## Wednesday, February 29, 2012

### Curses! Foiled again.

It's the damned calculus that gets me every time. I actually ran the integration on my antenna cross-section calculation, and I got the same answer I got the first time by just guessing. Which means I'm still out of whack with the rest of the world by a factor of 2. Hard to understand.

One the one hand, you don't get that close to the right answer unless your physics is pretty sound to begin with. There's little doubt that the discrepancy is just a matter of a math error on my part. The problem is: do you go back and obsess over it until you find the error, or do you accept that you're basically on the right track and move on to other things? You'd theoretically make more progress that way, by pressing ahead; but you never know when you might miss something important. It's a tough one.

The calculus was in fact rather picturesque, but it's not really the kind of thing I that lends itself to a blog treatment. However, I suppose I ought to sketch it out, just in case anyone is wondering. So here goes:

If you recall the superposition of the two wave systems, one circular and one planar, the question becomes: how fast doe the circular waves go out of phase with the plane waves, as you move away from the axis of symmetry? It shouldn't be that hard to convince yourself that the phase is quadratic in x, where x is the radial distance. (That's because a circle is basically the same as a parabola, if you take it over a shallow enough section of the curve. Which we can always do by backing off sufficiently far from the antenna.) The assumption is going to be that we're far enough away from the receiving antenna that the circular field is much smaller than the planar field, so the binomial approximation for power density will apply: (1-x)^2 = 1-2x, where x is the in-phase component of the circular wave.

Here it gets a little tricky: we still don't know what the phase difference of the two wave patterns is, taken along the axis of symmetry. I've drawn it as though the phase difference is zero and grows as you move outwards; but in fact, it might be anything. It's not clear exactly how it has to start off in order to maximize the power absorption. Fortunately, we can cover all possibilities by two special cases: sin and cosine. Either you start off exactly in phase, or you start of 90 degrees out of phase...or it's an intermediate case which you can make up by putting together those two special cases.

Then we just have to integrate the field over the cross-section. Don't forget the factor of 2*pi*xdx which you get for the circular rings when you set up the integrals: we then have to evaluate:

and we also have to check the quadrature case,
where k is a scaling factor to be determined by the physics of the situation. I actually ran these integrals in Excel and had two surprises: first, it was the sin integral that gave me the positive result. I always thought it was the cosine integral, but that one goes to zero. It means that along the axis of symmetry, far away from the antenna, the two fields are 90 degrees out of phase. For maximum power absorption, of course. The other weird thing was that doing it numerically, I clearly came up with a simple multiple of pi. It's funny in physics how the crazy integrals that come up happen to be analytically solvable as often as they are. So I looked again, and there is an obvious substitution of u=x^2. I guess anyone would have seen that. The integral solves pretty easily after all. It's the scaling factor that takes a bit of work.

But if you really think about it, you shouldn't be that quick to accept the formal mathematical solution. Just look at the function. It grows and grows, and oscillates faster and faster. If we graph it, it's pretty nasty looking:

Doesn't that grow like crazy? Yes and no. Physically, what's happening is that as you get farther from the axis of symmetry, the two wave patterns are going in and out of phase so rapidly that there is no net effect. It's like shining to flashlight beams across each other. In theory, you should calculate the power flow by adding the wave vectors everywhere and calculating the Poynting vector, which will be going crazy. But in practise, the effect is for all the micro-fluctuations to average out to zero. All the real physics happens in the first few lumps.

But how do we handle that mathematically? I actually did a problem like this once before, when I was adding up one of those crazy Ramanujan series. It came up in my calculation of the Casimir effect, and it was something like 1+2-3+4.... which adds up to 0.25. How? You cover it over with a very gentle Gaussian that preserves the low end and gradually suppresses the high-end fluctuation. I did exactly the same thing when I did this sin integral in Excel. You just keep adjusting the width of the Gaussian until the value stabilises, and then you know you're done. Actually, the fluctuations in this integral are almost like that other series, except instead of alternating integers, it's pretty much the square roots of the integers that alternate. (Because of the x-squared inside the sine.)

