Equipartition and the Larmor Formula

Today I’m going to do something that’s probably never been done before. I’m going to derive the Larmor Formula from without using Maxwell’s Equations. I’m not going to use anything except the Equipartition Theorem. Well, maybe I’m going to cheat a little. I’m going to use the Rayleigh-Jeans formula for the number of modes in a cavity. I’ve used this formula once before. It looks like this:

Here is how the derivation is going to proceed. I’m going to start with a box 30cm x 30cm x 30cm. Inside the box I’m going to construct a tiny, miniature tether ball. You know what a tether ball is. It’s a ball that you tie to the end of a pole, and then you whack the ball so it flies around the pole. Except I don’t think I want the string to wrap around the pole, so I’m going to have a rotating sleeve where the string is attached. So the ball can circle indefinitely without getting wrapped up.

I need to specify the exact parameters of the tether ball. It will weigh 10ˆ-27 kilograms, and the length of the string will be 25 nanometers. Oh yes, and the ball will be charged. I will give it a charge of exactly 10ˆ(-20) coulombs. I think we’re all set.

Now we go up to the tether ball and give it a good hard whack, so it starts spinning around the pole. Since it’s charged, it starts spraying electromagnetic radiation into the box. The radiation goes in all directions, bouncing off the walls; at the same time, the tether ball slows down, because it’s losing energy. Does it finally come to rest? No…because at some point, there is so much radiation in the box that sometimes instead of losing energy, the tether ball actually absorbs energy from the radiation. At some point, on average, the ball is absorbing just as much radiation as it is emitting. That’s called thermal equilibrium. And that’s what I’m going to calculate, right?

Not exactly. I’m going to do something a little different today. We know that the system has to come to equilibrium: what I’m going to do is make use of a peculiar fact about that equilibrium to calculate the radiative power of the tether ball, otherwise known as the Larmor Formula. It’s a very backwards way of doing it, but it ought to work.

What is the singular property of the system in equilibrium which will give us the key to the calculation? Simply this: that for a given frequency, the energy per mode  of the electromagnetic field is equal to the energy per mode of the mechanical system.

That’s it. It’s easy to state, and not so hard to interpret. Can it really be enough derive the Larmor Formula, which tells us the amount of radiation from an accelerating charge? Let’s give it a try.

I know what some of you are thinking: "It's wrong! The equipartition formula fails! It leads to the ultraviolet catastrophe!"  Calm down. I never said that every mode of the e-m field gets the same amount of energy. I said that the equivalence between electrical and mechanical mode energies holds true at any give frequency. That's true classically, and I'm going to show that it remains true in quantum mechanics.

I’m going to make one more adjustment to the system before we get started. Instead of using a piece of string to attach my tether ball to the pole, I think I’ll use a rigid massless rod, with a sleeve bearing on the end of it. So the tether ball is constrained to rotate around one axis only, at a fixed distance from the pole. I think that might simplify the calculation a bit.

So here goes: we whack the ball, it spins like crazy, it fills the box with radiation, and eventually we come to some equilibrium where the ball stops slowing down on average. Let’s suppose that when we get to this point, the ball has an average velocity of 16,000 meters/sec. It could be anything, after all; it depends on how hard we whacked the ball to get it started. Let’s say it’s 16,000 m/sec.

It’s not too hard to convert this to equivalent energy: we already know the mass of the ball, and if you put the numbers in it comes to 1.25 x 10^(-19) joules. It’s 125 nanomicromicrojoules. Or 125 nanopicojoules, if you prefer.

The Equipartition Theorem tells us that energy distributes itself equally among all modes. The spinning of the tether ball is exactly one mode: each standing wave in an electromagnetic cavity is also a mode. So each electromagnetic standing wave mode should have, on average, 125 npjoules of energy. 

How do you calculate the energy of an electric field? You take the square of the field, divide by the impedance of free space, multiply by the volume of the box and divide by the speed of light. Or something like that..I think you also have to dived by two for some reason vaguely related to RMS values. I find if I start with 1.0 millivolts/meter, it comes out just right…a nice round number, isn’t it? Of course, that’s no accident. I cooked up all my parameters so it would end up this way.

Now let’s see what else I’ve cooked up. I gave the tether ball an average velocity of 16,000 m/sec, on a radius of 25 nanometers. If I’m right, this gives me an average frequency of exactly 100 GigaHertz. This gives me a radiated wavelength of exactly 3 millimeters, which fits quite nicely into my 30-cm box.

This is starting to look like a calculation I did once before, exactly fifty-one weeks ago. I had a different angle on it back then: I was going to show that by calculating the equilibrium between the field and an atomic oscillator, you could prove the Equipartition Theorem. I did a lot of pretty loose approximations and came up within a factor of ten of the exact answer. Which was pretty good at the time, all things considered. But since I figured out the other day that I had been using the wrong approximation for the antenna-to-atomic-oscillator correspondence, I now had the tantalizing possibility of recovering a missing factor of five. Hence today’s calculation. But along the way I decided to get fancy and do it backwards: instead of demonstrating the equipartition theorem, I’m trying to use  the equipartition theorem to derive the radiation of an atomic oscillator.

What it all means is I’m finding a lot of corrections and adjustments in my original calculation. These things tend to equalize themselves in the long run when you’re just doing ballpark work, which is why I was OK with being within a factor of ten last year. But now I’m trying to do accurate work, and it’s stressing me out just a bit. Never mind. We must persevere.

We’re getting close to the real meat of the calculation right now. This is based on a brilliant insight which I figured out last year. The problem is to calculate the effective field at the oscillator. But haven’t we already done that? I cooked up the numbers so that my electric field would come to 1 millivolt per meter...what more do I need?

We’re not even close yet. That 1mV/meter is the field amplitude per mode. There are millions  of modes, all randomly adding up with different phases and frequencies. Yes, it would be simple if we just had to calculate the motion of a tuned oscillator driven by a 100-mHz field of known amplitude. That is so  not what we have to do.

The brilliant thing I figured out last year was that I could apply the paradigm of the drunkard’s walk to the driven oscillator. I can hardly begin to describe what a brilliant idea this is. (Yes, that’s the third time I’ve used the word brilliant. Does that get me points on the crackpot index?)

The crux of the idea is this: I can truncate my field distribution anywhere I want, and I’ll still get the same result. I have a formula for the number of modes in a cavity. I can take all the modes between 99 GHz and 101 GHz, or I can just take only the modes between 99.9 GHz and 100.1 GHz. It doesn’t matter: I’m going to get the same result when I add them up and apply them to the harmonic oscillator. I demonstrated how that works in this article from last year. 

Let’s pick up this calculation again when I come back tomorrow.