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I'm trying to get a more intuitive understanding of resonant inductive coupling. It's supposed be a more efficient way to transfer electrical energy wirelessly, because the coils are only coupled by near fields, and don't waste energy as far field radiation. The efficiency is supposed to increase as Q increases (the resistive component decreases).
LC circuit schematic

Ideally, an LC circuit has a Q of ∞. If you charge up the capacitor in an LC circuit, and then leave it floating in space away from anything else, the energy will cycle back and forth between the L and C. The voltage, current, electric field and magnetic field will all cycle sinusoidally (right?) With ideal components, and no resistive component, this will continue cycling back and forth forever.

Real LC circuits have some resistance, which wastes some of the energy as thermal radiation (in a frequency band related only to temperature), and the cycling eventually dies. I think they also have some other non-idealities that allow energy to escape as far-field electromagnetic radiation (at a frequency related to L and C), correct? What are these non-idealities? Are they independent of the resistive component?

If another identical LC circuit is brought near this oscillating circuit, will energy transfer from one circuit to the other due to the changing magnetic fields? Will this happen even with ideal components that don't have far field radiation? Is there no energy transfer if the coils are perpendicular and the axis of one is coplanar with the other, like this?: 

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Will the energy distribute itself evenly among the circuits, so they reach an equilibrium where each is circulating half of the original energy? Or will the energy oscillate back and forth between the two circuits, with one alternately being depleted or full?

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Real LC circuits have some resistance, which wastes some of the energy as thermal radiation, and the cycling eventually dies. I think they also have some other non-idealities that allow energy to escape as far field electromagnetic radiation, correct? What are these non-idealities? Are they independent of the resistive component?

Energy will also be lost due to electromagnetic radiation. The amount of energy loss depends on the size of the circuitry. In the limit of infinitely small inductors and capacitors the loss to radiation falls to zero. In general, you want the circuit to be smaller than the natural wavelength of the radiation it produces. For example, if the frequency is 1GHz the period is 1 nsec. With the speed of light $3\times 10^8$ m/s, the wavelength is $(3\times 10^8)/10^9 = 0.3$ meters so you want your circuitry to be much smaller than 0.3 meters.

In addition to far field radiation, all circuit components have many other non-idealities and some of these result in thermal losses. Mostly it's about inductors, capacitors, and wires having resistance as the other circuit elements do not create heat.

If another identical LC circuit is brought near this oscillating circuit, will energy transfer from one circuit to the other due to the changing magnetic fields?

Yes.

Will this happen even with ideal components that don't have far field radiation?

No. To get near field radiation you need to have an antenna and these are not included in ideal circuit components. In the limit of infinitely small circuit components the near field is infinitely small. (This is an instinct from many years of circuit design; I'll check it more carefully and remove this comment if I'm assured of the answer.)

Is there no energy transfer if the coils are perpendicular and the axis of one is coplanar with the other, like this?: | —

Correct.

Will the energy distribute itself evenly among the circuits, so they reach an equilibrium where each is circulating half of the original energy? Or will the energy oscillate back and forth between the two circuits, with one alternately being depleted or full?

This is a wonderful question! What you have is a pair of coupled oscillators. In physics we analyze these sorts of things by looking at what kind of activity would NOT alternate between the two circuits. These are called the "normal modes" and since there are two harmonic oscillators here, there will be two normal modes.

Let's look at the problem from a general point of view. Let $(1,a)$ and $(1,b)$ be the two normal modes where $a$ and $b$ are complex numbers. That is, I've normalized the normal modes to the first oscillator. These two normal modes each has an exponential (i.e. cosine plus sine) dependence. That is, the first would actually look like $(1,a)\cos(\omega_1 t)$ and the second with an unequal $\omega_2$.

To get all the energy into the second oscillator (at time t=0), requires taking equal and opposite amounts of the two modes. Thus the values would be $$(\cos(\omega_1 t) - \cos(\omega_2 t), a\cos(\omega_1 t)-b\cos(\omega_2 t))$$.