Like I said, the integral turned out to be not the hard part. It was getting the right scaling factor to line it up to the physics. I did my best; I'm not going to drag you through the details, but when it was all over, I was still out by a factor of 2. It's just one of those things.

## Tuesday, February 28, 2012

### Effective Cross-Section of a Dipole Antenna

Over the last couple of weeks I’ve been doing some pretty rough calculations to figure out the radiation from an oscillating charge. I’ve been in the ballpark, but today I think I’m like to get it right on the money. This is a calculation I figured out just over twenty years ago, and I’m pretty sure you won’t see it anywhere. I sketched it out last year in this article on the Crystal Radio, but I never did the numbers. That will be today’s project.

So far we’ve worked on the Larmor Formula, which gives you the radiation of an accelerating charge, and we’ve also tackled the radiation resistance of a short dipole. A charge in sinusoidal motion is just a special case of an oscillating charge, and I worked out a geometric argument which shows how to convert this case to an equivalent short dipole antenna. So these two formulas are really different versions of the same thing. Today I’m going to do a third version, which is also equivalent: the effective cross-sectional area of a small receiving antenna. If we know the power carried in the wave, this cross-section gives us the same information, in terms of absorbed power, as the radiation resistance. So if we know one, we know the other…in fact, we know all three.

The key to the calculation is this picture (see below), showing the interference of two wave patterns. The plane waves, moving from left to right, represent the incoming power from a distant transmitter. The circular waves, moving outwards, represent the response of the receiving antenna. What you have to understand is an antenna becomes a receiver by being a transmitter. The key is the shadow zone I’ve shown, where the two wave systems are close enough to be in phase with each other.

Everyone knows that a parabola is defined by the equation y = xˆ2, but not everyone knows this alternative definition: the locus of all points such that the difference of the distances to a given point (the focus) and a line not on that point (the directrix) is a constant. It’s called the “focus-directrix definition", and there are similar definitions for all the conic sections: circles, ellipses, and hyperbolas. For the antenna system shown in the diagram, the rough outline of the shadow zone is defined by the set of points where the phase difference of the two wave systems becomes 180 degrees, and according to the focus-directrix definition, that outline is a perfect parabola, Actually, it’s a paraboloid of rotation, if we look at it in three dimensions.

The calculation I’m about to do doesn’t rely on any details of electromagnetic theory, except two: I need to assume that the power in a wave is proportional to the square of its amplitude, and I will also assume that the radiated power depends only on that component of motion perpendicular to the angle of viewing. Putting these two assumptions together, for example, we would conclude that a short dipole puts out 50% of it’s peak output when viewed from a 45 degree angle. It’s a physically reasonable result, and it happens to be true.

We’re not going to worry about the details of tuning and matching: we’re going to focus on the resulting interference pattern. There’s a nice, quick-and-dirty way of ballparking the answer, and that is just to assume that inside the shadow zone, we have perfect interference; and outside, we don’t. It’s the cross-sectional area of the shadow zone that’s a bit dodgy: from the picture, you can see that it depends how far you are back from the receiving antenna, and there’s a square root relationship. In fact, if you are, say, seven wavelengths back of the antenna, the the cross-section of the paraboloid will be on the order of seven square wavelengths. And that will be enough to tell us the effective absorption cross-section of the antenna.

Here is how it goes. We back off reasonably far from the antenna, e.g. 20 wavelengths. The cross-section of the shadow is therefore on the order of 400  (oops) 20 square wavelengths. Now we ask: how much energy is being scattered by the receiving antenna? Suppose the incident field has unit amplitude, and let the power in the wave be one watt per square wavelength. What is the amplitude of the scattered field? Well, it can be almost whatever we want, because we control it by how we tune and match the antenna. Let’s say, for example, that we tune the antenna so the amplitude of the scattered field at a distance of 20 wavelengths is 1% of the incident field. That means the power density is1/10,000 watts/(squ. wavelength), since power goes as the square of amplitude. Multiplying this over the spherical surface at a radius of 20 wavelengths, we get a scattered power of… 4*pi*400/10000, or close to 0.51 watts.

