- #1
ergospherical
- 1,072
- 1,363
Then the equation of motion for a photon in the beam (moving parallel to the ##x## axis) is\begin{align*}
\ddot{y} = - \Gamma^y_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} = -[\Gamma^y_{00} + \Gamma^y_{xx} + 2\Gamma^y_{0x}](\dot{x}^0)^2
\end{align*}and similar for ##z##. In the linearised theory (##h_{\mu \nu} \equiv g_{\mu \nu} - \eta_{\mu \nu}##) we have ##h_{00} = h_{xx} = - h_{0x}## which after some algebra implies that the right hand side vanishes. This result should still hold in the non-linearised theory, but how can one prove this?
\ddot{y} = - \Gamma^y_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} = -[\Gamma^y_{00} + \Gamma^y_{xx} + 2\Gamma^y_{0x}](\dot{x}^0)^2
\end{align*}and similar for ##z##. In the linearised theory (##h_{\mu \nu} \equiv g_{\mu \nu} - \eta_{\mu \nu}##) we have ##h_{00} = h_{xx} = - h_{0x}## which after some algebra implies that the right hand side vanishes. This result should still hold in the non-linearised theory, but how can one prove this?