1-2+3-4+5....

I thought maybe I should tell you how I came up with this technique. It wasn't a discreet series, it was an integral I had to deal with. It came up when I was working on the problem of the crystal radio: what is the maximum theoretical power you can absorb from a plane wave using a small antenna?

The key to solving this problem is to realize you don't need to know (hardly) anything about antennas. Everything that happens comes as a result of wave interactions. You have an incoming plane wave, which causes currents to flow in your antenna. As a result, your antenna radiates spherically. (Of course its a modified spherical pattern because of the donut characteristic of the dipole field. But that's a minor detail.) The point is, looking at the system from above, you have the incoming waves from the left and the outgoing spherical waves from your antenna like so:

(I talked about this a couple of years ago in this blogpost.) You can see that the little antenna is scattering power in all directions, except in the shaded region you have the possiblility (if you can get the phase just right) to actually remove power from the incident wave. This can only happen effectively within the shaded region when the phase difference between the two systems is less than 180 degrees (or 1/2*lambda).

The really interesting thing is that since power goes as the square of the wave amplitude, you have two things going on. As the antenna current grows, the re-radiated power grows quadratically. But at the same time, within the shadow region, the absorbed power grows

*linearly*. Because if the incident field is 1 and the re-radiated field is dx, the resultant power is (1-dx)^2 where 2dx is the absorbed power (and you ignore the term in (dx)^2).
You know that if you have two quantities and one of them is growing linearly and the other is growing quadratically, that at first the linear term is going to dominate, but then the quadratic term is going to catch up. And in between there is an optimum. That optimum represents the maximum power you can absorb with a small antenna...and it has nothing to do with the size of the antenna. It's strictly a result of the interacting field patterns.

How do you achieve the optimum conditions? First, you have to get the phase relationship just right. That's called "tuning the antenna" and you do it with a variable inductor. Then, you have to get just the right current. That's called "impedance matching". We won't get into that today.

The point is...it all depends on comparing the re-radiated power to the absorbed power. The spherical power isn't too hard to calculate...in the worst case, we can look up the standard antenna formulas. But what about the absorbed power...how do we do that?

For starters, let's realize that if we back off far enough away from the antenna....100 wavelengths, or even 1000 wavelengths...then we can use the small-angle approximations like sin(x) = x. I know you math guys

*hate*that approximation, as I did the first time I saw it in Grade 11...but in physics you learn to*love*it. But that's a whole nother story.
You'll notice that "shadow region" we're working with is a paraboloid of revolution. That comes straight from the focus-directrix definition of a parabola as the locus of all points such that the difference of the two distances is a constant. And we said that within the first lobe, we can get destructive interference. But outside that lobe, we get

*positive*interference...the waves re-inforce. So that works against us...and so on and so forth to infinity, the interference pattern alternately giving us rings of positive and negative interference.
Also notice that the rings are not the same width. Because of the geometry, they get closer together as you get farther from the axis of symmetry. (It goes as sqrt(r)...that comes straight from putting successive integer values into the focus-directrix definition of the parabola.) Here then is a picture of the positive/negative interference pattern as though you projected it on a screen 1000 wavelengths behind the antenna:

Without getting bogged down in the details, this basically meant that to figure out the power absorbed by the antenna, I had to integrate a function of the form (again, taking advantage of the small-angle approximations)

It's helpful to know what these functions look like. Maybe you can imagine them, but here is the cos function, courtesy of Excel:

How do you integrate this from zero to infinity? By remembering the physics. Wave interference is a funny thing. If you shine two flashlight beams across each other, then there is all kinds of crazy interference patters going on where they beams cross. If you followed the power flow in detail, its going here and there and everywhere. But the overall result is that Beam A ends up where it was going, and Beam B goes its own way too. Like they never crossed. And essentially, that's the situation we have here. The crazy inteference peaks that get bigger and bigger...those represent power flowing from the incident plane wave and the small receiving antenna. But once you diverge from the axis of symmetry, there really can't be much going on that is of any physical significance. All those oscillating powers have to average out to zero. So let's just force them to do that. And the obvious way to do that mathematically is to multiply the integrand by a gently-decreasing Gaussian, something like this:

As you let k become arbitrarily large (so the envelope gets wider and wider) this integral should converge to a constant value...and in fact it does. I don't remember which one of them comes out non-zero...the sine or the cosine. Ultimately that just tells you the relative phase of the wave pattern. At some point I ran it in Excel and came up with a value. But actually, as I look at it now, I'm thinking that you can do this one exactly with integration by parts (with the obvious u=r^2 substitution). When I was doing this calculation, I think all I really wanted was a ballpark estimate of the power, so I just made the arbitrary assumption that interference was 100% inside the first nodal line (the shadow zone) and negligible outside. That turned out to be not too far off.
And that, my friends, is how I came up with the idea for using a Gaussian Envelope to tame ill-behaved integrals. Necessity was the mother of invention.

So how much power can you absorb with a crystal radio? According to the calculation, as it turns out, a rather huge amount of power. Remember, the size of the antenna is irrelevant: all you need is to have a perfectly tuned and perfectly matched receiver. In that case, you can absorb a quantity of power equivalent to everything flowing through an area approximately one-half wavelength in diameter. So for a typical AM radio station, that would be all the power flowing through a cross-section of

(200 m)x(200m)=40,000 sq. meters

But oddly enough, nature provides us with an example of the perfectly tuned and lossless antenna in the form of...the atom. Yes, an atom can behave as a tiny antenna, and in this case the equations really work. The absorption radius of a hydrogen atom in visible light is on the order of the wavelength of light, which represents a cross-sectional area around 10 million times the physical cross-section of a hydrogen atom.

When you take physics in high school, they tell you that it's impossible to explain the photo-electric effect using the wave theory of light because the energy in a wave isn't concentrated enough to knock an electron out of an atom. They even "prove" this by multiplying the power density of the light wave by the cross-sectional area of an atom. Therefore, they claim, light must be made of particles...the so-called "photons".

The people who make this argument (and they include physicists at the very highest level, who should know better) have never done the crystal radio calculation. The cross-section of the atom, like the cross-section of the radio antenna, has nothing to do with it. The same argument that supposedly debunks the wave theory of light also "proves" that a crystal radio can't work without a battery.

But it does.

POST-SCRIPT: Maybe I'm getting a little senile, but it turns out I went through most of these calculations two years ago, here and here. Either way, it's still stuff you won't find anywhere else.

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