Actually, the scattered power is a little less than that, because a short dipole does not radiate uniformly over a sphere. There’s something called the antenna gain, which we can actually calculate, but for now I’m going to assume you can look it up. The bottom line is you only radiate 2/3 as much as I calculated, so it’s actually 0.34 watts. But the real question is: where did that power come from? Answer: it was absorbed from the wave, and we should therefore see a weakening of the wave in the shadow zone.

And of course we do…it’s weakened by exactly 1%, because of the superposition of the two field patterns. Now, how much power do we take out of a wave at unit strength if we diminish its amplitude by 1%? Well, power goes as amplitude squared, so we actually reduce it by 2%; and this prevails over the cross-sectional area of the shadow, which we took to be 20 square wavelengths. You should be able to see that we have taken 0.4 watts out of the incoming wave.

What happened to the missing 60 milliwatts, I hear you ask? That’s the magic of an antenna. You have 60 milliwatts available to drive a load with, perhaps a small speaker.

Well, that’s convenient. Why don’t we gather more power, then? Instead of a paltry 1%, why not go for 2%? All you have to do is let twice as much current flow in your receiving antenna. If it’s just a question of changing resistors and tweaking a coil…

A nasty thing happens if you try to up your radiated amplitude to 2%. Radiated power goes as the square of the amplitude, so instead of pumping out 0.34 Watts, you’re now wasting 1.36 Watts in all directions. But the absorbed power, which happens in the shadow zone, is different. Instead of reducing the amplitude of the incident field in the shadow zone by one percent, giving you 98% power density, you now reduce it by two percent, giving you…96% power density. Your power absorption is only linear with the antenna current. You only absorb 0.8 watts, which isn’t enough to cover the wasted, re-radiated power, never mind driving a speaker.

Whenever you have two quantities, and one is linear and the other is quadratic, the linear quantity dominates first, and then the quadratic takes over. The receiving antenna works that way. At first, the more current you allow to flow, the more useful power you absorb. But then the wasted power starts to grow faster, and it dominates. There is a maximum useful power you can draw. I have drawn a graph of it here (I worked out the numbers so it agrees with the calcuations I've done so far:

You can see that the maximum useful power you can capture with this antenna is 0.22 Watts. SInce we are working with unit power density (1 watt/square wavelength) that also happens to be the amount of power flowing through 0.22 square wavelengths, and so we call that the absorption cross-section. Or we can also talk about the scattering cross-section, which is basically when you short-circuit the antenna and maximize the current. That’s of course four times as much power. (You can see why from basic circuit theory and load matching!) Either way, once you’ve got the cross-section, you’ve implicitly determined the radiation resistance. You just have to convert wave power to volts-per-meter, mutliply by the length of the antenna, and push your numbers around until everything agrees with everything else.

Actually, looking it up on Wikipedia I see they give an actual cross-section (they call it “effective aperture) of 0.11 square wavelengths, which is just half of what I got. It’s not surprising I’m out, since I used the coarsest imaginable approximation for the effective cross-section of the shadow zone. There’s an integration that gives me an exact value, and oddly enough, in this case I’m actually capable of doing the calculus. It’s just that the math tends to get in the way of the physics, and it’s so much easier to describe what you’re doing when you keep it simple.

Still, I promised I was going to try and get this one exact for a change, so maybe we’ll come back tomorrow and see if we can’t work in that missing factor of 2.

## 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 right...in 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.

## Friday, February 24, 2012

### Every good calculation should have a road map.

This is such a cool calculation. There is so much in here, it would be a real shame if the physics gets lost in the details. So before we continue, let's take a step back and recall what we're doing.

First, we're setting up a thermal equilibrium between a mechanical oscillator and the radiation field. You might think you need to have a statistical quantity of oscillators, but you don't. It's the same equilibrium if you have just one single charged oscillator. So I've created an artificial tether-ball in the middle of an empty box. Actually, it wouldn't have hurt to have filled the box with uncharged balls the same size as the tether-ball, and let them bounce about freely. We could have given them a temperature, and they would have imparted random motion to the constrained, charged ball. Since the tether-ball has only one degree of freedom, it would take on the same average energy as all the other balls. Sometimes it would have less energy, and in those circumstances the subsequent collision would by more likely to add energy to the tetherball. Sometimes it would have more than average energy, and...well, you get the idea. When it happens to have the average energy, it is no more likely to gain than to lose with the next collision...it is in mechanical equilibrium. The real point to understand is that at that moment, it is also  in equilibrium with the radiation field. You should be able to see why this has to be so. The bottom line is you only need one charged oscillator in the box to establish the equilibrium between matter and radiation.

