Sunday, February 27, 2011

Counting Modes

How many ways can you set up a standing mode of electromagnetic waves in a rectangular cavity? This question has been holding me up for a week now. There is a formula which I found on the hyperphysics webpage; it is basically the Rayleigh-Jeans analysis which gives the low-frequency limit of the black-body law. The formula must be right because it agrees with experimental results, assuming each mode gets its alloted energy of 1/2kT. My problem is I have my own way of counting modes which disagrees by a factor of 3, and I can't for the life of me see where.

There are a lot of ways you can be out by a factor of 2 or 4, but it is very hard to be out by a factor of three. Furthermore, I understand the basic concept of counting modes. I know how to do it in two dimensions for electromagnetics, and know how to do it in three dimensions for a gas. I thought I knew how to do it in three dimensions for solids, but now I'm not so sure. At first I thought it was the same as a gas, but now I realize that a solid, unlike a gas, can support shear waves as well as compressional waves.

The electromagnetic case in two dimensions is not so bad because you can start by taking waves in the xy plane and letting the polarization be all in the z direction. That lets you separate out the vector portion and leaves you with nothing but scalar waves to add up in the plane. It doesn't take much more than high-school trig identities to show that four sets of plane waves criss-crossing each other can be added up to equal a checkerboard pattern of standing waves.

That's how the hyperphysics explanation starts out, except it's in three dimensions. Here is what they say:

I really don't see that this can be correct. They call for the electric field to go to zero at the cavity walls, which is of course the correct condition for acoustic vibration amplitude in a gas. But electromagnetics? It's certainly not the condition for a waveguide. We need the electric field to have no tangential component, and we need the magnetic field to have no perpendicular component at the walls. Those are very different conditions than what they've described here. But never mind, let's go on.
We now have the rectangular cavity divided up into unit cells, and we want to count the modes up to a given wavelength. They do the thing of working in k-vector space and counting the lattice points within a spherical volume. I haven't explained it in detail but it's the same thing I do and they end up with the following formula:
It's essentially the number of unit cells, with an extra factor of two for what they say is polarization. I think that's also wrong; I would say the factor of two is necessary for the time displacement of one-quarter cycle; but that's a technical point. In any case I said I wouldn't be arguing about factors of two. It's the factor of three that's getting me.

What exactly does the electromagnetic field look like?? I first thought that it could be polarized in either the x, y, or z directions, giving me an extra factor of three. But then I realized that didn't make sense. In two dimensions you can add up all kinds of waves in the xy plane, keeping the polarization in the z plane. But there's no way you can take plane waves in three dimensions and add them up with all the same polarization. The polarizations is going to vary in different places... in other words, the field lines are going to curve.

How can you picture field lines curving through space and still meeting the required boundary conditions? Not the simplistic, incorrect boundary conditions described in the hyperphysics, link, but the correct e-m cavity conditions: perfectly conducting walls with the electric field perpendicular and the magnetic field tangential? After some hours staring at a box of Rice Krispies (as my typical rectangular cell) and drawing ellipses on different faces, this is what I came up with:
The green lines are the electric fields and the red ellipses are the magnetic fields. I am a bit troubled by a certain assymetry .... the unit cell has two "ellipsoids" of magnetic lines and one of electrical lines. (I've drawn the green electrical lines as though they form a torus, but that's bad artwork: the ellipses should be parallel.) Still, I think this is the correct configuration. It seems to work: as the magnetic loops collapse, the generate electric loops in generally the right places.

It's the count that bothers me. Of the six cell walls, four have magnetic loops on them and two are empty... in this case the side walls (the walls where they list the ingredients on the Rice Krispies box.) But I could have just as easily taken the front/back or the top/bottom faces to be the null faces. So there are three field configurations that fit this geometry. I can't think of any physical reason that shouldn't work.

So my mode count disagrees with Rayleigh-Jeans by a factor of three. And yet the Rayleigh-Jeans formula gives the correct energy for the field (at low frequencies), assuming each mode gets its alloted energy. It's a problem and I can't get around it.

For now I think I'm going to make use of the Rayleigh-Jeans formula and go ahead with my equilibrium calculations.

JAN 24 2013: It's taken me almost two years, but I found the flaw in my counting method. You can see what I did wrong if you click here.

Monday, February 21, 2011

What's the connection?

