# What does the electric field of a "spherical wave" look like?

by user299792458
Tags: electric, field, spherical wave
 P: 17 You often hear about spherical light waves. For example, something like a light bulb is said to emit waves which are more or less spherical. Lets assume this light bulb is perfectrly round and emits in all directions (as opposed to a real light bulb, which has that metal thing at the bottom and doesn't emit light in that direction). Supposedly, the light bulb I'm describing would emit spherical waves. But what I don't understand is, what would the electric field of these waves look like? I can understand spherically symmetric sound waves, but light is a transverse wave. The electric field vector has to be perpendicular to the direction that the wave is travelling. It seems to me that, in this case, the electric field lines would have to resemble lines of latitude or longitude (on a globe) or something else, but the electric field can not be spherically symmetric. So what does the solution of Maxwell's equations look like for this spherical wave? What do we mean when we say that this wave is spherically symmetric? I can't find this in my physics books and I also tried searching Google with no luck.
 P: 175 Well there are two different aspectes to your question. 1) a light bulb like object (assuming it didn't have any piece of it to block the light in any direction) would emit what's called isotropically in its power density vs. solid angle. It is an incoherent source emitting actually relatively few photons per time interval, but because it's a 'random' thermal symmetric isotropic emitter, it has a uniform probability of emitting energy packets in any given direction. So, on a long term average basis, there'd be a spherically symmetric light flux with equal illuminance everywhere on any larger enclosing sphere. The individual photons / wave packets that are a result of blackbody radiation from the hot surface, though, would not necessarily be without a specific direction and non-isotropic concentration of their fields. In that respect it wouldn't be so different than something like the sun, composed of a large number of atoms / ions / particles, each of which emits radiation essentially randomly and relatively independently of the others, but over all it'd be equiprobably directed over a long time. 2) A hypothetical coherent radiator that's either an infinitely tiny point source or some hypothetical extended source that's coherent might be modeled as emitting spherical wavefronts in the manner you describe, and quite simply such wave fronts would 'look like' a plane wave except with curvature to cover a sphere. If you conformally mapped a plant to a sphere or in analogy a line to a circle (and ended up with something like ripples on a pond surface -- transverse waving up and down but circularly symmetric wave-fronts), you'd have the idealized version of the coherent spherical emitter. The individual waves would still be transverse E/M but that doesn't preclude them from being smooth plane waves, and since a plane is equivalent to a sphere you could say the same for spherical waves. However I have no information that any real point source or extended emitter is truly an coherent isotropoc emitter when examined over short time-scales and levels of individual photons being emitted. Perhaps I'm wrong about that, QED isn't my specialty, though I know a bit about classical and quantum E/M theory. I've also heard it argued that, in fact, that truly spherical wavefronts cannot exist physically since though they could be permitted by the Maxwell equations in some senses, they'd reportedly necessarily violate divergence-free conditions in other aspects required for the Maxwell equations and physicality. I don't recall the exact argument / mathematics / author, but I could probably easily find a citation on request. I believe that author was criticizing the Maxwell's equations themselves in some respects, and that argument was just a small point in an overall thesis. The more important thing from a practical standpoint is that real physical quantum-isotropic emitters may not really exist on the quantum level, though you can certainly have isotropic probability emitters that are quantum or macroscopic in average. If I'm wrong about any of this, I'd love to hear more details about the situation from people that know more about coherent quantum electrodynamic statistics or topological aspects of Maxwell's theory.
 P: 175 oops I made lots of typos in a rush; please forgive that; I'll let them stand for now since I have to attend to other matters.
P: 657

## What does the electric field of a "spherical wave" look like?

Wow. That was quite a response, xez!

Here's an attempt at a simpler explanation: the wave solution applies to the magnitude of the E/M wave, that is, if you look at the time and spatial dependence of the magnitudes of the electric and magnetic field vectors, they will fit the form of a nice spherical wave. When you ask about the directions of those vectors, then you're getting into details of the emission process, and yes, I believe you can't have a truly spherically symmetric solution (as mathematicians put it, "you can't comb a hairy tennis ball"). Generally, however, the directions of the vectors are randomized (incoherent), so you really can think just in terms of the amplitudes (vector magnitudes) of the fields, being symmetric.
P: 17
 Quote by belliott4488 (as mathematicians put it, "you can't comb a hairy tennis ball").
I've never heard of that theorem (just looked it up in Wikipedia), but that's basically what I was thinking, that you can't have a truly spherically symmetric EM wave. It seems that even the magnitude of the electric field vectors can't be completely spherically symmetric? So I guess it varies randomly and when you average it over time you get spherical symmetry? Is that the case?
P: 2,050
 Quote by xez emitting actually relatively few photons per time interval
In what sense?
P: 175
 Quote by cesiumfrog In what sense?
e.g. the number of atoms in a macroscopic object versus
the total number of photons emitted per second.

