The Unapologetic Mathematician

Mathematics for the interested outsider

A Short Rant about Electromagnetism Texts

I’d like to step aside from the main line to make one complaint. In refreshing my background in classical electromagnetism for this series I’ve run into something that bugs the hell out of me as a mathematician. I remember it from my own first course, but I’m shocked to see that it survives into every upper-level treatment I’ve seen.

It’s about the existence of potentials, and the argument usually goes like this: as Faraday’s law tells us, for a static electric field we have \nabla\times E=0; therefore E=\nabla\phi for some potential function \phi because the curl of a gradient is zero.

What?

Let’s break this down to simple formal logic that any physics undergrad can follow. Let P be the statement that there exists a \phi such that E=\nabla\phi. Let Q be the statement that \nabla\times E=0. The curl of a gradient being zero is the implication P\implies Q. So here’s the logic:

\displaystyle\begin{aligned}&Q\\&P\implies Q\\&\therefore P\end{aligned}

and that doesn’t make sense at all. It’s a textbook case of “affirming the consequent”.

Saying that E has a potential function is a nice, convenient way of satisfying the condition that its curl should vanish, but this argument gives no rationale for believing it’s the only option.

If we flip over to the language of differential forms, we know that the curl operator on a vector field corresponds to the operator \alpha\mapsto*d\alpha on 1-forms, while the gradient operator corresponds to f\mapsto df. We indeed know that *ddf=0 automatically — the curl of a gradient vanishes — but knowing that d\alpha=0 is not enough to conclude that \alpha=df for some f. In fact, this question is exactly what de Rham cohomology is all about!

So what’s missing? Full formality demands that we justify that the first de Rham cohomology of our space vanish. Now, I’m not suggesting that we make physics undergrads learn about homology — it might not be a terrible idea, though — but we can satisfy this in the context of a course just by admitting that we are (a) being a little sloppy here, and (b) the justification is that (for our purposes) the electric field E is defined in some simply-connected region of space which has no “holes” one could wrap a path around. In fact, if the students have had a decent course in multivariable calculus they’ve probably seen the explicit construction of a potential function for a vector field whose curl vanishes subject to the restriction that we’re working over a simply-connected space.

The problem arises again in justifying the existence of a vector potential: as Gauss’ law for magnetism tells us, for a magnetic field we have \nabla\cdot B=0; therefore B=\nabla\times A for some vector potential A because the divergence of a curl is zero.

Again we see the same problem of affirming the consequent. And again the real problem hinges on the unspoken assumption that the second de Rham cohomology of our space vanishes. Yes, this is true for contractible spaces, but we must make mention of the fact that our space is contractible! In fact, I did exactly that when I needed to get ahold of the magnetic potential once.

Again: we don’t need to stop simplifying and sweeping some of these messier details of our arguments under the rug when dealing with undergraduate students, but we do need to be honest that those details were there to be swept in the first place. The alternative most texts and notes choose now is to include statements which are blatantly false, and to rely on our authority to make students accept them unquestioningly.

February 18, 2012 Posted by | Electromagnetism, Mathematical Physics | 28 Comments