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Confusion Field Tensor and derivation of Maxwell's equations

  1. Aug 29, 2012 #1
    Hi as I'm reading http://www.maths.tcd.ie/~cblair/notes/432.pdf at page 13 I see that he states that the covariant and contravariant field tensors are different. But how can that be? Aren't they related by

    [tex] F_{\mu \nu} = \eta_{\nu \nu'} \eta_{\mu \mu '} F^{\mu ' \nu '} ?[/tex]

    and is not the product of the metric tensor eta with it self the identity? I view this as a matrix product

    [tex] F = \eta \eta F'[/tex]

    and writing out either η or the metric g it seems like their product with eachother are the identity such that


    [tex] F=\eta \eta F' = I F'.[/tex]

    Where is my reasoning wrong?
    At page 14 he derived two of the maxwell equations from a lagrangian. But what about the other two? The author just states them. Are not these derivable from a lagrangian?
     
  2. jcsd
  3. Aug 29, 2012 #2
    This is not the product of two metric tensors. In the definition of the contravariant field tensor appears the product:
    $$\eta_{\nu\nu'}\eta_{\mu\mu'}$$
    while the product of two metric tensors should be (with explicit components):
    $$\left(\eta\eta\right)_{\nu\mu}=\eta_{\nu\nu'} \eta_{\nu'\mu}=\delta_{\nu\mu}$$
    just as any other matrix product.
     
  4. Aug 29, 2012 #3
    So I can't compute it using matrix products? How would you suggest computing it?
    Writing out the components one by one using the relation?
     
  5. Aug 29, 2012 #4
    If your problem is to compute $$F^{\mu\nu}$$ you can do it by components remembering that it is antisymmetric so you have to compute just 6 of them. Otherwise, if you want to calculate the contravariant tensor instead of the covariant one the problem is a very easy one as you just have to low the indexes and remember that
    $$B^i=-B_i \;\;\mbox{ and }\;\; E^i=-E_i$$

    I hope to remember well :tongue2:
     
  6. Aug 29, 2012 #5

    That does not seem right from

    http://www.maths.tcd.ie/~cblair/notes/432.pdf

    where only the E's not the B's has changed sign.
     
  7. Aug 29, 2012 #6
    I'm sorry you are right. You have to change the sign only to E's components. B's components don't do that because both the gradient and the A field gain a - sign, so the overall change is a + sign. I'm sorry :biggrin:
     
  8. Aug 29, 2012 #7

    Bill_K

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    Science Advisor

    In order for a tensor expression to be writeable as a matrix product, the contractions must occur between adjacent indices. That is,

    Cik = AijBjk

    can be written as a matrix product C = AB. In your case,

    Fμν = ημμ'ηνν'Fμ'ν'

    you have to rearrange the factors:

    Fμν = ημμ'Fμ'ν'ην'ν

    which is then a matrix product

    F = ηF'η
     
    Last edited: Aug 29, 2012
  9. Aug 29, 2012 #8
    Ah, thanks! That was a revelation :)
     
  10. Aug 29, 2012 #9
    The other two Maxwell equations, the homogenous equations, come from the Bianchi Identity that the field tensor must obey.
     
  11. Aug 29, 2012 #10
    But shouldn't all equations of motions be deriable from the action? Why are the homogeneous equations left out? And how is this identity derived?
     
  12. Aug 29, 2012 #11

    haushofer

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    Science Advisor

    It is an identity. In form notation one has F=dA, and so dF=0 because d^2=0.

    Identities are not equations of motion; they are satisfied identically. In this case because partial derivatives commute. I don't need an action to know that.
     
  13. Aug 29, 2012 #12
    The Field tensor which is the simplest, positive definite, topologically non-trivial gauge invariant object, is defined via the commutator of of two covariant derivatives acting on an object (see Weinberg II first chapter). Thus the field tensor obeys the bianchi identity by construction.

    There are geometric ways of saying this but I am not as familiar with that.
     
  14. Aug 31, 2012 #13
    It is perhaps more simply expressed in vector calculus notation. Once we have postulated the 4-potential [itex]A^\mu = (\phi, \textbf{A})[/itex] with [itex]\textbf{B} = \nabla \times \textbf{A}[/itex] and [itex]\textbf{E} = - \nabla \phi \ - \ \partial\textbf{A}/ \partial t[/itex], then

    [itex]\nabla . \textbf{B} \ = \ \nabla . \nabla \times \textbf{A} \equiv 0[/itex]

    and

    [itex]\nabla \times \textbf{E} \ = \ \nabla \times (- \nabla \phi - \partial\textbf{A}/ \partial t) \ = \ - \nabla \times (\nabla \phi) - \partial(\nabla \times \textbf{A})/ \partial t \ = \ - \partial\textbf{B}/ \partial t[/itex], because [itex]\nabla \times (\nabla F) \equiv \textbf{0}[/itex] for any scalar field F.

    The only condition required by above is that the partial derivative operators for the four dimensions commute with one another, ie [itex]\partial_\mu\partial_\nu \equiv \partial_\nu\partial_\mu[/itex].

    The identities [itex]\nabla . \nabla \times \textbf{V} \equiv 0[/itex] and [itex]\nabla \times (\nabla F) \equiv \textbf{0}[/itex] are easily proved by writing out these expressions in full with the individual coordinates of the vectors then simplifying the results. For example, the [itex]x[/itex] component of the latter is [itex]\partial_y \partial_z F - \partial_z \partial_y F \equiv 0[/itex].
     
    Last edited: Aug 31, 2012
  15. Aug 31, 2012 #14
    Thanks for all the brilliant answers! It's all sorted out now :)
     
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