In my last posting, I showed that you could get a convert the formula for electromagnetic power density into a real-time picture of actual sine waves, but you got different results depending on how you chopped up the spectrum and how wide you chose the bandwidth. Here are three pictures of typical waveforms you can get from a given spectrum. The pictures are highly simplified to illustrate the most essential features of the waves. The single cycle of a sine wave is really intended to suggest hundreds or millions of cycles in general:

It's not completely arbitrary. I've written in a formula at the bottom of the picture to show that the three waveforms must have a certain property in common. I've expressed this property as a mathematical formula because that's what people do, but the physical significance of this common property is what really matters. And it relates to ... the drunkard's walk!

If any of these wave trains are applied to a harmonic oscillator, they will tend to build up the vibration; but because the applied wave and the mechanical oscillator are in general out of phase with each other, the amplitude of oscillation will build up not steadily, but randomly. Like the drunkards walk.

What these three wave forms have in common is that if you apply them to a given oscillator over a period of time, they will cause the oscillation to grow at equal rates!

You don't even need to know much about the dynamics of a driven oscillator to see why this ought to be so. It goes back to the drunkard's walk. Let's compare the first two wave forms. The driving force in the second case is four times as great, applied over one quarter the time. That's why those two waveforms both give the same rate of growth.

The third case is interesting. Compared to the first waveform, it's the twice the impulse applied one quarter as often. It's like the drunkard who takes a giant two-meter step, but only once every four seconds. Figure it out ... he has the same rate of progress as the drunkard who takes regular one-meter steps every second.

So you can take your pick. Any one of these waveforms applied to a harmonic oscillator will cause it to grow at the same rate. What is the rate of growth? It appears to get slower and slower as the amplitude of oscillation builds up. But notice something funny. If we measure the size of the oscillation not by its amplitude, but by its total energy ... the growth rate becomes a constant. The energy of oscillation grows at a constant rate.

So why doesn't it build up without limit? Because as it oscillates, it becomes a transmitting antenna, and radiates outwards in all directions. There are formulas from antenna theory that will give us an exact number for this, but it's enough for now to undertand that the radiated power is proportional to the square of the amplitude. In other words, its proportional to the total power of the mechanical oscillator.

As the oscillator builds up, it emits power at a faster and faster rate. At some moment this becomes equal to the rate at which it is absorbing power. That is the point of equilibrium, and we can now calculate it.

The Drunkard's Walk

Yesterday I showed how you could take a piece of the energy spectrum and convert it into an actual waveform. You just approximate the energy distribution by a series of sine waves, pick an arbitrary bandwidth, and add up the waves. The picture you end up with depends on how finely you chop up the distribution into discrete waves, and how many of those waves you choose to add up.

In each case you get series of wave trains of a certain duration, amplitude, repetition rate. Some examples of what you might get from a given distribution are:

10 mV/m pulses 4 microsecond duration 100 kHz repetition rate
40 mV/m pulses 1 microsecond duration 100 kHz repetition rate
20 mV/m pulses 4 microsecond duration 25 kHz repetition rate

(These aren't the exact values I worked out yesterday, but they illustrate the general pattern: multiply column A by column B and the square root of column C to get a constant value.)

I then claimed that no matter what choices you make, you will get the exact same result when you apply the resulting waveform the the problem of finding the equilibrium of a mechanical resonator in that field. Now I'm going to show why it works. To understand the reason we must first consider a famous math problem known as the Drunkard's Walk.

The premise of the problem is that you have a drunk standing under a lamppost and he begins to wander aimlessly. Every second he takes one step, and each step is one meter long. The question is, how far from the lamppost will he be found after a given time?

It's not such an easy question, but the answer turns out to be simple: in this example it's just the square root of the number of steps. So after 100 seconds the drunk is expected to be found (on average) ten meters from the lamppost.

Why should this be? Since his steps are totally random, shouldn't he make absolutely no progress on average?

That's also true. His most likely progress in any given direction is zero, and is most likely endpoint is...right where he started. But his average distance away from the lamppost is still ten meters. Go figure.

It's actually not hard to see why he must, on average, continue to get farther and farther away. Suppose at some time he happens to be five meters away from the lamppost. His next step will place him somewhere on an imaginary circle exactly one meter from his present location.

If you look at that circle, you will see that most of it (more than half of it, at least) lies outside a five-meter radius from the lamp!

