Brian Bi

What is an intuitive explanation of the D'Alembertian?

Let's review first the Laplacian, $$\nabla^2$$. It has the intuitive interpretation of measuring the amount by which a function's value at a point differs from the average value of the function on the surface of a sphere centered at that point. (See, e.g., Intuitive interpretation of the Laplacian) For a harmonic function, for which the value at a point always equals the average value on a sphere centered at that point, the Laplacian vanishes. Otherwise, you might say the Laplacian measures how far the function deviates from being harmonic. If the Laplacian is positive then the point is "cold"; if it's negative then the point is "hot".

The d'Alembertian is the operator that appears in the electromagnetic wave equation,$\Box^2 \mathbf{E} = \mathbf{0}$ or, alternatively, $\Box^2 \mathbf{B} = \mathbf{0}$ or, in covariant form in the Lorenz gauge, $\Box^2 A^\alpha = 0$ As we know, solutions to the electromagnetic wave equations are waves that propagate at the speed of light (without dispersion) in any direction. So we know something important about the d'Alembertian: it vanishes for waves that travel at the speed of light . So we can say that the d'Alembertian measures how much the function deviates from being a (not necessarily monochromatic plane) wave that travels at the speed of light.

To try to get a bit more clarity on what this means, notice that the d'Alembertian can be written as $\Box^2 = \frac{\partial^2}{\partial(ct)^2} - \nabla^2$ We said the Laplacian is positive for a "cool" point. For the electric and magnetic fields you can think of that as meaning a wave trough. Likewise the Laplacian is negative for a wave crest. So when we're on a crest, the second time derivative has to be negative also in order for the d'Alembertian to vanish. This is indeed what happens on a wave crest (like $$\sin x$$ for $$x \in (0, \pi/2)$$). When we're on a trough, the second time derivative has to be positive.

So when the d'Alembertian is nonzero, it measures deviation from this wavelike behaviour of a crest accelerates downward and a trough accelerates upward. In electrodynamics this happens when we have sources. For instance, inside a uniformly charged sphere rotating around the z-axis, the magnitude of the z-component of the magnetic field increases as we approach the centre of the sphere. So interior points of the sphere are like "permanent crests" or "permanent troughs" in the z-component of the magnetic field. We can superimpose various kinds of electromagnetic waves on this configuration, but as the d'Alembertian is linear, it just "ignores" all the superimposed waves (on which it vanishes) and picks out the field due to the source, that is, the non-wavelike part. The result is nonzero. If we reversed the direction of rotation, we reverse the sign of the d'Alembertian too. In the case that we have a "permanent crest", the d'Alembertian will be positive, and in the case that we have a "permanent trough" the d'Alembertian will be negative.

(Note for the curious: here, $$\Box^2 \mathbf{B} = \mu_0(\nabla \times \mathbf{J})$$, if my math is correct.)

Sorry, I guess this isn't really that intuitive. I'm not sure if I can come up with a more intuitive explanation.

 Strictly speaking, there are also nonphysical solutions that do not represent electromagnetic waves, such as harmonic functions stationary in time. These are nonphysical because they blow up at infinity.