Thursday, February 17, 2011

About the Black Body Spectrum

I haven't posted for almost six months because I've been stuck on problems I can't solve. Today I solved a little problem so I'm back.

What happened is I stumbled across a physics essay contest just the other day at a site called It was the very final day for submissions so I whipped something together and got it in just before closing time. You can read it at

In the essay I take on the three main pillars of the case for the photon: the black body spectrum, the photo-electric effect, and the Compton effect. I call them the three pillars because for most physics students, that's how the case is made; there is a lot of history behind them, and even among people at the highest levels who should probably know better, these are the main arguments raised to debunk the wave theory of light. Anyhow, in my essay, I took a certain tack on the black body spectrum and dug myself into a bit of a hole. I had to finish it off in a hurry and in the end I wasn't happy with the way I dealt with it.

So here we are. Everybody knows that the black body spectrum was a huge problem for classical physics. But why exactly was it a problem? It boils down to something called the equipartition of energy. There is a "theorem" of thermodynamics that for a system of thermal equilibrium, energy will tend to be shared equally among all possible "modes". I put the word "theorem" in quotation marks because if it's a theorem, obviously you can't argue with it, even if (or especially if) you don't understand it. It just is.

I also put the word "modes" in quotation marks because I'm going to admit I don't know how a mode is defined. But in certain cases we can identify the modes without too much ambiguity, and in those cases the equipartition theorem can be a marvellous thing. For example, the diatomic gas (eg. H2 or N2): each molecule has three modes of translational energy (the x,y, and z directions) and just two modes of rotational energy (not three), because there is no energy involved in spinning about the long axis. Five modes altogether. The equipartition energy says they're all equal. Each atom gets a thermal energy of 1/2kT (where k is Boltzmann's constant and I don't know why it's defined with a factor of 1/2) and that tells you how much heat there is in a liter of gas. It doesn't even matter if you mix hydrogen and nitrogen, because the heat per molecule is still the same. Those are the modes.

And they give you the right physics if you measure the specific heat of different gasses. They're all the same hydrogen, nitrogen, or oxygen - if you take it by volume. And the reason that works is because equal volumes of gas contain equal numbers of molecules - the Ideal Gas Law.

Something funny happens if you take argon or helium. The specific heat comes out to only 60% of what you get for everything else. But this is the beauty of the equipartition theorem: those are now mono-atomic gases, so they don't have tumbling energy of rotation. Only three translational modes. Three out of five is 60%.

There is a worse problem though. We considered the hydrogen molecule to be a tiny rigid dumbbell flying through space, but that's not all it is. The bond between the two atoms, rigid as it seems, is also elastic to some extent. So it can theoretically vibrate, and that is one more mode to count. Actually, the people who really understand these things tell us it counts as two modes - one for kinetic and one for potential energy. (I wonder if it should be one for the sine wave and one for the cosine wave...but that's another question.) Either way, it comes to seven modes. But that's not what we get when we measure the specific heat. What gives?

That's a problem but it's just the tip of the iceberg. The real problem is when you count the modes of the electromagnetic field. Because they are also a part of the thermodynamic equilibrium, and they have to get their share of energy. Here's how you count them. You put the gas in a cubical metal box, for the sake of argument, to simplify the boundary conditions.
The electromagnetic modes are the standing waves. The first one fills the whole box. The second one, you divide the box in two and you get a standing wave with a nodal surface in the middle. There are three ways you can do this (the xy, the xz, and the yz planes) so there are three modes. Then you divide the box in three - oh, wait: you can divide it in two along one direction, and three, or five, or eleven along another direction... there are an awful lot of modes. Since there is no limit to how finely you can divide the volume, there are in fact an infinite number of modes.

Something is wrong, but surprisingly it's not completely wrong. It works up to a point.
You can organize the modes by frequency, since the lower order modes have the longest wavelength and hence the lowest frequency, and you find that the density of the modes (plotted against frequency) goes up as the square of frequency. And indeed, if we allow each of these modes to share in the energy - by giving them each an average of 1/2 kT - we get a radiation spectrum that agrees quite well with experiment.

Up to a point. If we keep going, the higher order modes suck out more and more energy, without limit. Since those higher frequency modes are in the ultraviolet spectrum and beyond, this is what they called the "ultraviolet catastrophe". It's a catastrophe for the theory.

It's not hard to see what is wrong with the theory. It's that pesky equipartition theorem. Why don't we just get rid of the problem by throwing out that theorem?

Apparently, because it's a "theorem", and you don't just throw out theorems. We have to admit that if it was good enough for people a lot smarter than you or me, it must be there for a reason. The equipartition theorem was going to stay put, no matter what the consequences.

And the consequences were devastating. To preserve that theorem, Max Planck was forced to throw away the wave theory of light. Ever since we have been living with that damned quantum mechanics, with all its paradoxes. In my next essay I'm going to have a second look at the equipartition theorem.

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