The blackbody radiation problem revisited: quantum statistics

The equipartition theorem – which states that the energy levels of the modes of any (linear) system, in classical as well as in quantum physics, are always equally spaced – is deep and fundamental in physics. In my previous post, I presented this theorem in a very general and non-technical way: I did not use any exponentials, complex numbers or integrals. Just simple arithmetic. Let’s go a little bit beyond now, and use it to analyze that blackbody radiation problem which bothered 19th century physicists, and which led Planck to ‘discover’ quantum physics. [Note that, once again, I won’t use any complex numbers or integrals in this post, so my kids should actually be able to read through it.]

Before we start, let’s quickly introduce the model again. What are we talking about? What’s the black box? The idea is that we add heat to atoms (or molecules) in a gas. The heat results in the atoms acquiring kinetic energy, and the kinetic theory of gases tells us that the mean value of the kinetic energy for each independent direction of motion will be equal to kT/2. The blackbody radiation model analyzes the atoms (or molecules) in a gas as atomic oscillators. Oscillators have both kinetic as well as potential energy and, on average, the kinetic and potential energy is the same. Hence, the energy in the oscillation is twice the kinetic energy, so its average energy is 〈E〉 = 2·kT/2 = kT. However, oscillating atoms implies oscillating electric charges. Now, electric charges going up and down radiate light and, hence, as light is emitted, energy flows away.

How exactly? It doesn’t matter. It is worth noting that 19th century physicists had no idea about the inner structure of an atom. In fact, at that time, the term electron had not yet been invented: the first atomic model involving electrons was the so-called plum pudding model, which J.J. Thompson advanced in 1904, and he called electrons “negative corpuscles“. And the Rutherford-Bohr model, which is the first model one can actually use to explain how and why excited atoms radiate light, came in 1913 only, so that’s long after Planck’s solution for the blackbody radiation problem, which he presented to the scientific community in December 1900. It’s really true: it doesn’t matter. We don’t need to know about the specifics. The general idea is all that matters. As Feynman puts it: it’s how “A hot stove cools on a cold night, by radiating the light into the sky, because the atoms are jiggling their charge and they continually radiate, and slowly, because of this radiation, the jiggling motion slows down.” 🙂

His subsequent description of the black box is equally simple: “If we enclose the whole thing in a box so that the light does not go away to infinity, then we can eventually get thermal equilibrium. We may either put the gas in a box where we can say that there are other radiators in the box walls sending light back or, to take a nicer example, we may suppose the box has mirror walls. It is easier to think about that case. Thus we assume that all the radiation that goes out from the oscillator keeps running around in the box. Then, of course, it is true that the oscillator starts to radiate, but pretty soon it can maintain its kT of energy in spite of the fact that it is radiating, because it is being illuminated, we may say, by its own light reflected from the walls of the box. That is, after a while there is a great deal of light rushing around in the box, and although the oscillator is radiating some, the light comes back and returns some of the energy that was radiated.”

So… That’s the model. Don’t you just love the simplicity of the narrative here? 🙂 Feynman then derives Rayleigh’s Law, which gives us the frequency spectrum of blackbody radiation as predicted by classical theory, i.e. the intensity (I) of the light as a function of (a) its (angular) frequency (ω) and (b) the average energy of the oscillators, which is nothing but the temperature of the gas (Boltzmann’s constant k is just what it is: a proportionality constant which makes the units come out alright). The other stuff in the formula, given hereunder, are just more constants (and, yes, the is the speed of light!). The grand result is:

Rayleigh's law

The formula looks formidable but the function is actually very simple: it’s quadratic in ω and linear in 〈E〉 = kT. The rest is just a bunch of constants which ensure all of the units we use to measures stuff come out alright. As you may suspect, the derivation of the formula is not so simple as the narrative of the black box model, and so I won’t copy it here (you can check yourself). Indeed, let’s focus on the results, not on the technicalities. Let’s have a look at the graph.

Rayleigh's law graph

The I(ω) graphs for T = T0 and T = 2T0 are given by the solid black curves. They tell us how much light we should have at different frequencies. They just go up and up and up, so Rayleigh’s Law implies that, when we open our stove – and, yes, I know, some kids don’t know what a stove is :-/ – and take a look, we should burn our eyes from x-rays. We know that’s not the case, in reality, so our theory must be wrong. An even bigger problem is that the curve implies that the total energy in the box, i.e. the total of all this intensity summed up over all frequencies, is infinite: we’ve got an infinite curve here indeed, and so an infinite area under it. Therefore, as Feynman puts it: “Rayleigh’s Law is fundamentally, powerfully, and absolutely wrong.” The actual graphs, indeed, are the dashed curves. I’ll come back to them.

The blackbody radiation problem is history, of course. So it’s no longer a problem. Let’s see how the equipartition theorem solved it. We assume our oscillators can only take on equally spaced energy levels, with the space between them equal to h·f = ħ·ω. The frequency f (or ω = 2π·f) is the fundamental frequency of our oscillator, and you know and ħ = h/2π, course: Planck’s constant. Hence, the various energy levels are given by the following formula: En = n·ħ·ω = n·h·f. The first five are depicted below.

energy levelsNext to the energy levels, we write the probability of an oscillator occupying that energy level, which is given by Boltzmann’s Law. I wrote about Boltzmann’s Law in another post too, so I won’t repeat myself here, except for noting that Boltzmann’s Law says that the probabilities of different conditions of energy are given by e−energy/kT = 1/eenergy/kT. Different ‘conditions of energy’ can be anything: density, molecular speeds, momenta, whatever. Here we have a probability Pn as a function of the energy En = n·ħ·ω, so we write: Pn = A·e−energy/kT = A·en·ħ·ω/kT. [Note that P0 is equal to A, as a consequence.]

Now, we need to determine how many oscillators we have in each of the various energy states, so that’s N0, N1, N2, etcetera. We’ve done that before: N1/N0 = P1/P0 = (A·e−2ħω/kT)/(A·eħω/kT) = eħω/kT. Hence, N1 = N0·eħω/kT. Likewise, it’s not difficult to see that, N2 = N0·e−2ħω/kT or, more in general, that Nn = N0·e−nħω/kT = N0·[eħω/kT]n. To make the calculations somewhat easier, Feynman temporarily substitutes eħω/kT for x. Hence, we write: N1 = N0·x, N2 = N0·x2,…, Nn = N0·xn, and the total number of oscillators is obviously Ntot = N0+N1+…+Nn+… = N0·(1+x+x2+…+xn+…).

What about their energy? The energy of all oscillators in state 0 is, obviously, zero. The energy of all oscillators in state 1 is N1·ħω = ħω·N0·x. Adding it all up for state 2 yields N2·2·ħω = 2·ħω·N0·x2. More generally, the energy of all oscillators in state n is equal to Nn·n·ħω = n·ħω·N0·xn. So now we can write the total energy of the whole system as Etot = E0+E1+…+En+… = 0+ħω·N0·x+2·ħω·N0·x2+…+n·ħω·N0·xn+… = ħω·N0·(x+2x2+…+nxn+…). The average energy of one oscillator, for the whole system, is therefore:

average energy

Now, Feynman leaves the exercise of simplifying that expression to the reader and just says it’s equal to:

Formula

I should try to figure out how he does that. It’s something like Horner’s rule but that’s not easy with infinite polynomials. Or perhaps it’s just some clever way of factoring both polynomials. I didn’t break my head over it but just checked if the result is correct. [I don’t think Feynman would dare to joke here, but one could never be sure with him it seems. :-)] Note he substituted eħω/kT for x, not e+ħω/kT, so there is a minus sign there, which we don’t have in the formula above. Hence, the denominator, eħω/kT–1 = (1/x)–1 = (1–x)/x, and 1/(eħω/kT–1) = x/(1–x). Now, if (x+2x2+…+nxn+…)/(1+x+x2+…+xn+…) = x/(1–x), then (x+2x2+…+nxn+…)·(1–x) must be equal to x·(1+x+x2+…+xn+…). Just write it out: (x+2x2+…+nxn+…)·(1–x) = x+2x2+…+nxn+….−x2−2x3−…−nxn+1+… = x+x2+…+xn+… Likewise, we get x·(1+x+x2+…+xn+…) = x+x2+…+xn+… So, yes, done.

Now comes the Big Trick, the rabbit out of the hat, so to speak. 🙂 We’re going to substitute the classical expression for 〈E〉 (i.e. kT) in Rayleigh’s Law for it’s quantum-mechanical equivalent (i.e. 〈E〉 = ħω/[eħω/kT–1].

What’s the logic behind? Rayleigh’s Law gave the intensity for the various frequencies that are present as a function of (a) the frequency (of course!) and (b) the average energy of the oscillators, which is kT according to classical theory. Now, our assumption that an oscillator cannot take on just any energy value but that the energy levels are equally spaced, combined with Boltzmann’s Law, gives us a very different formula for the average energy: it’s a function of the temperature, but it’s a function of the fundamental frequency too! I copied the graph below from the Wikipedia article on the equipartition theorem. The black line is the classical value for the average energy as a function of the thermal energy. As you can see, it’s one and the same thing, really (look at the scales: they happen to be both logarithmic but that’s just to make them more ‘readable’). Its quantum-mechanical equivalent is the red curve. At higher temperatures, the two agree nearly perfectly, but at low temperatures (with low being defined as the range where kT << ħ·ω, written as h·ν in the graph), the quantum mechanical value decreases much more rapidly. [Note the energy is measured in units equivalent to h·ν: that’s a nice way to sort of ‘normalize’ things so as to compare them.]

400px-Et_fig2

So, without further ado, let’s take Rayleigh’s Law again and just substitute kT (i.e. the classical formula for the average energy) for the ‘quantum-mechanical’ formula for 〈E〉, i.e. ħω/[eħω/kT–1]. Adding the dω factor to emphasize we’re talking some continuous distribution here, we get the even grander result (Feynman calls it the first quantum-mechanical formula ever known or discussed):

formula 2So this function is the dashed I(ω) curve (I copied the graph below again): this curve does not ‘blow up’. The math behind the curve is the following: even for large ω, leading that ω3 factor in the numerator to ‘blow up’, we also have Euler’s number being raised to a tremendous power in the denominator. Therefore, the curves come down again, and so we don’t get those incredible amounts of UV light and x-rays.

Rayleigh's law graph

So… That’s how Max Planck solved the problem and how he became the ‘reluctant father of quantum mechanics.’ The formula is not as simple as Rayleigh’s Law (we have a cubic function in the numerator, and an exponential in the denominator), but its advantage is that it’s correct. Indeed, when everything is said and done, indeed, we do want our formulas to describe something real, don’t we? 🙂

Let me conclude by looking at that ‘quantum-mechanical’ formula for the average energy once more:

E〉 = ħω/[eħω/kT–1]

It’s not a distribution function (the formula for I(ω) is the distribution function), but the –1 term in the denominator does tell us already we’re talking Bose-Einstein statistics. In my post on quantum statistics, I compared the three distribution functions. Let ‘s quickly look at them again:

  • Maxwell-Boltzmann (for classical particles): f(E) = 1/[A·eE/kT]
  • Fermi-Dirac (for fermions): f(E) = 1/[AeE/kT + 1]
  • Bose-Einstein (for bosons):  f(E) = 1/[AeE/kT − 1]

