Geodesic Equation in Spacetime
The following is a solution to TTM book 4, lecture 5, exercise 1:
Given the metric $g_{\mu \nu}(X)$, show that the Euler-Lagrange equation, to minimize the action along a trajectory in space-time:
\[\begin{equation} \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot X^m} = \frac{\partial \mathcal L}{\partial X^m} \label{el} \end{equation}\]where the Lagrangian $\mathcal L$ is:
\[\mathcal L = -m \sqrt{-g_{\mu \nu}(X) \frac{dX^\mu}{dt}\frac{dX^\nu}{dt}}\]is equivalent to the definition of a geodesic given by:
\[\begin{equation} \frac{d^2 X^\mu}{d\tau^2} = -\Gamma_{\sigma \rho}^\mu \frac{dX^\sigma}{d\tau}\frac{dX^\rho}{d\tau} \label{geodesic} \end{equation}\]which says that the tangent vector to the trajectory in space-time stays constant.
Starting from equation 4 of book 4, lecture 5 (and abbreviating $g_{\mu \nu}(X)$ as $g_{\mu \nu}$):
\[\begin{align*} d\tau &= \sqrt{-g_{\mu \nu} dx^\mu dx^\nu} \\ &= \sqrt{-g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt} dt^2} \\ &= \sqrt{-g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}} dt \\ \therefore \frac{d\tau}{dt} &= \sqrt{-g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}} \end{align*}\]Which we can use for $\color{red} \frac{\partial \mathcal L}{\partial X^m}$:
\[\begin{align*} {\color{red} \frac{\partial \mathcal L}{\partial X^m}} &= \frac{m}{2} \frac{\partial_m [g_{\mu \nu} \frac{dX^\mu}{dt}\frac{dX^\nu}{dt}]}{\sqrt{-g_{\mu \nu} \frac{dX^\mu}{dt}\frac{dX^\nu}{dt}}} \\ &= \frac{m}{2} \frac{\partial_m [g_{\mu \nu} \frac{dX^\mu}{dt}\frac{dX^\nu}{dt}]}{\frac{d\tau}{dt}} \\ &= \frac{m}{2} \frac{\partial_m [g_{\mu \nu} \frac{dX^\mu}{d\tau} \frac{d\tau}{dt} \frac{dX^\nu}{d\tau}\frac{d\tau}{dt}]}{\frac{d\tau}{dt}} \\ &= \frac{m}{2} \partial_m [g_{\mu \nu} \frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau}] \frac{d\tau}{dt} \\ \end{align*}\]And also for $\color{red} \frac{\partial \mathcal L}{\partial \dot X^m}$ (with the product rule for tensors):
\[\begin{align*} {\color{red} \frac{\partial \mathcal L}{\partial \dot X^m}} &= \frac{m}{2} \frac{\frac{\partial}{\partial (\frac{d X^m}{dt})}[g_{\mu \nu} \frac{dX^\mu}{dt}\frac{dX^\nu}{dt}]}{\frac{d\tau}{dt}} \\ &= \frac{m}{2} \frac{g_{\mu \nu}(\frac{\partial}{\partial (\frac{d X^m}{dt})}[\frac{dX^\mu}{dt}]\frac{dX^\nu}{dt} + \frac{\partial}{\partial (\frac{d X^m}{dt})}[\frac{dX^\nu}{dt}]\frac{dX^\mu}{dt})}{\frac{d\tau}{dt}} \\ &= \frac{m}{2} \frac{(g_{m \nu}\frac{dX^\nu}{dt} + g_{\mu m}\frac{dX^\mu}{dt})}{\frac{d\tau}{dt}} \\ &= \frac{m}{2} \frac{(g_{m \nu}\frac{dX^\nu}{d\tau} \frac{d\tau}{dt} + g_{\mu m}\frac{dX^\mu}{d\tau}\frac{d\tau}{dt})}{\frac{d\tau}{dt}} \\ &= \frac{m}{2}(g_{m \nu}\frac{dX^\nu}{d\tau} + g_{\mu m}\frac{dX^\mu}{d\tau}) \end{align*}\]Allowing us to take the time derivative (using the product rule for tensors twice) in $\eqref{el}$:
\[\begin{align*} {\color{red} \frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot X^m}} &= \frac{m}{2} \frac{d}{dt}(g_{m \nu}\frac{dX^\nu}{d\tau} + g_{\mu m}\frac{dX^\mu}{d\tau}) \\ &= \frac{m}{2} \frac{d}{d\tau} \frac{d\tau}{dt}(g_{m \nu}\frac{dX^\nu}{d\tau} + g_{\mu m}\frac{dX^\mu}{d\tau}) \\ &= \frac{m}{2} (\frac{d}{d\tau}[g_{m \nu}\frac{dX^\nu}{d\tau}] + \frac{d}{d\tau}[g_{\mu m}\frac{dX^\mu}{d\tau}]) \frac{d\tau}{dt}\\ &= \frac{m}{2} (g_{m\nu} \frac{d^2 X^\nu}{d \tau^2} + \partial_\lambda g_{m\nu}\frac{dX^\lambda}{d\tau}\frac{dX^\nu}{d\tau} + g_{\mu m} \frac{d^2 X^\mu}{d\tau^2} + \partial_\lambda g_{\mu m}\frac{dX^\lambda}{d\tau}\frac{dX^\mu}{d\tau}) \frac{d\tau}{dt}\\ \end{align*}\]At this point, we are ready to set $\color{red} \frac{\partial \mathcal L}{\partial X^m} = \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot X^m}$ to give us $\eqref{el}$:
\[\begin{align*} \require{cancel}\cancel{\color{blue}\frac{m}{2}} \partial_m g_{\mu \nu} \frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau} \cancel{\color{blue}\frac{d\tau}{dt}} &= \\ \cancel{\color{blue}\frac{m}{2}} (g_{m\nu} \frac{d^2 X^\nu}{d \tau^2} + \partial_\lambda g_{m\nu}\frac{dX^\lambda}{d\tau}\frac{dX^\nu}{d\tau} + g_{\mu m} \frac{d^2 X^\mu}{d\tau^2} &+ \partial_\lambda g_{\mu m}\frac{dX^\lambda}{d\tau}\frac{dX^\mu}{d\tau}) \cancel{\color{blue}\frac{d\tau}{dt}}\\ \end{align*}\]And combine like terms:
\[\begin{align} \partial_m g_{\mu \nu} \frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau} - \partial_\lambda g_{m\nu}\frac{dX^\lambda}{d\tau}\frac{dX^\nu}{d\tau} - \partial_\lambda g_{\mu m}\frac{dX^\lambda}{d\tau}\frac{dX^\mu}{d\tau} &= g_{m\nu} \frac{d^2 X^\nu}{d \tau^2} + g_{\mu m} \frac{d^2 X^\mu}{d\tau^2} \label{liketerms} \end{align}\]For the rhs of $\eqref{liketerms}$, use the fact that $g$ is a symmetric rank 2 tensor (i.e. $g_{m \mu} = g_{\mu m}$) and that dummy variables are exchangeable to combine terms:
\[\begin{align*} g_{m\nu} \frac{d^2 X^\nu}{d \tau^2} + g_{\mu m} \frac{d^2 X^\mu}{d\tau^2} &= g_{m\nu} \frac{d^2 X^\nu}{d \tau^2} + g_{m \mu} \frac{d^2 X^\mu}{d\tau^2} \\ &= g_{m\nu} \frac{d^2 X^\nu}{d \tau^2} + g_{m \nu} \frac{d^2 X^\nu}{d\tau^2} \\ &= 2 g_{m \nu} \frac{d^2 X^\nu}{d\tau^2} \end{align*}\]For the lhs of $\eqref{liketerms}$, replace the dummy variable $\lambda$ with $\mu$ and $\nu$ respectively to factor out $\frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau}$:
\[\begin{align*} -\frac{1}{2}(\partial_\mu g_{m\nu}\frac{dX^\mu}{d\tau}\frac{dX^\nu}{d\tau} + \partial_\nu g_{\mu m}\frac{dX^\nu}{d\tau} & \frac{dX^\mu}{d\tau} -\partial_m g_{\mu \nu} \frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau}) \\ = -\frac{1}{2}(\partial_\mu g_{m\nu} + \partial_\nu g_{\mu m} -&\partial_m g_{\mu \nu} ) \frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau} \end{align*}\]Giving an intermediate result for $\eqref{el}$:
\[\begin{align} -\frac{1}{2}(\partial_\mu g_{m\nu} + \partial_\nu g_{\mu m} -\partial_m g_{\mu \nu} ) \frac{dX^\mu}{d\tau} \frac{dX^\nu}{d\tau} &= g_{m \nu} \frac{d^2 X^\nu}{d\tau^2} \label{elsoln} \end{align}\]The Christoffel symbol as given in equation 17, book 4, lecture 3 is:
\[\begin{align*} \Gamma_{m n}^t = \frac{1}{2}g^{r t}(\partial_n g_{r m} + \partial_m g_{r n} - \partial_r g_{m n}) \end{align*}\]Or in terms of $\eqref{geodesic}$:
\[\begin{align} \Gamma_{\sigma \rho}^\mu = \frac{1}{2}g^{r \mu}(\partial_\rho g_{r\sigma} + \partial_\sigma g_{r \rho} - \partial_r g_{\sigma \rho}) \label{christoffel} \end{align}\]
Now, we need to simplify $\eqref{elsoln}$ using $\eqref{christoffel}$ to give the result in $\eqref{geodesic}$. First, we use $r=m, \mu=\sigma, \nu=\rho$, and a transpose of $g_{\mu m}$ to match the variable names in the result:
\[\begin{align*} -\frac{1}{2}(\partial_\rho g_{r\sigma} + \partial_\sigma g_{r \rho} -\partial_r g_{\sigma \rho}) \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau} &= g_{r \rho} \frac{d^2 X^\rho}{d\tau^2} \\ \end{align*}\]And lastly, using the metric inverse $g^{r \mu}$ and the identity $g_{mn}g^{np} = \delta_{m}^p$ from equation 34 of book 4, lecture 2:
\[\begin{align*} -\frac{1}{2}(\partial_\rho g_{r\sigma} + \partial_\sigma g_{r \rho} -\partial_r g_{\sigma \rho}) \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau} &= g_{r \rho} \frac{d^2 X^\rho}{d\tau^2} \\ -\frac{1}{2}g^{r \mu}(\partial_\rho g_{r\sigma} + \partial_\sigma g_{r \rho} -\partial_r g_{\sigma \rho}) \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau} &= g^{r \mu} g_{r \rho} \frac{d^2 X^\rho}{d\tau^2} \\ -\Gamma_{\sigma \rho}^\mu \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau} = \delta_{\rho}^{\mu} \frac{d^2 X^\rho}{d\tau^2} \\ -\Gamma_{\sigma \rho}^\mu \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau} = \frac{d^2 X^\mu}{d\tau^2} \\ \require{enclose}\enclose{box}{\color{red}\frac{d^2 X^\mu}{d\tau^2} = -\Gamma_{\sigma \rho}^\mu \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau}} \end{align*}\]Or, using $t^\lambda = \frac{dX^\lambda}{d\tau}$ and following the equations in book 4, lecture 4:
\[\begin{align*} \frac{d^2 X^\mu}{d\tau^2} &= -\Gamma_{\sigma \rho}^\mu \frac{dX^\sigma}{d\tau} \frac{dX^\rho}{d\tau} \\ \frac{d t^\mu}{d\tau} + \Gamma_{\sigma \rho}^\mu t^\sigma t^\rho &= 0 \\ dt^\mu + \Gamma_{\sigma \rho}^\mu t^\sigma dX^\rho &= 0 \\ \require{enclose}\enclose{box}{\color{red}Dt^\mu = 0} \end{align*}\]Which says that the tangent vector ($t^\mu$) to the trajectory in space-time stays constant (covariant change in the tangent vector from one point to another does not change). Or, that we can parallel transport $t^\mu$ along a space-time trajectory.