If you look at the small circle you see that the segment of arc (in green) outside the five-meter radius is longer than the segment (in purple) inside five meters. This is true wherever he starts from; and so, on average, the drunk keeps getting farther from the lamppost.
How much farther? That's a tough one. But there is a little trick that will let us guess the answer. The most "neutral" thing the drunk can do is, arguably, to take a step perpendicular to the radius of the big circle. We can then use the Law of Pythagoras to see where he ends up. 5^2 plus 1^2 equals 26, so his distance from the lamppost is the square root of 26. Since his previous location at a distance of 5 represented the square root of 25, we see that this little trick gives us exactly the answer that we wanted...the distance goes as the square root of the number of steps.
We're going to need this trick when we go back to the harmonic oscillator.

Sunday, February 20, 2011

Harmonic Oscillator: The problem is solved

Okay. Now I've figured out something good. This is a problem that has plagued me for years and now I've solved it.

I've been talking about the equilibrium between the radiation field and the molecular oscillator. We said that the oscillator gets its energy from collisions with other molecules and pumps energy into the radiation field. Of course that's only half the story. The radiation field also pumps energy into the molecular oscillator. The problem is to calculate the equilibrium point, and now I've figured out how to do it.

Why is this such a difficult problem? It's difficult in every possible way. But probably the greatest difficulty of all is that the fields are random. So we have an oscillator vibrating at 1000 MHz. It woud be fine and dandy if we simply had a radiation field at 1ooo MHz with a strength of 100 milliwatts per square meter. That we could work with. But that is not what we have. No, what we have is a radiation distribution of 100 mW/m^2 .....per MegaHertz! So what do we work with? Do we consider only the radiation at exactly 1000 MHz? No, because there is zero radiation at any exact frequency! We have to sample it over a certain finite bandwidth or we get nothing. So do we consider all radiation between 999 and 1001 MHz? That gives us a total power density of 200mW/m^2? But we could just as well have considered all the radiation between 996 and 1004 MHz, giving us a total power density of 800 mW/m^2. Which one should it be??? It's all very confusing.

What I've figured out today is doesn't matter! We're free to make either choice, and we get the same result. This seems crazy, but I'm going to show you why it works. The first step is to describe the power distribution. Mathematically, it's very tricky do deal with continuous distributions, but we can always approximate them by discrete sums. I've drawn a sketch of a graph showing a continuous frequency distribution in the vicinity of 1000 MHz, represented by a sum of equal sine waves spaced 1 MHz apart. (The graph says hertz, but that's a mistake. It should be labeled MHz!)

By the way, there's nothing special about MegaHertzes. We could also have chosen to represent our field by taking sine waves every 250 kiloHertz. Those are the skinny little lines between the heavy black lines. This is a more accurate breakdown. If we choose the 250 kHz separation, the sine waves are only half as big, because there are more of them. I know...there are four times as many, so shouldn't they be a quarter the height? No, because power goes as the square of the amplitude. I've chosen the correct height to give the same power in either representation.

Now let's see what this power distribution looks like from the point of view of the oscillator. The oscillator is really only sensitive to power close to its frequency of oscillation. How close? That's a tough one, but let's try for starters just including power between 999 and 1001 MHz ... a bandwith of 2 MHz. There's a little circled sketch in the previous graph showing exactly how I'm calculating this - I've taken half the sine waves from the side bands, and the full wave in the middle. (If you took first-year calculus I think it's what they call Simpson's Rule.) The resulting wave looks like this:

It rises and falls over a span of one microsecond. This pattern repeats itself a million times a second.
That's one way we can do it. But who said we had to restrict our bandwidth to 2 MHz? I've redone the calculation below, taking in a bandwith of 8 MHz between 996 and 1004 MHz. Again, I've used Simpson's Rule for the endpoints. Here again is a schematic of the distribution:

The resulting waveform is shown below. (By the way, these pictures are for real. I actually programmed nine sine waves into an Excel spreadsheet, added them up, and plotted the graph.)

It looks very different. The total time interval is still one microsecond, but the energy pulse is jammed into the middle of the frame, and the peak is much higher. In fact, the math shows you get four times the amplitude over one quarter the time interval. Is that the same total power? No, because power goes as the square of the amplitude. It's four times as much power. But that makes sense, after all...we took four times the bandwidth over a constant power density.

