How do I derive the Covariant Derivative for Covectors? (Lower index)

[LoadedAnvils]

Hello everyone!

I'm trying to learn the derivation the covariant derivative for a covector, but I can't seem to find it.

I am trying to derive this:

[itex]\nabla_{α} V_{μ} = \partial_{α} V_{μ} - \Gamma^{β}_{αμ} V_{β}[/itex]

If this is a definition, I want to know why it works with the definition of the covariant derivative of a vector:

[itex]\nabla_{α} V^{μ} = \partial_{α} V^{μ} + \Gamma^{μ}_{αβ} V^{β}[/itex]

Why is the Christoffel symbol negative for a covector? Can anyone explain this to me?

Thank you all so much.

 

[George Jones]

Try working out [itex]\nabla_\alpha \left( V^\mu V_\mu \right)[/itex] a couple of different ways:

1) Using the product rule for covariant derivatives;

2) using the fact that [itex]V^\mu V_\mu[/itex] is a scalar.

 

[LoadedAnvils]

This is one of the leads I found here, but I don't know how to progress.

Right now, I am here:

[itex]V^{μ} \nabla_{α} V_{μ} = \partial_{α} (V^{μ} V_{μ}) - V_{μ} \partial_{α} V^{μ} - V_{μ} \Gamma^{μ}_{αβ} V^{β}[/itex]

Here are some questions I have:

I understand that [itex]V^{μ} V_{μ}[/itex] is a scalar, so what is the partial derivative of it? I tried to do it with zero if it is independent of α but it doesn't look like it leads to the covariant derivative.

Also, how do I remove the [itex]V^{μ}[/itex] on the LHS? Do I multiply by [itex]V_{μ}[/itex]?

 

[George Jones]

I understand that [itex]V^{μ} V_{μ}[/itex] is a scalar, so what is the partial derivative of it?

Use the product rule for partial derivatives.

Also, how do I remove the [itex]V^{μ}[/itex] on the LHS? Do I multiply by [itex]V_{μ}[/itex]?

You don't have to remove it; you need to relabel indices in the last term on the right side.

 

[LoadedAnvils]

Use the product rule for partial derivatives.

I don't know what two terms I would use in the product rule. Once I separate the vector and covector I can't use the partial derivative anymore (or so I think). Can you explain to me how I would use the product rule in this case?

You don't have to remove it; you need to relabel indices in the last term on the right side.

I see what you mean, however, without knowing the partial derivative I don't know what I can do with this yet.

 

[George Jones]

I don't know what two terms I would use in the product rule. Once I separate the vector and covector I can't use the partial derivative anymore (or so I think). Can you explain to me how I would use the product rule in this case?

I am not sure what you mean. More explicitly: use the product rule for partial derivatives on [itex]\partial_\alpha \left( V^\mu V_\mu \right)[/itex].

 

[LoadedAnvils]

Oh, now I get it!

For those reading this in the future, this is what I got:

[itex] \partial_{α} (V^{μ} V_{μ}) = V^{μ} \partial_{α} V_{μ} + V_{μ} \partial_{α} V^{μ} [/itex]

Thus,

[itex] V^{μ} \nabla_{α} V_{μ} = V^{μ} \partial_{α} V_{μ} + V_{μ} \partial_{α} V^{μ} - V_{μ} \partial_{α} V^{μ} - V_{μ} \Gamma^{μ}_{αβ} V^{β} [/itex]

So [itex]V_{μ} \partial_{α} V^{μ}[/itex] cancels out, and we have [itex] V^{μ} \nabla_{α} V_{μ} = V^{μ} \partial_{α} V_{μ} - V_{μ} \Gamma^{μ}_{αβ} V^{β} [/itex].

We can rewrite [itex]V_{μ} \Gamma^{μ}_{αβ} V^{β}[/itex] as [itex]V^{β} \Gamma^{μ}_{αβ} V_{μ} [/itex].

Since [itex]μ[/itex] and [itex]β[/itex] are arbitrary indices, we can swap them around.

Thus, [itex] V^{β} \Gamma^{μ}_{αβ} V_{μ} = V^{μ} \Gamma^{β}_{αμ} V_{β} [/itex], and

[itex] V^{μ} \nabla_{α} V_{μ} = V^{μ} \partial_{α} V_{μ} - V^{μ} \Gamma^{β}_{αμ} V_{β} [/itex].

Eliminating [itex]V^{μ}[/itex], we reach the desired result:

[itex]\nabla_{α} V_{μ} = \partial_{α} V_{μ} - \Gamma^{β}_{αμ} V_{β} [/itex]