# Index Concordia

## Maxwell wearing differential forms

Posted in Classical Electrodynamics, Geometry, Relativity by Index Guy on July 16, 2007

[The title of this post is not intended to offend the memory of the great Scottish physicist J. C. Maxwell.]

Let us try to write the four Maxwell equations using the language of differential forms, exterior differentiation and the Hodge star duality operation. We work over the flat four-dimensional Minkowski spacetime with flat metric $\eta_{\mu\nu}$.

We start with some definitions. The exterior derivative $\text{d}$ of a $p$-form $A$ is a $(p + 1)$-form with components

$(\text{d}A)\displaystyle_{\mu_{1} \mu_{2} ... \mu_{p+1}} = (p+1)\partial_{[\mu_{1}}A_{\mu_{2}...\mu_{p+1}]}.$

The square brackets denote anti-symmetrization. Let us also agree that the Hodge star duality operator $*$ acts on a $p$-form and makes it a $(n - p)$-form, with $n$ being the dimension of our manifold. The components of the new form are given by

$(*A)_{\mu_{1}...\mu_{n-p}} = \displaystyle\frac{1}{p!} \epsilon^{\nu_{1}...\nu_{p}}_{\mu_{1}...\mu_{n - p}} A_{\nu_{1}...\nu{p}}.$

The $\epsilon$ tensor is the completely Levi-Civita antisymmetric symbol times the square root of the absolute value of the determinant of the spacetime metric tensor. In Minkowski spacetime this factor is unity and the Levi-Civita symbol is the same as the Levi-Civita tensor.

Recall that the four Maxwell’s can be written with index notation as

$\partial_{\alpha}F^{\alpha \beta} = J^{\beta}$ and $\partial_{[\alpha}F_{\beta \gamma]} = 0.$

In this notation we combine the electric and magnetic fields with the field strength

$F_{\alpha \beta} = \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha},$

and the electric charge and current densities go into the four-current $J^{\beta}$. It is no surprise that electromagnetic field strength can be written as a 2-form. This is true since the homogeneous Maxwell equations can be written as the exterior derivative of the field strength. The vanishing of this derivative in Minkowski space implies that the field strength is the exterior derivative of some 1-form $A$. This 1-form is called the electromagnetic potential. Similarly one can write the four-current as a 1-form. So we have so far that

$\text{d}F = 0 \implies F = \text{d}A.$

This gives us the first set of the Maxwell equations written with differential forms. For the second set we use the Hodge star to calculate $*F$, the dual field strength. The first thing we can say is that $*F$ will be a 2-form. If we were to exterior-differentiate $*F$ we would obtain a 3-form. Notice also that just like the four-potential $A^{\mu}$ can be written as 1-form, we can also take the four-current $J^{\mu}$ and express it as a 1-form. The Hodge dual of this 1-form will be then the 3-form $*J$. So now we know about two 3-forms (besides $*A$), we could aim at getting an expression that involves them. Of course, since we know that the remaining Maxwell equations have the source terms, we can predict that it should have the dual of the current 1-form. We postulate that the Maxwell equations written in the language of differential forms look like

$\text{d}F = 0$ and $\text{d}(*F) = *J.$

Now we proof that the second of these terms gives the two inhomogeneous Maxwell equations. We start by writing explicitly the components of the differential forms zoo defined above. But first let us name

$M = *J,$ $G = *F$ and $B = \text{d}G.$

We now need to show that $B = M$. This forms have components given by

$M_{\alpha \beta \gamma} = \epsilon_{\thickspace \medspace \alpha \beta \gamma}^{\nu}J_{\nu}$, $G_{\mu \nu} = \frac{1}{2}\epsilon_{\quad \mu \nu}^{\alpha \beta}F_{\alpha \beta}$ and $B_{\alpha \beta \gamma} = 3\thinspace\partial_{[\alpha}G_{\beta \gamma ]}.$

Let us perform some index gymnastics. We start by lowering the indices on the Levi-Civita tensor,

$M_{\alpha \beta \gamma} = \epsilon_{\nu \alpha \beta \gamma}J^{\nu}$ and $G_{\mu \nu} = \frac{1}{2}\epsilon_{\alpha \beta \mu \nu}F^{\alpha \beta}.$

Now let us write explicitly the anti-symmetrized $B$ tensor.

$2\thinspace B_{\alpha \beta \gamma} = \partial_{\alpha}G_{\beta \gamma} - \partial_{\alpha}G_{\gamma \beta} + \partial_{\gamma}G_{\alpha \beta} - \partial_{\gamma}G_{\beta \alpha} + \partial_{\beta}G_{\gamma \alpha} - \partial_{\beta}G_{\alpha \gamma}.$

So what is the anti-symmetrization of two $G$ tensors? This is just

$G_{\mu \nu} - G_{\nu \mu}= \frac{1}{2}\left[ \epsilon_{\quad \mu \nu}^{\alpha \beta} - \epsilon_{\quad \nu \mu}^{\alpha \beta}\right]F_{\alpha \beta} = 2\thinspace G_{\mu \nu}.$

With this we can write

$B_{\alpha \beta \gamma} = \partial_{\alpha}G_{\beta \gamma} + \partial_{\beta}G_{ \gamma \alpha} + \partial_{\gamma}G_{\alpha\beta }.$

Now we write the $G$ tensors in terms of the field strength,

$2 \thinspace B_{\alpha \beta \gamma} = \epsilon_{\mu \nu \beta \gamma } \partial_{\alpha} F^{\mu \nu} + \epsilon_{\mu \nu \gamma \alpha}\partial_{\beta} F^{\mu \nu} + \epsilon_{\mu \nu \alpha \beta }\partial_{\gamma} F^{\mu \nu}.$

Now we contract on both sides with a Levi-Civita tensor. We will do this term by term, so my poor readers do not get overwhelmed with details (or maybe for my sake…). Each term is getting two indices contracted, for example the last one gives

$\epsilon^{\alpha \beta \gamma \sigma}\epsilon_{\alpha \beta \mu \nu}\partial_{\gamma}F^{\mu \nu} = 4 (\partial_{\mu}F^{\mu \sigma} - \partial_{\nu}F^{\sigma \nu}) = 8 \thinspace \partial_{\mu}F^{\mu \sigma} = 8 \thinspace J^{\sigma}.$

In the last step we have identified and used Maxwell’s equations to introduce the current four-vector. We have also ignore the minus-one factor that comes from me metric’s signature. This factor will cancel in the end. The other two terms are the same, so we have

$\epsilon^{\alpha \beta \gamma \sigma}B_{\alpha \beta \gamma} = 12\thinspace J^{\sigma}.$

Finally, we contract again with another Levi-Civita tensor. The factor of 12 will cancel, and it will give us the components of the $M$ tensor. Thus this shows that $B = M$. This shows that Maxwell’s equation can be written as

$\text{d}F = 0$ and $\text{d}(*F) = *J.$

This still gives the same physics as the Maxwell equations when written in vector form or in four-vector notation (or even quaternions, just like Maxwell originally wrote them). We can argue that differential form notation and covariant notation reduce the number of equations from 4 to 2. The truth is that the 4 equations still remain, since one has to remember how to write the field strength in component form to get the more-measurement-friendly electric and magnetic fields. This is just a notion of what is practical; if we adopt the concept of the field strength as the only entity and then say that we can measure certain components of it, then it is perfectly fine. So geometry killed the electric and magnetic fields.