# Some aspect of covariant derivative of point particle energy-momentum tensor

My question is related to Derivation of the geodesic equation from the continuity equation for the energy momentum tensor

I need to understand one step in derivation.

Let's consider the Energy-momentum tensor of point particle: $$\begin{equation}\label{1} T^{\mu\nu}(x) = \frac{m}{\sqrt{-g(x)}}\int d\tau \frac{dX^{\mu}}{d\tau}\frac{dX^{\nu}}{d\tau}\delta^{(4)}(x - X(\tau)) \end{equation}$$

We want to find a covariant derivative of $$T^{\mu\nu}$$. For arbitrary symmetric tensor, the covariant derivative is: $$\begin{equation}\label{2} \nabla_{\mu} T^{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\partial \left( \sqrt{-g} T^{\mu\nu}\right) }{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda}T^{\mu\lambda} \end{equation}$$

And for our case, let's consider derivative $$\frac{1}{\sqrt{-g(x)}} \frac{\partial \left( \sqrt{-g(x)} T^{\mu\nu}\right) }{\partial x^{\mu}}$$:

$$\begin{multline} \frac{1}{\sqrt{-g(x)}} \frac{\partial \left( \sqrt{-g(x)} T^{\mu\nu}\right) }{\partial x^{\mu}} = \\ = \frac{1}{\sqrt{-g(x)}} m \int d\tau \frac{dX^{\mu}}{d\tau}\frac{dX^{\nu}}{d\tau}\frac{\partial }{\partial x^{\mu}} \left[ \delta^{(4)}(x - X(\tau))\right] = \\ = - \frac{1}{\sqrt{-g(x)}} m \int d\tau \frac{dX^{\mu}}{d\tau}\frac{dX^{\nu}}{d\tau}\frac{\partial}{\partial X^{\mu}} \left[ \delta^{(4)}(x - X(\tau)) \right] = \\ = - \frac{1}{\sqrt{-g(x)}} m \int d\tau \frac{dX^{\nu}}{d\tau}\frac{d}{d\tau}\left[ \delta^{(4)}(x - X(\tau))\right] = ?\\ \end{multline}$$

What correct property of $$\delta$$-function should I use for the next step? Intergrating by parth I think is no completely correct.

• Why do you think integration by parts is not correct? – Quantumness Jun 2 at 20:16
• I think I get $\frac{dX}{d\tau}\delta(x- X(\tau))$ outside integral, which is undefined. – Sergio Jun 2 at 20:23
• Integration by parts with derivatives of the delta function should give you derivatives of the other part of the integrand, as the delta function is 0 at any nonzero point so you don't have delta's outside the integral. – Quantumness Jun 2 at 22:17
• @Quantumness out of integral $\int udv = uv - \int vdu$ we have $uv = \left.\frac{dX}{d\tau}\delta(x- X(\tau))\right|_{-\infty}^{\infty}$, so, you assert that it is equal to zero? Am I right? – Sergio Jun 3 at 19:00
• @Quantumness I define $x$ as arbitrary point of space-time, and $X$ as a coordinare of particle. – Sergio Jun 3 at 22:06

## 2 Answers

OP is right. Integration by parts leads to boundary terms at the initial and final point of the geodesic worldline. These become creation & annihilation source terms for the energy-momentum continuity equation. I updated my Phys.SE answer accordingly.

This expression is defined only if the right is 0 (as it should be in GR). The problem is that the product of generalized functions is poorly defined (non-associative), therefore, multiplying this expression on any function is impossible, but, from another side, it should be understood only when integrated with any carrier

• We are far from being used Einstein's equations, we are still in the derivative stage – Sergio Jun 2 at 20:35