How to derive the geodesic equations, via Euler Lagrange, from $L=\frac12 g_{ij}\dot x^i\dot x^j$?

Question

I’m trying to use the Euler Lagrange equations to derive the geodesic equations. I’ve assumed a lagrangian:

$$ L = {1\over 2} g_{ij}\dot x^i \dot x^j $$

So one of the terms of the equation requires:

$${\partial L\over \partial x^k} = {1\over 2} {\partial\over \partial x^k}\left( g_{ij}\dot x^i \dot x^j \right) $$

Some references I’ve seen are saying that this is equal to:

$$ {\partial L \over \partial x^k} = {1\over 2}{\partial g_{ij}\over \partial x^k}\dot x^i \dot x^j $$

So I thought we would need to use the product rule on these terms, but it seems that:

$$ {\partial \dot x^i \over \partial x^j } = 0 $$

Can anybody explain why this should be true, what am I missing? Thanks in advanced.

Possible duplicates: https://math.stackexchange.com/q/580858/11127 and links therein. — Qmechanic, Feb 10 '19 at 20:59
I suppose its not an exact duplicate, but it was very helpful; so thank you :) I appreciate that! — user2662833, Feb 11 '19 at 00:35

score 1 · Accepted Answer · answered Feb 10 '19 at 08:44

For the purpose of the Euler–Lagrange equations, the variables $x^i$ and $\dot x^i$ are considered as independent. In fact, I think it would be pedagogically better to call them (for example) $x^i$ and $v^i$ instead. Then you have a function $L(x,v)$, and $\partial L/\partial x^i$ means nothing but the usual partial derivative: vary one $x^i$, keeping the other $x^i$ and all the $v^i$ constant. After you've computed that derivative, you substitute $v^i=\dot x^i$.

Thanks! That makes it clearer. – user2662833 Feb 11 '19 at 00:35 — user2662833, Feb 11 '19 at 00:35

How to derive the geodesic equations, via Euler Lagrange, from $L=\frac12 g_{ij}\dot x^i\dot x^j$?

1 Answers1