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Commutator of two covariant derivatives

  1. Aug 9, 2015 #1
    Hello all,

    I'm trying to calculate a commutator of two covariant derivatives, as it was done in Caroll, on page 122. The problem is, I don't get the terms he does :-/

    If ##\nabla_{\mu}, \nabla_{\nu}## denote two covariant derivatives and ##V^{\rho}## is a vector field, i need to compute ##[\nabla_{\mu}, \nabla_{\nu}]V^{\rho}##. Covariant derivative is defined as

    [tex]\nabla_{\mu} = \partial_{\mu} + \Gamma^{\nu}_{\mu\lambda}[/tex]

    Putting it into the definition of the commutator, one can write

    [tex]
    \begin{align}
    [\nabla_{\mu}, \nabla_{\nu}]V^{\rho} &= \nabla_{\mu} \nabla_{\nu} V^{\rho} - \nabla_{\nu}\nabla_{\mu} V^{\rho} \nonumber \\ &=\partial_{\mu} (\nabla_{\nu} V^{\rho})+\Gamma^{\rho}_{\mu\sigma}(\nabla_{\nu}V^{\sigma})-\Gamma^{\lambda}_{\mu\nu}\nabla_{\lambda}V^{\rho}+(\mu \leftrightarrow \nu) \nonumber \\ &=\ldots \nonumber
    \end{align}
    [/tex]

    What gives me problems is the 3rd term in the 2nd row. I don't know where this third term comes from. The expansion is even more problematic, Caroll expands these three terms into 7:

    [tex]\ldots\partial_{\mu}\partial_{\nu}V^{\rho}+(\partial_{\mu}\Gamma^{\rho}_{\nu\sigma})V^{\sigma}+\Gamma^{\rho}{\gamma\sigma}\partial_{\mu}V^{\rho}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda}V^{\rho}-\Gamma^{\lambda}_{\mu\nu}\Gamma^{\rho}_{\lambda\sigma}V^{\sigma}+\Gamma^{\rho}_{\mu\sigma}\partial_{\nu}V^{\sigma}+\Gamma^{\rho}_{\mu\sigma}V^{\lambda}-(\mu \leftrightarrow \nu)\ldots[/tex]

    Any ideas would be very much appreciative.

    :-)
     
  2. jcsd
  3. Aug 9, 2015 #2

    Orodruin

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    The action of the first covariant derivative is on a type (1,1) tensor. As such, you must include one term with a Christoffel symbol for both the covariant and the contravariant index of that tensor.

    This seems to be part of your problem. Your definition of the contravariant derivative both contains too many indices on one side and seems like it is coming from the action on a contravariant vector only. You need to look up the general definition.
     
  4. Aug 9, 2015 #3
    Caroll defines the covariant derivative as follows:

    [tex]\nabla_\mu V^{\nu} = \partial_{\mu}V^{\nu}+\Gamma^{\nu}_{\mu\sigma}V^{\sigma}[/tex]

    (formula 3.2 on page 93)

    :when i wrote the formula in my first post, i omitted the vector field, just gave a definition on what ##\nabla_{\mu}## is. that's why i don't really understand your comment. can you clarify a bit further?
     
  5. Aug 9, 2015 #4

    Orodruin

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    You will note that in this definition the indices are contracted between the vector field and the Christoffel symbol. It is therefore not possible to simply remove the vector field from the definition.

    Which part do you have problems with? The fact that ##\nabla_\mu V^\nu## is a type (1,1) tensor or how the covariant derivative acts on arbitrary tensors? The latter should be defined a little bit later in Carroll.
     
  6. Aug 9, 2015 #5
    true, my mistake about that one.

    now when i reread the first comment again, i see your point :-) thanks. one more question: are indicies in the 2nd and 3rd term correct? now that i look at it, it seems they aren't but i'm not sure.
     
  7. Aug 9, 2015 #6

    Orodruin

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    Yes.
     
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