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?

 

[sardar]

for any help or insights you can provide!

Hello,

The geodesic equation is derived from the conservation of energy-momentum in general relativity. This equation describes the path of a particle moving under the influence of gravity. Eq. 19.1 is the conservation of energy-momentum equation, which states that the covariant derivative of the energy-momentum tensor is equal to zero. This equation can be rewritten as Eq. 19.3, which is the geodesic equation, by using the definition of the energy-momentum tensor and the Christoffel symbols.

To get to the equation before 19.4, you can use the definition of the covariant derivative and the fact that the Christoffel symbols are symmetric in their lower indices. This will allow you to simplify the equation and get to the desired form.

I hope this helps and provides some insight into the derivation process. Let me know if you have any further questions.