So here we simply substitute ħω for E, which makes sense, as the Planck-Einstein relation tells us that the energy of the particles involved is, indeed, equal to E = ħω . Below, you’ll find the graph of these three functions, first as a function of E, so that’s f(E), and then as a function of T, so that’s f(T) (or f(kT) if you want).

graph energy graph temperature

The first graph, for which E is the variable, is the more usual one. As for the interpretation, you can see what’s going on: bosonic particles (or bosons, I should say) will crowd the lower energy levels (the associated probabilities are much higher indeed), while for fermions, it’s the opposite: they don’t want to crowd together and, hence, the associated probabilities are much lower. So fermions will spread themselves over the various energy levels. The distribution for ‘classical’ particles is somewhere in the middle.

In that post of mine, I gave an actual example involving nine particles and the various patterns that are possible, so you can have a look there. Here I just want to note that the math behind is easy to understand when dropping the A (that’s just another normalization constant anyway) and re-writing the formulas as follows:

  • Maxwell-Boltzmann (for classical particles): f(E) = e−E/kT
  • Fermi-Dirac (for fermions): f(E) = e−E/kT/[1+e−E/kT]
  • Bose-Einstein (for bosons):  f(E) = e−E/kT/[1−e−E/kT]

Just use Feynman’s substitution xeħω/kT: the Bose-Einstein distribution then becomes 1/[1/x–1] = 1/[(1–x)/x] = x/(1–x). Now it’s easy to see that the denominator of the formula of both the Fermi-Dirac as well as the Bose-Einstein distribution will approach 1 (i.e. the ‘denominator’ of the Maxwell-Boltzmann formula) if e−E/kT approaches zero, so that’s when E becomes larger and larger. Hence, for higher energy levels, the probability densities of the three functions approach each other indeed, as they should.

Now what’s the second graph about? Here we’re looking at one energy level only, but we let the temperature vary from 0 to infinity. The graph says that, at low temperature, the probabilities will also be more or less the same, and the three distributions only differ at higher temperatures. That makes sense too, of course!

Well… That says it all, I guess. I hope you enjoyed this post. As I’ve sort of concluded Volume I of Feynman’s Lectures with this, I’ll be silent for a while… […] Or so I think. 🙂

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Planck’s constant (II)

My previous post was tough. Tough for you–if you’ve read it. But tough for me too. 🙂

The blackbody radiation problem is complicated but, when everything is said and done, what the analysis says is that the the ‘equipartition theorem’ in the kinetic theory of gases ‘theorem (or the ‘theorem concerning the average energy of the center-of-mass motion’, as Feynman terms it), is not correct. That equipartition theorem basically states that, in thermal equilibrium, energy is shared equally among all of its various forms. For example, the average kinetic energy per degree of freedom in the translation motion of a molecule should equal that of its rotational motions. That equipartition theorem is also quite precise: it also states that the mean energy, for each atom or molecule, for each degree of freedom, is kT/2. Hence, that’s the (average) energy the 19th century scientists also assigned to the atomic oscillators in a gas.

However, the discrepancy between the theoretical and empirical result of their work shows that adding atomic oscillators–as radiators and absorbers of light–to the system (a box of gas that’s being heated) is not just a matter of adding additional ‘degree of freedom’ to the system. It can’t be analyzed in ‘classical’ terms: the actual spectrum of blackbody radiation shows that these atomic oscillators do not absorb, on average, an amount of energy equal to kT/2. Hence, they are not just another ‘independent direction of motion’.

So what are they then? Well… Who knows? I don’t. But, as I didn’t quite go through the full story in my previous post, the least I can do is to try to do that here. It should be worth the effort. In Feynman’s words: “This was the first quantum-mechanical formula ever known, or discussed, and it was the beautiful culmination of decades of puzzlement.” And then it does not involve complex numbers or wave functions, so that’s another reason why looking at the detail is kind of nice. 🙂

Discrete energy levels and the nature of h

To solve the blackbody radiation problem, Planck assumed that the permitted energy levels of the atomic harmonic oscillator were equally spaced, at ‘distances’ ħωapart from each other. That’s what’s illustrated below.