Now we invoke the equipartition theorem, which says that the energy per mode of the radiation field has to equal the energy per mode of the mechanical system. This is the same principle that supposedly leads to the ultraviolet catastrophe, but we really don't need to worry about that here. What people forget about this principle is that it is strictly true only in the vicinity of a specific frequency. There might be different average mode energies at  200 Megahertz versus 200 Gigahertz, but at each frequency the electrical mode energy will equal the mechanical mode energy. The idea that a mechanical oscillator at 200 Gigahertz must have the same energy as one at 200 Megahertz is a misapplication of the equipartition principle.

Having said all that, for the purposes of the present calculation it is rather a moot point. We can choose our numbers so that the frequencies are all within the Rayleigh-Jeans regime where equipartition prevails in the broader sense...that is, all frequencies have the same mode energy regardless. It ought to be noted, however, that the validity of this calculation will ultimately not depend on any such assumption.

The equilibrium will be independent of any specific properties of the oscillator. That means we can give it any arbitrary properties we like. In our case, we have chosen to configure it as what we call a rigid rotor, with a certain arbitrary mass, charge, and radius of rotation. One degree of freedom, which defines its frequency and energy at equilibrium. The radiation field must then be at equilibrium with the rigid rotor, or the tether-ball as we call it.

But how can our charged tether-ball be at equilibrium if it is spinning about, radiating energy like crazy? There is only one way out: it must be also absorbing energy at the same time, and it must be absorbing energy at the same average rate as it is radiating energy. If we can possibly figure out how much energy the tether-ball is absorbing, we will automatically know how much it is radiating. And wouldn't that be something. You tie a charged ball to a string, and whirl it about your head, and now you can calculate how much power you're radiating. Without looking up any formulas. You can actually figure it out.

I'm not saying we won't use any formulas. We'll use mostly basic formulas that everyone knows, like the force on a charged body and the energy of an electric field. We'll do some stuff with Fourier Series that you can ultimately verify with high-school level trig identities. And we'll use one formula that you could theoretically figure out by yourself, but each time I try to do it I end up off by a factor of three, so I caved in and copied it from the internet. That's the famous Rayleigh-Jeans formula for counting the modes of the electromagnetic field in a rectangular box. The point is that none of these formulas tell you how much radiation you get from an accelerating charge. That's the result we're going to come up with, and ultimately we'll get it by doing nothing more than enforcing the condition of equilibrium between the electromagnetic field and the mechanical oscillator.

It's a very funny trick, and to be honest I'm still not really sure why it works.  Part of the secret is that there is a fundamental asymetry between absorption and emission of radiation. Emission of radiation is a huge conceptual mystery: the charge is giving off energy as it oscillates, which means you are doing work to shake the charge: but what is the force which is reacting against the shaking charge, the force which you are working against? I'm not sure anyone really knows.

The absorption of energy is different. An oscillating electric field comes along and pushes on the charge, so the charge starts to oscillate in response to the field. It's pretty straightforward. Actually, even here there is a hidden mystery: we agree that the charge moves, gaining energy, but how does this process remove energy removed from the electromagnetic field? That's a story for another day. The point is that unlike the case of radiation,  for absorption we have a pretty reasonable calculation.

It's a reasonable calculation, but it's far from a walk in the park. There are two pretty huge problems facing us. First, the electromagnetic field. What is it? Why, it's just equal to the average energy of the rigid rotor, which is an arbitrary number we said we can just pull out of our ass. No, that's not quite right: we said that energy would be the average energy per mode of the electromagnetic field. We still have to add up all the modes to get the total field.

Add up all the modes? But there are billions of modes, and they are all all kinds of crazy frequencies. Why not just take the ones that are at the same frequency as the tether-ball? Because there is not one mode out of all those billions that has the exact frequency we want. There are all kinds of modes very close, but how close is close enough? It's a horrible mess.

