# Differentiating the Lagrangian to find geodesic equations?

I'm stuck pretty much at the first hurdle trying to follow the derivation of the geodesic equations from the Lagrangian $L\left(\dot{x}^{c},x^{c}\right)\equiv\frac{1}{2}g_{ab}\left(x_{c}\right)\dot{x}^{a}\dot{x}^{b}$ in Foster and Nightingale's A Short Course in General Relativity.

Differentiating the Lagrangian they give $$\frac{\partial L}{\partial\dot{x}^{c}}=\frac{1}{2}g_{ab}\delta_{c}^{a}\dot{x}^{b}+\frac{1}{2}g_{ab}\dot{x}^{a}\delta_{c}^{b},$$ which I can see, because $\frac{\partial\dot{x}^{a}}{\partial\dot{x}^{c}}$ is 1 only when $a=c$ (otherwise equals zero), and $\frac{\partial\dot{x}^{b}}{\partial\dot{x}^{c}}$ is 1 only when $b=c$ (otherwise equals zero). However, they then go on to say that the above simplifies to $$\frac{\partial L}{\partial\dot{x}^{c}}=g_{cb}\dot{x}^{b}.$$

How do they do that? Why do the indices change on the metric? I'm a self-taught plodder, so please don't worry about making your answers too simple.

The metric times the Kronecker delta gives

$$g_{ab} \delta^a_c = g_{cb}$$

Since the Kronecker delta tells us to replace the $a$ indice with $c$. We do this for both terms in your equation,

$$\frac{\partial L}{\partial \dot{x}^c} = \frac{1}{2} g_{cb} \dot{x}^b + \frac{1}{2} g_{ac} \dot{x}^a$$

and then rename the dummy indices (the indices that are summed over) so that both terms have the same form. Adding them gives

$$\frac{\partial L}{\partial \dot{x}^c} = g_{cb}\dot{x}^b$$

• Yes, the Kronecker delta renames the index. I see it now. Thanks. – Peter4075 May 10 '15 at 11:25