Sunday Function

You know what's gonna happen? I'll tell you what's gonna happen. Troops are now forming behind the line of trees. When they come out, they'll be under enemy long-range artillery fire. Solid shot. Percussion. Every gun they have. Troops will come out under fire with more than a mile to walk. And still, within the open field, among the range of aimed muskets. They'll be slowed by that fence out there, and the formation - what's left of it - will begin to come apart. When they cross that road, they'll be under short-range artillery. Canister fire. Thousands of little bits of shrapnel wiping the holes in the lines. If they get to the wall without breaking up, there won't be many left. A mathematical equation... But maybe, just maybe, our own artillery will break up their defenses. There's always that hope. That's Hancock out there, and he ain't gonna run. So it's mathematical after all. If they get to that road, or beyond it, we'll suffer over fifty percent casualties. But, Harrison, I don't believe my boys will reach that wall.
- Lt. Gen. James Longstreet, as portrayed in Gettysburg

We can't figure out exactly what the equation was, and it would be morbid to try. But in less bloody contexts, physics requires a solution of a problem along essentially these same lines. In radiation shielding you might want to know what the rate of attrition of x-rays is in lead. As those photons make their Pickett's Charge through the lead jacket you wear at the dentist's office when taking a x-ray, it would be nice to know just how many of them are stopped before reaching your cellular DNA.

Initially we can think of it in terms of coin flipping. If you have a million coins and flip them all, you'll end up with about half heads and half tails. Think of the heads as those x-ray photons that happen to hit a lead atom and he absorbed in that first millimeter, and think of the tails as those that don't. The next millimeter is thus like another coin flip - half will survive, leaving a quarter of a million. And so on and so forth until by random chance either all the photons have been absorbed or you've reached the end of the lead with the surviving photons. The survival of any individual photon is a matter of pure chance, but the survival rates of all the photons in the aggregate is statistically clear. If you give the distance required to absorb half the photons a name - say, the letter T - then you can write our Sunday Function. It represents the fraction of surviving photons as a function of the distance x traveled:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png
i-7b7142f1efa09b47c3b8242f89bcb9f4-graph.png

Here I've plotted the function with T = 1. Every time you go another distance T to the right, you lose another half of the function's value. This sort of decay doesn't just occur in shielding of course, it's also the equation describing radioactive half life, the current in a circuit with a discharging capacitor, and any number of other applications. The function falls off very rapidly. By the time T = 20, you've lost more than 99.9999% of the value of the original function.

Conceptually convenient though it is, using the number 2 as the base of our exponent is mathematically cumbersome. When doing derivatives and integrals we'd be accumulating factors of 0.693147 all over the place. We want to avoid this if at all possible. And why not? Exponentials are exponentials, maybe we can find a constant (call it by the Greek letter tau) for e^-x/tau that matches 2^-x/T. Let's try:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

Take the natural logarithm of both sides:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

Solve:

i-8d7070ce28f0af28047c7a5fb8246d53-4.png

The natural logarithm of 2 is less than 1, so tau is larger than T. Other than that the two constants work much the same way: T is the amount of distance (or time, or whatever) that it takes to reduce the function value by a factor of 2. Tau is the amount required to reduce the function value by a factor of e, which is roughly 2.718.

If half the x-rays make it one centimeter, T = 1 centimeter and correspondingly tau = 1.4427 centimeters. About 36.8% of the rays would make it that far. Fewer than 5% would make it past a distance 3*tau. Exponents fall off quickly, as we mentioned.

This Sunday Function is a little more utilitarian than usual, and in fact we're going to use this tomorrow to help solve a physics problem. (This SF is also a day late, but I have been traveling so we'll just chalk it up to time dilation. Apologies all around!)

More like this

Heavy metal shielding has an asterisk. If by any means big fat 1.022+ MeV photons arise you will get pair formation when one grazes a heavy nucleus. Pair formation makes more gammas by annihalation. Alpha- or beta-ray bremsstrahlung can do it as well as direct input. (And if you are an asstronaught, cosmic ray target spallation will cook you medium rare with all sorts of secondary radiation).

Lots of concrete (low Z) is very competitive with lead, tungsten, and depleted uranium if volume and opacity (heavy metal methacrylate transparent beta-shields) are not issues.

I've been wondering about this for a while. How do you build a detector for super high energy gamma rays? You need them to be absorbed to get a signal, so you'd need a super dense detector crystal, right?

Maybe you could use a detector optimized for lower energies and just shield it with a couple feet of lead or concrete. That just seems wasteful, though.

By SimplyHarmonic (not verified) on 03 Aug 2009 #permalink

As Uncle-Al hints, that equation is rarely useful in shielding calculations.

It works fine if you are losing photons to photoelectric interactions. At higher energies, you need to worry about Compton scattering, then you have scattered photons to worry about.

If your X-ray source is a linear accelerator, you also may have pair production.

@SimplyHarmonic what energy do you want to know about?
I work in radiotherapy, we have 15MV X-ray beams (modal energy around 5MeV). Imagers at this energy tend to have lowish-Z metal build up plates. This makes secondary electrons (mostly Compton interactions) that travel a few mm (in the metal). The actual detector is an amorphous silicon panel. (They used to use fluorescent panels, mirrors and cameras)

By ColonelFazackerly (not verified) on 04 Aug 2009 #permalink

I was wondering about 1GV+ photons. Stuff from the astronomical Heavy Hitters, so to speak.

By SimplyHarmonic (not verified) on 04 Aug 2009 #permalink

1GeV. Not sure what interactions you get at that energy. Quite a lot of variety, I guess. I expect the products will have a long range, and the resolution will be poor as a result. I guess you already knew all that, though...

By ColonelFazackerley (not verified) on 05 Aug 2009 #permalink

To put it a different way from the Colonel and Uncle Al, exponential attenuation is appropriate for pencil beams ("narrow-beam attenuation"). In that situation, we may reasonably assume that any interaction results in the photon being removed from the beam. Note that the photon is not necessarily absorbed, but that in being scattered, it is no longer present in the beam that we are examining. (Out of sight, out of mind?)

For an x-ray source, this equation is (I think) not a bad approximation. It isn't going to work well for the shielding situation considered in the 60-Co posts.

By Mister Troll (not verified) on 15 Aug 2009 #permalink