Geodesic Equation from conservation of energy-momentum

[PLuz]

Hi everyone,

While reading http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html 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

 

[Bill_K]

The stress-energy tensor is correctly defined as T^μν = 2 δL/δg_μν. So as he says, "the particle’s energy-momentum tensor, obtained by functional differentiation of S_particle with respect to g_αβ(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, g_μν(z)z^μz^ν with respect to g_αβ(x). (The z's are constant.)

He gives a hint: "the parallel propagators appear naturally by expressing g_μν(z) as g^α_μ(z,x) g^β_ν(z,x) g_αβ(x)." When we vary this with respect to g_αβ(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.)

 

[PLuz]

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:\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.

So the first term in the integral gives (ignoring the fraction) 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)

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

and there shouldn't be a 2 there...

Does anybody see my mistake?

 

[PLuz]

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 g^\beta_\nu \dot{z}^\nu= \dot{z}^\beta but that is only true if \dot{z}^\nu is parallel transported along de curve which is the goal of the proof...Anyone?