Equally space energy levels

Now, I don’t want to make too many digressions from the main story, but this En = nħω0 formula obviously deserves some attention. First note it immediately shows why the dimension of ħ is expressed in joule-seconds (J·s), or electronvolt-seconds (J·s): we’re multiplying it with a frequency indeed, so that’s something expressed per second (hence, its dimension is s–1) in order to get a measure of energy: joules or, because of the atomic scale, electronvolts. [The eV is just a (much) smaller measure than the joule, but it amounts to the same: 1 eV ≈ 1.6×10−19 J.]

One thing to note is that the equal spacing consists of distances equal to ħω0, not of ħ. Hence, while h, or ħ (ħ is the constant to be used when the frequency is expressed in radians per second, rather than oscillations per second, so ħ = h/2π) is now being referred to as the quantum of action (das elementare Wirkungsquantum in German), Planck referred to it as as a Hilfsgrösse only (that’s why he chose the h as a symbol, it seems), so that’s an auxiliary constant only: the actual quantum of action is, of course, ΔE, i.e. the difference between the various energy levels, which is the product of ħ and ω(or of h and ν0 if we express frequency in oscillations per second, rather than in angular frequency). Hence, Planck (and later Einstein) did not assume that an atomic oscillator emits or absorbs packets of energy as tiny as ħ or h, but packets of energy as big as ħωor, what amounts to the same (ħω = (h/2π)(2πν) = hν), hν0. Just to give an example, the frequency of sodium light (ν) is 500×1012 Hz, and so its energy is E = hν. That’s not a lot–about 2 eV only– but it still packs 500×1012 ‘quanta of action’ !

Another thing is that ω (or ν) is a continuous variable: hence, the assumption of equally spaced energy levels does not imply that energy itself is a discrete variable: light can have any frequency and, hence, we can also imagine photons with any energy level: the only thing we’re saying is that the energy of a photon of a specific color (i.e. a specific frequency ν) will be a multiple of hν.

Probability assumptions

The second key assumption of Planck as he worked towards a solution of the blackbody radiation problem was that the probability (P) of occupying a level of energy E is P(EαeE/kT. OK… Why not? But what is this assumption really? You’ll think of some ‘bell curve’, of course. But… No. That wouldn’t make sense. Remember that the energy has to be positive. The general shape of this P(E) curve is shown below.

graph

The highest probability density is near E = 0, and then it goes down as E gets larger, with kT determining the slope of the curve (just take the derivative). In short, this assumption basically states that higher energy levels are not so likely, and that very high energy levels are very unlikely. Indeed, this formula implies that the relative chance, i.e. the probability of being in state E1 relative to the chance of being in state E0, is P1/Pe−(E1–E0)k= e−ΔE/kT. Now, Pis n1/N and Pis n0/N and, hence, we find that nmust be equal to n0e−ΔE/kT. What this means is that the atomic oscillator is less likely to be in a higher energy state than in a lower one.

That makes sense, doesn’t it? I mean… I don’t want to criticize those 19th century scientists but… What were they thinking? Did they really imagine that infinite energy levels were as likely as… Well… More down-to-earth energy levels? I mean… A mechanical spring will break when you overload it. Hence, I’d think it’s pretty obvious those atomic oscillators cannot be loaded with just about anything, can they? Garbage in, garbage out:  of course, that theoretical spectrum of blackbody radiation didn’t make sense!

Let me copy Feynman now, as the rest of the story is pretty straightforward:

Now, we have a lot of oscillators here, and each is a vibrator of frequency w0. Some of these vibrators will be in the bottom quantum state, some will be in the next one, and so forth. What we would like to know is the average energy of all these oscillators. To find out, let us calculate the total energy of all the oscillators and divide by the number of oscillators. That will be the average energy per oscillator in thermal equilibrium, and will also be the energy that is in equilibrium with the blackbody radiation and that should go in the equation for the intensity of the radiation as a function of the frequency, instead of kT. [See my previous post: that equation is I(ω) = (ω2kt)/(π2c2).]