I was stuck on this point until I made an amazing discovery that changed everything. It's that business with Fourier Series that I mentioned earlier. You add up a bunch of modes, starting with the ones closest frequency to your target, and moving outwards as you go. And then you just stop. You truncate your series. Where do you truncate it? Anywhere you want. It turns out you're going to get the same final answer for the energy of the tetherball regardless of where you truncate your series. It's hard to believe, but I show why it works here.

The Fourier calculation converts your random, distributed field into a uniform chain of pulse trains of a specific size, frequency, and duty cycle. Depending on where you cut off your frequency band, you get different parameters for those three values of the pulse train. That's a bad sign: it means you're not calculating anything that is physically real in a measurable sense. But here's the catch: when you apply any one of those pulse trains to your tether-ball, you get the exact same final result. So who cares if they're real or not? How can you doubt that the result you get is consistent with the true physical result? (Or at least a very good approximation to it.)

But let's not get ahead of ourselves. The result of the Fourier calculation is a train of pulses - frequency bursts, actually. We apply that pulse train to the tether-ball, and it twirls about its post. How fast does it go? It turns out we can figure that out using the theory of the drunkard's walk. Everybody knows that the expactation of the distance traveled by the drunkard from the lamppost goes as the square root of the number of steps. But for an amplitude of oscillation, which is what we are dealing with here, the energy of oscillation is proportional to the square of the amplitude. Since it turns out that the amplitude of oscillation corresponds to the distance of the drunkard from the lamppost, the conclusion is that the energy of the oscillator grows linearly with time. Each pulse train adds an equal amount of energy to the oscillator!

What oscillator am I talking about? I thought we were going to drive a tether ball. That's not a harmonic oscillator, it's a rigid rotor. Well, I have to admit I'm cheating a bit here. The first time I did this calculation I applied it to a harmonic oscillator. When I came back to it the other day, I decided the rigid rotor would be a better case to work out, for a couple of reasons. I have an easier time restricting the degrees of freedom. I have a very convenient formula (the Larmor Formula) to check my work against, whereas with the harmonic oscillator I was having trouble getting it to fit into the antenna formula for the short dipole. And not least, I like the image of the tether-ball. I think it's evocative.

The problem with the tether ball is it's hard to get it started. To get it up to speed, you have to drive it through the whole spectrum of frequencies. That's a problem. The nice thing about the harmonic oscillator is that you're working with the same frequency through the whole range of amplitudes. So I end up returning to the old reliable mass-on-a-spring after all. It's kind of cheating, but you can ultimately justify it. After all, you could imagine some kind of variable-tether mechanism whereby the ball is let out gradually as it revs up, maintaining the same frequency the whole time. But more importantly, once you actually get up to speed, the equilibrium conditiions are exactly the same for the tether-ball and the mass-on-a-spring. So with apologies, I'm going to start the system off as a harmonic oscillator. It's a bit of a mixed metaphor, but it works.

Here's how it works. You have this oscillating charged mass, and you hit it with one of these standard pulse trains that you've worked out. This gives a certain impulse to speed up the oscillation...or does it? The problem is you don't know the relative phase. Isn't it just as likely that your pulse train arrives 180 degrees out of phase, so it slows down  the oscillator instead of speeding it up? Or any other phase in between? In fact, you have no idea what the outcome of the interaction is going to be.

That's why it's just like the drunkards walk. He takes a step in a random direction, and you have absolutely no way of knowing whether he's getting closer or farther from the lamppost. It turns out to be slightly more probable that he will end up farther, but it's totally random. Except, that is, in one particular circumstance: the very first step. You know that if he starts out right under the lamppost, that after the first step he will be...exactly one step away from the lamppost. And the harmonic oscillator works exactly the same way!

We have to apply one of our standard frequency bursts to our mass-on-a-spring when it is perfectly at rest. In that case, and in that case only, can we calculate the outcome of the interaction. We can calculate the resulting amplitude of oscillation, and also the energy of the oscillation. With subsequent frequency bursts, the amplitude will grow more and more slowly, like the drunkard's walk. But the energy, since it is the square of the amplitude, will grow...on average...linearly with each frequency burst. Each frequency burst adds, on average, the same energy to the oscillator. And we know how big the bursts are, and how far apart they are.

Does the oscillator grow without limit? No it doesn't. Because as it oscillates, it begins to radiate. The problem is, we don't yet know the laws of radiation for a harmonic oscillator, or a rigid rotor, or any kind of accellerating charge. Well, that's not quite right. We ought to know on general grounds that the radiation is quadratic with the amplitude of oscillation; and as goes the amplitude, so go the velocity and accelertation. We will guess that the formula for radiation should be in some way proportional to the square of the acceleration.

So what are we missing? We need to calculate how much energy our frequency burst puts into a stationary oscillator. That's going to be pretty much an F=ma calculation; it shouldn't be an obstacle. Since we know how often the pulse trains are coming, we know the rate at which energy is being absorbed: in fact, it's being absorbed at a constant rate. How about the rate at which energy is being radiated? From what we just said, it's clearly proportional to the total energy of the oscillator, which is growing at a linear rate. In other words, at some point, the radiation must overtake the absorption. That is the point of equilibrium, and we already know where that point is!

At the point of equilibrium, we now know the amplitude of oscillation, and we also know how fast it is radiating. The formula which relates these two quantities is nothing other than the Larmor Formula, and our data must agree with it. All we need is to divide quantity A by quantity B to derive the constant of propotionality which links them. That will be our job for tomorrow.

### 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.

## Thursday, February 23, 2012

### The equivalent antenna of an atom

Here’s the problem we were working on when I left off yesterday. I wanted to use the classical antenna equations to calculate the amount of radiation from a hydrogen atom. Specifically, I was looking at the s-p transition, at the point where the atom was just 1% excited. I already showed that the classical calculation gives you the same radiated power as you calculate Copenhagen-style with the Einstein A coefficient and Fermi’s Golden Rule. I got it to come out exactly right using the Larmor Equation for the radiation from an accelerating charge. I was just having some trouble getting the same result from the antenna equations.

It might seem like a small point, but it turns out to have huge consequences for me, which I’ll get to at the end of this post. But let’s start off by remembering where we were yesterday. The Schroedinger equation of .995*s + 0.1*p (that’s an atom excited to the extent of 1%) gives me an oscillating dipole moment of 8 picometer-electrons. I did some rough geometry to equate this to a dipole antenna 32 picometers in length. I wasn’t very explicit about how I got that, but it’s a fairly straightforward geometrical argument using moments. Here is a picture of how it looks:
The first picture shows the Larmor Atom, which is pretty close to the actual atom except it's oscillating in two dimensions instead of one. You can see that I've drawn an equivalent antenna with the same dipole moment; I hope it's clear why the first one has a radius of 8 picometers, and the second one has a length of 32 picometers. They both have the same oscillating dipole moment.

So if that's my equivalent antenna, it should be pretty easy to use the antenna equations. I showed yesterday that for the frequency in question (2.5*10^15 Hz) you get a radiation resistance of about 12 micro-ohms. So all you need is to put the current in and calculate i-squared-r. But that's easy. The current is just the charge times the frequency, right?

Wrong! That's what was screwing me up so bad. It's actually really tricky to figure out the equivalent current. I finally worked it out like this:

This graph represents a sinusoidal current with unit amplitude. It delivers a total charge of 2 coulombs in pi seconds. The frequency is 1/2*pi and the charge is 2. If I said that the current was just the charge times the frequency (which is what I said it would be yesterday!) I would calculate a current of 1/pi. But that means I'd be out by a factor of pi!

It's worse than that because I also forgot that the antenna formulas are written in terms of RMS currents...the electrical engineers's standard. Which throws me out by another factor of sqrt(2). It turns out that the two errors are in opposite directions, so the compound error works out to 2.2. Since current is squared in the power formula, this gives me a factor of 5. Which is exactly the amount I was out by.

There's no reason to torture you by walking you through the numbers: you can do it yourself if you really care. The point is it came out perfectly. The antenna calculation agreed with the Larmor formula, which agreed with the Fermi Golden Rule calculation. So everything was fine after all.

