Bianchi identity with F^ab F^cd

samedi 1 mars 2014

The Reimann curvature tensor has the following symmetry resulting from a Bianchi identity[tex]R_{abcd}+R_{acdb}+R_{adbc}=0[/tex]

The derivative of the electromagnetic field tensor also yields some of Maxwell's equations from a Bianchi identity[tex]\partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha } = 0[/tex]

Unfortunately, while I can do tensor manipulations at the index level and calculate results in some coordinate system, when articles start manipulating things index free as "forms" and deriving results, I get lost. So while I recognize the above as coming from Bianchi identities, it is only because I have heard them referred to as such, but beyond checking by hand (painfully to verify the above) I cannot actually "derive" them from some Bianchi identity.



Now the question:



Does [itex]F^{ab} F^{cd}[/itex] have some kind of index permuting symmetry? And if so, is it due to a Bianchi identity?

(Below, there is some evidence a result can be obtained with the Levi-Civita permutation symbol, so the answer may be yes, but I am unsure how to derive it.)



Also, while it may not be pertinent, this object does have some of the symmetries of the Riemann tensor.

Defining [tex]X^{abcd} = F^{ab} F^{cd}[/tex] it has the symmetries [tex]X^{abcd} = X^{cdab}[/tex] [tex]X^{abcd} = - X^{bacd} = - X^{abdc}[/tex]



I saw in a discussion on science2.0 with someone trying to derive [itex]X^{abcd} X_{bcda} [/itex] and they found

[tex]X^{abcd} X_{bcda} = \frac{1}{2}(F^{ab}F_{ab})^2 + \frac{1}{4}(F^{ab}G_{ab})^2[/tex]

Where [itex]G_{ab}[/itex] is the dual electromagnetic field tensor

[tex]G_{ab}=\frac{1}{2}\epsilon_{abcd}F^{cd}[/tex]



The way in which it was derived was pretty messy, but it looks okay to me. (If you are curious its in the comments of a blog post by Doug Sweetser who used to post here a lot. http://ift.tt/1ofkm1z ) I'm wondering if there is some Bianchi like identity that would make the above "obvious" somehow.





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