Geodesic equation via conserved stress tensor

[haushofer]

Hi,

I have a question which was raised after reading the article "Derivation of the string equation of motion in general relativity" by Gürses and Gürsey.

The geodesic equation for point particles can apparently be obtained as follows.

First one takes the stress tensor of a point particle,


T^{\mu\nu} = \frac{m}{\sqrt{-g}} \int \frac{dz^{\mu}}{ds}\frac{dz^{\nu}}{ds}\delta^{4}(x - z(s))ds


Then one uses that it is covariantly conserved,


\nabla_{\mu}T^{\mu\nu} = 0


Because \sqrt{-g} is a density with weight +1, one has the identity


\nabla_{\mu}(\sqrt{-g} T^{\mu\nu}) = \partial_{\mu} (\sqrt{-g} T^{\mu\nu}) +
\sqrt{-g} \Gamma^{\nu}_{\mu\rho}T^{\mu\rho}


Then the idea is to integrate this over a "tubular spacetime region" \Omega (I suppose they mean a tubular region around the spacetime path of the particle) and one should then obtain


\int [\frac{d^2 z^{\rho}}{ds^2} + \Gamma^{\rho}_{\mu\nu}\frac{dz^{\mu}}{ds}\frac{dz^ {\nu}}{ds}] ds = 0


The question is: how is this done? The second term I can see arising, but how to handle the first one? This first term involves


\int_{\Omega} \frac{\partial}{\partial x^{\mu}} [\int\frac{dz^{\mu}}{ds}\frac{dz^{\nu}}{ds}\delta^{ 4}(x - z(s))ds] d^4 x


which somehow should give

+ \int \frac{d^2 z^{\rho}}{ds^2} ds
.

Does anyone have an idea? Partial integrations give me the wrong answer somehow, but perhaps I'm overlooking something silly :)

[arkajad]

I am not sure how they do it, but essentially the following trick is probably being used:

\partial_\mu\int\,\frac{dz^\mu}{ds}\ldots =\int\partial_\mu\,\frac{dz^\mu}{ds}\ldots=\int \frac{dz^\mu}{ds}\partial_\mu\ldots=\int \frac{d}{ds}\ldots

An alternative version (though more general, too general for your purpose) is here (http://arkadiusz-jadczyk.org/papers/conservation_laws.pdf) (the G&G paper being Ref. 2)

[haushofer]

Thanks Arkajad. That would explain the overal plus sign. My confusion is now that if you do your trick, the derivative operator (wrt x) is acting on the delta function. You seem to use the delta function to convert this derivative wrt x into a derivative wrt z, but for me that seems rather fishy. Could you elaborate on that?

Of course one has identities for expression in which a derivative is acting on a delta function, but that also doesn't seem the right way.

[arkajad]

OK. Will try to manage the details. Do not have the G&G derivation with me, so I will try to figure it out all by myself.

[haushofer]

Well, their "derivation" is pretty much what was in my openingspost :P But I would be happy to hear your details!

[arkajad]

So, I thought about it and I am not buying G+G derivation, although it may be expanded so, that it will work. I would follow another method.

So we know we should have \nabla_\mu T^{\mu\nu}=0, thus
\xi_\nu\nabla_\mu T^{\mu\nu}=0

for any \xi_\nu

Thus \nabla_\mu(\xi_\nu T^{\mu\nu})=T^{\mu\nu}\nabla_\mu\xi_\nu

Now integrate both sides over the tubular domain over the path, assuming \xi_\nuvanishes at the boundary (but is arbitrary inside). The LHS vanishes by the Stokes theorem. If follows that the integral of the RHS must be zero. however small the region is. So, lets us just restrict ourselves to the integration over the path itself.
Take
T^{\mu\nu}(s)=m\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}
We have
\int \nabla_\mu\xi_\nu \frac{dx^\mu}{ds}\frac{dx^\nu}{ds}ds=0
But
\nabla_\mu\xi_\nu\frac{dx^\mu}{ds}=\frac{D\xi_\nu} {ds}
where the D denotes the covariant derivative along the path. So we have:

\int\frac{D}{ds}(\xi_\nu) \frac{dx^\nu}{ds}ds=0

Now integrate by parts taking into account the fact that \xi_\nu vanishes at the ends. We get
\int\xi_\nu \frac{D}{ds}\dot x^\nu ds=0

But \xi_\nu is arbitrary inside the path. Thus
\frac{D}{ds}\dot x^\nu =0
Which is the geodesic equation.

