It wouldn't need to be salt.

Basically I was initially thinking about a mechanical transmitter, essentially just taking two equal opposite charges and fixing them to the opposite ends of a pole. Then you spin the pole around it's center (like a baton twirler) and it will emit some (mostly dipole) radiation.

Take an identical setup and fix the centers of the two poles together so we have a cross and spin that in the plane of the cross (picture a tire iron), and we should have a decent quadrapole moment and thus some quadrapole radiation (supposing you could spin it fast enough).

This made me consider iterations of this until you've a ring of alternating charges side by side spinning in a circle. Naturally I thought of table salt (some ionic crystal or another). So classically I get that it should radiate, but since it's a bound quantum state I'm not sure? Also, the strongest multipole moment I imagine would be some ridiculously large number for such a setup (probably making the radiation exceedingly weak and hard to detect?).

Sorry no pictures or equations, I'm tired

Anyway, for those interested I was initially considering how the moment of inertia of our initial pole would end up being dependent upon rotation speed.

  • $\begingroup$ My feeling is that (1) the radiation is too week for detection and (2) each dipole of salt molecule is not spinning coherently, the em wave probably destructive interfere with one another. $\endgroup$ – K_inverse Dec 4 '18 at 12:16
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    $\begingroup$ This is not an answer since I don't know the answer, but have you tried estimating how fast it should fall off? If there's $10^{23}$ ions then you're going to have radiation going something like $(a/r)^{10^{23}}$ where $a$ is the lattice spacing I think. $\endgroup$ – jacob1729 Dec 4 '18 at 12:20
  • $\begingroup$ @jacob1729 I was thinking I could model it as a bunch of superimposed dipole antennas placed at small angles to one another. From that perspective certain frequencies (ie rotation speeds), might either reinforce or partly cancel different charge spacings I was thinking (I'll have to check it out). $\endgroup$ – R. Rankin Dec 4 '18 at 12:25
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    $\begingroup$ To put it simpler: no, salt is electrically neutral. $\endgroup$ – my2cts Dec 4 '18 at 12:25
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    $\begingroup$ @my2cts and so is the initial baton, by that logic that wouldn't radiate either, but that definitely shouldn't be the case. The vanishing of the monopole plays no part in whether it radiates. But that's the interesting thing. If it doesn't radiate, what's the length scale.where it stops $\endgroup$ – R. Rankin Dec 4 '18 at 12:30

First, note that you are using a classical approximation for both the composition of a salt and the nature of electromagnetic radiation. This approximation breaks down both at very small length scales (where the distribution of the electron clouds within the salt becomes important) and at very low intensities of radiation (where the discrete nature of electromagnetic radiation becomes important).

Also, note that even in the classical approximation, you cannot consider the radiation of an individual charge in isolation from its environment. In the radiation zone (namely, at distances much larger than both the spacing between charges and the wavelength derived from the frequency of oscillations), the contribution of every charge in the sample is important (since they're all at nearly the same distance from a point in the radiation zone), and the total radiation is the sum of the oscillations in the field from all of the charges.

If you pick a charge in the middle of a salt, you can find an opposite neighboring charge, which means you have an electric dipole (and, in particular, the monopole moment for this distribution is zero). For this dipole, you can find a neighboring dipole pointing in the opposite direction, which means you have an electric quadrupole (and, in particular, the monopole and dipole moments for this distribution are zero). For this quadrupole, you can find a neighboring oppositely-oriented quadrupole, which means you have an electric octopole (and in particular, the monopole, dipole, and quadrupole moments for this distribution are zero). You can continue this process until you reach the end of the salt crystal. For a salt crystal consisting of $N$ ions, you may have a nonzero electric $2^{\lfloor \log_2 N\rfloor}$-pole moment, while all lower moments are zero.

It turns out that in the radiation zone, the radiation from an electric $2^\ell$-pole is suppressed by a factor of $1/(1+2\ell)!!$ relative to an electric monopole (source: https://en.wikipedia.org/wiki/Multipole_radiation). For a macroscopic crystal with $N=10^{24}$ ions, this corresponds to $\ell=79$, which means that the salt may have a nonzero $2^{79}$-pole moment, whose radiation is suppressed by a factor of $3/159!!\approx10^{-141}$ relative to your dipole. This means the radiation is certainly undetectable and would break the classical approximation even if it was detectable.

  • $\begingroup$ That's kind of what I thought, also as @jacob1729 mentioned the lattice parameter would be featured to a large power. Thank you. $\endgroup$ – R. Rankin Dec 5 '18 at 0:12
  • $\begingroup$ Regarding the breakdown of classical radiation formula in small charges, what does that imply about a single electron (or just a small group) in a cyclotron. Does it still emit cyclotron radiation? $\endgroup$ – R. Rankin Dec 5 '18 at 0:24
  • $\begingroup$ @R.Rankin The lattice parameter is not featured to a large power at all. The lattice parameter is nowhere in the above argument whatsoever, except for setting the length scale where the radiation zone starts and being large enough that the classical approximation for the salt's composition still holds. The suppression is purely a function of the number of ions present in the crystal. If you somehow slightly shrank or lengthened the lattice spacing in the crystal, the suppression in the radiation zone would be the same. $\endgroup$ – probably_someone Dec 5 '18 at 0:45
  • $\begingroup$ @R.Rankin The argument that jacob1729 mentioned holds only for static fields, not for radiation. Contributions to radiation can only die off as $1/r^2$ with distance (as any higher power in inverse distance doesn't carry any energy to infinity, and lower powers are not possible). The suppression occurs in the coefficients that multiply the various multipole contributions to the radiation, and it is these coefficients that are suppressed by $1/(1+2\ell)!!$, where $\ell$ is purely a function of the number of ions (as their arrangement is set by the crystal structure). $\endgroup$ – probably_someone Dec 5 '18 at 0:49
  • $\begingroup$ @R.Rankin The breakdown in the classical approximation does not occur for small charges, nor did I say that it did. The reason that it occurs for small length scales, specifically in a salt crystal, is that the positive and negative ions are actually composed of a central nucleus surrounded by a delocalized electron cloud, which must be treated using quantum mechanics. Single electrons are point particles, so there is no internal length scale that matters except the one set by the frequency of the oscillations. $\endgroup$ – probably_someone Dec 5 '18 at 0:55

In order to radiate, a system must have an oscillating dipole (or higher multipole) moment. Salt does not consist of discrete molecules, but rather of positive and negative ions arranged so that the moments cancel out very accurately in macroscopic crystals. Some other substances (e.g., quartz) do consist of oriented dipoles, and you might think that crystals would present bound surface charges and dipole moments analogous to permanent magnetic moments. (You could call them electrets.) However, the surface charges attract free charges that neutralize them, at least when in equilibrium. Such substances turn out to be piezoelectric instead.

  • $\begingroup$ But microscopically, there are charges, which are accelerating. Why should not they radiate? $\endgroup$ – Archisman Panigrahi Dec 4 '18 at 12:57
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    $\begingroup$ @ArchismanPanigrahi For the same reason that an ordinary electric dipole ceases to radiate as the dipole moment $\vec{p}\to 0$ (i.e. when you place one charge essentially directly on top of the other). The radiation at macroscopic distances is not determined by whether charges are accelerating, but rather, what the moments of the charge distribution are doing, and the moments cancel quite nicely on a macroscopic scale. $\endgroup$ – probably_someone Dec 4 '18 at 13:13

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