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There is a lot of documentation about magnetic and electric fields being perpendicular in plane waves, I'd like to know if these two fields are perpendicular also in spherical wave. Is it possible to find an analogous formula to the one used for plane waves? $$\vec E=\vec B \times \vec c$$ (where $\vec c$ is a vector with intensity the speed of light and directed along the propagation direction)

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  • $\begingroup$ Do Maxwell's equations even have a spherically symmetric solution? Sound (pressure) waves certainly do, but it's not so clear that electromagnetic waves do. $\endgroup$ – Roger Wood Sep 16 at 18:40
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The only way for a vector field to have strict spherical symmetry is for it to be purely in the radial direction. For, if it had a non-radial component then that component would have to be preserved under rotations, but you cannot construct a vector field which has that property everywhere on the surface of a sphere. I provide a proof below. (This is closely related to, but not exactly the same as, the hairy ball theorem.) So the only type of vector field which has strict spherical symmetry is a purely radial one, such as a Coulomb field. Such a field cannot be an electromagnetic wave. So it is not possible to have exactly a spherical electromagnetic wave (i.e. one with no change at all under rotations).

You can have a wave which in the limit $r \rightarrow \infty$ has spherical wavefronts and is transverse, but I suppose the question is not about that limit, since it amounts to adopting a plane wave approximation for each part of the spherical wavefront.

You can have an oscillating field which has spherical wavefronts, where a wavefront is a locus of a fixed value of the phase of the oscillation. Such a field is not exactly transverse everywhere.

A proof of the claim (I just made up this proof; I'm adding it to see if anyone likes it or tells me it is not good enough.)

Take a sphere, and put a vector $\bf E$ at some point P on it. Let's define the 'equator' of our sphere to be the great circle running through P and parallel to $\bf E$ there. Now rotate the sphere through 90 degrees, carrying P and $\bf E$ up to the north pole. The vector is pointing in a direction we shall agree to call $x$.

Now return to the initial condition, and this time rotate the ball by 90 degrees about an axis through the poles, thus carrying P around the equator, and $\bf E$ with it. Then rotate again, carrying P up to the north pole. Now $\bf E$ is sitting at the north pole and pointing in a direction $y$, at right angles to the direction we got in the first rotation. But if we had been able to paint a vector field onto our sphere such that it had spherical symmetry, then these two transformations should both give no net effect on the whole sphere, and therefore both should carry $\bf E$ to a direction at the pole which would be the same in both cases. But it is not the same, so we have a contradiction, and the false step was the assumption that a vector field could be painted on the sphere in a spherically symmetric way.

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  • $\begingroup$ Anything to do with the 'hairy ball theorem'? $\endgroup$ – user45664 Sep 17 at 17:25
  • $\begingroup$ @user45664 yes it is related, though not exactly the same; I just added an extension which mentions this $\endgroup$ – Andrew Steane Sep 17 at 18:24
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Write Ignatowsky's equations (erroneously referred to as Jefimenko's equations in [1]) in the form as follows [2]:

$$ \mathbf{B} = \frac{1}{c}\int d^3\mathbf{x}'\frac{([\mathbf{J}]\times\hat {\mathbf{n}})}{R^2} +\frac{1}{c^2}\int d^3\mathbf{x}'\frac{([\dot{\mathbf{J}}]\times\hat {\mathbf{n}})}{R} \tag{4}\label{4} $$ $$ \mathbf{E} = \int d^3\mathbf{x}' \frac{[\rho] \hat {\mathbf{n}}}{R^2} +\frac{1}{c}\int d^3\mathbf{x}'\frac{([\mathbf{J}]\cdot\hat {\mathbf{n}})\hat {\mathbf{n}}}{R^2} +\frac{1}{c^2}\int d^3\mathbf{x}'\frac{([\dot{\mathbf{J}}]\times\hat {\mathbf{n}})\times \hat {\mathbf{n}}}{R} \tag{5}\label{5} $$

The brackets $[]$ mean retarded time. While assuming that the current density $\mathbf{J}$ is localized in space for large $R=|\mathbf{x}-\mathbf{x}'|$ only the terms whose integrand is proportional to $1/R$ will matter representing the radiation field, while terms having $1/R^2$ is the near field; so the radiation field is: $$\mathbf{B} \approx \frac{1}{c^2}\int d^3\mathbf{x}'\frac{([\dot{\mathbf{J}}]\times\hat {\mathbf{n}})}{R} \approx -\hat {\mathbf{n}}\times \frac{1}{c^2}\frac{1}{R}\int d^3\mathbf{x}'[\dot{\mathbf{J}}] \\ \mathbf{E} \approx \frac{1}{c^2}\int d^3\mathbf{x}'\frac{([\dot{\mathbf{J}}]\times\hat {\mathbf{n}})\times \hat {\mathbf{n}}}{R} \approx \frac{1}{c^2}\frac{1}{R}\hat {\mathbf{n}}\times \left(\hat {\mathbf{n}}\times \left(\int d^3\mathbf{x}'[\dot{\mathbf{J}}]\right)\right)$$ from which it is obvious that in the radiation field $\mathbf{E} \perp \mathbf{B} \perp \hat {\mathbf{n}}$ where $\hat {\mathbf{n}}$ is the unit vector in the direction of propagation.

[1]: https://en.wikipedia.org/wiki/Jefimenko%27s_equations

[2]: Kirk T. McDonald, The relation between expressions for time-dependent electromagnetic fields given by Jefimenko and by Panofsky and Phillips, American Journal of Physics 65 (11) (1997), 1074-1076

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  • $\begingroup$ Good answer, but maybe we need clarification from the questioner. I take a spherical wave to be one that has exact spherical symmetry. For example, the field from a sinusoidally varying point-charge or monopole is perfectly spherical and can certainly be sensed at a distance, but it drops off as 1/R^2, not 1/R, as in eqn (5). $\endgroup$ – Roger Wood Sep 16 at 21:14
  • $\begingroup$ @RogerWood this derivation shows that the orthogonality of $E$ and $B$ does not depend on the propagating wave being spherical, it depends only the source $\mathbf{J}$ being localized (having a finite extent). Of course when that is the case the wave is asymptotically becoming spherical as you move away from the source. $\endgroup$ – hyportnex Sep 16 at 21:20
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    $\begingroup$ @RogerWood as long as charge is conserved, $\nabla\cdot \mathbf{J}=- \partial \rho / \partial t$, the near field of an oscillating charge drops off as $1/R^2$ while in the far field it drops off as $1/R$. $\endgroup$ – hyportnex Sep 16 at 21:24
  • $\begingroup$ I don't see how any vector quantity can have spherical symmetry around a source unless it is directed radially. This includes E and H and the J. If the J associated wih the source is uniformly directed radially, again it doesn't seem like there will be any far field. Isn't this the same problem that you can't make an isotropic antenna that has the same sensitivity in all directions? $\endgroup$ – Roger Wood Sep 16 at 22:04
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    $\begingroup$ @RogerWood what you are saying is in essence Poincare's theorem or sometimes referred to as the "hairy ball" theorem: en.wikipedia.org/wiki/Hairy_ball_theorem $\endgroup$ – hyportnex Sep 16 at 22:28

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