect

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    Just getting rid of the square root is wrong, as other posters have mentioned. However, there is a cute trick, known as an einbein, that allows one to accomplish the same end legitimately.

    See for instance http://arxiv.org/pdf/hep-ph/9708319v1.pdf

    Maybe more later if I have time.
     
  15. Jan 7, 2015 #14
    I meant the first integral you're right.
    I checked my notes: I did not used the Lagrange equation to derive the geodesics. I studied the variation of the action ##\delta S## with respect to the coordinates . It turns out I have indeed set ##dS=d\tau## during the computation.
    I have to say, using the Euler-Lagrange equations is much more handy!



    Then that explain how I got the equations of motion in the Schwarzchild metric! Thanks!



    I am not sure what you mean by that. Is it that in the first integral, ##d\tau## is just any variable in the integral ? In other words ##d\tau## is any infenitesimal space-time interval, as long as it follows the integration path ? I might as well write ##dP,dD,dT## etc.? (like ##dx,dy,dz,dt## mean the same thing in a integral, they are just the "integration variables").
    And since I am integrating along the trajectory in space time it turns out that ##d\tau## is the proper time during the integration ?

    I hope this makes sense and that I am not too far from the actual meaning of ##d\tau, dS##, thank you very much to all
     
    Last edited: Jan 7, 2015
  16. Jan 7, 2015 #15

    Orodruin

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    ##d\tau## in the first integral is a priori only a curve parameter and does not have to be an affine parameter for the proper time. It is only after setting ##d\tau = dS## that you can interpret it as proper time.
     
  17. Jan 7, 2015 #16
    Hum so:
    ##dS## is the infinitesimal displacement in spacetime along the true trajectory
    ##d\tau## is any curve parameter

    And I can always choose ##d\tau=dS## ?
     
  18. Jan 7, 2015 #17

    Orodruin

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    Yes. You can always make such a reparameterisation. It is a matter of solving the differential equation ##dS/d\tau = 1##.
     
  19. Jan 7, 2015 #18
    I think I'm getting the meaning of all this
    You've been very patient thank you very much!
     
  20. Jan 7, 2015 #19

    Matterwave

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    I'd like to point out though, that usually the notation ##d\tau## is for a proper time. For a non-proper time parametrization one uses usually ##dt##. People will be confused if you use ##d\tau## to mean an arbitrary curve parametrization.
     
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