Friday, February 18, 2011

Avoiding Equipartition

In the last posting, I raised the tantalizing possibility that we could squirm out of the straitjacket caused by the equipartition theorem for mechanical energy, and in doing so save the world from the ultraviolet catastrophe. Let's see if this really works.

First of all, do we really need equipartition on the mechanical level in order to create equipartition on the electromagnetic level? There are good reasons to think that we do. If we put a handful of atoms into a box and let them bounce off the walls forever, it's not hard to imagine that they exchange energies in such a way that every possible combination of velocites exists at one time or another. Because in a typical "glancing" collision, the energies of the outgoing particles are basically randomized. Electgromagnetic energy is different. If we set up standing waves inside a box at different frequencies, they will basically stay that way forever (we are taking the walls of the box to be perfect conductors). There is no mixing of frequencies and therefore no tendency to find an equilibrium. We count on the presence of charged matter to tranfer electrical energy from one mode to another.

It's worth noticing that you don't need an awful lot of charged matter to mediate this equilibrium. Suppose in a box of helium gas there is one and only one special molecule that is capable of vibrating at 100 GigaHertz; and suppose further that the collisions are "hard" enough so that mode of vibration is in fact driven to its "equipartition value" of 1/2kT. If that molecule is electrically neutral there is nothing more to say. But if it happens to have some charge distribution whereby the net average motion of the positive charge is different from the negative charge - then that molecule becomes an antenna at 100 GHz. And that's all we need. Once those conditions are met, the electromagnetic field must be driven to its equilibrium value. At least, that is, in the vicinity of 100 GHz; there may be no energy at 90 or 150 GHz but at 100 GHz, the average electric field energy will approach the value given by Planck's formula for black body radiation.

Here's the catch: it will also approach the value given by the much-maligned Rayleigh-Jeans formula! (Remember, that's the formula that leads to the ultraviolet catastrophe.) Because wherever the mechanical equilibrium follows the Law of Equipartition, the Rayleigh-Jeans formula also holds true.

You can add a second vibrating molecule and it won't upset the equilibrium. Because on average, the little molecular antenna will be absorbing as much as it is emitting. You can add a hundred or a million more, and the equilibrium will still establish itself at the same level.

It doesn't even matter how efficient the antenna is. You can re-arrange the charge distribution so that for the same mechanical oscillation, there is much less electrical action. It doesn't matter! The equilibrium will be at the same level, although it may take much longer to get there. Mind you, there is also a bandwidth effect to consider. For a very efficient oscillator, the radiation field may establish itself over a reasonably wide range of frequencies, e.g. between 98 and 102 GHz; as you make the individual oscillator less efficient, this bandwidth will get smaller and smaller until
ultimatley you only get get the expected Rayleigh-Jeans field intensity within a few kiloHertz of the midpoint of the band. The point is, it's the same field intensity no matter how many oscillators or how efficient they are.

Another way of looking at is to count the standing wave modes. If we have a finite box, then the modes are discrete, not continuous; so they may be counted. A very weak oscillator may fill up only a handful of modes very close to 100 GHz, but they will all be filled right up to 1/2kT. A more efficient oscillator will fill many more modes, but only to the same level of 1/2kT. It has to be that way; otherwise you could connect the two boxes by a fibre optic cable that passes energy only at or very near 100GHz, and the result would be a continuous flow of energy from one box to the other. That would be a violation of the Second Law of Thermodynamics.

We started this argument by supposing that the collisions were sufficiently rigid to ensure that the vibrational modes were fully excited. What if they aren't? In the last posting we pointed out how these modes are not typically excited for diatomic gasses like H2 or N2, and that the reasons were not hard to argue classically: that the actual collision occurs over a characteristic time span depending on the strength (mostly electrostatic) of the interatomic forces; and that if the duration of the collision spans several cycles of the vibrational period, then the driving force becomes so smeared out as to be inefffective.

We can see that this effect must take place for any vibrating species sooner or later. First drop the temperature in half. We get half the kinetic energy, half the vibrational energy, and half the radiation energy. That's OK. Drop the temperature again. Same thing. But there's a catch. The vibrational frequency doesn't change as you drop the temperature, so not everythings works according to scale. Eventually the helium atoms are moving so slowly that they can't effectively drive the vibration. That's when the radiation density drops below the Rayleigh-Jeans level.

Is this a true thermodynamic equilibrium, or is it just a case of imperfect materials deviating from the ideal laws? It's an important distinction. Because on the one hand, there is the case where different vibrators have different efficiencies and the equilibrium for a particular gas depends on what molecule you've used for the vibrator. On the other hand, there is the possibility that the efficiency whereby mechanical energy is converted to electromagnetic energy turns out to be independent of the specific mix of molecules in a particular experiment. Which one is it?

The clue which will help us settle this question is reversibility. Let's suppose the whole thing is material dependent, so if you put a water molecule (H2O) into a box helium gas at 300 degrees K and a deuterium sulfide molecule (D2S) into another box (let's say that the bond stiffness scales perfectly according to the mass ration of hydrogen to deuterium so we get exactly the same vibrational frequency - it could happen!), we should in general, once we're below the Rayleigh-Jeans cutoff, expect the radiation spectrum to be different for the two boxes.

The problem is that for each box, the mechanical equilibrium and the electromagnetic equilibrium are independent of each other. If we take the charge away from the H2O molecule or the D2S, it vibrates just the same in equilibrium with its box of helium. But if we take away the helium gas altogether, and just let the H2O molecule or the D2S molecule float around by itself, it still has to be in equilibrium with its radiation field. And if it's just a matter of specific properties of different molecules, those radiation fields will in general be different.

Here is the consequence. Set up the two boxes side by side, with a thermally conductive wall separating them. Connect a fiber optic cable from one box to another to let radiation pass between them. Suppose the H2O box (A) has twice the radiation density as the D2S (B) box. Then electomagnetic energy will flow from A to B, where it will pump up the D2S molecule.

So the D2S builds up to the level of the H2O, until the radiation is equalized in both boxes. But then you have to ask: what about the helium in the D2S box? If it stays as it was, without responding to the increased D2S vibration, then you get a new balance; and you have to conclude that the former "equilibrium" wasn't really equilibrium after all; it was just a matter of the failure to transfer energy from the kinetic modes to the vibrational mode. Now that you've provided a mechanism (basically via the H2O molecule) the true equilibrium is established, and it's the Rayleigh-Jeans equilibrium.

The other possibility is messed up: that the D2S does transfer some of its extra energy back into the helium and now you have more kinetic energy in box B as compared to A. So this sets up an imbalance whereby thermal energy flows by conduction through the wall between the chambers in one direction, while electromagnetic energy circulates back in the other direction through the fiber optic cable. That's crazy.

So you have to conclude that thing where a vibrational mode fails to get fully excited does not actually represent a true equilibrium. It's just case where the system gets stuck in a state where it cannot truly maximize its entropy, simply because a pathway is blocked. The true equilibrium, including both mechanical and electromagnetic, is still where Rayleigh-Jeans predicts.

And the only reason the ultraviolet catastrophe doesn't occur is that there is simply no mechanism available that you can put together from the 92 elements of the periodic table.

Or is it? That is the question I'll take up in my next posting.

No comments: