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Geodesic Equation from conservation of energy-momentum

  1. Sep 26, 2012 #1
    Hi everyone,

    While reading this reference I bumped into a result. Can anyone get from Eq.19.1 to Eq.19.3?

    I've also been struggling to get from that equation to the one before 19.4 (which isn't numbered)...anyone?

    Thank you very much
     
  2. jcsd
  3. Sep 26, 2012 #2

    Bill_K

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    The stress-energy tensor is correctly defined as Tμν = 2 δL/δgμν. So as he says, "the particle’s energy-momentum tensor, obtained by functional differentiation of Sparticle with respect to gαβ(x)".

    Well, calculus tells us for any A, δ√A = (1/2√A) δA, which explains the √ thing in the denominator. All that remains is to vary the argument, gμν(z)zμzν with respect to gαβ(x). (The z's are constant.)

    He gives a hint: "the parallel propagators appear naturally by expressing gμν(z) as gαμ(z,x) gβν(z,x) gαβ(x)." When we vary this with respect to gαβ(x), all that happens is that the last factor drops out, and we are left with just the two parallel propagators in the numerator. (The z's are still there.)
     
  4. Sep 27, 2012 #3
    Yes I agree with him and obviously with what you wrote. It's my mistake, I should have been more specific. What I don't understand is from where does de dirac delta appear? Because as you said from his definition of the action, differentiating we should only have the other terms, right?


    And the other equation?taking the the covariante derivative of Eq.19.3 should give us something like this:[tex]\nabla_{\beta}T^{\alpha \beta}=\int_{\gamma}\frac{1}{\sqrt(-g_{\mu \nu}\dot{z}^\mu \dot{z}^\nu)}g^{\alpha}_{\mu}g^{\beta}_{\nu}\nabla_{\beta}(\dot{z}^\mu \dot{z}^\nu)\delta_{4}(x,z) + g^{\alpha}_{\mu}g^{\beta}_{\nu}\nabla_{\beta}(\dot{z}^\mu \dot{z}^\nu) \nabla_{\beta}(\frac{1}{\sqrt(-g_{\mu \nu}\dot{z}^\mu \dot{z}^\nu)})\delta_{4}(x,z) d\lambda[/tex].

    So the first term in the integral gives (ignoring the fraction) [tex]g^{\alpha}_{\mu}\dot{z}^\mu g^{\beta}_{\nu}\nabla_{\beta}(\dot{z}^\nu)+ g^{\alpha}_{\mu} g^{\beta}_{\nu}\dot{z}^\nu \nabla_{\beta}(\dot{z}^\mu)[/tex]

    I can argue that the tensor is symmetric in [itex]\alpha[/itex] and [itex]\beta[/itex] and in [itex]\mu[/itex] and [itex]\nu[/itex] (right?) and then I end up with
    [tex]2\frac{D}{d \lambda}(g^{\alpha}_{\mu}\dot{z}^{\mu})[/tex]

    and there shouldn't be a [itex]2[/itex] there...

    Does anybody see my mistake?
     
    Last edited: Sep 27, 2012
  5. Sep 27, 2012 #4
    Finnaly I found from where the dirac delta comes from. One has to write the action in terms of a Lagrangian density. But I still can't get the expression before Eq.19.4...

    Actually I believe there might be some inconsistency in the calculation done by Poisson. The fact that in the end he ends up with the derivative along the curve means that has used at some point [itex]g^\beta_\nu \dot{z}^\nu= \dot{z}^\beta[/itex] but that is only true if [itex]\dot{z}^\nu[/itex] is parallel transported along de curve wich is the goal of the proof...


    Anyone?
     
    Last edited: Sep 27, 2012
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