## Wednesday, March 2, 2011

### Let's run the numbers

I've proposed that a mechanical oscillator will be in equilibrium with an external field when that field has the same energy per mode as the mechanical oscillator. Now I have to show how the numbers work. I'm going to sacrifice some accuracy and rigor in order to present the calculation in a way that is easy (?) to follow.

There was a stumbling block for me when I tried to count modes and ended up with a result different from the Rayleigh-Jeans formula by a factor of three. I couldn't resolve the discrepancy so I'm just going to use the number from Rayleigh Jeans. Here is that formula again:

I'm going to choose a frequency of 100 GHz and make a box 30 cm on a side. You can see that I have a wavelength of 3 mm so exactly one hundred waves fit into the box lengthwise.

I'm going to add up all the modes between 99 and 101 GHz, so I need to know how many modes there are. The formula given here is for wavelength, not frequency, so I'm going to let lamda range from 2.97 mm to 3.03 mm, which should come to just about the same thing. Letting pi=3 and using some shortcuts, I'm getting 240 000 modes, or about a quarter of a million. You can see if I space the waves out equally over the frequency interval they are about 8 kHz apart.

So we have a quarter million sine waves between 99 and 101 GHz and we want to add them up and see what they look like. Remember, I could have just as easily taken one tenth of that bandwidth and added up only the waves between 99.9 and 100.1 GHz. The beauty of this analysis is that the equilibrium would come out to the same point either way, as I explained in a previous post. So I am at liberty to choose any arbitrary numbers. There are people who would say "let's just work in symbols and let the formulas do the work" but I think it's easier to follow with the actual numbers.

I'm going to let each sine wave have an intensity of one millivolt per meter. There are two things we have to do next: add up a quarter of a million sine waves to get the composite waveform, and figure out the energy in each mode. Let's do the energy calculation first. It's the square of the field divided by the impedance of free space (377 ohms) and the speed of light. It comes to 10^-17 joules. I'm ignoring things like a factor of two somewhere.

EDIT: Oops. I actually ignored the size of the box, which matters a lot. What I got was energy per unit volume. The actual mode energy therefore comes to around 4 x 10^-19 joules.

Now let's add up the sine waves. I'm again ignoring details like the fact that they are all going in different random directions. These calculations are basically a Fourier series which becomes in the limiting case a Fourier transform. There are a quarter of a million waves all in phase at the exact center frequency. So the peak amplitude is 250 volts/meter. (Remember the mode amplitude was 1 mV/m). I have a total bandwidth of 2000 MHz with a space of just 8 kHz between waves. I explained how to add these things up in a previous post. You start by taking the middle three waves... in AM radio, that would be the carrier and the two sidebands. It's pretty well known that this gives you an 8 kHz signal modulated at 100 GHz. To simplify the math, we will think of this as a series of 100 GHz wave trains 62.5 microseconds in duration (let's just call it 60), spaced 125 microseconds apart (a 50% duty cycle.)

Now we will start adding more side bands. When we add a second pair of sidebands, the pulse train gets shorter ... 30 microseconds instead of 60 ... but the "window" of 125 microseconds stays the same. We get a 25% duty cycle. The repetition rate stays the same but the duty cycle gets shorter.

This is how it works each time we add a pair of sidebands, and we must add 120,000 such pairs. The end result is a series of pulse trains 500 picoseconds long, separated by an interval of 125 microseconds. With a repetition rate of 8 kHz.

Let's apply this pulse to an atomic oscillator with a reasonable mass and charge: let's say, 10^-27 kg and 10^-20 coulombs. It's of course a sinusoidal force and there is a spring to consider but I'm going to ignore all that and just calculate the effect of a straight force applied to an inertial mass. I'm going to be out by a factor of 2 or 4 or something, but that's OK. The final velocity of the particle will be (force) x (time) / (mass). The force is just (field) x (charge). So the calculation gives me

(250 V/m) x (10^-20 Coulombs) x (500 x 10^-12 sec) / (10^-27 kg) = 1.25 m/sec

So the energy of oscillation, from 1/2 mv^2, is close to 10^-27 joules. That's from a single impulse.

Let's now remember how the drunkard's walk works. The distance increases as the square root of the number of steps; so the square of the distance is linear in the number of steps. Likewise with the harmonic oscillator. Each impulse adds to the amplitude in a randomly oriented phase, but the square of the amplitude (the energy) increases linearly. There are, we may recall, 8000 impulses per second; so the energy starting from rest increasees at a rate of 8 x 10^-24 joules per second. The question is: what will limit this buildup of energy?

