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Quick question on notation.

  1. Mar 10, 2007 #1

    cristo

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    Does anyone know what this means: [tex]\nabla_{[a}F_{bc]}[/tex]? I know that [tex]F_{(ab;c)}=\frac{1}{3}(F_{ab;c}+F_{bc;a}+F_{ca;b})[/tex], and presume that the first expression can be written thus [tex]F_{[bc;a]}[/tex], but am not sure what it means!

    Can anyone help?
     
  2. jcsd
  3. Mar 10, 2007 #2
    Any time you see square brackets around indices it means that you permute the indices, with even permutations receiving a plus sign and odd permutations receiving a minus sign. Thus,

    [tex]\nabla_{[a}F_{bc]} = \frac{1}{3!}\sum_{\pi\in S(3)}\textrm{sign}(\pi)\nabla_{\pi(a)}F_{\pi(b)\pi(c)}[/tex]

    where [itex]S(3)[/itex] is the symmetric group of order three, [itex]\pi[/itex] is a permutation, and sign[itex](\pi)[/itex] equals one for an even permutation of the elements and minus one for an odd permutation of the elements. In your case you can expand out the above definition to obtain

    [tex]\nabla_{[a}F_{bc]} = \frac{1}{3!}(\nabla_aF_{bc} + \nabla_bF_{ca} + \nabla_cF_{ab} - \nabla_aF_{cb} - \nabla_bF_{ac} - \nabla_cF_{ba})[/tex]
     
    Last edited: Mar 10, 2007
  4. Mar 10, 2007 #3

    cristo

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    Ahh ok, that makes sense. Thanks for the quick reply! In this case F is the electromagnetic field tensor, and so is antisymmetric. Would I be right in assuming that in this case the equation becomes [tex]\nabla_{[a}F_{bc]} = \frac{1}{3}\left(\nabla_aF_{bc} + \nabla_bF_{ca} + \nabla_cF_{ab}\right)[/tex]
     
  5. Mar 10, 2007 #4
    Yes. If [itex]F_{ab}[/itex] are taken as the components of the Maxwell tensor then [itex]\nabla_{[a}F_{bc]}=0[/itex] is essentially the Bianchi identity for the electromagnetic field.

    Another way to think about it is to notice that [itex]\nabla_{[a}F_{bc]}=0[/itex] is precisely the same statement as [itex]d\mathbf{F}=0[/itex] where [itex]\mathbf{F}=d\mathbf{A}[/itex] is the Maxwell two-form. The identity [itex]d\mathbf{F}=0[/itex] is guaranteed since for any [itex]p[/itex]-form field [itex]\mathbf{A}[/itex] one has [itex]d\cdot(d\mathbf{A})=0[/itex].
     
    Last edited: Mar 10, 2007
  6. Mar 10, 2007 #5

    cristo

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    That's a good way to think about it. Thanks a lot for your help, shoehorn!
     
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