Thus we let N0 be the number of oscillators that are in the ground state (the lowest energy state); N1 the number of oscillators in the state E1; N2 the number that are in state E2; and so on. According to the hypothesis (which we have not proved) that in quantum mechanics the law that replaced the probability eP.E./kT or eK.E./kT in classical mechanics is that the probability goes down as eΔE/kT, where ΔE is the excess energy, we shall assume that the number N1 that are in the first state will be the number N0 that are in the ground state, times e−ħω/kT. Similarly, N2, the number of oscillators in the second state, is N=N0e−2ħω/kT. To simplify the algebra, let us call e−ħω/k= x. Then we simply have N1 = N0x, N2 = N0x2, …, N= N0xn.

The total energy of all the oscillators must first be worked out. If an oscillator is in the ground state, there is no energy. If it is in the first state, the energy is ħω, and there are N1 of them. So N1ħω, or ħωN0x is how much energy we get from those. Those that are in the second state have 2ħω, and there are N2 of them, so N22ħω=2ħωN0x2 is how much energy we get, and so on. Then we add it all together to get Etot = N0ħω(0+x+2x2+3x3+…).

And now, how many oscillators are there? Of course, N0 is the number that are in the ground state, N1 in the first state, and so on, and we add them together: Ntot = N0(1+x+x2+x3+…). Thus the average energy is

formula

Now the two sums which appear here we shall leave for the reader to play with and have some fun with. When we are all finished summing and substituting for x in the sum, we should get—if we make no mistakes in the sum—
energy

Feynman concludes as follows: “This, then, was the first quantum-mechanical formula ever known, or ever discussed, and it was the beautiful culmination of decades of puzzlement. Maxwell knew that there was something wrong, and the problem was, what was right? Here is the quantitative answer of what is right instead of kT. This expression should, of course, approach kT as ω → 0 or as → .”

It does, of course. And so Planck’s analysis does result in a theoretical I(ω) curve that matches the observed I(ω) curve as a function of both temperature (T) and frequency (ω). But so what it is, then? What’s the equation describing the dotted curves? It’s given below:

formula blackbody

I’ll just quote Feynman once again to explain the shape of those dotted curves: “We see that for a large ω, even though we have ωin the numerator, there is an e raised to a tremendous power in the denominator, so the curve comes down again and does not “blow up”—we do not get ultraviolet light and x-rays where we do not expect them!”

Is the analysis necessarily discrete?

One question I can’t answer, because I just am not strong enough in math, is the question or whether or not there would be any other way to derive the actual blackbody spectrum. I mean… This analysis obviously makes sense and, hence, provides a theory that’s consistent and in accordance with experiment. However, the question whether or not it would be possible to develop another theory, without having recourse to the assumption that energy levels in atomic oscillators are discrete and equally spaced with the ‘distance’ between equal to hν0, is not easy to answer. I surely can’t, as I am just a novice, but I can imagine smarter people than me have thought about this question. The answer must be negative, because I don’t know of any other theory: quantum mechanics obviously prevailed. Still… I’d be interested to see the alternatives that must have been considered.

Post scriptum: The “playing with the sums” is a bit confusing. The key to the formula above is the substitution of (0+x+2x2+3x3+…)/(1+x+x2+x3+…) by 1/[(1/x)–1)] = 1/[eħω/kT–1]. Now, the denominator 1+x+x2+x3+… is the Maclaurin series for 1/(1–x). So we have:

(0+x+2x2+3x3+…)/(1+x+x2+x3+…) = (0+x+2x2+3x3+…)(1–x)

x+2x2+3x3… –x22x3–3x4… = x+x2+x3+x4

= –1+(1+x+x2+x3…) = –1 + 1/(1–x) = –(1–x)+1/(1–x) = x/(1–x).

