How to derive the Euler Lagrange equation for geodesics?

Question

In my book, it says a geodesic is associated to the functional $\int_0^l |\gamma'|^2$ , with a metric g.

It then jumps to $\ddot{\gamma}^k + \Gamma^k_{ij}\dot{\gamma}^i\dot{\gamma}^j = 0$ where $\Gamma^k_{ij} := \frac{1}{2}g^{kl}\left( \frac{\partial g_{jl}}{\partial x^i} + \frac{\partial g_{il}}{\partial x^j} - \frac{\partial g_{ij}}{\partial x^l}\right).$

I'm trying to find out how it gets there.

My functional is $L(t, \gamma, \dot \gamma) = g_{ij}(\gamma(t)) \dot\gamma ^i \dot\gamma ^j$

From wikipedia Euler-Lagrange, $\frac{\partial L}{\partial \gamma ^k} - \frac{d}{dt} \frac{\partial L}{\partial \dot \gamma ^k} = 0$

but this only leads me to $\partial_k g_{ij} \gamma' ^i \gamma' ^j - \frac{d}{dt} (g_{kj} \sum_j \gamma '^j + g_{kk}\gamma '^k)=0 $

There are lots of more developments to if I derive this last thing with regard to t (since $g$ is $g(\gamma(t))$). Am I on the right direction or is there something I'm missing?

Answer 1

Let $(M, g=g_{ij}dx^idx^j)$ be a Riemannian manifold. In particular, this implies that $(g_{ij})$ is a symmetric matrix. We consider as the functional the kinetic energy \begin{equation*} \begin{split} K:\mathbf{T}M &\rightarrow\mathbb{R} \\ (x,\dot{x}) & \mapsto K(x,\dot{x}) =\frac{1}{2}g_{ij}(x)\dot{x}^i\dot{x}^j \end{split} \end{equation*} and we want to compute its extremal curves. To do so, $K$ must satisfy Euler-Lagrange equations at this curves: \begin{equation*} \frac{\partial K}{\partial x^k}-\frac{d}{dt}\frac{\partial K}{\partial \dot{x}^k}=0, \end{equation*} for $k=1,\ldots,n$ and where $x^i=x^i(t)$ y $\dot{x}^i=\dot{x}^i(t)$. It is easy to show that (just see the case $k=1$ to understand it) \begin{equation*} \frac{\partial K}{\partial \dot{x}^k}=g_{ik}\dot{x}^i. \end{equation*} On the other hand \begin{equation*} \frac{d}{dt}\frac{\partial K}{\partial \dot{x}^k}=\frac{\partial g_{ik}}{\partial x^j}\dot{x}^i\dot{x}^j+g_{ik}\ddot{x}^i. \end{equation*} By convinience, we can express the first term of the right-hand side as \begin{equation*} \frac{\partial g_{ik}}{\partial x^j}\dot{x}^i\dot{x}^j=\frac{1}{2}\frac{\partial g_{ik}}{\partial x^j}\dot{x}^i\dot{x}^j+\frac{1}{2}\frac{\partial g_{ik}}{\partial x^j}\dot{x}^i\dot{x}^j=\frac{1}{2}\frac{\partial g_{ik}}{\partial x^j}\dot{x}^i\dot{x}^j+\frac{1}{2}\frac{\partial g_{jk}}{\partial x^i}\dot{x}^i\dot{x}^j, \end{equation*} where we have interchanged the indexes $i$ and $j$ in the last term. Finally, the first term of the Euler-Lagrange equation is \begin{equation*} \frac{\partial K}{\partial x^k}=\frac{1}{2}\frac{\partial g_{ij}}{\partial x^k}\dot{x}^i\dot{x}^j. \end{equation*} Thus, Euler-Lagrange equations are \begin{equation*} g_{ik}\ddot{x}^i+\frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^j}+\frac{\partial g_{jk}}{\partial x^i}-\frac{\partial g_{ij}}{\partial x^k}\right)\dot{x}^i\dot{x}^j=0,\quad k=1,\ldots,n. \end{equation*} The coefficients in $\dot{x}^i\dot{x}^j$ are denoted as \begin{equation*} \Gamma_{kij}=\frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^j}+\frac{\partial g_{jk}}{\partial x^i}-\frac{\partial g_{ij}}{\partial x^k}\right). \end{equation*} With this notation, Euler-Lagrange equations can be written as \begin{equation} \label{3_eq_EEL_K_1especie} g_{ik}\ddot{x}^i+\Gamma_{kij}\dot{x}^i\dot{x}^j=0,\quad k=1,\ldots,n \end{equation} or in a matrix form as \begin{equation*} (g_{ij})\begin{pmatrix} \ddot{x}^1\\ \vdots\\ \ddot{x}^n \end{pmatrix}=-\begin{pmatrix} \Gamma_{1ij}\dot{x}^i\dot{x}^j\\ \vdots\\ \Gamma_{nij}\dot{x}^i\dot{x}^j \end{pmatrix}. \end{equation*} Now, if we define \begin{equation*} (g^{hk})=(g_{ij})^{-1},\quad\Gamma_{ij}^k=g^{kh}\Gamma_{hij}, \end{equation*} we get \begin{equation*} \begin{pmatrix} \ddot{x}^1\\ \vdots\\ \ddot{x}^n \end{pmatrix}=-(g^{hk})\begin{pmatrix} \Gamma_{1ij}\dot{x}^i\dot{x}^j\\ \vdots\\ \Gamma_{nij}\dot{x}^i\dot{x}^j \end{pmatrix}=-\begin{pmatrix} \Gamma_{ij}^1\dot{x}^i\dot{x}^j\\ \vdots\\ \Gamma_{ij}^n\dot{x}^i\dot{x}^j \end{pmatrix} \end{equation*} or \begin{equation} \label{3_eq_EEL_K_2especie} \ddot{x}^k+\Gamma_{ij}^k\dot{x}^i\dot{x}^j=0,\quad k=1,\ldots,n. \end{equation}