## 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.