Note the tricky bit: if x = e−ħω/kT, then eħω/kis x−1 = 1/x, and so we have (1/x)–1 in the denominator of that (mean) energy formula, not 1/(x–1). Now 1/[(1/x)–1)] = 1/[(1–x)/x] = x/(1–x), indeed, and so the formula comes out alright.

Planck’s constant (I)

If you made it here, it means you’re totally fed up with all of the easy stories on quantum mechanics: diffraction, double-slit experiments, imaginary gamma-ray microscopes,… You’ve had it! You now know what quantum mechanics is all about, and you’ve realized all these thought experiments never answer the tough question: where did Planck find that constant (h) which pops up everywhere? And how did he find that Planck relation which seems to underpin all and everything in quantum mechanics?

If you don’t know, that’s because you’ve skipped the blackbody radiation story. So let me give it to you here. What’s blackbody radiation?

Thermal equilibrium of radiation

That’s what the blackbody radiation problem is about: thermal equilibrium of radiation.

Huh? 

Yes. Imagine a box with gas inside. You’ll often see it’s described as a furnace, because we heat the box. Hence, the box, and everything inside, acquires a certain temperature, which we then assume to be constant. The gas inside will absorb energy and start emitting radiation, because the gas atoms or molecules are atomic oscillators. Hence, we have electrons getting excited and then jumping up and down from higher to lower energy levels, and then again and again and again, thereby emitting photons with a certain energy and, hence, light of a certain frequency. To put it simply: we’ll find light with various frequencies in the box and, in thermal equilibrium, we should have some distribution of the intensity of the light according to the frequency: what kind of radiation do we find in the furnace? Well… Let’s find out.

The assumption is that the box walls send light back, or that the box has mirror walls. So we assume that all the radiation keeps running around in the box. Now that implies that the atomic oscillators not only radiate energy, but also receive energy, because they’re constantly being illuminated by radiation that comes straight back at them. If the temperature of the box is kept constant, we arrive at a situation which is referred to as thermal equilibrium. In Feynman’s words: “After a while there is a great deal of light rushing around in the box, and although the oscillator is radiating some, the light comes back and returns some of the energy that was radiated.”

OK. That’s easy enough to understand. However, the actual analysis of this equilibrium situation is what gave rise to the ‘problem’ of blackbody radiation in the 19th century which, as you know, led Planck and Einstein to develop a quantum-mechanical view of things. It turned out that the classical analysis predicted a distribution of the intensity of light that didn’t make sense, and no matter how you looked at it, it just didn’t come out right. Theory and experiment did not agree. Now, that is something very serious in science, as you know, because it means your theory isn’t right. In this case, it was disastrous, because it meant the whole of classical theory wasn’t right.

To be frank, the analysis is not all that easy. It involves all that I’ve learned so far: the math behind oscillators and interference, statistics, the so-called kinetic theory of gases and what have you. I’ll try to summarize the story but you’ll see it requires quite an introduction.

Kinetic energy and temperature

The kinetic theory of gases is part of what’s referred to as statistical mechanics: we look at a gas as a large number of inter-colliding atoms and we describe what happens in terms of the collisions between them. As Feynman puts it: “Fundamentally, we assert that the gross properties of matter should be explainable in terms of the motion of its parts.” Now, we can do a lot of intellectual gymnastics, analyzing one gas in one box, two gases in one box, two gases in one box with a piston between them, two gases in two boxes with a hole in the wall between them, and so on and so on, but that would only distract us here. The rather remarkable conclusion of such exercises, which you’ll surely remember from your high school days, is that:

  1. Equal volumes of different gases, at the same pressure and temperature, will have the same number of molecules.
  2. In such view of things, temperature is actually nothing but the mean kinetic energy of those molecules (or atoms if it’s a monatomic gas).

