# Covariant conservation of the energy-momentum tensor

In order to derive the geodesic equation of motion from the covariant conservation of the energy-momentum tensor we have to do the following procedure:

$$T^{\mu\nu}_{\space\space\space\space;\mu}= \partial_\mu T^{\mu\nu}+\Gamma^{\mu}_{\space\space\sigma\mu}T^{\sigma\nu}+\Gamma^{\nu}_{\space\space\sigma\mu}T^{\mu\sigma}$$

$$T^{\mu\nu}_{\space\space\space\space;\mu}= \frac{1}{\sqrt{-g}} \partial_\mu (\sqrt{-g} T^{\mu\nu})+ \Gamma^{\nu}_{\space\space\sigma\mu}T^{\mu\sigma}.$$

The first term is zero since the energy momentum tensor is divergenceless. Using the definition for the energy-momentum tensor:

$$T^{\mu\nu}(x)= \frac{m}{\sqrt{-g}} \int u^{\mu}u^{\nu} \delta^4(x-z(\tau))d\tau$$

where: $$u^{\mu} = \frac{dz^\mu(\tau)}{d\tau}$$.

1. When taking the derivative $$\partial_\mu (\sqrt{-g} T^{\mu\nu})$$ it doesn't act in the $$u^{\mu}u^{\nu}$$ terms, only on the delta function and I dont understand why.

2. Another thing is when we integrate by parts there is a boundary term of the form: $$-\frac{m}{\sqrt{-g}} u^{\nu} \delta^4 (x-z(\tau)) \ \Big|_a^b$$ where it is being evaluated at the proper time limits that I labeled as $$a$$ and $$b$$. The other two terms combine in order to give the geodesic equation. Why do we ignore the boundary term?

1. The particle velocity $$u^{\nu}(\tau)=\dot{z}^{\nu}(\tau)$$ is independent of the spacetime coordinate coordinate $$x^{\mu}$$.