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The concept of a spherical electromagnetic wave is a nice fiction, which is sometimes called upon in introductory optics textbooks, but it runs into a deep topological problem in the form of Brouwer's hairy ball theorem, which essentially states that

if $\mathbf f:\mathbb S^2\to\mathbb R^3$ is a continuous function that assigns a vector $\mathbf f(p)$ in $\mathbb R^3$ to every point $p$ on a sphere such that $\mathbf f(p)$ is always tangent to the sphere at $p$, then there is at least one $p$ such that $\mathbf f(p) = 0$,

or, in other words, "you cannot comb a hairy sphere". As regards electromagnetism, this means that a linearly-polarized spherical wave cannot be isotropic, because Gauss's law requires the radiation component to be transverse, and the hairy-ball theorem then requires it to have zeroes in its angular intensity distribution.

The usual argument with e.g. stars and the like is that the radiation they emit is not coherent, which sidesteps this limitation. However, a recent answer pointed out that if you relax the requirement that the polarization be uniform, it becomes possible to have coherent spherical waves with an isotropic intensity distribution by letting them take elliptical or circular polarizations over some directions, and that is plenty interesting in its own right. However, this is not a point that is made very often, so I would like to see explicit forms of how this can be done.

More specifically, I would like to see explicit exact solutions of the vacuum Maxwell equations on full space minus a sphere, i.e. $\{\mathbf r\in\mathbb R^3:\|\mathbf r\|>a\}$, which are (i) monochromatic, (ii) outgoing spherical waves, and (iii) have constant intensity over every sphere centered at the origin. If there is an explicit decomposition of this wave into e.g. a sum of two orthogonal linear polarizations with different intensity distributions, that would be good to see as well.

I believe the references in the previous answer, including this, can be worked into that form with some ease, but I'm not particularly fussed about the existence of easy implementations with straight-wire antennas or anything like that - I would value more solutions that are exact everywhere over solutions that have the desired properties in an asymptotic sense. That said, if there is a current distribution on the sphere itself that will give the desired solutions exactly (in the sense of this answer), then that's certainly interesting.

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  • $\begingroup$ Several comments removed. Please don't use comments for answers. $\endgroup$ – Qmechanic Feb 15 '17 at 10:30
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Qmechanic Feb 16 '17 at 18:53
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The solution proposed by Matzner and outlined in the other answer is only approximate, not exact. The problem is, how much of an approximate solution it really is.

The original idea in Int. J. Antenn. Propag. 2012, 187123 (2012) was to obtain a field with a spherically symmetric field intensity in the asymptotic far-field region. Once such a solution is known, it is in principle possible to find a finite current distribution that produces the same field, and a corresponding distribution on a spherical shell was indeed calculated.

The proposed far-field solution has a simple separable form and reads $$ \vec{E}(\vec{x}, t) = \left(-i\omega\frac{\mu_0 k}{4\pi} e^{i\omega t} \right) \frac{e^{-ikr}}{kr} \vec{A}(\theta, \phi) = C(t) \vec{E}_0(r, \theta, \phi) $$ with $$ \vec{E}_0(r, \theta, \phi) = f(kr) \vec{A}(\theta, \phi) $$ and \begin{align} A_x(\theta, \phi) & = \exp\left(-i\frac{\pi}{4}\cos\theta\right)\;,\;\;\; A_y = 0\\ A_z(\theta, \phi) & = i\;\frac{\cos\phi}{\sin\theta} \left[ \exp\left(i\frac{\pi}{4}\cos\theta\right) - i \cos\theta \exp\left(-i\frac{\pi}{4}\cos\theta\right) \right] \end{align} But it is not hard to see that it is only a solution to first order in $1/r$ and only for $k\ll1$.

  1. The field intensity is claimed to be spherically symmetric because the magnitude of the tangential component of $\vec{A}(\theta, \phi)$ is indeed spherically symmetrical, as pointed out in the other answer.

