Sunday, February 26, 2012

Deriving the Larmor Formula

I promised to derive the Larmor Formula and yesterday I drew up an outline of how the calculation is supposed to go. It’s based on the equilibrium between mechanical and electromagnetic energy in a cavity. Reviewing the outline, we can see that these are the key steps:

1. Set up a rigid rotor inside the cavity with arbitrary parameters of mass, length, and charge. Let it come to thermal equilibrium inside the cavity with an arbitrary amount of energy. By the Equipartition Theorem, that will also be the energy per mode of the electromagnetic field.

2. Pick an arbitrary size for the cavity and use the Rayleigh-Jeans formula to count up how many cavity modes there are in the vicinity of the rotor frequency, as already defined by its energy and dimensions. Pick an arbitrary bandwidth and only count the number of modes within that bandwidth.

3. From the mode energy of the field and the size of the cavity, calculate the electric field amplitude of the mode. Add up all the modes within our arbitrary bandwidth as though they were a Fourier Series, with the amplitudes in phase at an arbitrary point. The result will be a pulse train of frequency bursts with a characteristic amplitude, burst length, and repetition rate.

4. Imagine a harmonic oscillator with the same mass as our rigid rotor, and the same characteristic frequency. Apply a single frequency burst, as calculated in Step 3, to this oscillator at rest, and calculate the resulting amplitude of oscillation.

5. From the amplitude determined in (4), calculate the equivalent energy. By analogy with the Drunkard’s Walk, this is the average amount of energy gained from each pulse burst. Therefore, the rate of power absorption is just this amount of energy multiplied by the repetition rate (from Step 3).

6. The system is in balance when the rate of absorption is equal to the rate of emission. Assume that the emission rate is proportional to the square of (acceleration)*(charge). From all the parameters determined so far, calculate the constant of proportionality. Compare the result to the Larmor Formula.
Shall we begin? I already did the first couple of steps the other day. Let’s pick up where we left off.

1. The rigid rotor is a single mass on the end of a swivel arm attached to a post, so it is constrained to spin about one axis only. We gave it these arbitrary parameters:
Mass = 10ˆ(-27) kg
Charge = 10ˆ(-20) Coulomb
Radius of gyration = 25 nanometers
Frequency at Equilibrium = 100 GigaHertz
From these parameters, we can also derive the energy:
Energy at Equilibrium = 1.25 x 10ˆ(-19) joules (or 125 nanopicojoules).

2. We put our rigid rotor inside a cubical cavity 30 cm on a side. From our frequency, we can calculate a wavelength of 3 millimeters, so the cavity accommodates exactly 100 wavelengths each way. Now we use the Rayleigh-Jeans formula to count up how many modes there are between 99 GHz and 101 GHz. This is actually a calculation we did last year, and we counted up approximately 250,000 modes. But this year we’re going to notice that only some of those modes are effective in driving the rotor. Considering three orthogonal axes, it is clear that two out of three orientations will have no driving force on the rotor. So we will divide by three and work with only 80,000 modes.

3. From the size of the cavity and the energy of the standing wave, we can calculate the peak electric field amplitude of a typical mode. Correct me if I’m wrong, but I make it 1.0 millivolt/meter. That’s not exactly an accident: I picked the rotor energy to make it come out that way. Now comes the Fourier magic. We know that we have to add up 80,000 sine waves, centered on a frequency of 100 GHz. Spreading these modes out over our chosen bandwidth (2 GHz), we find that the average spacing between modes is 25,000 Hz. 

The calculation is actually easier than it looks. We start off just by focussing on the middle frequency and it’s two closest sidebands:

I’ve cut the sidebands in half so this picture represents the very familiar AM radio calculation for a modulated sine wave. We know what the result looks like: 
It’s a modulated sine wave. For purposes of description, I’m going to pretend it’s an “equivalent”square wave, so I can use the language of duty cycle and repetition rate. The nice thing about the calculation is that the basic repetition rate isn’t going to change as we add more sidebands. The additional sidebands only shape the pulse within that framework. It’s not hard to see that the peak amplitude grows by two millivolts each time you add another pair of sidebands; so the height is proportional to the number of sidebands. What about the burst duration? It turns out to be just inversely proportional to the number of sidebands. You can see that it has to be that way, because that’s the only way the total energy of the burst will be proportional to the number of sidebands. (I explained how to do these Fourier series last year.) The result is a train of pulse bursts with the following parameters:

Amplitude:  80 volts/meter
Duration:  500 picoseconds
Repetition Rate: 25,000 per second

4. The next thing we have to do is apply one of these pulse bursts to a harmonic oscillator initially at rest.  I’m going to assume that in this situation the force is always applied with maximum efficiency, that is: it is always in synch with the velocity.With this assumption it is fairly easy to ballpark how much momentum it pumps into the oscillator: it is just ½ *(force)* (time). (The factor of ½ is due to the pulsating character of the driving force. The force is of course given by the (electric field)*(charge); put the numbers in, and I get a momentum of of 2*10ˆ(-28) Newton-seconds.

5. How much energy did that pulse burst impart to the oscillator? Using the formula pˆ2/2m, this converts to an equivalent energy of... 2*10ˆ(-29) joules. Since there are 25,000 pulse bursts per second, it means the oscillator will be absorbing power at a rate of  of 0.5 * 10ˆ(-24) joules per second, or one-half pico-pico-watt.

Remember that funny property of the random walk: the amplitude-squared grows at a constant rate. Similarly with the driven oscillator: the random phase of the incoming wave burst corresponds to the random direction of the drunkard’s footstep, so in this case it is the energy of the oscillator that grows at a constant rate…a rate of 0.5 ppWatts.

6. Here’s where I mix metaphors just a little. I’ve been building up the oscillations as though I have a harmonic oscillator…a mass-on-a-spring. But now that I’m there, I’d rather apply it to the case of the rigid rotor. At the point of equilibrium, it’s the same frequency and the same energy in either case. The way it works is that you assume the radiated power is proportional to (square of the charge) x (square of the acceleration). This is a totally straightfoward assumption that is justified by all kinds of physical reasoning. All we need to do is figure out the constant of proportionality. The charge we know: 10ˆ(-20) coulombs; the acceleration is easy to get from omega-squared-r. Put them together and square them and I get very nearly 10ˆ(-8) in SI units.  Recalling that we are putting out a radiative power of one-half pico-pico-watts, this means the constant of proportionality is…5 x 10ˆ(-17).

How does this compare with the Larmor Formula? It’s pretty close. Recalling the formula with the permittivity of free space and all that, we see that all the constants multiplied together come to…2.2 x 10ˆ(-16). I seem to be within a factor of 4 of the exact answer.

I have to say that’s not a bad outcome. With all the approximations I made and the short-cuts I took, there are so many places I could have slipped in an extra factor of 2…which incidentally would have given me an error of 4x on the power. So all things considered, I’m really lucky I came out as close as I did. I could  have easily gone back and patched it up after the fact, to make it look like I got it perfect, but I think it’s better to leave it as it is.

Finally, we shouldn’t forget that having basically derived the Larmor Formula, we’ve also effectively derived the radiation resistance of a short dipole…since I worked out the physical  correspondence between those two formulas last week. I know I did a derivation for the antenna resistance last year by letting sheets of current flow on a sphere, but that was actually a pretty sketchy calculation. Now we’ve done it theory, at least. Regardless of calculation errors, the method obviously works in principle.

It turns out that I have one more method for calculating the radiation resistance of an antenna; it’s very different, and as I don’t seem to get tired of saying these days, it’s also a very cool calculation. I’m going to post it one of these days.

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