Also in the sense of the "photon density" of photons per
square meter per second at a modestly extended distance
from the emitter.

Let's see:
[eV/J] = 1/1.60217733*10^-19 = 6.24150636*10^18.

So if a 100W incandescent bulb was 10% efficient at
emitting visible light photons, that'd be 10 Joules/second
of photons from the bulb.

If the average photon energy emitted is 2eV, that'd be
3.12*10^18 * 2eV average photons / Joule, and

3.12*10^19 photons per second total, which over a
10m * 10m * 10m room = 10^12 square mm would
be 3.12*10^7 photons per second per square millimeter
which is already rather sparse and certainly an easily
physically countable discrete rate of them.

Even looking at fairly bright stars or other astronomical
objects in a fairly sizable telescope you end up getting
relatively small total numbers of photons per second
collected in your instrument, and the 'image' being more
of a statistical accumulation of lots of discrete values
rather than something that's "present" in any continuous
simultaneous sense.
P: 657
 Quote by user299792458 I've never heard of that theorem (just looked it up in Wikipedia), but that's basically what I was thinking, that you can't have a truly spherically symmetric EM wave. It seems that even the magnitude of the electric field vectors can't be completely spherically symmetric? So I guess it varies randomly and when you average it over time you get spherical symmetry? Is that the case?
Why do say that the magnitude can't be spherically symmetric? The magnitude of a vector field is a scalar field, and there's no reason why that can't be spherically symmetric, is there? Think of spherical shells whose color varies periodically.
 P: 2,050 The problem, belliott, is (vector) polarisation (which breaks the spherical symmetry). Of course, for a light bulb (not a coherent source) you might be right.
 Mentor P: 10,699 The light emitted by a light bulb isn't a single simple wave, spherical or otherwise, because a light bulb isn't a single simple source. It contains bazillions of sources (individual atoms) radiating in short bursts at random times and random polarizations. I don't know for sure what form the wave emitted from an individual atom is, but I suspect it might be like the wave emitted by an oscillating electric dipole. I couldn't find (with a quick Google search) a diagram illustrating both direction and magnitude for either the electric or magnetic field from such a dipole, but maybe someone else knows where to find one. As I recall, the electric field amplitude and direction both vary with the angle from the dipole axis. However, for something like a light bulb, the individual sources' dipole axes are oriented randomly with respect to each other, so the net electric field amplitude is spherically symmetric for practical purposes, and the direction of the field at any given point varies rapidly and randomly with time.
P: 657
 Quote by cesiumfrog The problem, belliott, is (vector) polarisation (which breaks the spherical symmetry). Of course, for a light bulb (not a coherent source) you might be right.
Yes, I know. That's why I earlier said that the approximation of a spherical wave is applied only to the magnitude. If you want to get into the vector components of the field, you have to get into the details of the production mechanism. If this is a coherent field, such as a polarized wave, then it's not spherically symmetric, but it's likely not coming from a point source, either.

I don't think anyone really speaks of spherically symmetric wave fronts when the specific directions of the vector fields are relevant. You tend to speak of plane waves in such cases.
 P: 2,050 Would you normally speak at all of wave fronts when the source is incoherent?
P: 657
 Quote by cesiumfrog Would you normally speak at all of wave fronts when the source is incoherent?
Um ... well, I thought that's what the spherical waves were. We already established that a coherent wave couldn't be spherically symmetric, so any spherical wave makes sense only if you're talking about a wave with unspecified or randomized vector directions. Would you not call that a wave front? Maybe not ...
P: 1,433
 Quote by belliott4488 Um ... well, I thought that's what the spherical waves were. We already established that a coherent wave couldn't be spherically symmetric...
A coherent wave can still be unpolarised, and thus spherically symmetric.

Claude.
P: 657
 Quote by Claude Bile A coherent wave can still be unpolarised, and thus spherically symmetric. Claude.
Yes, you're right, of course. I've been speaking too loosely - I'll just be quiet, now.
 Sci Advisor P: 1,433 I didn't mean to sound harsh! Claude.
P: 17
 Quote by Claude Bile A coherent wave can still be unpolarised, and thus spherically symmetric. Claude.
Isn't a spherically symmetric transverse wave inconsistent with the "hairy ball" theorem that has been mentioned here?
 Sci Advisor P: 1,433 For a spherical, unpolarised wavefront, each point on the spherical wavefront is indistinguishable from any other and any orientation (direction) is indistinguishable from any other. I would say that spherical symmetry is preserved in this case. Claude.

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