    But for the given cartesian components the radial component $A_r(\theta, \phi)$ is not null, and not spherically symmetric. It reads in fact \begin{align} A_r(\theta, \phi) & = A_x \sin\theta \cos\phi + A_y \sin\theta\sin\phi + A_z \cos\theta \\ & = \frac{\cos\phi}{\sin\theta}\left[ \exp\left(-i\frac{\pi}{4}\cos\theta\right) + i\cos\theta \exp\left(i\frac{\pi}{4}\cos\theta\right)\right] \\ & \neq 0 \end{align} Note: In the subsequent paper on Isotropic Radiators the radial component of $\vec{E}\;$ is explicitly set to 0, and the intensity is calculated again using the traversal components.

  2. The more pressing problem has to do with the form of the radial factor. Since $f(kr) = \exp(-ikr)/kr$ is itself a spherically symmetric solution to the Helmholtz equation $$ \left(\Delta + k^2 \right)f(kr) = 0 $$ the nontrivial contributions from $\vec{A}(\theta, \phi)$ must satisfy $$ \frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_x(\theta, \phi) \right) = 0\\\frac{f(kr)}{r^2} \;{\hat L}^2 \left( A_z(\theta, \phi) \right) = 0 $$ where ${\hat L}^2$ is none other than the square angular momentum operator.

    If read literally, this is an eigenvalue equation of ${\hat L}^2$ for a null eigenvalue, with the only exact solution $\sim Y_0^0(\theta, \phi) = const$. So any $A_x(\theta, \phi)$ and $A_z(\theta, \phi)$ may be approximate solutions as long as the presence of the $1/r^2$ factor makes their contribution smaller than some agreed upon tolerance limits.

  3. Same goes for the transversality condition $\nabla \cdot \vec{E} = 0$, which in spherical coordinates reads $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \;E_r \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \;E_\theta \right) + \frac{1}{r\sin\theta} \frac{\partial}{\partial \phi} E_\phi = 0 $$ For the particular solution sought here it becomes, after slight rearrangement, $$ k \times \Big\{ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)^2 f(kr) \right] A_r(\theta, \phi) + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \\ + \frac{f(kr)}{(kr)\sin\theta} \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big \} = 0 $$ The radial term then gives $$ \frac{1}{(kr)^2} \frac{\partial}{\partial (kr)} \left[ (kr)\; \exp\left(-i\; kr\right) \right] = \frac{f(kr)}{kr} - if(kr)\;, $$ and ignoring a factor of $\exp(-ikr)/\sin\theta\;$, $$ -\frac{ik}{(kr)}\; A_r(\theta, \phi) \sin\theta + \frac{k}{(kr)^2} \Big\{ A_r(\theta, \phi) + \frac{\partial}{\partial\theta} \left[ \sin\theta \;A_\theta(\theta, \phi) \right] + \frac{\partial}{\partial \phi} A_\phi(\theta, \phi) \Big\} = 0 $$ For this to be exactly satisfied both the leading first term and the one in the big curly brackets must vanish separately (correspond to different powers or $(kr)$). But since this doesn't happen for the solution at hand, we must conclude that transversality only holds approximately in the limit $kr\gg1$ and/or $k \rightarrow 0$.

So what can be done about the desired exact monochromatic solution with a spherically symmetric intensity?

The existence of such solutions when field polarization is elliptical and varies from point to point is not such a new idea, see this paper from IEEE Transactions on Antennas and Propagation, Nov. 1969, 209 (eprint). However, finding an exact solution is still a tricky task.

Formally, it amounts to solving for the electric field as a divergenceless solution to a Helmholtz equation, $$ \left(\Delta + k^2 \right) \vec{E} = 0 \;, \;\;\; \nabla \cdot \vec{E} =0 $$ under the additional requirement that the field intensity $\vec{E}^2$ be spherically symmetric.

One standard strategy is to consider a multipole expansion with scalar radial factors given by spherical Bessel functions and vectorial directional factors as linear superpositions of spherical harmonics. But then imposing the spherical intensity condition brings up a hornet's nest of Clebsch-Gordon coefficients. Or, potentially, some really neat irrep argument.

Anyone up to it?

