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Here is a question that I've had for $\geq$ 30 years but only now am in a position to ask properly. Electromagnetic waves are "transverse" which according to the textbooks means that the amplitude (in this case E and B fields) are perpendicular to the direction of propagation.

Now, if we have a point source then the amplitude vectors should furnish a vector field defined on a small ball around that source but according to the aforementioned hairy-ball theorem such cannot exist in a continuous fashion. What is the way out of this?

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The answer is that physical sources generally do not emit isotropically; instead, the archetypical radiator emits in a dipole pattern with the emission concentrated along a plane, with two zeros in its intensity distribution.

That said, it is possible to produce so-called isotropic radiators.

  • One way is to have two orthogonal polarizations with complementary intensity distributions and then add them incoherently, i.e. having two sources without a definite phase relationship on the two, so they don't actually form a vector field. This is how light from a star, even when monochromatized, gets to have an isotropic intensity distribution.

  • The more interesting way is to do this coherently, by exploiting the fact that monochromatic EM radiation can quite happily accommodate circular polarizations. For more details, see How do coherent isotropic radiators evade the hairy-ball theorem?.

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  • $\begingroup$ Correct; and here is a related fact. If you try to be more specific about a "point source", then you need to say in what way it is oscillating. If it is a point charge oscillating to and fro along a line, then it is not spherically symmetric. If it is a spherical ball of charge whose radius is oscillating, then it does not emit any radiation! It is a rather beautiful fact that the field around such a ball is just the static Coulomb field. A similar result holds for gravity in General Relativity. $\endgroup$
    – Andrew Steane
    Commented Nov 13, 2018 at 16:05
  • $\begingroup$ @Andrew Steane "If it is a spherical ball of charge whose radius is oscillating, then it does not emit any radiation! It is a rather beautiful fact that the field around such a ball is just the static Coulomb field" Is there a reference on this? $\endgroup$
    – user45664
    Commented Jan 26, 2019 at 18:41
  • $\begingroup$ @Emilio Pisanty Does this apply to scalar radiators or longitudinal waves? $\endgroup$
    – user45664
    Commented Jan 26, 2019 at 18:52
  • $\begingroup$ @user45664 No, nothing here applies to either of those. $\endgroup$
    – Emilio Pisanty
    Commented Jan 26, 2019 at 18:58
  • $\begingroup$ @user45664 The result follows directly from Gauss' law and spherical symmetry. Gauss' law says the total flux outwards is equal to the charge enclosed---it doesn't matter if the charge is moving around. Then the spherical symmetry says the field in one direction is same as in all other directions, at any given radius. So what can the field do? It can only stay constant if the charge moves in a spherically symmetric way such as expand or contract. $\endgroup$
    – Andrew Steane
    Commented Jan 26, 2019 at 18:59
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The long distance component of electromagnetic radiation is dipolar, not monopolar. That means that you need to choose a direction for the dipole emitters before writing a generic wave solution. Only scalar fields can have monopolar terms, like pressure for example.

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  • $\begingroup$ Yes, but EM radiation is not a vector field, it's an ellipse field. You can switch from linear to elliptical and circular polarization in a way that keeps the intensity (a scalar) monopolar. $\endgroup$
    – Emilio Pisanty
    Commented Oct 31, 2017 at 19:35

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