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"

 

[paranormal]

Hello, it seems like the equation you are trying to reduce is a form of the Maxwell's equations in electromagnetism. The '*' symbol could possibly represent a cross product, but as you mentioned, the partial derivative is not a vector so it doesn't make sense in this context.

To reduce the equation into the Jacobi identity form, we can start by expanding the partial derivative on the left side of the equation using the product rule:

\partial_\mu {*} F^{\mu \nu} = \partial_\mu (F^{\mu \nu}) + F^{\mu \nu} \partial_\mu

Since the equation is equal to zero, we can set the first term on the right side to zero, leaving us with:

F^{\mu \nu} \partial_\mu = 0

Now, we can use the definition of the Jacobi identity, which states that for any three vectors A, B, and C, the identity can be written as:

A \cdot (B \times C) + B \cdot (C \times A) + C \cdot (A \times B) = 0

In our case, we can replace the vectors A, B, and C with the partial derivatives and the field tensor F, giving us:

\partial_\lambda (F^{\mu \nu}) \cdot (\partial_\mu \times \partial_\nu) + \partial_\mu (F^{\mu \nu}) \cdot (\partial_\nu \times \partial_\lambda) + \partial_\nu (F^{\mu \nu}) \cdot (\partial_\lambda \times \partial_\mu) = 0

Now, we can recognize that the cross product of two partial derivatives is equal to the negative of the cross product of the two derivatives in the opposite order. This allows us to simplify the equation to:

\partial_\lambda (F^{\mu \nu}) \cdot (\partial_\nu \times \partial_\mu) - \partial_\mu (F^{\mu \nu}) \cdot (\partial_\nu \times \partial_\lambda) + \partial_\nu (F^{\mu \nu}) \cdot (\partial_\lambda \times \partial_\mu) = 0

Finally, using the definition of the field tensor, we can write this equation as:

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