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\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}.

\end{equation}

Let contract ##T^{\mu\nu}##:

\begin{equation}

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

\end{equation}

Lets get covariant derivative

\begin{equation}

\nabla_{\nu} T = \nabla_{\nu} \left(m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right) = \partial_{\nu} \left(m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right) = m\int d \tau \partial_{\nu} \left(\frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right) = 0.

\end{equation}

Does it wright ##\nabla_{\nu} T = 0##? Or I'm wrong?