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The energy-momentum tensor of a free particle with mass ##m## moving along its worldline ##x^\mu (\tau )## is

\begin{equation}

T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2}

\end{equation}

The covariant derivative of tensor gives

\begin{equation}

\nabla_{\mu} T^{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\partial \left( \sqrt{-g} T^{\mu\nu}\right) }{\partial y^{\mu}} + \Gamma^{\nu}_{\mu\lambda}T^{\mu\lambda}

\end{equation}

And a derivation of geodesics equation from energy-momentum tensor of point particle in GR ##\nabla_{\mu} T^{\mu\nu} = 0## looks like:

\begin{align}

0 &~~~~~~=~ \frac{1}{\sqrt{-g(y)}}\partial^{(y)}_{\mu} \left(\sqrt{-g(y)}T^{\mu\nu}(y)\right)

+\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr

&~~~~~~=~\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(y)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr

&~~~~~~=~\underbrace{-\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau ~\dot{x}^{\nu} \frac{d}{d\tau}\delta^4(y\!-\!x(\tau ))}_\text{this term we integrating by parts}

~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr

&\stackrel{\text{int. by parts}}{=}~\frac{m}{\sqrt{-g(y)}}\underbrace{ \dot{x}^{\nu}\delta^{(4)}(y−x(τ))}_\text{shoul be =0! Why?} + \cr

&~~~~~\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau\underbrace{\left[\ddot{x}^{\nu}+ \Gamma^{\nu}_{\mu\lambda}(x(\tau))\dot{x}^{\mu}\dot{x}^{\lambda} \right]}_\text{=0 geodesics equation}\delta^4(y\!-\!x(\tau )) .

\end{align}

Can any one explain where did the term ##uv=\dot{x}^{\nu}\delta^{(4)}(y−x(τ))## go after integration by parts ##\int udv=uv−\int vdu## in the last term (5)?

\begin{equation}

T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2}

\end{equation}

The covariant derivative of tensor gives

\begin{equation}

\nabla_{\mu} T^{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\partial \left( \sqrt{-g} T^{\mu\nu}\right) }{\partial y^{\mu}} + \Gamma^{\nu}_{\mu\lambda}T^{\mu\lambda}

\end{equation}

And a derivation of geodesics equation from energy-momentum tensor of point particle in GR ##\nabla_{\mu} T^{\mu\nu} = 0## looks like:

\begin{align}

0 &~~~~~~=~ \frac{1}{\sqrt{-g(y)}}\partial^{(y)}_{\mu} \left(\sqrt{-g(y)}T^{\mu\nu}(y)\right)

+\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr

&~~~~~~=~\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(y)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr

&~~~~~~=~\underbrace{-\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau ~\dot{x}^{\nu} \frac{d}{d\tau}\delta^4(y\!-\!x(\tau ))}_\text{this term we integrating by parts}

~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr

&\stackrel{\text{int. by parts}}{=}~\frac{m}{\sqrt{-g(y)}}\underbrace{ \dot{x}^{\nu}\delta^{(4)}(y−x(τ))}_\text{shoul be =0! Why?} + \cr

&~~~~~\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau\underbrace{\left[\ddot{x}^{\nu}+ \Gamma^{\nu}_{\mu\lambda}(x(\tau))\dot{x}^{\mu}\dot{x}^{\lambda} \right]}_\text{=0 geodesics equation}\delta^4(y\!-\!x(\tau )) .

\end{align}

Can any one explain where did the term ##uv=\dot{x}^{\nu}\delta^{(4)}(y−x(τ))## go after integration by parts ##\int udv=uv−\int vdu## in the last term (5)?

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