# Geometric Version of Maxwell Equation related to Tensor Dual

## Homework Statement

From Misner, Thorne and Wheeler's text Gravitation (MTW), exercise 3.15:
Show that, if F is the EM field tensor, then $\nabla \cdot *F$ is a geometric, frame-independent version of the Maxwell equation $F_{\alpha,\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0$
(and similarly for the current equations)
$*F$ is the dual of $F$

## Homework Equations

The Maxwell equation $F_{\alpha,\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0$
The dual of a vector $J$ is defined $*J_{\alpha\beta\gamma}=J^\mu \epsilon_{\mu\alpha\beta\gamma}$
where $\epsilon$ is the Levi-Civita symbol.
The dual of a second-rank antisymmetric tensor F is defined by $*F_{\alpha\beta}=F^{\mu\nu}\epsilon_{\mu\nu\alpha\beta}$
The dual of a third-rank antisymmetric tensor $B$ is defined by $*B_\alpha=B^{\lambda\mu\nu}\epsilon_{\lambda\mu\nu\alpha}$

## The Attempt at a Solution

$(\nabla \cdot *F)_\beta=(*F)_{\alpha\beta,\alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}$
$=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu,\alpha}$
In the second expression, the first $\alpha$ should be in the upper position. In the last expression, the final $,\alpha$ should be in the lower position. But I don't know how to get Physics Forum to accept that. On my local machine I declare \usepackage{tensor} and say F\indices{...} but that doesn't seem to work on the forum.

I know that the Maxwell equation as given above is actually 64 equations, mostly redundant. I In MTW $F$ has two indices for the components, which leads me to think $*F$ is second-rank. But when I take the divergence of the second-rank tensor I get a vector as shown in my attempt at a solution. This seems to yield four equations rather than 64.

In equation (3.3) of MTW, it has the expression $F(u)$ which makes me think F is a first-rank tensor. I am unclear what is the rank of $F$ in this problem.

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robphy
Homework Helper
Gold Member
$(\nabla \cdot *F)_\beta=(*F)^{\alpha}{}_{\beta,\alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}$
Code:
(\nabla \cdot *F)_\beta=(*F)_{\alpha\beta,     \alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}
(\nabla \cdot *F)_\beta=(*F)^{\alpha}{}_{\beta,\alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}
$=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu}{}_{,\alpha}$
Code:
=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu,\alpha}
=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu}{}_{,\alpha}
Since $F_{ab}=\nabla_a A_b - \nabla_b A_a$, it's a two-form.
In n-dimensional space[time], its Hodge dual $*F$ has (n-2) indices. So, in (3+1)-dimensions, its Hodge dual also has 2 indices.

In Eq. (3.3), $d{\bf p}/{dt}=e{\bf F}({\bf u})$ is a 4-vector equation for the Lorentz Force.
In abstract index notation, this is $d{\bf p}_b/{dt}=e{\bf F}_{ab}{\bf u}^a$

Look at Box 3.3 on "Index Gymnastics" on page 85.

Thanks for the help on Latex coding of tensor indices. Much appreciated!

Still wondering, though, about $\nabla \cdot *F$. If $*F$ has two indices, shouldn't $\nabla \cdot *F$ have just one index? I seem to need three.

nrqed
Homework Helper
Gold Member
Thanks for the help on Latex coding of tensor indices. Much appreciated!

Still wondering, though, about $\nabla \cdot *F$. If $*F$ has two indices, shouldn't $\nabla \cdot *F$ have just one index? I seem to need three.
What do you mean by "I seem to need three"? In your expression given in your original post in the "attempt of a solution", what you wrote has only one uncontracted index (the alpha's are contracted).

The problem asks me to demonstrate that it reduces to maxwells equation $F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha}=0$.

That has three free indices.

robphy
Homework Helper
Gold Member
The problem asks me to demonstrate that it reduces to maxwells equation $F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha}=0$.

That has three free indices.
Have you seen "Eq. 3.49b"
MTW said:
Show that Maxwell's "magnetic" equations
$F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha}=0$
can be rewritten in the form
$F_{[\alpha\beta,\gamma]}=0$.
Have you seen Exercise 3.13 on the Levi-Civita tensor ?
https://en.wikipedia.org/wiki/Levi-Civita_symbol

In n-dimensions, the dual of a 3-form (with 3 indices) has (n-3) indices.

(Probably overkill for this problem... but may be a useful reference: https://www.physicsforums.com/threads/algebraic-proofs-of-levi-civita-symbol-identities.919269/ )

Yes, I worked exercise 3.13.

My problem is that $\nabla \cdot *F=0$ seems to produce a formula with only one free index, whereas the exercise asks me to demonstrate that it reduces to an equation with three indices, i.e. $F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0$

I seem to be missing your point, I am sorry.

robphy
Can you form the dual of $F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}$?
$*(F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha})^\mu=\frac{1}{2}F_{\alpha\beta,\gamma}\epsilon^{\alpha\beta\gamma\mu}$
$=-\frac{1}{2}F_{\alpha\beta,\gamma}\epsilon^{\alpha\beta\mu\gamma}=-\frac{1}{2}\frac{\partial}{\partial x^\gamma}F_{\alpha\beta}\epsilon^{\alpha\beta\mu\gamma}=-\frac{1}{2}(\nabla \cdot *F)^\mu=0$