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Suppose in a system of $n$ dynamic degrees of freedom $q_i$, $i=1,\dots,n$, there is only kinetic energy: $$L=\frac{1}{2}\sum_{a,b}g_{ab}(q_i)\dot{q_a}\dot{q_b}$$

where $g_{ab}(q_i)=g_{ba}(q_i)$ are the components of a symmetric $n\times n$ matrix. Show that the Euler-Lagrange equations for this system are $$\ddot{q_a}+\sum_{b,c}\Gamma^{a}_{bc}\dot{q_b}\dot{q_c}=0$$ Where $$\Gamma^{a}_{bc}=\sum_{d}\frac{1}{2}g_{ad}^{-1}\Big(\frac{\partial g_{bd}}{\partial q_c}+\frac{\partial g_{cd}}{\partial q_b}-\frac{\partial g_{bc}}{\partial q_d}\Big)$$ I know there is a similar question in mse but that lacks some explanations so I cannot fully understand. May anyone help?

End points
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1 Answers1

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In your notation,the Euler-Lagrange (EL) equations of your lagrangian are $$ \frac{\partial L}{\partial q^a}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^a}=0; $$ where each $q:I\subset \mathbb{R}\to M$ are smooth curves valued the space of orbits of your physical system. For example if you are studying a pendulum the latter has to be $S^1$, since the motion of the point is constrained on the circonference. Generally speaking you can however assume that $M$ is any open subset of $\mathbb{R}^n$. Then the EL eq. yields \begin{equation}\label{eq_1} \frac{\partial g_{ab}}{\partial q^c}\dot{q}^a\dot{q}^b- \frac{d}{dt}\left(2g_{ac}\dot{q}^a \right)=\frac{\partial g_{ab}}{\partial q^c}\dot{q}^a\dot{q}^b- 2g_{ac}\ddot{q}^a - 2\frac{\partial g_{ac}}{\partial q^b}\dot{q}^b\dot{q}^a =0. \end{equation} Here we are using Einstein notation where repeated indices, one up and one down, mean summation (e.g. $g_{ab}\dot{q}^a\dot{q}^b=\sum_{1\leq a,b\leq n}g_{ab}\dot{q}^a\dot{q}^b$). Next we use the index symmetry of the term $\dot{q}^a\dot{q}^b=\dot{q}^b\dot{q}^a$, thus we can write $$ 2\frac{\partial g_{ac}}{\partial q^b}\dot{q}^b\dot{q}^a= \frac{\partial g_{bc}}{\partial q^a}\dot{q}^a\dot{q}^b+\frac{\partial g_{ac}}{\partial q^b}\dot{q}^a\dot{q}^b, $$ substituting in the EL eq. and changinng the sign yields \begin{equation} g_{ac}\ddot{q}^a + \frac{1}{2}\bigg(-\frac{\partial g_{ab}}{\partial q^c}+ \frac{\partial g_{bc}}{\partial q^a}+\frac{\partial g_{ac}}{\partial q^b}\bigg)\dot{q}^a\dot{q}^b =0. \end{equation}

Baol
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