But then it hit me: exactly one year ago, I was trying to do a similar calculation where I equated the mode energy of the electric field in a cavity to the mode energy of a mechanical oscillator. It was a very cool calculation but no matter what I did, my numbers were still out by a factor of ten! Could it be the same error,  and if so, was it in the right direction? I'd hate to go and fix it up only to find that now I was out by a factor of fifty or a hundred.

When we return, I'm going to check this out.

## Wednesday, February 22, 2012

### The antenna calculation

Yesterday I said I couldn’t get the radiation emission from hydrogen to come out correctly, using classical antenna theory. Does this mean that atoms do not behave like classical antennas? Possibly. The other possibility is that I was messing up the calculation. Which could it be?

Eventually, in desperation, looked up the formula for the radiation from a charged body moving in a uniform circular orbit. It’s called the Larmor Formula, and it’s a little more general than that: it claims to give the radiation for any acclerating motion. I adapted the atomic equations to the case of uniform circular motion, and I got exactly the right amount of radiation. The semi-classical picture agrees with the Copenhagen picture.

So why wouldn’t the antenna equations work? That’s a tough one, but I think I have the answer. The proper thing for me to do now would be to show you how its done correctly, but I can’t bring myself to do that. For some reason, I feel I have to first show you how I did it wrong. A though you were to somehow be in danger of making the same mistake as I did. So here goes:

Everybody knows that the radiation resistance of a half-wave dipole antenna is 73 ohms. Right? I mean, we have to start somewhere. I actually did a calculation a couple of months ago to justify those 73 ohms. It wasn’t really a very good calculation, but it kind of at least gets you in the ballpark. I actually have a much better way of doing it, where you can actually see the physical basis for what's happening, but again it's just a ballpark calculation. The three-dimensional calculus is just too hairy, so I have to take some shortcuts. Someday I'll write it up and post it.

But for today, I'm actually going to just use a formula we can find on the internet. It's not the formula for the half-wave dipole, it's the more general formula for the short dipole, where by "short" we mean much shorter than the half-wave. Since atoms are on the order of anstroms, and they emit light in the optical or (at worst) ultraviolet spectrum, this qualifes them as "short" dipoles. And as such, we can use this formula for their radiation resistance:

You can check this formula for the case of lambda = 1/2: you see it gives you a value of about 50 ohms, which is a little on the low side. That's because the half-wave dipole is a little more focussed than the short dipole. It makes sense.

Now we can plug our numbers in and see what to expect for the power output. First, compare the radiated wavelength to the atomic dimension. We already noted that the wavelength is 120 nanometers, and the dipole moment is 8 picometers. Actually, that's not quite the right figure to use for the ratio: that's the one-sided  moment, so we should double it up for full distance the charge travels. And that's not all either. What is the dipole moment of a short electrical antenna? Is it equal to its full length? Not really. In a small dipole, the current distribution is such that you get a uniform distribution of charge, equal and opposite, on the two legs of the antenna. The center-of-mass of an equal distribution is half-way along the length: so the dipole antenna which has the same moment as the atom should actually be twice as long. All together, I get a length of 32 picometers. I think this is right.

From the ratio of these two lengths, using the formula you should calculate a radiation resistance of
about 12 micro-ohms. Now we just have to figure out the current and plug everything into the formula "power equals i-squared-r".  Are you ready?

### The Semi-Classical Calculation

Yesterday, I set up a sample of hydrogen atoms in an excited state and calculated the radiation output according to the standard Copenhagen picture. Or, at least, I walked you through Richard Fitzpatrick's lecture on the University of Texas website. I took the sample to be 1,000,000 atoms, with 1% of them in the 2p state. (I ignored the possibility that some of the excited atoms should have been in the 2s state, but I don't think that matters. Except it would have screwed up my temperature calculation.) Using these numbers, I calculated a power output of 10 microwatts.

Then I said I was going to redo the same calculation, this time treating the atoms as tiny classical antennas. Since everybody knows that Maxwell's Equations don't apply to atomic systems, we should get nonsense. Let's see what actually happens.