So depending on how we divide and truncate our power distribution, we get very different electromagnetic waveforms as applied in real time to the mechanical oscillator. It gets even more complicated if we approximate our distribution with sine waves every 250 kHz (the skinny lines in the graph). For example, taking in eight bands of power between 999 and 1001 MHz, we get a pulse train lasting one microsecond out of four, and twice the height calculated in our first example. Twice the power for one quarter the time interval? That's the same net energy .... that's a good sign at least. But again it's a very different looking wave.
But here's what I just figured out .... these waves all have exactly the same effect on a mechanical oscillator! It doesn't matter which approximation or how much bandwidth we choose. We're going to find it gives the same average boost to a random oscillator. It means we can go ahead with the equilibrium calculation. That will be the topic of my next post.

What was Planck thinking?

I ended my last posting by speculating that the only way to fix things was to find some way of recalculating the entropy. Because no matter how many practical reasons I could find as to why the high-frequency vibrational modes should not be activated in "real" matter, the fundamental problem remains: we can still imagine an ideal gas made of perfect billiard balls joined by flexible rods, and for this theoretical gas, the entropy calculation agrees with the practical outcome: all the modes, vibrational included, must be excited to their full complement of energy. And any "real" gas must have the same equilibrium as this "ideal" gas.

Isn't this something like what Planck did? By quantising the energy modes of the electromagnetic resonances, he fixed things so the high frequency modes went away. Because when you calculate the entropy of these quantised modes, it gives the desired outcome. But there's something very wrong with this approach. If Planck allows the gas itself to behave as an ideal collection of springs and ball bearings, then both thermodynamics and practical reasoning tell us that the high-frequency vibrations must exist. And it doesn't help to recalculate the entropy of the radiation field, because if those molecules are vibrating, then they will (and must) radiate power into those forbidden frequencies.

One hundred years after the fact, it's hard to know exactly what people were thinking back then. Planck had conjured up this miraculous formula that worked perfectly, but nobody (Planck included) knew quite what to make of it. It seems like people were focused on the thermodynamics of the electromagnetic field, so people figured out that by quantising the modes of vibration, and allowing only certain values of the energy at any given frequency, they could make the entropy come out "right" - in other words, consistent with the experimental facts. But what about the entropy of the mechanical oscillators? Wasn't that still a problem?

It would be a number of years before Einstein would propose that the mechanical oscillations should follow similar rules of quantisation. In the meantime, one has to assume that people were content to allow the mechanical vibrations to have their place in the theory, but to somehow insist, against all logic, that they would not contribute to the intensity of the radiation field. To make this work, they had to suspend the laws of electromagnetism! That is the origin of the idea that energy could only be emitted or absorbed in discrete lumps.

There is a much better way out of this dilemma, and that is to attack the problem at its source: the mechanical vibrations. Because if the unwanted mechanical vibrations are supressed, then the electromagnetic field follows suit and we avert the ultraviolet catastrophe. I have already shown how there are many practical reasons to suppose why those vibrations shouldn't happen in "real" atoms. The problem is with the entropy: the entropy calculation shows that those vibrations must assume their alloted share of the energy; and the entropy calculation is backed up by the theoretical example of the imaginary ideal gas made of springs and billiard balls, where you can verify by mechanical reasoning that it works as claimed - that equipartition must prevail.

What I am saying is that if we solve the problem at the mechanical level, we avoid the problem with the electromagnetic field. You can keep the laws of Maxwell and there is no need for quantized lumps of energy. In my next post, I'm going to delve more deeply into just what is required to solve the problem at the mechanical level.

Saturday, February 19, 2011

Where are we now?

I never intended to use this blog for random, thinking-out-loud speculations. But I'm afraid that is what it's come down to. I started out three days ago thinking I would make a clear case for the semi-classical picture of the black-body spectrum, but the farther I go the more messed up it gets.

So where are we now?

I've put forward the suggestion that maybe the true black-body spectrum is the old Rayleigh-Jeans distribution, with its ultraviolet catastrophe; and proposed that the reason we avoid the catastrophe is that there is simply no mechanism available in the form of atomic collisions that can effectively drive the high-frequency molecular oscillations needed in turn to excite the high-frequency electromagnetic standing waves. In all modesty I must say it's an intriguing theory, but in the end I can't quite bring myself to believe in it.