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  • $\begingroup$ I'm not sure how this squares with the last section of Matzner (2012), which presumably has exactly the desired solutions, as eqs. (31-32); these are a specialization of Jackson's eq. (9.122) and are therefore full solutions of the Maxwell equations. On the other hand, the series-of-vector-spherical-harmonics isn't the most manageable expression in the world. $\endgroup$ – Emilio Pisanty Feb 17 '17 at 20:25
  • $\begingroup$ Eqs.(31)-(32) are indeed the multipole expansions for the general solution to the source-free Maxwell's eqs. The catch is in Eqs.(37)-(41): the expansion coefficients derived from the source currents are not limited to $l=0$, and the exact solution definitely does not separate into one radial and one directional factor, unlike the form assumed in Eqs.(18)-(20). The latter is supposed to be the far-field ($kr >>1$), low-frequency ($k\rightarrow 0$) limit. Anything else is algebraically exact. $\endgroup$ – udrv Feb 18 '17 at 3:48
  • $\begingroup$ My problem is that it's kinda too strong a limit, and the exact solution from the multipole expansion is only isotropic in intensity in this same limit. As for spherical harmonics, what do you think the ${\bf X}_{lm}$ factors hide? $\endgroup$ – udrv Feb 18 '17 at 3:48
  • $\begingroup$ I'm not sure that separability is strictly necessary, though you're right that it would be nice. A separable solution will have the same polarization regardless of distance, but a non-separable one can still be isotropic in intensity. I'll have to think some more about the spherical harmonics - I'm not that familiar with the vector versions. $\endgroup$ – Emilio Pisanty Feb 18 '17 at 13:38
  • $\begingroup$ I only mentioned separability because the asymptotic solution in Eqs.(18)-(20) is separable, whereas the multipole expansion of the exact solution is definitely not. Otherwise, at least for now, I see no reason to expect separability either. And Jackson uses vector spherical harmonics because of the way the divergence condition is implemented, but in principle the whole thing can be rewritten using the scalar ones. $\endgroup$ – udrv Feb 18 '17 at 15:49
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Fields that probably satisfy those requirements are given in

H. Matzner and E. Levine. Can Radiators Be Really Isotropic? Int. J. Antenn. Propag. 2012, 187123 (2012).

The claim is: $$ \mathbf{E} = \frac{i\omega \mu_0}{4 \pi r} e^{i(kr-\omega t)}\mathbf{A}(\theta,\phi) $$

where $\mathbf{A}(\theta,\phi)$ has altitudinal and azimuthal components $$ A_\theta(\theta,\phi) = i \cos(\phi) e^{-i\frac{\pi}{4} \cos \theta} $$ and $$ A_\phi(\theta,\phi) = - \sin (\phi) e^{+i\frac{\pi}{4} \cos \theta} $$ respectively. The conditions (i) and (iii) from the question are clear (and relying on $ |A_\theta|^2 + |A_\phi|^2) $ , we still should check (ii)). I will check later whether the wave equation is really fulfilled.

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  • $\begingroup$ It appears that Matzner's PhD thesis (Moment method and microstrip antennas, Weizmann Institute of Science, 1993), also contains a fair bit of information on the topic. $\endgroup$ – Emilio Pisanty Feb 15 '17 at 16:41
  • $\begingroup$ Are you sure that the field is divergence-free? $\endgroup$ – Raziman T V Feb 17 '17 at 11:06
  • $\begingroup$ @Raziman T V Very good question. The paper seems to silently imply this, but one has to check. Also, although one freely uses complex E fields, nature is real. So we also have to see how the real field gets around Birkhoff. $\endgroup$ – lalala Feb 17 '17 at 12:26
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How do coherent isotropic radiators evade the hairy-ball theorem?

By having a toroidal topology.

The concept of a spherical electromagnetic wave is a nice fiction

It is not a fiction.

which is sometimes called upon in introductory optics textbooks, but it runs into a deep topological problem in the form of Brouwer's hairy ball theorem...

No it doesn't.

or, in other words, "you cannot comb a hairy sphere".

But you can comb a hairy torus such that it doesn't have a cowlick: "A hairy doughnut (2-torus), on the other hand, is quite easily combable."

enter image description here Public domain image by The Evil Midnight Uploader, see Wikipedia.

