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A Trying to understand covariant derivative on tensors

  1. Jun 7, 2017 #1
    Hello there,

    Recently I encountered a type of covariant derivative problem that I never before encountered:

    $$
    \nabla_\mu (k^\sigma \partial_\sigma l_\nu)
    $$

    My goal: to evaluate this term

    According to Carroll, the covariant derivative statisfies ##\nabla_\mu ({T^\lambda}_{\lambda \rho}) = {{\nabla(T)_\mu}^\lambda}_{\lambda \rho}\ \ (\dagger)## and also ## \nabla(T \otimes S) = (\nabla T)\otimes S + T \otimes (\nabla S) \ \ (\ddagger)##. We know that ##\partial \sigma## forms a basis, so therefore ##k^\sigma \partial_\sigma ## is a (1,0) tensor
    $$
    \nabla_\mu (\underbrace{k^\sigma \partial_\sigma}_{\equiv T} l_\nu) = \nabla_\mu (T l_\nu) \stackrel{(\dagger)}{=} \nabla(T l)_{\mu \nu} = [\nabla(T l)]_{\mu \nu} \stackrel{(\ddagger)}{=} [\nabla(T) \otimes l + T\otimes \nabla(l)]_{\mu \nu}
    $$
    But how do I continue from there? I want to distribute back the ## \mu \nu## and replace ## \otimes## with regular multiplication, but I dont know any algebra rules which applies, also my textbook doesn't help. I know that the tensors are multilinear, but its not quite what I need. How do I continue to evaluate ##\nabla_\mu (k^\sigma \partial_\sigma l_\nu)##?
     
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  3. Jun 7, 2017 #2

    Paul Colby

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    Is ##k^\sigma\partial_\sigma## a scalar operator when applied to the tensor ##l_\nu##? I would think not?
     
  4. Jun 7, 2017 #3
    Great, I don't know. I am trying to understand how this all works. Would you like to explain some more?
     
  5. Jun 7, 2017 #4

    Paul Colby

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    This is the part I get in trouble with. The operator, ##\partial_\sigma##, I read as regular partial differentiation. When applied to a scalar ##\partial_\sigma \phi## yields a 4-vector. In fact ##\partial_\sigma = \nabla_\sigma## when acting on a scalar. If I wrote ##\nabla_\nu(k^\sigma\nabla_\sigma)l_\mu## I would just distribute chain rule the way one normally does with covariant derivatives.

    ##\nabla_\mu(k^\sigma\nabla_\sigma)l_\nu = (\nabla_\mu k^\sigma)(\nabla_\sigma l_\nu)+k^\sigma (\nabla_\mu\nabla_\sigma l_\nu)##
    Now if I replace ##\nabla_\sigma = \partial_\sigma + \Gamma^._{\cdots}## where the ##\Gamma##'s are whatever that's needed to make things covariant, I could solve for the thing you are looking to evaluate.
     
  6. Jun 7, 2017 #5

    strangerep

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    The simplest way is to treat the argument as a (covariant) vector with index ##\nu## and apply the rules found in the "examples" section of this Wiki page. I.e., $$\nabla_\mu (k^\sigma \partial_\sigma l_\nu) ~=~\partial_\mu (k^\sigma \partial_\sigma l_\nu) ~-~ \Gamma^\alpha_{~\mu\nu} k^\sigma \partial_\sigma l_\alpha ~.$$If you're a masochist, you could in fact evaluate it using the Leibniz product rule as
    $$\nabla_\mu (k^\sigma \partial_\sigma l_\nu) ~=~ k^\sigma \nabla_\mu ( \partial_\sigma l_\nu) ~+~ (\nabla_\mu k^\sigma) ( \partial_\sigma l_\nu) ~=~ \dots$$ This is indeed equivalent to "treating the argument as a (covariant) vector with index ##\nu##".
    (I've just now double-checked this, but if you want see my work you'll have to show me your attempt first. :oldbiggrin:)

    If you're a truly psychotic masochist, you could even treat it as the covariant derivative of a 3-index quantity, provided you follow the rules carefully, and preserve ordering when using the Leibniz rule.

    HTH.
     
    Last edited: Jun 8, 2017
  7. Jun 7, 2017 #6

    Orodruin

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    Adding to this, I think it is worth pointing out that the argument is not a covariant vector. To make it covariant, you would need ##k^\sigma\nabla_\sigma l_\nu## instead. As written, I am very sceptical regarding the origins of the expression itself. Of course, it might be a term in a larger covariant expression.
     
    Last edited: Jun 8, 2017
  8. Jun 13, 2017 #7

    stevendaryl

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    There is a problem with the expression [itex]k^\sigma \partial_\sigma l_\nu[/itex]. That isn't actually a tensor. This can be seen in two different ways:
    1. [itex]l_\nu[/itex] is a one-form, and you can't apply a partial derivative to a one-form. Partial derivatives can only apply to scalar fields.
    2. If you think of [itex]\partial_\sigma[/itex] as a basis vector, then [itex]k^\sigma \partial_\sigma[/itex] is just the vector [itex]\vec{k}[/itex] written in terms of components. So the expression [itex]k^\sigma \partial_\sigma l_\nu[/itex] is a vector acting on a one-form. How can a vector act on a one-form? Well, typically using covariant (or directional) derivatives: [itex]k^\sigma \nabla_\sigma l_\nu[/itex] makes sense, but [itex]k^\sigma \partial_\sigma l_\nu[/itex] doesn't.
     
  9. Jun 13, 2017 #8

    haushofer

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    Shouldn't the title of this thread read "trying to understand covariant derivatives on non-tensors"? :P
     
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