Question about Geodesic Equation Derivation using Lagrangian

[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.

 

[paranormal]

The reason why the partial derivative of {g}_{γβ} with respect to \dot{x}^{α} is zero is because the metric tensor {g}_{γβ} does not depend on the velocity terms \dot{x}^{α}. The metric tensor is a function of the position coordinates x^{α} and it describes the geometry of the space in which the particle is moving. In the Lagrangian formulation, the partial derivative \frac{\partial {L}}{\partial \dot{x}^{α}} represents the momentum of the particle in the direction of \dot{x}^{α}. Since the metric tensor does not depend on the velocity terms, it means that the momentum in that direction is conserved and therefore the partial derivative is zero.

To further understand this, let's consider the simpler case of a particle moving in a flat Euclidean space with no curvature. In this case, the metric tensor is simply the Kronecker delta {\delta}_{γβ} and it is constant, meaning it does not depend on the position or velocity terms. In this case, the Lagrangian would be {L} = {\delta}_{γβ} \dot{x}^{γ} \dot{x}^{β} and the partial derivative \frac{\partial {L}}{\partial \dot{x}^{α}} would also be zero. This is because the momentum in any direction is conserved in a flat space.

In summary, the reason why the partial derivative of the metric tensor with respect to the velocity terms is zero is because the metric tensor is a function of the position coordinates and not the velocity terms. This is a fundamental property of the metric tensor and is crucial in the derivation of the geodesic equation.