- #1
sergiokapone
- 302
- 17
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}.
\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?
\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?