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 that...it 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.

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.

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.

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 fxqi.org. 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 http://fqxi.org/community/forum/topic/927

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.

Free Energy: The rest of the story

In my next-to-last post on this subject, I left the question hanging in the air: how does a reaction proceed when the Gibb's Free Energy is positive? I am now going to try to answer this.

The reaction in question is the decomposition of methane in the presence of CO2:

CH4 + CO2 ===> 2CO + 2H2

As I recall I determined that at STP conditions (25 C and 1 atm) the Free Energy was definitely unfavorable. I think it worked out to something on the order of 90 kiloJoules. There was a complication however: the reaction takes place in a nuclear reactor at a temperature of 300 C. The effect of this is to bring the Free Energy down quite a bit (because the entropy term is favored by the smaller mole fragments on the product side of the equation) but not nearly enough to make it go negative. Let's say it came to around 65 kJ.

The funny thing is that these numbers lead to all kinds of consequences in terms of measuring the actual gas concentrations. To see what happens we have to convert the Gibbs Free Energy to the corresponding Equilibrium Constant. The equilibrium constant k is given by the formula:


k = exp(-(Gibbs Free Energy) /RT)

where RT is the gas constant product from the formula PV=nRT. It's easy to remember that RT comes to exactly 2.2 kJ because one litre of air at STP occupies just 22 litres.

Now it gets interesting. The equilibrium constant in the first case (room temperature) works out, on the ratio of 90/2.2, to around e^-40, or dividing by 2.3 you get 10^-17. At the higher temperature, however, the numbers come out a little less extreme, at around 10^-12. These are all still very tiny numbers. How could they make much difference?

It turns out that the difference is critical in the case of interest. In a previous blog I showed how to use the equilibrium constant to calculate the concentrations of a reaction. Now lets apply it to the trace gasses in the nuclear reactor. We assume that reactor coolant is leaking into the CO2 purge gas stream through a tiny crack in a pressure tube. The coolant, a complex hydrocarbon, will typically break down into methane-like fragments. Let us suppose the initial concentration of methane in CH4 is 20 ppm, or 2 * 10^-5.

How can we balance the equilibrium equation in this case? We will write the equation as


k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

For a reactor temperature of 300 degrees, we have k = 10^-13 and we need to make the concentrations in the above equation work out to that value. By toying with numbers, we can see that if the initial concentration of methane is 90% dissociated, so that only 2 ppm remains, that the CO and H2 concentrations must each be close to 40 ppm: we then get for the numerator a value of 256 *10^-20, which requires the denominator to be 256 * 10^-8, or just over 2 ppm. That's where the equation becomes balanced.

The situation is quite different, however, when the gas returns to room temperature. Now the equilibrium constant is 10^-17. The math is a little hard to write out in ASCII, but it turns out if only 10% of the methane dissociates, you get a concentration of 4 ppm for both the CO and the H2. Multiply out the terms and the equation balances at these concentrations. You have 18 ppm methane.

These numbers describe exactly what was seen to be happening in the leak detection system! When gas was initially sampled at the detector, it showed definite methane readings in the tens of parts per million. But after drawing for several minutes on a sample line, the reading "settled down" to zero, reassuring everyone that there was no cause for concern. What was actually happening was that the initial reading came from the room temperature gasses that had been sitting in the tubes: the methane that had initially decomposed at high temperatures had recombined into its original form. But as you kept drawing on the sample line, you then got to the fresh gasses which had just decomposed at reactor temperatures. Even if the sample point was at room temperature, the gasses had not yet had enough time to recombine, so the methane reading was basically zero.

I started off telling this story because almost thirty years later, it would have critical consequences in my development of the theory of Quantum Siphoning. I was involved in an online discussion of the quantum nature of the photographic process, and my opponents had thrown in my face the fact that the photo-reduction of silver bromide to silver was thermodynamically non-spontaneous. It was only because of my previous experience in analyzing the decomposition of trace methane in CO2 that I was able to see the possibility of doing the same thing for trace concentrations of silver in a silver bromide crystal. This insight was to have critical consequences for my ability to see the way to Quantum siphoning.

There was one other physics problem from my younger days that gave me an unusual perspective on processes involve waves and the interchange of energy. That was the problem of the crystal radio. But that's a story for another day.

There is a final postscript to the story of the nuclear reactor that needs to be told. Having figured out what was the problem with the leak monitoring system, I naturally wanted to tell people all about it; in no small measure, to show people how smart I was. I called a meeting of the interested parties. The department manager was the chair of the meeting. I began my presentation but almost immediately the manager interrupted me. "Before we go too far, I want to ask one question". He then turned to the reactor superintendant and said: "Even if what Marty says is right, would it make any difference to how you operate the reactor?".

"Absolutely not," replied the superintendant. The manager then turned to me and said, "Well, I guess there's no need to continue with this meeting". And that was it.

I didn't object because at the time I was actually impressed at what a good job the manager had done: he had gotten right down to the essentials without wasting a lot of time over theoretical issues. I was young and I wanted to be a team player, so I accepted the result.

It wasn't until I started writing this series of articles several weeks ago that it struck me: I had been set up! The manager and the superintendant had gotten together and figured out how to shut me up. It's dirty trick that people use when they want to dismiss what you're saying without letting you make your point: they start off with "even if what you say is true...". Of course, if what I was saying were true, it should have made a difference. They just didn't want to admit it.