Geodesic Equation from energy-momentum conservation

Question

I've been reading the excelent review from Eric Poisson found here. While studying it I stumbled in a proof that I can't make... I can't find a way to go from Eq.(19.3) to the one before Eq.(19.4) (which is unnumbered).

I've been able to do some progress (which I present below), but can't get the right answer... Please somebody help me. It's getting really frustating...

Thank you

Given the energy-momentum tensor$$T^{\alpha\beta}\left(x\right)=m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\left(x,z\right)g^{\beta}{}_{\nu}\left(x,z\right)\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)d\lambda,}$$

one can take it's divergence\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda+m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\beta}\left[\delta_{4}\left(x,z\right)\right]d\lambda.}} \end{alignedat}

But using Eq.13.3 of the reference one finds that the divergence of the energy-momentum tensor is also given by\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]g^{\beta}{}_{\nu}\delta_{4}\left(x,z\right)d\lambda-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}} \end{alignedat}

Which means that$$m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\beta}\left[\delta_{4}\left(x,z\right)\right]d\lambda=-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}}$$

so it must be zero (please correct me if I'm wrong).

Then, using Eqs.(5.14) and (13.3), the divergence of the energy-momentum tensor is simply\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\frac{D}{d\lambda}\left[\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda}+\\ & \,\,\,+m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\beta}\left[g^{\beta}{}_{\nu}\dot{z}^{\nu}\right]\delta_{4}\left(x,z\right)d\lambda.} \end{alignedat}

If what I've done is correct then comparing with the reference's result the last term must be zero. Can anyone think why?

I thought that, since the covariant derivative is taken in the point $x$ then $g^{\beta}{}_{\nu}\nabla_{\beta}\dot{z}^{\nu}$ is zero but then, what forbids $g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\nu}\nabla_{\beta}\dot{z}^{\mu}$ to be equally zero?

Answer 1

I will abuse notation a little, but I hope you don't find it terrible. After all, I only abuse notation: not children!

Trick 1: Parametrize by Proper Length. We will pick for our affine parameter $\lambda=s$ the proper length. Then the stress energy tensor becomes $$\tag{1}T^{\alpha\beta}(x)=m\int_{\gamma}u^{\alpha} u^{\beta}\frac{\delta^{(4)}\bigl(x,z(s)\bigr)}{\sqrt{|g|}}\,\mathrm{d}s$$ where $u^{\alpha}=\mathrm{d}x^{\alpha}/\mathrm{d}s$ and $g=\det{g_{\mu\nu}}$.

Trick 2: Covariant Derivative Trick. We can write $$\nabla_{\mu}f^{\mu}=\frac{1}{\sqrt{|g|}}\partial_{\mu}(\sqrt{|g|}f^{\mu})$$ for arbitrary $f^{\mu}$.

Exercise: Using Poisson's notation (13.2), we have $$\delta(x,x') = \frac{\delta^{(4)}(x-x')}{\sqrt{|g|}} = \frac{\delta^{(4)}(x-x')}{\sqrt{|g'|}}$$ and thus using our Covariant Derivative trick, find $$\nabla_{\mu}\delta(x,x')=???$$

This will tell you that $$\int u^{\alpha}u^{\beta}\nabla_{\alpha}\delta(x,x')\,\mathrm{d}s = \mbox{boundary terms}$$ and thus we can ignore it.

Remark 1. You are in error writing \begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}} \end{alignedat} This should have been a simple application of the product rule. That is, the minus sign should be a plus sign.

Remark 2. Why should we expect the right hand side of $\nabla_{\beta}T^{\alpha\beta}=0$? Well, because using Einstein's field equation it's $\nabla_{\beta}G^{\alpha\beta}$ and this is identically zero.

This is why we set $$\tag{2}m\int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$ ...which is precisely the geodesic equation for a point particle as discussed in Poisson's article section 3.

Edit We can rewrite (2) since ${g^{\alpha}}_{\beta}={\delta^{\alpha}}_{\beta}$ is the Kronecker delta. So $$\tag{3}m\int_{\gamma}\nabla_{\nu}\left[\frac{\dot{z}^{\alpha}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\beta}}}\right]\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$ But if we pick the arclength as the parameter, this becomes simply $$\tag{4}m\int_{\gamma}\nabla_{\nu}(u^{\alpha}u^{\nu})\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$ Great, but really is $$\tag{5}\nabla_{\nu}(u^{\alpha}u^{\nu})=0?$$ Lets recall for a geodesic using arclength parametrization we have $$u_{\mu}u^{\mu}=1\implies u_{\mu}\nabla_{\nu}u^{\mu}=0.$$ Thus (5), when contracted by a non-negative vector (say $u_{\alpha}$) becomes \begin{alignedat}{1}u_{\alpha}\nabla_{\nu}(u^{\alpha}u^{\nu}) &= u_{\alpha}\underbrace{u^{\nu}\nabla_{\nu}u^{\alpha}}_{=0} + \underbrace{u_{\alpha}u^{\alpha}}_{=1}\nabla_{\nu}u^{\nu}\\ &=\nabla_{\nu}u^{\nu}\end{alignedat} But this is a continuity-type equation (and if you use trick 2, it really resembles electromagnetism's continuity equation!).

Now we can go back, and by inspection we find $$\nabla_{\nu}(u^{\alpha}u^{\nu})=\left(\begin{array}{c}\mbox{Geodesic}\\ \mbox{Equation}\end{array}\right)+\left(\begin{array}{c}\mbox{Continuity}\\ \mbox{Equation}\end{array}\right).$$ This is, of course, $\nabla_{\mu}T^{\mu\nu}$. Why should we expect it to be zero?

Well, if the Einstein field equations hold, then $$G^{\mu\nu}-\kappa T^{\mu\nu}=0$$ and moreover $$\nabla_{\mu}(G^{\mu\nu}-\kappa T^{\mu\nu})=0.$$ However, $\nabla_{\mu}G^{\mu\nu}=0$ identically thanks to geometry.