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Consider a small LC circuit which is oscillating at some rate. The wavelength is VERY long compared to the diameter of the inductor. (100's of meters)

Beyond the radiansphere, where the EM waves are detached, they are photons of that frequency correct?

Inside the radiansphere, where they are not yet detached, what are they?

Is there a good reference I can read up on for details of what's happening inside the radiansphere?

I'm aware this is not an abrupt transistion.

user103218
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    What does the word "detached" mean here? – probably_someone Feb 20 '20 at 22:30
  • In the far field, outside the radiansphere, the waves no longer couple back to the source, and are said to be detached. – user103218 Feb 20 '20 at 22:49
  • What does “no longer couple back to the source” mean mathematically? I am not familiar with this way of thinking. – G. Smith Feb 20 '20 at 22:52
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    Are you taking a physics course or an EE course? Sometimes engineers think about EM differently from physicists. – G. Smith Feb 20 '20 at 22:54
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    The "detachment" isn't nearly as clean as you appear to be imagining it, then. The radiansphere defines the surface at which the induction and radiation terms are the same magnitude; the induction term isn't zero outside the radiansphere, it's just less than the induction term. If by "coupling back to the source" you refer to this induction term, then there is still significant coupling back to the source for some distance outside the radiansphere. – probably_someone Feb 20 '20 at 22:55
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    From my point of view, an EM wave never “becomes” a photon. But the EM wave is related to the probability of detecting a photon. – G. Smith Feb 20 '20 at 22:59
  • I would agree with G. Smith. Photons and EM waves are both models of radiation that are useful in different contexts. Radiation can be represented as either one under almost any circumstances, but most of the time one model is vastly more useful than the other. In my experience, usually the dividing line is some combination of the frequency of the radiation and the intensity (low intensity and high frequency = better represented as photons, high intensity and low frequency = better represented as EM waves). – probably_someone Feb 20 '20 at 23:07
  • I'm not taking any courses at the moment, but as I've said for a long time, "When I stop learning, get a shovel". :) I'm just trying to understand this area. It seems that what happens around the radiansphere is very interesting. – user103218 Feb 21 '20 at 17:08

2 Answers2

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This is rather a comment on the "VERY long" wavelength. About the "inside the radiantsphere" you think about a liquid and the inner tension at which a drop can leave the liquid.

Let us not start with your LC-circuit, but only with a resistor. The resistor is connected to a DC power source. After switching on the source, the temperature of the resistor increases. This happens because the inner (atomic) structure of the resistor hinders the forward movement of the electrons. The electrons on their way through the resistor many times lose kinetic energy and radiate this energy in the form of photons. The same happens in the wires, by the way.

The wavelength of these photons is in the range of infrared. A powerful source heats a robust resistor even to a temperature at which radiation (visible light) can be seen. In general, the wavelength depends on the free path length in which the electrons can accelerate (until they collide with the next atom) and on the acceleration rate obtained from the potential difference of the source (the voltage).

How does this phenomenon differ for an alternating current? The potential difference of the source now follows a sine function. As a result, the resistance R of the resistor increases. In general, the energy losses increase with the increasing frequency of the source. This is quite natural, because in addition to the bouncing of the electrons, they are accelerated back and forth by the alternating source and emit photons during these accelerations.

As long as the frequency of the source is low, the periodic emission of photons is overshadowed by the thermal noise of the bouncing electrons. At higher frequencies, the skin effect is the main process and the electrons on the surface of the wire emit photons into the environment. With a receiver you can measure the frequency of the source from a distance.

Now it is questionable whether the photons of a modulated radiation have the frequency of the source. Or does only the intensity of the photon emission follow the frequency of the source and every single photon still has its own frequency in the infrared range? How long is the wavelength of each photon? In the range of infrared or hundred of meters?

HolgerFiedler
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  • Really nice answer. Here is an interesting one: https://physics.stackexchange.com/questions/579543/is-it-possible-to-bend-light-without-changing-its-color – Árpád Szendrei Sep 14 '20 at 16:24
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If your circuit was oscillating since the beginning of time: Many photons are generated simultaneously, and the collection of these packets of multiple photons being produced at a constant rate produce the E-field. (As a technical point, it's actually the coherence between these possibilities that produce a nonzero E-field, but you can think of it like their "collective" properties produce the classical E-field.)

If your circuit had a starting "ramp-up" and "ramp-down" time: Your photons are less likely to occur during the ramp-up and ramp-down, so you get a wavefunction in time. (And then you can answer "when were the photons created" by specifying that the photons wavefunction in time was created when the circuit was "ramped-up")

Steven Sagona
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