Divine Plagiarism

So I was confused about what voltage was. Yeah, yeah, potential difference and all. I know the book definition. I know the units. But I wanted to see it on a quantum level. Here’s another boring, crazy theory. Don’t worry, I have no intentions of quitting my day job.

So voltage is like ‘smash.’ ‘Cause you can have the same current, and it seems like it has more ‘smash’ when its at a higher voltage. And ‘smash’ is momentum. (I know it’s not elegant like a proof. I don’t like proofs. And I don’t get paid to prove my theories. ‘Push stick forward, trees get bigger; Pull stick back, trees get smaller.’) So that is what it seems like to me. Therefore, in atrociously bad methodology, I’ll use whatever data is convenient to support my pre-arrived-at conclusion (which everyone does anyways, no matter how many regressions they use.) That’s what I learned in the Social Sciences.

So ‘ya gotta decide what kind of atom you’re gonna use if you wanna see voltage on a quantum level. (Wanted to see how much bad grammar I could use in a sentence and still say the word quantum.) You’d think a metal would work, but the problem with that is the very thing that makes it a metal. With the loose affinity of outer orbital electrons for their respective nuclei, you throw an electron with a lot of smash into a metal, and it’ll just knock all the other electrons in sequence, until the whole electron cloud pukes out a ‘tron with just about as much smash on the other side of the field (minus transmission losses due to resistance.) It’s like a thick straw full of marbles. Put a marble in one side, it will just knock out another marble out the other side with about as much momentum. So we won’t use a metal now. But we’ll come back to it. Let’s use a noble gas instead, like in a florescent light. There’s a pretty cool instant where the electron jumps up before collapsing back down and emitting a photon.

So to our inert gas and florescent lamp. Three fundamental parts of our electricity equation: V=IR. So we’ll drop one time term out of the equation for the time being. Instead of current (charge per time,) we’ll just use charge. And to keep it as simple as possible on the quantum level, we’ll make that charge be just one electron’s worth. So what actually happens in the light? An electron smashes into the noble gas, knocks an electron up a few quantum shells (conservation of energy and all,) which comes crashing back down, emitting a photon. Hence light. The voltage is burned when the photon is emitted. But let’s figure out how that happens. Go to the instant when the electron is kicked up some quantum shells. (Note: I’m talking deterministically about this stuff, when I know its probabilistic, non-zero probabilities and all. I’m assuming a zillion atoms so I get a nice mean and standard dev. Just easier to talk about that way.) So we’ve got a ’tron which has leapt up a number of quantum shells (or a fractional number if we’re using averages.) Let’s bring in the laws of orbital motion. In order for something to have constant radial velocity and a wider orbit, it must have higher linear velocity. So in that leap, there is a positive delta in linear velocity on the electron. Holding mass constant, that means the momentum of the electron has increased. But voltage is potential difference, and in order to have potential differences, you have to be resisting something (the null state has to be something other than the current state, and there has to be something trying to pull to the null.) Consider holding a book off the floor. You have a potential difference, which stores energy, because the null state of the book is on the floor, and gravity is resisting you. If you let go of the book, the potential difference is swapped for the energy it was holding. Back to the atom. Electrons orbit in shells (back to deterministic vs. stochastic, acknowledging Heisenberg, etc, just using big numbers,) so their potential difference is the distance between their null state (normal orbital) and their excited state (leap orbital.) So this explains voltage, which must be then a function of the momentum of electrons, and hence a function of aggregate quantum leaps. And this brings us to our third term: resistance. This one is simpler. Gravity resisted our book. You need resistance to have potential difference, and this is why there is an R in the equation. And R pulls against the electrons, which makes sense given its placement next to the current term. Its pulling against electro-weak (Electromagnetism.) The negatively charged electrons are drawn to the positively charged nucleus. So these are our three terms.

Putting back in the time term we took out earlier, we find that 1) Current is electrons per unit time (we already knew that), 2) Voltage is aggregate electron momentum per unit time, or change in electron momentum (expressed in a summation of quantum leaps,) and 3) Resistance is the pull of electromagnetism on electrons per unit time. Going back to the metal from the noble gas, a lot of our deterministic assumptions don’t work as well, with the cloud of electrons and all. But momentum is still expressed in the same way. So in an infinite sample size, we would still see voltage stored in quantum leaps, even though those quantum leaps would be transferred very rapidly, until they hit something which could resist them, where they would drop their voltage.

This brings us to the third law of thermodynamics and the idea of superconductors. It is motion hence and imperfect momentum transfer which introduces resistance (and hence heat) into conductors. If our marbles in a tube analogy was perfect, then all the electron momentum would be transferred (hence no voltage loss, no resistance (because there would be no time for magnetism to pull on the offending ‘tron.) and no heat buildup.) And here we return to the perfect crystal lattice at absolute zero in the third law. There would be no resistance, as momentum would come out the other side at the instant it went in one side. Of course, this can only occur in theory, or at least not in this universe or in any that could have any meaningful connection to ours. A useful analogy would be ice. When you step on ice, it melts and becomes slippery. Your interaction in the system is what introduces the resistance (or lack thereof.) So a perfect conductor would be like ice which doesn’t melt when you step on it. Where your involvement in the system changes it as little as possible. Because it takes energy to change the other system, and the measure of your losses is the measure of the changes in the transport system (your voltage drop is making the heat in the wires.) So if you were to hit a patch of ice, and come out the other side at the same speed, you have perfect ice. (Of course, unmelted ice isn’t slippery, and it takes your pressure, hence friction, hence losses to melt the ice. So its not a perfect analogy.) That was fun. See ‘ya.