Yes, I might be able to help you a little. I'm not a professor or anything but I do know a little bit.
Reactance (X) is analogous to resistance but only shows up when the voltage is changing.
Capacitors and inductors have opposite signed reactance. That is, when putting a voltage to a coil it will resist the flow of current and has a positive reactance, however putting that same voltage to a capacitor a current will immediately flow to fill the plate with electrons and we can say the capacitor has a negative reactance.
Also, with regard to the frequency, capacitors and inductors behave oppositely. Capacitors have very little problem passing current of high frequency but at low frequency the plates would fill with electrons and the flow would cease as the plates would reach capacity rather briefly. Inductors have a smaller reactance to lower frequencies on account the greater time for the voltage to push it's way though the coil and really get some current flowing.
Since capacitive reactance decreases with frequency, X=1/ωC (? = radians per sec. = 2*pi*f), and inductive reactance increases with frequency, X=ωL, in a LC circuit there will be one resonant frequency where they are equal. Since they are of opposite sign their reactances will cancel each other out and all you would see is the resistance.
Note; You have probably heard the word "impedance". Impedance(Z) = Resistance(R) + Reactance(X). Impedance is the total of the resistance to current flow for that circuit. Again, resistance happens anytime, even with DC. Reactance happens only with AC. Impedance is the combination of the two.
This is where Q comes in. Generally the "Quality" of capacitors are rather large and can be ignored as it's harder to make a high "Quality" inductor it is the limiting factor. So, Q=L/R. If the inductance (L) is large and the resistance (R) small the Q will be large.
Speaking of Q, besides resistance there is another thing that can decrease the Q and that is self capacitance. Self capacitance will short circuit your inductive reactance. Instead of the signal having to push the whole way through the inductor and really build up some momentum, like an inductor should do, if the coil has any self capacitance the signal will immediately propagate through the coil, as much as the self capacitance will let it, and a small current will immediately show up on the other side of the inductor. That decreases how much the inductor looks like a good inductor, decreases the inductive reactance and kills the Q of the circuit.
I'm sure you know that a higher Q let's the circuit ring for longer. Like a high quality bell, when struck it rings for a long time. Unlike a bell made of lead giving a muffled "thud" when struck. Your car suspension is low Q by design, springs (capacitor), mass of car (inductor), shock absorbers (large resistor). Otherwise your car would still be bouncing from the first pothole you hit, not to mention all the others along the way, when you arrive at your destination.
Q is also the magnifying factor of a resonant circuit. That basically means the voltage and or current inside the resonant circuit will equal the input voltage multiplied by Q. If a circuit had a Q = 100 and the input voltage is 10 volts after a while the voltage inside the circuit will build up to 100 * 10V = 1,000 Volts.
As for "matching" an inductor to your water capacitor, I would say; Unless you are trying to do something supper special like trying to match the length of your coil to your operating frequency (which wouldn't work anyway because the frequency is going to change even with the bubbles in the tubes) your might be able to throw any coil on there. I imagine you would only need to maintain a phase lock i.e. PLL (Phase Lock Loop). PLL's are what modern radios are built on so there is lot's of documentation, theory, examples, etc.
As for Stan's "resonant cavity", I really believe that he is talking about the cavity in the capacitor of the resonant electrical circuit. It may well be a way to extract the hidden energy of the tuned circuit.
I should explain about "hidden" or "invisible" energy. In a tuned circuit all the energy you add stays inside the circuit. Because the current and the voltage have the same phase there is no way to see any voltage difference on the outside of the circuit. Not being able to see voltage means no current will flow. That energy will stay inside the circuit because it has the path of least resistance right inside the circuit. It is happy as can be. No need to venture outside.
Stans "resonant cavity" is sitting right in the middle of the tuned circuit. As you add energy it will be stored in the circuit and just like hammering a bell, it gets louder and louder. If the electric field was strong enough to break down the water then that could be one way to tap into that "hidden" energy.
As for a physical force, that is absolutely true. If you know any thing about chemistry you probably know it is based on electricity. Anion and cations are electric charges.
Quarts crystal, as you may know, actually changes shape when under the influence of an electric field. It will shrink in length while growing in width. Water, being a polar molecule, will do the same thing. It will physically change shape when under the influence of an electric field. Granted, it would be harder to see it because it's a liquid.
I hope I've described things enough to prompt some new questions.
By the way, I hope you don't mind if I use the content of this message to post on the forum. I don't explain things too often so when I do I'd like to get it out there for everyone to critique. (Or I'm just lazy and don't like to type too much.)
