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I Two Conserved Quantities Along Geodesic

  1. Jul 25, 2017 #1
    Hi Everyone!

    I have done three years in my undergrad in physics/math and this summer I'm doing a research project in general relativity. I generally use a computer to do my GR computations, but there is a proof that I want to do by hand and I've been having some trouble.

    I want to show that the Lagrangian and the Hamiltonian are conserved along the geodesic. I could be wrong about that, but I actually think they might be equal to each other in this case.

    The Lagrangian is: $$L = 1/2 gabVaVb = VbVb$$
    and the Hamiltonian is [itex] H = 1/2 gabVaVb = VbVb[/itex]

    where V is the geodesic and equals q dot.

    Where do I go from here? Can I just say [itex]A =VbVb [/itex] and then take the Lie derivative of A along V, or should I take a covariant derivative of A? I really need some help. Please show the steps that you would take to prove this.

    Also, how long did it take you guys to get comfortable with this mathematics? I've worked through the first half of the Schutz GR book, if you have any other recommendable resources, let me know!

    Thanks in advance!
    Last edited: Jul 25, 2017
  2. jcsd
  3. Jul 25, 2017 #2


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    I haven't worked this out. The approach that comes to mind as to where you go next from what you wrote is to write down the Euler-Lagrange equations. This should give you the equations of motion, and if the Lagrangian is correct, then you should get as a result the standard geodesic equations:

    $$\frac{d^2 x^a}{ds^2} + \Gamma^a{}_{bc} \frac{dx^a}{ds} \frac{dx^b}{ds}= 0$$

    You can identify ##V^a = \frac{dx^a}{ds}##.

    (see wiki on geodesic equations, for instance).

    I'm not quite sure about getting two conserved quantities or what they are in the general case, you don't seem to be making at this point any assumption that ##g_{ab}## is a static and/or stationary metric. The origin of conserved energy and angular momentum along a geodesic in the Schwarzschild metric comes from the fact that the metric is static, for instance, but this may just be irrelevant to what you are doing, I don't know.

    I'd also want to do a Legendre transformaton to confirm that the Hamiltonian is as you say it is. (See wiki for Legendre transformation too, if it's not familiar).
  4. Jul 25, 2017 #3
    Thank you! I should have stated that the metric does NOT change with time. My real question is how to show that those values are conserved. Can I take the Lie Derivative of VaVa with respect to V and see if it is 0?
  5. Jul 25, 2017 #4


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    I don't use Lie derivatives that much. I probably should - but I don't. I believe one can say that having a metric "not change with time" means that the Lie derivative of the metric along some vector field is zero. This vector field is called the Killing vector field, and in a coordinate system where the metric coefficients do not change, is represented by ##\partial / \partial t##.

    So this would be the place to start, I think - you're essentially finding the Kiling vectors of the static metric, though perhaps those terms are not familiar to you? Hopefully the above, which points out that said Killing vectors can be described as ##\partial / \partial t## and also as the vector field along which the lie derivative of the metric tensor vanishes will get you started.

    My biggest concern here is that you're familiar with the interpretation of partial derivative operators as vectors / vector fields.
  6. Jul 25, 2017 #5
    Thank you again! I think I figured it out. I don't know why latex is not working for me. Here's my
    solution. I am quite familiar with the terms metric and Killing Vectors, in fact Killing vectors are symmetries of the metric. Pervect, what do you do your research in?

  7. Jul 26, 2017 #6


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    It's much simpler :-). Start from
    $$L=\frac{1}{2} g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}.$$
    It's not explicitly dependent on the parameter (which is affine automatically, as we'll show right now) and thus variational calculus tells you that with the canonical momentum
    $$p_{\mu}=\frac{\partial L}{\partial \dot{q}^{\mu}}$$
    the "Hamiltonian"
    $$H=p_{\mu} \dot{q}^{\mu}-L=\text{const},$$
    and since ##g_{\mu \nu}## depends only on the ##q^{\alpha}## but not on the ##\dot{q}^{\alpha}## you simply have
    $$p_{\mu}=g_{\mu \nu} \dot{q}^{\nu}.$$
    This implies that
    and this means that ##\lambda## is automatically an affine parameter.

    The good thing with that is that with this form of the Lagrangian for the geodesics you can treat the timelike (massive particles' trajectories in the given spacetime) and lightlike (mass less particles, aka "naive photons") geodesics on the same footing. For timelike geodesics you can always set ##\lambda=\tau## (where ##\tau## is the proper time of the particle), while for lightlike (null) geodesics ##\lambda## is an arbitrary affine parameter, defined only up to a scale, which however is irrelevant for all the physics you can do with null geodesics.
  8. Jul 26, 2017 #7
    Thank you Vanhees71! This is a wonderful and easy to follow proof.

    I don't regret what I came up with, because I learned (and re-learned) a lot, but your proof is far better! Thanks again!

    Case closed!
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