As regards electromagnetism, this means that a linearly-polarized spherical wave cannot be isotropic, because Gauss's law requires the radiation component to be transverse, and the hairy-ball theorem then requires it to have zeroes in its angular intensity distribution.

Electromagnetic waves are comprised of photons anyway. Those photons remain coherent. There are no photons radiating outwards in an all-round spherical fashion. So what you're asking about is hypothetical anyway. However given the scenario, the torus fixes the issue you raise.

The usual argument with e.g. stars and the like is that the radiation they emit is not coherent, which sidesteps this limitation. However, a recent answer pointed out that if you relax the requirement that the polarization be uniform, it becomes possible to have coherent spherical waves with an isotropic intensity distribution by letting them take elliptical or circular polarizations over some directions, and that is plenty interesting in its own right. However, this is not a point that is made very often, so I would like to see explicit forms of how this can be done.

You "inflate" your torus. The more you inflate it, the more spherical it gets. See Adrian Rossiter's torus animations. Or draw two adjacent circles to represent a cross-section through the torus, then using the same centres, draw bigger and bigger circles. In the limit your two circles are congruent:

enter image description here

More specifically, I would like to see explicit exact solutions of the vacuum Maxwell equations on full space minus a sphere, i.e. $\{\mathbf r\in\mathbb R^3:\|\mathbf r\|>a\}$, which are (i) monochromatic, (ii) outgoing spherical waves, and (iii) have constant intensity over every sphere centered at the origin.

You're asking for too much. Particularly when it comes to point iii. There are no point particles.

If there is an explicit decomposition of this wave into e.g. a sum of two orthogonal linear polarizations with different intensity distributions, that would be good to see as well.

Work backwards.

I believe the references in the previous answer, including this, can be worked into that form with some ease, but I'm not particularly fussed about the existence of easy implementations with straight-wire antennas or anything like that - I would value more solutions that are exact everywhere over solutions that have the desired properties in an asymptotic sense. That said, if there is a current distribution on the sphere itself that will give the desired solutions exactly (in the sense of this answer), then that's certainly interesting.

I think it's more interesting than perhaps you realise.

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    $\begingroup$ Your method does not work: if you inflate the torus whilst keeping a fixed internal diameter, either (i) it never becomes a sphere (if the cross-section remains circular), so it does not become isotropic, or (ii) the vector field will develop a discontinuity (in the case of the 'spindle torus'), which makes it incompatible with the Maxwell equations. If you cannot exhibit solutions of the Maxwell equations to back up your claims, you're not doing electromagnetism - plain and simple. $\endgroup$ – Emilio Pisanty Feb 15 '17 at 13:37
  • $\begingroup$ @Emilio Pisanty : don't keep a fixed internal diameter. Draw bigger and bigger circles like I said. Keep doing it, and in the limit your two circles are congruent. Which means they're one and the same circle. PS: I am most definitely doing electromagnetism. Do note my last line. If there's any more electromagnetic questions you'd like to ask, I'd be only to happy to offer an answer. $\endgroup$ – John Duffield Feb 15 '17 at 13:50
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    $\begingroup$ No, that still doesn't work. You don't really specify whether your torus is self-intersecting, but: (i) if it is self-intersecting, then again you develop a discontinuous field, which is not a solution of the Maxwell equations; or (ii) if it's not self-intersecting, then it will either retain its donut shape or, at best, approach this shape, which is still not isotropic. That said, if you do in fact claim that there are electromagnetic fields that do not obey the Maxwell equations, this is a good time to clarify that. $\endgroup$ – Emilio Pisanty Feb 15 '17 at 14:00
  • $\begingroup$ @Emilio Pisanty : of course it's self-intersecting. In the limit it's all intersecting. See the drawing I've added to my answer above, and take it to the limit. As for whether it's a solution of the Maxwell equations, like I said above you're asking for too much. Which is presumably why you have no other answers. $\endgroup$ – John Duffield Feb 15 '17 at 14:30
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    $\begingroup$ OK, thanks for clarifying the shapes. And, just to double down on that one - you are in fact claiming that being a solution of the Maxwell equations is not a requirement for electromagnetic fields to exist physically? $\endgroup$ – Emilio Pisanty Feb 15 '17 at 14:33

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