I'm going to admit that I sweated bullets doing this calculation. I probably did it at least twenty times and got a different answer each time. The antenna calculation just wouldn't come out right. The best I could do was still out by a factor of ten. In desperation, I looked up the formula for the radiation from an accelerating charge. It's called the Larmor Formula, and I found it on the Wolfram Alpha website:

This gives us a formula to work with: but what numbers do we put in the formula? Well, there's not too much to worry about; we've got a couple of constants, which are provided in the table; we've got the charge on the electron, which everyone knows is 1.6x10^-19. The only other thing we need is the acceleration.

That's the hard part. What is the acceleration of the electron, according to the Schroedinger picture? Fortunately, Professor Fitzpatrick has already done all the hard work for us. I said there were two alternative pictures of the physics. In the Copenhagen picture, one percent of the hydrogen atoms are in the excited state. In the Schroedinger picture, each of the atoms is in a mixed state to the extent of one percent excited. To calculate the dipole moment of such a mixed state, you do the bra-ket thing with x in the middle. Actually, the order of operations doesn't really matter in this calculation (as it sometimes does in quantum mechanics). You can just think of this as the square of the amplitude (that's the product of the bra and ket states) integrated against x to get the dipole moment. The calculation looks like this:

What exactly are we seeing here? I said that the atoms were one percent excited, but here I'm showing the P state with a ten percent amplitude. Is that a mistake? No, because don't forget you square the amplitude to get the density. And why don't I have any s^2 or p^2 terms in my result? Because there is no dipole moment for a pure state: because the s^2 and p^2 states are symmetric (even) about the origin, the integral against x (an odd function) gives you zero. No, the calcuation reduces to the dipole coupling of s versus p.

But this is the calculation which the Professor has already done for us. Recall from yesterdeay:

He's actually given us the dipole moment for all three p states in terms of the Bohr Radius of 5.3*10^-11 m. (It's the third value which pertains to our case, the one with the extra factor of root 2.) Working out all those powers, it comes to about 75%, or 0.4 Angstroms. We also mustn't forget the factor of 20% which came out of the 1% excitation: so the dipole moment in our case actually comes to 8 x 10^-12 meters. (Multiplied by the charge of the electron, of course.) It's funny how much dipole moment you actually get by mixing in just 1% of the p state, but there it is.

Most importantly, let's notice that this dipole moment is an oscillating dipole moment. Because the s and p states precess through time at different exponential frequencies. So based on the difference in those two frequencies, what is initially a positive dipole moment becomes, after one-half cycle, a negative moment. It oscillates.

We've now got almost all the numbers we need to do the Larmor calculation. We still need the acceleration of the charge. We've got the oscillation amplitude - that's the 8 picometers - and we've got the frequency from yesterday - that was 2.5 x 10^15 Hz. Now I'm going to do something pretty slick. Instead of working out the acceleration due to sinusoidal motion, I'm going to pretend I'm working with uniform cirular motion instead. That's not what the atom is doing: in fact, it will throw me off by a factor of two, because the energy output of the circular motion is just the superposition of two orthogonal harmonic oscillators. No problem...I'll just divide by two at the end. The nice thing is I get to use the omega-squared-r formula for circular motion that I still remember from high school physics. Remembering the factor of 2-pi to change from hertz to radian frequency, I get an acceleration of 2*10^21 m/sec^2.

Putting all the numbers together, I get a power output of...20 picowatts. (You can check this if you like by plugging the numbers into the applet on the Wolfram site.) Divide by two to convert circular motion into simple harmonic motion, multiply by one million for the number of atoms in the sample, and I get, incredibly, exactly the same power output that I calculated yesterday. The Copenhagen picture with its quantum leaps gives exactly the same result as the Schroedinger picture with its tiny oscillating dipoles.

Not convinced? There's one more calculation that really needs to be done to bring this argument full circle. I said originally that I was going to use antenna theory to do the calculation, and when I tried, the numbers wouldn't come out. I kept getting two microwatts instead of ten. How do you get a factor of five for an error? It didn't make sense, and it was driving me crazy. But now I've worked it out so the antenna calculation comes out right. That's a story for my next blogpost.