Answer: the radiative losses. We simply need to calculate what size of antenna will radiate with a power of 8 x 10^-24 watts. At the point where the atomic oscillations build up to that amplitude, the system will be in equilibrium.

We will use the classical formula for the radiation resistance of a half-wave dipole: it is

200 (L/Lamda)^2

where lamda is the wavelength and L is the antenna length. You can see this gives the result of 50 ohms for the half-wave dipole; it is well known to radio amatures that the correct value is 73 ohms, but this diverges somewhat in the limiting case. From the assumed values of charge and frequency, and recalling the nursery rhyme "twinkle, twinkle little star, power equals I-squared-R", we get the following condition:

{(10^11 Hz) x (10^-20 Coulombs)} ^2 x (resistance) = 8 x 10^-24 Watts

We therefore require a radiation resistance of 8 x 10^-6 ohms, which from the formula gives us an antenna length of 6 x 10^-7 meters. We must now ask the question: what is the mechanical energy of this oscillator?

It's not hard to figure out a velocity of close to 10^5 m/sec for the oscillator; squaring this and multiplying by 1/2 times the mass, we get an energy of 5 x 10^-18 joules.

We're out by a factor of ten from what we set out to show: that the electromagnetic mode energy is equal to the mechanical energy. It should be realized that we took an awful lot of short-cuts and neglected a whole bunch of things, so it's not surprising that we might be out by an order of magnitude. I may be biased, but I'm going to judge this result to be "close enough".

The proper thing would be to redo the whole calculation with symbols instead of arbitrary values and show that no matter what parameters you choose, the result is that the energy of the mechanical oscillator comes out equal to the energy of a single mode of the electromagnetic field. I can hardly doubt that the result must come out correctly, but I'm not about to do the work to prove it. What I find hardest to understand is that there are people who would prefer to do it symbolically from the get-go, rather than run a numerical example first to make sure everything is lined up right.

There's one last thing that still bothers me. I have almost no doubt that the identity must hold, and the calculation strongly suggests that this is so. What I don't have is a simple way of understanding why it must be so. I think there ought to be a way of seeing that it has to work without actually grinding through the numbers.

### Where does the equilibrium come from?

In my very first blogpost just over a year ago, I talked about how I really felt I understood classical physics even if I wasn't capable of every single calculation. The example I took was Maxwell's calculation of the equilibrium distribution of velocities in a gas. I said that even if I didn't know how to calculate the distribution, I could "see" how it more or less had to come about.

The irony in this example is that over the last two weeks, I've been agonizing over how to work out the exact same equilibrium, except as applied to the classical radiation field. And my problem was not merely one of how to do the calculation. It was the basic physics that had me baffled. Just why and how does an equilibrium come about in the first place? I know that an oscillating atom can either absorb or emit radiation. But I had no definite way of seeing why the tendency to absorb or radiate should depend on the magnitude of the ambient field. Rather than saying that there had to be an equilibrium point, I could have just as well argued, for example, that the amount of outward radiation would always excede the amount of absorption.

I went down a number of dead end roads before solving the problem. Mostly I was trying to analyze the ambient field as the sum of an in-phase and and out-of-phase component to the atomic vibration. To the extent that the fields are in quadrature (out-of-phase) there is no interaction, so the atomic oscillator is always emitting. Where does the absorption come into effect, to counterbalance the emission? As the fields go in and out of phase, they are either in a leading or lagging relationship. One is absorbing, the other is emitting. But if the phases are basically random, the quantities should apparently just cancel out. So there is no equilibrium between absorption and emission.

So then I looked for mechanisms whereby the ambient field would "drive" the atomic oscillator slightly off frequency, pulling the two fields into synchronism just like a motor when it is connected to a 60-Hz power line. I tried in vain to make this model work and simply gave up.

Then I had two essential inspirations that gave me the solution of the problem. The first was the drunkard's walk. I realized that even when the impulse is in completely random directions, there is still a net tendency for outward progress. This was the first part of the puzzle. Then I came up with a way of analyzing random fields whereby I could convert a power spectrum into a time-varying electric field. (See most "Harmonic Oscillator: The problem is solved" ). The amazing thing about this analysis is that no matter what I chose for an arbitrary frequency cutoff or discrete resolution, I ended up with waveforms, all of them different from each other, but all of which gave the same result when used in a "random walk" analysis of the harmonic oscillator.

At last I could say I understood the essential physics of the equilibrium process. Now, when you really understand something, you ought to be able to do some calculations with it; and that's what I'm going to do next. It's not the kind of thing where the numbers are going to come out exact, because I've made some approximations along the way; and to some extent, I'm allowing myself to be a little sloppy. The point of this kind of calculation is to see if you are at least in the ballpark, and I think I've been able to do that much.