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- Thread starter PLuz
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Bill_K

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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

He gives a hint: "the parallel propagators appear naturally by expressing g

- #3

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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?

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?

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- #4

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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?

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?

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