[Altabeh]

Thanks Arkajad. That would explain the overal plus sign. My confusion is now that if you do your trick, the derivative operator (wrt x) is acting on the delta function. You seem to use the delta function to convert this derivative wrt x into a derivative wrt z, but for me that seems rather fishy. Could you elaborate on that?

Of course one has identities for expression in which a derivative is acting on a delta function, but that also doesn't seem the right way.

The part involving the derivative of delta function simply vanishes because

introducing u^\kappa=x^\kappa- z^\kappa we have

\int_{\Omega} [\int\frac{dz^{\mu}}{ds}\frac{dz^{\nu}}{ds}\frac{\p artial}{\partial x^{\mu}} \delta^{ 4}(u^\kappa)ds] d^4 x=

\int_{\Omega} [\int\frac{dz^{\mu}}{ds}\frac{\partial u^\kappa }{\partial x^{\mu}} \frac{\partial}{\partial u^{\kappa}} \delta^{ 4}(u^\alpha)dz^{\nu}] d^4 x=

\int_{\Omega} [\int\frac{dz^{\mu}}{ds}\frac{\partial u^\kappa }{\partial x^{\mu}} \frac{\partial}{\partial u^{\kappa}} \delta^{ 4}(u^\alpha){dz^{\nu}} ] d^4 x=

\int_{\Omega} [\int\frac{dz^{\mu}}{ds}\frac{\partial u^\kappa }{\partial x^{\mu}} \frac{\partial}{\partial u^{\kappa}} \delta^{ 4}(u^\alpha){dz^{\nu}} ] d^4 x=

\int_{\Omega} [\int\frac{dz^{\mu}}{ds} \frac{\partial u^\kappa }{\partial x^{\mu}} \frac{\partial}{\partial u^{\kappa}} \delta^{ 4}(u^\alpha){dz^{\nu}} ] d^4 x=

\int_{\Omega} [\int\frac{dz^{\mu}}{ds}\frac{\partial u^\kappa }{\partial x^{\mu}} \frac{\partial}{\partial u^{\kappa}} \delta^{ 4}(u^\alpha){dz^nu} ] d^4 x=-\int_{\Omega} [\int \frac{\partial}{\partial u^{\kappa}} (\frac{dz^{\mu}}{ds}\frac{\partial u^\kappa }{\partial x^{\mu}}) \delta^{ 4}(u^\alpha){dz^\nu} ] d^4 x.

Here we use the fact the tube is located around the point z^{\mu} and that the delta function over the boundaries of the region \Omega would be zero (use Gauss theorem). It is then a very easy exercise to continue from the last equation to get to zero.

AB

[haushofer]

But then I see two explanations which contradict each other; if the first term involving the derivative on delta vanishes, one does not obtain the geodesic equation.

[haushofer]

So, I thought about it and I am not buying G+G derivation, although it may be expanded so, that it will work. I would follow another method.

So we know we should have \nabla_\mu T^{\mu\nu}=0, thus
\xi_\nu\nabla_\mu T^{\mu\nu}=0

for any \xi_\nu

Thus \nabla_\mu(\xi_\nu T^{\mu\nu})=T^{\mu\nu}\nabla_\mu\xi_\nu

Now integrate both sides over the tubular domain over the path, assuming \xi_\nuvanishes at the boundary (but is arbitrary inside). The LHS vanishes by the Stokes theorem. If follows that the integral of the RHS must be zero. however small the region is. So, lets us just restrict ourselves to the integration over the path itself.
Take
T^{\mu\nu}(s)=m\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}
We have
\int \nabla_\mu\xi_\nu \frac{dx^\mu}{ds}\frac{dx^\nu}{ds}ds=0
But
\nabla_\mu\xi_\nu\frac{dx^\mu}{ds}=\frac{D\xi_\nu} {ds}
where the D denotes the covariant derivative along the path. So we have:

