Why is ∂g_γβ/∂ẋ^α zero in the geodesic equation derivation?

[LoadedAnvils]

I'm trying to derive the Geodesic equation, $\ddot{x}^{α} + {Γ}^{α}_{βγ} \dot{x}^{β} \dot{x}^{γ} = 0$.

However, when I take the Lagrangian to be ${L} = {g}_{γβ} \dot{x}^{γ} \dot{x}^{β}$, and I'm taking $\frac{\partial {L}}{\partial \dot{x}^{α}}$, I don't understand why the partial derivative of ${g}_{γβ}$ with respect to $\dot{x}^{α}$ is zero.

I've been looking for a derivation that explained this step but I'm having no luck.

Anyone care to explain?

 

[andrien]

metric tensor is function of coordinates only,not derivatives of it.So it is zero.

 

[DimReg]

I'm trying to derive the Geodesic equation, $\ddot{x}^{α} + {Γ}^{α}_{βγ} \dot{x}^{β} \dot{x}^{γ} = 0$.

However, when I take the Lagrangian to be ${L} = {g}_{γβ} \dot{x}^{γ} \dot{x}^{β}$, and I'm taking $\frac{\partial {L}}{\partial \dot{x}^{α}}$, I don't understand why the partial derivative of ${g}_{γβ}$ with respect to $\dot{x}^{α}$ is zero.

I've been looking for a derivation that explained this step but I'm having no luck.

Anyone care to explain?

Although you might think that since $\dot{x}$ is related to x, the derivative should be zero, we actually consider the two to be separate variables. This is because in the lagrangian formalism, you are considering functions, not positions (you are techinically varying over paths). To see how they are independent, consider a function $f(t) = x$. Since all we know is that it equals x at t, then the first derivative is not fixed (since the derivative is the slope of the tangent, think of how many lines you can draw through a single point). Since we are considering all functions, we don't actually know anything about the first derivative, so we have to treat it as an independent parameter.

In that case, you'll notice that the metric is not written to have explicit dependence on $\dot{x}$, which is for a reason, because the metric is chosen to not have any dependence on $\dot{x}$. Physically, this is because the distance between two points doesn't depend on the speed that you are going, so the metric field shouldn't depend on $\dot{x}$

 

[LoadedAnvils]

Thank you both. I can now continue with the derivation.