The Unapologetic Mathematician

Mathematics for the interested outsider

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 Posted by | Electromagnetism, Mathematical Physics | 2 Comments