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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.

Sergio
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    Why do you think integration by parts is not correct? –  Jun 02 '19 at 20:16
  • I think I get $\frac{dX}{d\tau}\delta(x- X(\tau))$ outside integral, which is undefined. – Sergio Jun 02 '19 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. –  Jun 02 '19 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 03 '19 at 19:00
  • I believe so, but it depends on the specific values of $x$ and $X$ (how are you defining them?): as long as the difference is not zero the delta function will be zero. –  Jun 03 '19 at 21:35
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    @Quantumness I define $x$ as arbitrary point of space-time, and $X$ as a coordinare of particle. – Sergio Jun 03 '19 at 22:06
  • And I think, instead use of term "integrating by parts" it is simple to use "definition of delta-function derivative" which is $\int u d\delta = \int \delta du$ – Sergio Jun 03 '19 at 22:32

2 Answers2

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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.

Qmechanic
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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

Pavel Nakaznoy
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