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Lie Derivatives

  1. Mar 19, 2010 #1
    I've been trying to get a grasp on Lie Derivatives. I understand that we can represent a lie derivative acting on a vector as a commutator. What do I do when I act a lie derivative on a tensor? Can I still just write out the commutator?
     
  2. jcsd
  3. Mar 20, 2010 #2

    haushofer

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    I don't think so, because it only makes sense to take commutators of vector fields. The Lie derivative can be calculated in a "coordinate free way" using pullbacks/pushforwards, or in a coordinate way by calculating

    [tex]
    T'(x) - T(x)
    [/tex]

    induced by the infinitesimal coordinate transformation

    [tex]
    x \rightarrow x + \xi
    [/tex]
     
  4. Mar 20, 2010 #3
    Thanks for your reply! Can you give an example of either of these methods?
     
  5. Mar 21, 2010 #4

    haushofer

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    I can give the rough idea :) Let's pick a tensor T

    [tex]
    T = T_{\mu\nu} dx^{\mu} \otimes dx^{\nu}
    [/tex]

    and perform the coordinate transformation

    [tex]
    x^{\lambda} \rightarrow x^{\lambda'} = x^{\lambda} - \xi^{\lambda}
    [/tex]

    We know that for our T we have

    [tex]
    T_{\mu' \nu '}(x') = \frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\nu }}{\partial x^{\nu '}}T_{\mu\nu}(x) \ \ \ \ \ \ \ \ (1)
    [/tex]

    For our coordinate transformation we have that

    [tex]
    \frac{\partial x^{\mu}}{\partial x^{\mu '}} = \delta^{\mu}_{\mu '} + \frac{\partial }{\partial x^{\mu '}}\xi^{\mu}
    [/tex]

    You can plug this into (1). But the Lie derivative of our tensor T with respect to [itex]\xi[/itex] is given by (in components)

    [tex]
    T_{\mu ' \nu '}(x) - T_{\mu\nu}(x)
    [/tex]

    You can calculate this by performing a Taylor expansion,

    [tex]
    T(x) = T(x' + \xi) = T(x') + \xi^{\mu}\partial_{\mu}T(x')
    [/tex]

    and recognizing that, to first order in xi,

    [tex]
    \xi^{\mu}\partial_{\mu}T(x') = \xi^{\mu}\partial_{\mu}T(x)
    [/tex]

    It's been a while that I did these kind of calculations explicitly, but I think you should be able to do it. It maybe helps to first do it for a scalar field, then for a vector or covector, and so on. Notice that the signs depend on how you define your coordinate transformation with either a plus or minus sign! For a general tensor T there always will be a leading term

    [tex]
    \xi^{\mu}\partial_{\mu}T(x)
    [/tex]

    The other terms will be due to the "pulling back of the coordinate", but that will become clear if you read something about pushforwards and pullbacks; the GR text by Carroll or Nakahara's geometry book are very useful. Hope this helps :)
     
  6. Mar 21, 2010 #5

    haushofer

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    By the way, Nakahara also calculates explicitly a Lie derivative of I believe a vector field in the "coordinate free formalism", using pullbacks and pushforwards, so you can check his geometry book to see how that goes.
     
  7. Mar 25, 2010 #6

    haushofer

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    Did you manage to do the calculation? :)
     
  8. Mar 25, 2010 #7
    I did, but I did so by defining a new commutator. Turns out that you actually can do all lie derivatives as commutators so long as you properly define what it means for a tensor to act on a vector (I proved this and, if people are interested, I can post something more explicit later.) After you do that, you can write out VT-TV.
     
  9. Mar 28, 2010 #8

    haushofer

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    Mmm, I must say this doesn't ring a bell instantaneously, so I would be quite curious how one does that :)
     
  10. Apr 1, 2010 #9
    Sorry it's taken me a while to respond. I've been pretty busy with grad school. At any rate, we want to define something like:

    [tex]L_V T=[V,T][/tex]

    or

    [tex]L_V T=VT-TV[/tex]

    for all tensors T and any vector V where L is the lie differential operator. Now, ordinarily, the lie derivative acting on a tensor is defined to be:

    [tex]L_V T=V^\mu \partial_\mu T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-

    (\partial_\lambda V^\mu_1)T^{\lambda \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-

    (\partial_\lambda V^\mu_2)T^{\mu_1 \lambda ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-...

    +(\partial_\nu_1 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\lambda \nu_2 ... \nu_l}

    +(\partial_\nu_2 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \lambda ... \nu_l}+...
    [/tex]

    Obviously, [tex]V^\mu \partial_\mu T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}[/tex] is just the action of V on T. Thus, our job is half done -- we have a natural way to define VT. What about the action of the tensor on the vector -- TV? Well, we can just define:

    [tex]TV=

    (\partial_\lambda V^\mu_1)T^{\lambda \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}+

    (\partial_\lambda V^\mu_2)T^{\mu_1 \lambda ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}+...

    -(\partial_\nu_1 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\lambda \nu_2 ... \nu_l}

    -(\partial_\nu_2 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \lambda ... \nu_l}-...
    [/tex]

    And now we can write:

    [tex]L_V T=VT-TV[/tex]

    With a little computation, you can show that this satisfies all of the desirable properties.

    [Note: the only way that I could make the indices appear in the right order was to include periods between the upstairs indices and the downstairs indices. If any one knows how to make the indices go in the right order without doing that, let me know; it would be really helpful.]
     
    Last edited: Apr 1, 2010
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