Help with Reducing an Equation into Jacobi Identity Form

[kreil]

Homework Statement

Reduce the equation \partial_\mu {*} F^{\mu \nu} = 0 into the following form of the Jacobi Identity:

\partial_\lambda F_{\mu \nu} + \partial_\mu F_{\lambda \nu} + \partial_\nu F_{\lambda \mu} = 0

The Attempt at a Solution

I can't figure out what the '*' is supposed to be. My first thought was that it was a typo and is meant to signify a dot product, but the partial derivative is not a vector, so I don't see how this could be the case.

At any rate, this problem seems straightforward but I could use some help getting started.

Thanks for your thoughts.

 

[tiny-tim]

hi kreil!

(have a curly d: ∂ and a mu and an nu: µν )

the clue is in the alteration of the position of the indices …

this is ∂_µ(*F^µν) ​

(see http://en.wikipedia.org/wiki/Hodge_dual" )

 

[kreil]

Thanks Tim, I figured it was an operator of some sort but my notes/book were not immediately helpful in figuring out what it was.

In a previous problem we used the following form for F,

F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu

I'm having trouble seeing how the dual operator works on this form of F. Assuming a more general form, is the following true?

*F^{\mu \nu} = \left ( F^\mu e_\mu + F^\nu e_\nu + F^\lambda e_\lambda \right ) e_\mu e_\nu e_\lambda = F_{\nu \lambda} + F_{\mu \lambda} + F_{\mu \nu}

 

[tiny-tim]

hi kreil!

i've never really grasped the index way of doing it

i always think of it this way …

i see F_µν as a wedge product E_xx∧t + E_yy∧t + E_zz∧t + B_xy∧z + B_yz∧x + B_zx∧y

then for "curl" i wedge-multiply by ∂∧ = ∂_xx∧ + ∂_yy∧ + ∂_zz∧ + ∂_tt∧,

and use the equivalence x∧y∧z → *t, y∧z∧t → *x, z∧x∧t → *y, x∧y∧t → *z

(the jacobi identity is written in triple-wedges x∧y∧z etc)

i seem to remember there's a fairly good description of this in (surprisingly) Misner Thorne and Wheeler "Gravitation"