### Surprises

We physicists tend to use ordinary English words, like “energy” or “field” in unconventional, technical ways. Mostly — even to a layman — it is recognizable, from the context, when a word like “energy” is being used in its technical, rather than its colloquial sense.

But, as I was reminded recently, that’s not always true, and can lead to some confusion. In particular, the use of the adjective “naïve” differs significantly from its colloquial meaning. In physics, a “naïve calculation” is one that omits some important effect, or makes some unwarranted simplification that, when properly understood, may significantly change the answer.

Which, in turn, reminds me of one of my favourite book, Rudolf Peierls’s *“Surprises in Theoretical Physics”* (and its sequel, *“More Surprises…”*). It’s a compendium of examples from various branches of Physics, where the naïve calculation yields the wrong answer, followed by a lucid explanation of *what* went wrong, and how a correct calculation fixes the problem.

To give you a taste, here’s a summary of one of my favourite “surprises,” that I sometimes talk about when I teach E&M.

Why isn’t a free electron gas paramagnetic?

In a uniform magnetic field, the trajectories of electrons are helices, which project to circles, when projected to the plane perpendicular to the applied magnetic field. Let $v_\perp$ be the projected velocity. The radius of the circle is $r = \frac{m v_\perp}{e B}$ and the magnetic moment should be $e v_\perp r$, or $\mu = \frac{m v_\perp^2}{B}$ (If you want the magnetic susceptibility, you would multiply this by the number density of electrons.)

That was easy. But, after a moment’s reflection, you realize that it’s complete *garbage*. This answer is independent of the electron charge, and it’s *inversely* proportional to the strength of the magnetic field. So it diverges for $B\to 0$!

What went wrong?

The first answer, as understood by H.A. Lorentz, and by Niels Bohr, is that you really need to think about a finite region. If you pick some rectangle in the plane, those orbits that lie entirely *outside* the rectangle do not contribute to the magnetic susceptibility of this rectangular region. Those orbits that lie entirely *inside* the rectangle, contribute paramagnetically, as described.

However, we made a mistake by neglecting the orbits which cross the boundary of the rectangle. We should only consider those portions of these orbits which lie *inside* the rectangle. These contribute a net current in the opposite sense (clockwise, instead of counterclockwise) from the electrons, whose orbits lie entirely inside the rectangle. This is a small current, but it encloses a large area (nearly the whole rectangle). The upshot, if you do it carefully, is that this boundary current precisely cancels out the effect of the electrons whose orbits lie inside the rectangle.

This is a somewhat gnarly calculation, and the sensitive dependence on the boundary conditions makes you think that we’re going about it the wrong way.

There’s a more robust calculation that yields the same result. The susceptibility
$\mu = - \frac{\partial F}{\partial B}$
where $F$ is the Helmholtz free energy. In Boltzmann Statistics,
$Z = e^{-\beta F} = \int e^{-\beta H(\mathbf{r},\mathbf{p})} d^3 \mathbf{r}d^3\mathbf{p}$
where
$H = \frac{1}{2m}(\mathbf{p}- e \mathbf{A})^2 + V(\mathbf{r})$
and $V(\mathbf{r})$ is the potential energy (which, say, confines the electrons to this rectangular region). The point is that, whereas our previous calculation was highly sensitive to the boundary conditions, calculating the free energy *first*, and *then* differentiating with respect to $B$ is more robust, as the free energy is much less sensitive to the boundary conditions.

Indeed, just by changing variables of integration from $\mathbf{p}$ to $\mathbf{q}=\mathbf{p}-e \mathbf{A}$, we see, immediately, that free energy is independent of $B$, and hence the susceptibility vanishes.

Peierls then goes on to discuss Landau’s famous extension of this story to the quantum case. Still neglecting electron spin, one finds (using either Boltzmann statistics, as above, or Fermi-Dirac statistics)

per electron.

So, now we see that, neglecting spin, the free electron gas in the quantum case, is *diamagnetic*. The magnitude of the above result is $1/3$ the contribution due to electron spin, which is paramagnetic. So, including spin, we expect the electron gas to be *paramagnetic*.

Hmmm. So that poses another puzzle. Why are some metals *diamagnetic*?

In a periodic potential, the effective mass of the electrons in the conduction band will, in general, not be equal to the free electron mass, $m$. This shifts the energies of the Landau levels. But the contribution of the spin still depends on the intrinsic magnetic moment, which depends on $m$. For $m_{\text{eff}}\ll m$, as in Bismuth, the contribution computed by Landau above wins, and one finds a large diamagnetic susceptibility.

Not all of the “surprises” have as many twists and turns as this one, but it’s all great stuff.

## Re: Surprises

This is interesting, although I sense a cultural barrier (between math and physics) that prevents me from appreciating the paradox, and therefore also its resolution.

It does remind me of my own pet favorite E&M problem, which I invented on my own when I learned that the momentum in an electromagnetic field is E cross B. Suppose that you have coil with a constant B inside. Suppose further that you place between two capacitor plates that generate a perpendicular E. (Perhaps coil is even square so that the plates are flush with the surface of the coil.) Then inside the coil there is a constant E cross B, while outside of the coil B is negligible. Where did this momentum come from?