# Euler Lagrange and Geodesics

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: math.stackexchange.com/q/580858/11127 and links therein. – Qmechanic Feb 10 at 20:59
• I suppose its not an exact duplicate, but it was very helpful; so thank you :) I appreciate that! – user2662833 Feb 11 at 0:35

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 $$v^i=\dot x^i$$.