I am trying to prove, using the explicit formula for the energy-momentum tensor of a particle, that its 4-divergence is zero. However, while attempting to do this, I got stuck. Here's why:
I know that the energy-momentum tensor for a particle can be written as:
$$T^{\mu \nu} = p^\mu(t) p^\nu(t) \frac{\delta^3(x - x(t))}{p^0(t) \sqrt{-g(\vec{x}, t)}}$$
where $ g $ denotes the determinant of the metric.
My first doubt is whether this expression should be evaluated at a fixed time $ t $, or more generally for a fixed value of the affine parameter used to describe the worldline of the particle.
What I thought was to compute the 4-divergence to check if it is indeed zero:
$$\nabla_{\mu} T^{\mu \nu} = \partial_{\mu} T^{\mu \nu} + \Gamma^{\mu}_{\mu \rho} T^{\rho \nu} + \Gamma^{\mu}_{\nu \rho} T^{\mu \rho}.$$
Then, the partial derivative of the first term should reduce to:
$$\partial_{\mu} T^{\mu \nu} = \frac{p^\mu(t) p^\nu(t)}{p^0(t)} \partial_{\mu} \left( \frac{\delta^3(x - x(t))}{\sqrt{-g(\vec{x}, t)}} \right).$$
Since:
$$p^{\mu}(t) = \frac{d x^{\mu}(t)}{dt}$$
evaluated at some fixed point $ t $, it is just a number, so it should be valid to take it outside the partial derivative. This is my second doubt.
For the third member, I thought that I could rewrite the Christoffel symbols as derivatives with respect to $ t $ of $ p^{\mu} $, since I am assuming that my particle is free-falling, and therefore it follows a geodesic in spacetime. Thus, the geodesic equation is satisfied.
So, what I am asking is just a confirmation of these two doubts: first, regarding the expression for the energy-momentum tensor for a particle that I provided, and second, about the partial differentiation with respect to the coordinates.
I know there is a faster way to check this, using the Einstein equation. Since the Einstein tensor has a 4-divergence of zero, this implies that the energy-momentum tensor also has a 4-divergence of zero. However, I would like to try a more computational approach.
