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Einstien notation question

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Is [tex] \partial ^ {\nu} F_{\mu \nu} + m^2 A_{\mu} [/tex] the same expression as [tex] \partial _{\nu} F^{\mu \nu} + m^2 A^{\mu} [/tex]?

    What form of the metric do I need to hit them with to show that?

    Also is

    [tex] \partial ^{\nu} \partial_{\mu} A_{\nu} = \partial_{\mu} \partial ^{\nu} A_{\nu} [/tex]

    ?

    2. Relevant equations



    3. The attempt at a solution

    EDIT: I changed the second expression so it makes sense.
     
    Last edited: Oct 12, 2007
  2. jcsd
  3. Oct 12, 2007 #2

    nrqed

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    No. The second expression does not make sense to start with since in the F term the two mu are downstairs. Did you mean F with with indices upstairs? If so, the tewo expressions could still not be equal since one would be a covariant vector and the other one a contravariant vector.



    Two tensors may only be equal if they have the same number of covariant and contravariant indices.
     
  4. Oct 12, 2007 #3
    Yes. That's the danger of copying and pasting. I changed it so it makes sense.
     
  5. Oct 13, 2007 #4

    CompuChip

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    Ad 1) You wrote down the covariant and contravariant form of the same expression. You can convert between them by just writing metric tensors in between, e.g.
    [tex] m^2 A^{\mu} = m^2 g^{\mu\nu} A_\nu [/tex]
    and similar for the first term. But the expressions are not the same, so do not expect the metric to drop out in the end.

    Ad 2) Your question is if
    [tex] g^{\nu\lambda} \partial_{\lambda} \partial_{\mu} = \partial_{\mu} g^{\nu\lambda} \partial_{\lambda} A_{\nu},[/tex]
    that is, do partial derivatives commute?
    AFAIK they do, and the identity at least holds when the metric is constant (that is, [itex]g_{\mu\nu}[/itex] does not explicitly depend on the coordinate w.r.t which you are differentiating).
     
    Last edited: Oct 13, 2007
  6. Oct 13, 2007 #5
    But if they are both set equal to 0 then the metric will drop out and they are indeed precisely the same equations, right?
     
  7. Oct 13, 2007 #6

    nrqed

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    Ah, if the expression is equal to zero, then it's a special case. In general, one cannot say that a covariant expression is equal to its contravariant equivalent. But if the covariant expression is equal to zero then its contravariant counterpart will be equal to zero, indeed.
     
    Last edited: Oct 13, 2007
  8. Oct 13, 2007 #7
    OK. Thanks.
     
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