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Tensors: switching between mixed and contravariant components

  1. Jun 30, 2014 #1
    I'm working on the electromagnetic stress-energy tensor and I've found this in a book by Landau-Lifshitz:

    [itex]
    T^{i}_{k} = -\frac{1}{4\pi} \frac{\partial A_{\ell}}{\partial x^{i}} F^{k\ell}+\frac{1}{16\pi}\delta^{k}_{i} F_{\ell m} F^{\ell m}
    [/itex]

    Becomes:

    [itex]
    T^{ik} = -\frac{1}{4\pi} \frac{\partial A^{\ell}}{\partial x_{i}} F^{k}_{\ell}+\frac{1}{16\pi}g^{ik} F_{\ell m} F^{\ell m}
    [/itex]

    I was wondering how this work? [itex]F^{ik}[/itex] is the electromagnetic field tensor, [itex]A_{\ell}[/itex] is the potential of the field.
     
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  3. Jun 30, 2014 #2

    Orodruin

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    Your first equation cannot be correct, you have switched the free covariant and free contravariant indices on the different sides of the equation. Let me assume that the LHS is instead ##T^k_i##. To obtain the second equation from the first, multiply on both sides with ##g^{in}##:
    $$
    T^{kn} \equiv g^{in} T^k_i =
    -\frac{1}{4\pi} g^{in} F^{k\ell} (\partial_i A_\ell)
    +\frac{1}{16\pi} g^{in}\delta^k_i F_{\ell m} F^{\ell m}
    $$
    Now use ##g^{in}## to raise the ##i## index and use ##A_ \ell B^\ell = A^\ell B_\ell## in the first term as well as ##g^{in} \delta^k_i = g^{kn}## in the second to rewrite this as
    $$
    T^{kn} =
    -\frac{1}{4\pi} F^{k}_{\phantom k \ell} (\partial^n A^\ell)
    +\frac{1}{16\pi} g^{kn} F_{\ell m} F^{\ell m}.
    $$
    Beware! Writing ##F^k_{\ell}## is dangerous. You must keep track of which index was lowered somehow since ##F## is anti-symmetric (which is why I have added the space in the subinex). Renaming ##n \to i## on both sides gives you the relation.

    Note that I have used the notation ##\partial_i = \frac{\partial}{\partial x^i}## and ##\partial^i = \frac{\partial}{\partial x_i}##.
     
  4. Jun 30, 2014 #3
    Ah yes sorry I made a typo, I should have read my post over more carefully.

    Right, I figured it had something to do with [itex]g^{in}[/itex] but I wasn't quite sure, I always forget that I can use dummy indices to multiply through with these things.

    I'm still not clear on how [itex]g^{in}[/itex] raises the [itex]i[/itex] index.

    I see how the [itex]\ell[/itex] index was swapped, but I'm not sure about the [itex]i[/itex] in the first term.

    Thank you for your help.

    Edit: Sorry about not keeping track with the mixed indices, I don't even understand them well enough to do that.
     
  5. Jun 30, 2014 #4

    Orodruin

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    This is essentially by definition. If we have a contravariant vector with components ##A^k##, then ##g_{ik} A^k## are the components of a covariant vector, we can call it ##A_i##. In the same way we can relate the components of a covariant vector with those of a contravariant one by ##g^{km}##. The procedure is consistent due to ##g^{km}g_{ik} = \delta^m_i##. The same goes for raising and lowering indices of higher order tensors.

    [Edit] For higher order tensors, you raise one individual index using one insertion of the metric. Since the indices generally are not equivalent, it is necessary to keep track of the index order in the original tensor unless it symmetric.
     
  6. Jun 30, 2014 #5
    Right but what does a mixed tensor even mean?
     
  7. Jun 30, 2014 #6

    Orodruin

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    A mixed tensor is a tensor which is not contravariant or covariant, i.e., its components has both contravariant and covariant indices. If you are familiar with the nomenclature "tensor of type ##(n,m)##", a mixed tensor is a tensor of type ##(n,m)##, where both ##n## and ##m \geq 1##. See also http://en.wikipedia.org/wiki/Mixed_tensor
     
  8. Jun 30, 2014 #7
    So under specific transformations it acts in different ways?
     
  9. Jun 30, 2014 #8

    Orodruin

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    Under any transformations, some indices of a mixed tensor will transform covariantly and others contravariantly.
     
  10. Jun 30, 2014 #9
    I'd just go read the Wikipedia articles on this but I have and they were slightly confusing.

    What is the importance of the distinction between the two?
     
  11. Jun 30, 2014 #10

    TSC

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    My first tensor book is Tensor Calculus by Kay. Very good for introduction. There are many which starts from even a simpler baseline.
    The second one is Synge and Schild.
    Try them!
    (note: isn't this suppose to be in the calculus page?)
     
  12. Jun 30, 2014 #11

    Matterwave

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    Thinking about it in terms of transformations of components, the components of a (n,m) tensor transform this way:

    $$T'^{~i...}_{~~~~n...}=T^{a...}_{~~~q...}\frac{\partial x'^i}{\partial x^a}...\frac{\partial x^q}{\partial x'^n}$$

    The contravariant indices transform with contravariant factors (primes above) and covariant indices transform with covariant factors (primes below).

    Thinking about it in more geometric terms, a (n,m) tensor is a multi-linear function of n one-forms and m vectors into real numbers. It accepts n one-forms as arguments and m vectors as arguments and returns one single number:

    $$T(\tilde{\omega},...,\vec{q},...)=T^{i...}_{~~n...}\omega_i...q^n... $$
     
  13. Jun 30, 2014 #12
    Right so thanks to TSC for the recommendation.

    Matterwave:

    I have no idea what a one form is. I feel like in my 3rd year of physics I should know?
     
  14. Jun 30, 2014 #13

    Matterwave

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    A one form is simply the dual to vectors. It is a linear function of vectors into numbers. In the old language it is called a covariant vector.
     
  15. Jul 1, 2014 #14
    Do you have any book recommendations on this subject? I don't know what dual means really.
     
  16. Jul 1, 2014 #15

    TSC

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    There are 2 approaches to learning tensors. The component approach and the component-free approach.
    In general for physics and engineering students, the component approach is easier to start with. The one-form is the component-free approach.
    The 2 books I recommend is the component approach. It is better, I think, for your purpose.(covariant electromagnetism)
     
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