To answer your question, we look at the 2nd part of the above. That is, I've arranged to match the phases and sum to zero for the first part at time t=0. At some later time will the second part also match phase and sum to zero?

The answer is that this can only happen if the magnitudes of $a$ and $b$ are the same. So in general, the answer to your question is that a long term oscillation which sees all the energy in one of the circuits will not, in general, eventually get to a point where all the energy transfers to the other circuit.

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  • $\begingroup$ Is there a geometry that maximizes near-field coupling while minimizing far-field radiation? Or is a good antenna for one inherently a good antenna for the other? $\endgroup$ – endolith Jun 7 '11 at 4:11
  • $\begingroup$ "since there are two harmonic oscillators here" - in this usage for energy transfer, the resonators are identical, or at least, have the same resonant frequency. $\endgroup$ – endolith Jun 7 '11 at 4:23
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    $\begingroup$ Geometry that maximizes near field coupling would be parallel coils, just like you already thought. The "two harmonic oscillators" can be quite arbitrary. If they are identical, then yeah, the normal frequencies will have the same magnitudes for $a$ and $b$. In fact, $a=-b$ I think. The nice applet works best when you make the coupling very small. And you analogy is correct, I think. $\endgroup$ – Carl Brannen Jun 7 '11 at 19:54
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    $\begingroup$ @Carl How do you justify the claim that radiation losses vanish with diminishing component size? My intuition says that in this limit, the capacitor and inductor will reduce to electric and magnetic dipoles, respectively, but will be radiating just fine. $\endgroup$ – Leo Alekseyev Jun 22 '11 at 9:53
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    $\begingroup$ @Leo; It's because the strength of, for example, an electric dipole, goes to zero as its size goes to zero. That is, what you call an "electric dipole" in order to maintain some specific strength as the distance between the poles decreases, must have a corresponding increase in the charge. If you keep the charge constant, the strength of the dipole goes to zero as the charges approach each other. $\endgroup$ – Carl Brannen Jun 22 '11 at 20:35
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You have asked many questions. Let me answer them one by one.

  1. If ideally there is no resistance then the only loss will be core loss. Whatever (except free space) is within the inductor undergoes hysteresis and because of that some energy will be lost. Also there will be some loss in the core due to eddy currents in the core.

  2. Stray magnetic field loss. Some energy will be lost due to whatever stray magnetic field present there.

  3. If another LC circuit is brought near the original circuit then it will act as a transformer and some energy will transferred from one inductor to another by transformer action. The usual losses like core loss, hysteresis and stray magnetic field losses (except copper loss) will be there.

  4. If the inductor of the second circuit is kept at perpendicular position to the first one then if the magnetic circuit between the two inductor closes itself through the coils then energy will be transferred as usual.

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  • $\begingroup$ Oh, you're right. I meant perpendicular and... symmetrical. I'll try to explain. $\endgroup$ – endolith May 3 '11 at 2:59
  • $\begingroup$ In my example, there is no core except free space. These are ideal components made of perfect conductors, etc. What do you mean by magnetic field loss? $\endgroup$ – endolith May 3 '11 at 3:01
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    $\begingroup$ If there is free space then there will be no core loss and eddy current loss. The stray loss is due the leakage of flux which may be intercepted by nearby conductor like the support structure etc. and for ideal cases it is also zero. $\endgroup$ – user1355 May 3 '11 at 4:01
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    $\begingroup$ If all fluxes are parallel to the whole part of the second coil and there is only free space then no transformer action can take place and energy will not be transferred. $\endgroup$ – user1355 May 3 '11 at 4:06
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    $\begingroup$ Dear @sb1, it's all great but could you please answer the main question - in non-engineering terms that often already suppose an answer without asking why - whether the energy just disappears in the form of electromagnetic waves and how much? How much do we have to try to "catch it" by surrounding the circuit? $\endgroup$ – Luboš Motl May 3 '11 at 4:20

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