Understanding radiation and coupling of LC circuits 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).

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?: 
| —

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?
 A: 
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.
A: You have asked many questions. Let me answer them one by one.


*

*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.

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

*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.

*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.
A: The charge of the inductor and capacitor will determine the loss. Off resonance, you look at heat and some radiation into open air, which might be picked up by other resonant inductors/capacitors by the inverse square law.
