Spherical Waves and the Hairy Ball Theorem

Boy howdy do we love spheres in physics. Sure we might tell you that the reason involves deep truths in topology, and symmetry, and group theory, and all that mathematical arcana, and in fact there's a lot of truth to that. But if we're completely honest, or at least if I'm completely honest, I have to admit that I love spheres because they're easy. All that lovely deep symmetry tends to produce enormous simplifications in whatever actual calculations we happen to do involving spheres.

Hence, our love for pretending everything is a sphere, or at least close enough for govermnemt work. There's even a name for taking this particular approximation method to an extreme - the spherical cow model.

In optics, we work with spherical waves and isotropic radiators and various other means of pretending that light waves are cleanly and coherently originating from a point source. Of course you wouldn't expect such an idealization to perfectly describe any physical device in the laboratory, but in theory it's fine. Right?

Let's say you have an electric field which is spherically symmetric. That is, the field is of the form:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

I.e., the magnitude of the field is uniform at any distance r from the center, and is directed radially outward. Faraday's law states that:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

The curl of a radially-symmetric vector field is zero, so the right-hand-side of that equation is zero. Thus a radially symmetric electric field necessarily implies that the magnetic field B is static. You can go through exactly this same argument for the B field using Ampere's law, and you'll find that a radially-symmetric magnetic field necessarily implies a static E field. Thus if you want both fields to have radial symmetry, they both have to be static. That means Maxwell's theory limits our spherical cow models to situations which are constant in time, which is a rather substantial limitation.

Ok, but what about spherically propagating waves? Light waves propagate perpendicular to the E and B fields, so perhaps we can have a light wave propagating radially outward. The E and B fields would be tangent to the outgoing spherical waves, skirting the problem that you can't have non-static E and B fields both pointing radially outward.

Unfortunately we run into a brick wall there too. There's a theorem in mathematics which states given a vector field defined on a sphere such that every vector is tangent to that sphere, the vector field must be zero on at least one point on that sphere. This theorem is called the hairy ball theorem. I'm not making this up. Basically if you have a basketball covered in fur, there's no way to comb the entire thing smoothly. You'll always have at least one cowlick.

i-bb92d1318c2acf2b7a583e351b26e53b-Hairy_ball.png

But if either one of the E or B fields is zero at some point, the Poynting vector ExB will be zero too, meaning no light is being radiated from that point. Thus a spherical light wave is impossible too, not just in practice but in theory as well. Sometimes that isn't a problem. We might be interested only in some region in which the assumption of spherical symmetry works just fine. Still, we are constrained and if we want to maintain connection with physical reality we can't let nice spherical cow assumptions get the best of us.

We'll finish this off with some comments on some possible counterexamples:

First, what about a steel ball-bearing which has been heated red hot, or a spherical star for that matter? Sure enough, both systems have spherically-symmetric time-averaged intensity distributions. But that doesn't make them sources of spherical waves. Every point on the surface of those objects is an independent source of radiation, and the E and B fields vary wildly at any given point near the surface. This wild variation will give a uniform time-averaged Poynting vector, but the fields and instantaneous intensity are far from uniform.

Second, what about a single atom in free space which has been raised to an excited state via a laser? Eventually it'll radiate by spontaneous emission, and since spontaneous emission is completely isotropic the field generated by the radiating atom has to be spherically symmetric. But since we're dealing with a single atom, quantum mechanics indicates that the atom emits a single photon, and that photon will assuredly be detected at exactly one point on a detecting sphere which surrounds the atom, breaking the symmetry.*

The moral of the story? How about this: even if you buy the (spherical) cow, the milk isn't necessarily free.

*I tend more and more toward the Willis Lamb view that the photon-as-a-particle picture is badly misleading, so we should really treat this glib explanation with more care. But in the spirit of the spherical cow, this handwaving argument is probably good enough for the moment.

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While I realize you're simplifying for the sake of this blog post, I think the "continuous" part of the spherical vector field is crucial enough that you should have included it in the bit about the hairy ball. Without specifying a continuous field, a counterexample is trivially constructed. (Which, of course, I'll leave as an exercise for the reader.)

By Randy Owens (not verified) on 13 Aug 2011 #permalink

I think this is worked around in practice by the concept of time averaged Poynting vectors. You can still have a spherical wave, but different parts of the wave will be zero at different times at the same radius from the center. The time average power transfers should still be the same though.

By ObsessiveMathsFreak (not verified) on 02 Sep 2011 #permalink

So, is this the closing post of "Built on Facts"? I'd personally like it if it weren't but I certainly understand the demands of real life. But, if you're all finished, maybe put up a post to that effect and I'll (sadly) remove the link from my blog roll.

Are you still out there Matt?

Theory that the earth is not spherical but is flat.

Since light travels in a straight line called ray a "spherical object"could only be seen from one direction showing us a "concave surface".
Again the rays which bounce of that "concave surface" would not have reached the eyes,since the rays hitting the surface would have dispersed at angles that wouldn't reach the eye.

By David Essiam (not verified) on 31 Jan 2012 #permalink