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

  1. May 1, 2013 #1
    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.
     
  2. jcsd
  3. May 1, 2013 #2

    George Jones

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    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.
     
  4. May 1, 2013 #3
    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]?
     
  5. May 1, 2013 #4

    George Jones

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    Use the product rule for partial derivatives.

    You don't have to remove it; you need to relabel indices in the last term on the right side.
     
  6. May 1, 2013 #5
    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 see what you mean, however, without knowing the partial derivative I don't know what I can do with this yet.
     
  7. May 1, 2013 #6

    George Jones

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    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].
     
  8. May 1, 2013 #7
    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]
     
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