I have recently been reading up on GR and I'm currently deriving the Geodesic Equation using the principle of least action.

When solving the Euler-Lagrange equation for $L=\frac{m}{2}g_{ij}(x)\dot{x}^i\dot{x}^j$i'm happy with step $$\frac{\partial L}{\partial \dot{x}^i}=mg_{ik}\dot{x}^k$$ What I don't understand is when we take the derivative w.r.t time. $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}^i}\right)=m\frac{\partial g_{ik}}{\partial{x_j}}\dot{x}^j\dot{x}^k+mg_{ik}\ddot{x}^k$$

Where does this first term come from? I understand that the metric is dependent on $x$ and so by extension $t$, but why is the derivative w.r.t $x^j$ and where does $\dot{x}^j$ come from?

(Sorry for any formatting mistakes, I wrote this on mobile.)


1 Answer 1


This is just the multivariable chain rule. It might be easier to see if you ignore the indices on $g_{ik}$ and just write it as $f$: $$\frac{\text{d} f}{\text{d} t} = \frac{\partial f}{\partial x^j} \frac{\text{d}x^j}{\text{d}t}$$


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