\int\frac{D}{ds}(\xi_\nu) \frac{dx^\nu}{ds}ds=0

Now integrate by parts taking into account the fact that \xi_\nu vanishes at the ends. We get
\int\xi_\nu \frac{D}{ds}\dot x^\nu ds=0

But \xi_\nu is arbitrary inside the path. Thus
\frac{D}{ds}\dot x^\nu =0
Which is the geodesic equation.
That's indeed a nice derivation; I would say that strictly one should take

T^{\mu\nu}(s)=m\int\frac{dz^\mu}{ds}\frac{dz^\nu}{ ds}\delta^{4}(x-z(s))ds

but for the argument that doesn't matter ;)

[arkajad]

That's indeed a nice derivation; I would say that strictly one should take

T^{\mu\nu}(s)=m\int\frac{dz^\mu}{ds}\frac{dz^\nu}{ ds}\delta^{4}(x-z(s))ds

but for the argument that doesn't matter ;)

Indeed. I have adjusted a little bit the argument mixing two different approaches - the one with the tube (tht you were discussing) and the other one that starts directly with a singular distribution of matter and concentrates on the supporting set (that I prefer). In this second approach one essentially derives the form of energy-momentum tensor along the path to get the formula I have used and that is akin to the one with the delta function.

But such a mixing is not dangerous in this case.

[haushofer]

Ok, thank you very much for your help! Altabeh, I'm still trying to understand your derivation, but your claim that that whole term should just vanish sounds a bit strange to me :)

[Altabeh]

Ok, thank you very much for your help! Altabeh, I'm still trying to understand your derivation, but your claim that that whole term should just vanish sounds a bit strange to me :)

I'm not really sure about it, but integration by parts appears to be helpful because then the first term vanishes by Gauss theorem and the second term is just

-\int_{\Omega} [\int\frac{dz^{\mu}}{ds}\frac{dz^{\nu}}{ds}\frac{\p artial}{\partial x^{\mu}} \delta^{ 4}(u^\kappa)ds] d^4 x .

Now what is with the contradiction?

AB

[haushofer]

If I understand your post correctly, the first term should vanish as a whole, instead of giving the \ddot{z}-term of the geodesic equation. Or am I misunderstanding you?

[Altabeh]

If I understand your post correctly, the first term should vanish as a whole, instead of giving the \ddot{z}-term of the geodesic equation. Or am I misunderstanding you?

Oh you got to be misunderstanding me! The first term not in the Leibniz expansion but in the integrated (by parts) equation of the derivative of delta function in the integrand as a whole vanishes. Yet we have two terms left: first a term from integration by parts and second the derivative of \frac{dz^{\mu}}{ds}\frac{dz^{\nu}}{ds}. Am I right?

AB

[Altabeh]

Oh if you have noticed I've made a mistake in my penult post here. In fact you can read the right term left from integration by parts in my first post:

-\int_{\Omega} [\int \frac{\partial}{\partial u^{\kappa}} (\frac{dz^{\mu}}{ds}\frac{\partial u^\kappa }{\partial x^{\mu}}) \delta^{ 4}(u^\alpha){dz^\nu} ] d^4 x.

Sorry for the inconvenience.

AB

[haushofer]

A small kick, and perhaps a silly question. With this procedure one gets


T^{\mu\nu} = \frac{m}{\sqrt{-g}} \int \frac{dz^{\mu}}{ds}\frac{dz^{\nu}}{ds}\delta^{4}(x - z(s))ds


But this is not the usual energy momentum tensor for a dust; instead of the usual Einstein equations


G_{\mu\nu} = \kappa \rho u^{\mu}u^{\nu}


one now seems to get


G_{\mu\nu} = \kappa \rho u^{\mu}u^{\nu} \frac{\delta^D (x-x(s)))}{\sqrt{g}}


This is not the usual Einstein equation sourced by a dust. The delta on the RHS tells me to evaluate the energy-momentum tensor along the world-line, which is fine. But the appearance of the sqrt{g} puzzles me. How should I then interpret this?

[haushofer]

Is it just that one throws away the factor


\frac{\delta^D (x-x(s)))}{\sqrt{g}}


and concludes that the energy momentum tensor is thus given by


T^{\mu\nu} = \rho u^{\mu}u^{\nu}

?