The equipartition theorem is not derived on the basis of any particular model of how collisions work. It is hard-core thermodynamics, based on maximizing the entropy of a system. For any mechanical collection of objects, including projectiles, oscillators, and rotators, so long as the energy is somehow quadratic with the motion (as it is in these cases)...the derivation shows that entropy is maximized when the energy is shared equally between all modes. I think I'm summarizing this correctly.

So under my proposal, the system would come to rest in a state of less than maximum entropy. This is a problem, if we believe the entropy calculation is correct. The problem is still perplexing: it is true that the entropy calculation does not rely on any specific mechanism to acheive its result, but should it also not be in physical agreement with a real mechanism which actually does exist? And I have put forward a plausible mechanism to show why the vibrational modes should remain low. What does it all mean?

Suppose I want to achieve the impossible, and redo the entropy calculation itself; suppose I then find that for a real gas, the entropy is maximized by those states I've described with the weak oscillators. I've still got a huge problem. Because my state has to be in thermodynamic equilibrium with a theoretical state made of classically "ideal" oscillators that follow Rayleigh-Jeans. And while I might have a faint hope that the two systems could be in mechanical equilibrium, there is no way they can still be in equilibrium when I include the electromagnetic field. It's a real problem.

Maybe there is a missing term in the entropy calculation which accounts for frequency. I'm damned if I can figure out how that would work.

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.

Thursday, February 17, 2011

What's Wrong with the Equipartition Theorem

Why should all the modes get the same energy in the first place? It's an important question, and let's start by looking at a case where it works.

In a diatomic gas like hydrogen, we consider the molecules to be tiny rigid dumbbells. Let's assume that they start off zipping every which way, but that none of them are spinning. It's easy to see that this situation cannot last long. Any kind of glancing collision will result in some energy being converted into rotational motion. As an extreme case, we can show a situation where the conversion is 100%. Imagine two molecules speeding towards each other with each of their axes perpendicular to the direction of motion. Two of the atoms collide - not four, in general - and at the moment of collision the other two atoms, at opposite ends of their respective dumbbells, are unaffected. If the two atoms collide exactly head on, they will each instantly reverse direction - because that is what billiard balls do.

The kinetic energy of this system has been converted from 100% translational to 100% rotational. You can see it must be so because at the moment the molecules begin to separate, their component atoms have equal and opposite velocities. Therefore, they are spinning, and their average forward motion is zero. It's not hard to imagine that eventually the motion in a gas is randomly distributed between all its different possible forms.

But what about the vibrations? Shouldn't a good clean hit between stiff molecules set each of them ringing? Indeed, that is what equipartition tells us, and in an ideal world that's just what would happen. But as pointed out in the last essay, the measuremen of specific heats shows that the vibrational modes are not in fact active for common diatomic gasses.

Why not? I think it's pretty obvious that the stumbling block is going to be that "good clean hit". Atoms are not dumbbells. When two atoms collide there is no meeting of hard surfaces. As they approach they influence each other with electric forces. It is a gradual process...very fast on our time scales, but compared to the vibrational frequency of a hydrogen molecule - well, it's easy to imagine that it might be very much slower indeed.

It's not a billiard ball collision and that's why the vibrational mode is not excited. So maybe there is no need to worry about the discrepancy in specific heats. In fact, when the temperatures get high enough, so the velocities are high enough, so the collisions take place on a more compressed time scale...voila, the molecules start to vibrate after all.

What's more, if we can explain away the vibrations, maybe we can explain away the ultraviolet catastrophe. The vibrating hydrogen molecule happens to be electrically neutral, so it doesn't influence the cavity modes one way or the other. But to drive the standing waves inside the metal box, you need some means of converting the thermal energy of the molecules into electromagnetic energy. You do that by getting charges to vibrate. Hydrogen doesn't work so well, but any composite molecule like CN or NO will do. They are polarized to some extent, so when they vibrate they become tiny antennas, pumping electromagnetic energy into the field.

But if they don't vibrate, then there is no high-frequency energy, and there is no ultraviolet catastrophe. The idea that they would have to vibrate was based on a literal reading of the equipartion theorem. We've made the case that there is no mechanism for equipartition when the collisions are "soft", as they must be in the real world. Have we thenb solved the problem that was the ruin of classical physics at the end of the nineteenth century?

That's something we'll take up in our next posting.

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.