# The Unapologetic Mathematician

## Maxwell’s Equations in Differential Forms

To this point, we’ve mostly followed a standard approach to classical electromagnetism, and nothing I’ve said should be all that new to a former physics major, although at some points we’ve infused more mathematical rigor than is typical. But now I want to go in a different direction.

Starting again with Maxwell’s equations, we see all these divergences and curls which, though familiar to many, are really heavy-duty equipment. In particular, they rely on the Riemannian structure on $\mathbb{R}^3$. We want to strip this away to find something that works without this assumption, and as a first step we’ll flip things over into differential forms.

So let’s say that the magnetic field $B$ corresponds to a $1$-form $\beta$, while the electric field $E$ corresponds to a $1$-form $\epsilon$. To avoid confusion between $\epsilon$ and the electric constant $\epsilon_0$, let’s also replace some of our constants with the speed of light — $\epsilon_0\mu_0=\frac{1}{c^2}$. At the same time, we’ll replace $J$ with a $1$-form $\iota$. Now Maxwell’s equations look like:

\displaystyle\begin{aligned}*d*\epsilon&=\mu_0c^2\rho\\{}*d*\beta&=0\\{}*d\epsilon&=-\frac{\partial\beta}{\partial t}\\{}*d\beta&=\mu_0\iota+\frac{1}{c^2}\frac{\partial\epsilon}{\partial t}\end{aligned}

Now I want to juggle around some of these Hodge stars:

\displaystyle\begin{aligned}*d*\epsilon&=\mu_0c^2\rho\\d(*\beta)&=0\\d\epsilon&=-\frac{\partial(*\beta)}{\partial t}\\{}*d*(*\beta)&=\mu_0\iota+\frac{1}{c^2}\frac{\partial\epsilon}{\partial t}\end{aligned}

Notice that we’re never just using the $1$-form $\beta$, but rather the $2$-form $*\beta$. Let’s actually go back and use $\beta$ to represent a $2$-form, so that $B$ corresponds to the $1$-form $*\beta$:

\displaystyle\begin{aligned}*d*\epsilon&=\mu_0c^2\rho\\d\beta&=0\\d\epsilon&=-\frac{\partial\beta}{\partial t}\\{}*d*\beta&=\mu_0\iota+\frac{1}{c^2}\frac{\partial\epsilon}{\partial t}\end{aligned}

In the static case — where time derivatives are zero — we see how symmetric this new formulation is:

\displaystyle\begin{aligned}d\epsilon&=0\\d\beta&=0\\{}*d*\epsilon&=\mu_0c^2\rho\\{}*d*\beta&=\mu_0\iota\end{aligned}

For both the $1$-form $\epsilon$ and the $2$-form $\beta$, the exterior derivative vanishes, and the operator $*d*$ connects the fields to sources of physical charge and current.

February 22, 2012 -

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