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B Euler-Lagrange equation for calculating geodesics

  1. Aug 22, 2016 #1
    Hello I am little bit confused about lagrange approximation to geodesic equation:
    So we have lagrange equal to L=gμνd/dxμd/dxν
    And we have Euler-Lagrange equation:∂L/∂xμ-d/dt ∂/∂x(dot)μ=0
    And x(dot)μ=dxμ/dτ. How do I find the value of x(dot)μ?
     
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  3. Aug 22, 2016 #2

    stevendaryl

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    I'm not sure I understand your question. The Euler-Lagrange equations produce a second-order differential equation for [itex]x^\mu[/itex]. The solution has two undetermined constants, which can be taken to be [itex]x^\mu[/itex] and [itex]\dot{x^\mu}[/itex] at some particular moment.

    It works much better if you use LaTex for mathematics. The "Lagrangian" for geodesic motion can be taken to be:

    [itex]L = \frac{1}{2} m g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}[/itex]

    where [itex]s[/itex] is proper time. Then

    [itex]\frac{\partial L}{\partial (\frac{dx^\lambda}{ds})}= \frac{1}{2} g_{\mu \nu} m (\delta^\mu_\lambda \frac{dx^\nu}{ds} + \delta^\nu_\lambda \frac{dx^\mu}{ds}) = m g_{\lambda \nu} \frac{dx^\nu}{ds}[/itex] (The factor [itex]m[/itex] is irrelevant, except that it makes [itex]\frac{\partial L}{\partial (\frac{dx^\lambda}{ds})}[/itex] into the relativistic 4-momentum, [itex]P_\lambda[/itex]).

    [itex]\frac{\partial L}{\partial x^\lambda} = \frac{1}{2} \frac{\partial g_{\mu \nu}}{\partial x^\lambda} m \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}[/itex]

    So the Euler-Lagrange equations give:
    [itex]\frac{d}{ds} (m g_{\mu \nu} \frac{dx^\nu}{ds}) = m \frac{\partial g_{\mu \nu}}{\partial x^\lambda} \frac{dx^\mu}{ds}\frac{dx^\nu}{ds}[/itex]
     
  4. Aug 22, 2016 #3

    vanhees71

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    That's not entirely true since, if ##s## is "proper time", i.e., defined by ##\mathrm{d} s^2=g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}## and then ##L=m/2=\text{const}##.

    What's correct is that for an arbitrary worldline parameter the geodesic equation is derived from the variational principle with
    $$L=-m \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
    Then you have to take the variation, leading to the Euler-Lagrange equations of motion. Then you can take ##\lambda=s##, and what comes out is that you can as well use your Lagrangian but you have to impose the constraint that ##s## is proper time, because only then the two variational principles yield the same result.

    You can formaly derive this also by implying that ##\lambda=s## by imposing the contraint ##g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}-1=0## via the Lagrange-multiplier method.
     
  5. Aug 22, 2016 #4

    stevendaryl

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    What are you saying is not true? That [itex]\frac{\partial}{\partial (\frac{d x^\lambda}{ds})} L = P_\lambda[/itex]?

    This is one of those tricky issues, where there is a distinction between what is true by definition, and what is true because of the equations of motion. If you start with the "Lagrangian" [itex]L = \frac{1}{2} m g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}[/itex], then the equations of motion imply that [itex]L = \frac{1}{2} m[/itex], but obviously, you can't let that be true by definition, or you can't derive the equations of motion.
     
    Last edited: Aug 22, 2016
  6. Aug 22, 2016 #5

    vanhees71

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    Ok, let me formulate my previous posting differently: The correct geodesic equation follows for a general parameter from (and only from) the "square-root form". Only after evaluating the equations of motion you can simplify the equations by choosing the parameter arbitrarily. That this is allowed is due to the explicit parameter invariance of the action (valid for the "square-root form" only!). It turns out that there's a short cut by using your form of the action when keeping in mind that after taking the variation you have to use the proper time ##s## as the parameter. That's true only for this choice of the parameter (and thus only valid for non-null geodesics!). To see this formally, you can use the method of a Lagrange multiplier for the (holonomous) constraint ##g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}-1=0##, leading to the action with the Lagrangian
    $$L=-\sqrt{\dot{x}^{\mu} \dot{x}^{\nu} g_{\mu \nu}}-\Lambda(g^{\mu \nu} \dot{x}_{\mu} \dot{x}_{\nu}-1).$$
    Here ##\Lambda## is the Lagrange parameter and ##\lambda## the general parameter of the worldline, ##\dot{x}^{\mu} = \mathrm{d} x^{\mu}/\mathrm{d} \lambda##.
    Now since the Lagrangian doesn't depend on ##\lambda## explicitly, the "Hamiltonian"
    $$H=p_{\mu} \dot{x}^{\mu}-L \quad \text{with} \quad p_{\mu}=\frac{\partial L}{\partial \dot{x}^{\mu}}$$
    is conserved, which implies (after doing the explicit calculation and implying the constraint) that ##\Lambda=\text{const}##. This shows that you can use your Lagrangian (modulo an arbitrary constant factor) under the constraint that ##\lambda=s##, i.e., ##g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=1##.
     
  7. Aug 22, 2016 #6

    stevendaryl

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    Right. I'm only claiming that using the quadratic form for [itex]L[/itex] does lead to the geodesic equations of motion (with an affine parametrization). Starting with the square-root form makes that more obvious (well, it makes it obvious if you define a geodesic to be a path making the proper time stationary; if you define a geodesic as a path that is autoparallel, then neither form is obvious).
     
  8. Aug 22, 2016 #7
    But why do you take s instead of τ?
     
  9. Aug 23, 2016 #8

    vanhees71

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    Well, some authors use ##s## and some ##\tau## for proper time. Sometimes, if you keep ##c## in the equations by not using natural units, you have ##s=c \tau##.
     
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