So we can actually measure temperature in terms of the kinetic energy of the molecules of the gas, which, as you know, equals mv2/2, with m the mass and v the velocity of the gas molecules. Hence, we’re tempted to define some absolute measure of temperature T and simply write:

T = 〈mv2/2〉

The 〈 and 〉 brackets denote the mean here. To be precise, we’re talking the root mean square here, aka as the quadratic mean, because we want to average some magnitude of a varying quantity. Of course, the mass of different gases will be different – and so we have 〈m1v12/2〉 for gas 1 and 〈m2v22/2〉 for gas 2 – but that doesn’t matter: we can, actually, imagine measuring temperature in joule, the unit of energy, including kinetic energy. Indeed, the units come out alright: 1 joule = 1 kg·(m2/s2). For historical reasons, however, T is measured in different units: degrees Kelvin, centigrades (i.e. degrees Celsius) or, in the US, in Fahrenheit. Now, we can easily go from one measure to the other as you know and, hence, here I should probably just jot down the so-called ideal gas law–because we need that law for the subsequent analysis of blackbody radiation–and get on with it:

PV = NkT

However, now that we’re here, let me give you an inkling of how we derive that law. A classical (Newtonian) analysis of the collisions (you can find the detail in Feynman’s Lectures, I-39-2) will yield the following equation: P = (2/3)n〈mv2/2〉, with n the number of atoms or molecules per unit volume. So the pressure of a gas (which, as you know, is the force (of a gas on a piston, for example) per unit area: P = F/A) is also equal to the mean kinetic energy of the gas molecules multiplied by (2/3)n. If we multiply that equation by V, we get PV = N(2/3)〈mv2/2〉. However, we know that equal volumes of volumes of different gases, at the same pressure and temperature, will have the same number of molecules, so we have PV = N(2/3)〈m1v12/2〉 = N(2/3)〈m2v22/2〉, which we write as PV = NkT with kT = (2/3)〈m1v12/2〉 = (2/3)〈m2v22/2〉.

In other words, that factor of proportionality k is the one we have to use to convert the temperature as measured by 〈mv2/2〉 (i.e. the mean kinetic energy expressed in joules) to T (i.e. the temperature expressed in the measure we’re used to, and that’s degrees Kelvin–or Celsius or Fahrenheit, but let’s stick to Kelvin, because that’s what’s used in physics). Vice versa, we have 〈mv2/2〉 = (3/2)kT. Now, that constant of proportionality k is equal to k 1.38×10–23 joule per Kelvin (J/K). So if T is (absolute) temperature, expressed in Kelvin (K), our definition says that the mean molecular kinetic energy is (3/2)kT.

That k factor is a physical constant referred to as the Boltzmann constant. If it’s one of these constants, you may wonder why we don’t integrate that 3/2 factor in it? Well… That’s just how it is, I guess. In any case, it’s rather convenient because we’ll have 2/3 factors in other equations and so these will cancel out with that 3/2 term. However, I am digressing way too much here. I should get back to the main story line. However, before I do that, I need to expand on one more thing, and that’s a small lecture on how things look like when we also allow for internal motion, i.e. the rotational and vibratory motions of the atoms within the gas molecule. Let me first re-write that PV = NkT equation as

PV = NkT = N(2/3)〈m1v12/2〉 = (2/3)U = 2U/3

For monatomic gas, that U would only be the kinetic energy of the atoms, and so we can write it as U = (2/3)NkT. Hence, we have the grand result that the kinetic energy, for each atom, is equal to (3/2)kT, on average that is.

What about non-monatomic gas? Well… For complex molecules, we’d also have energy going into the rotational and vibratory motion of the atoms within the molecule, separate from what is usually referred to as the center-of-mass (CM) motion of the molecules themselves. Now, I’ll again refer you to Feynman for the detail of the analysis, but it turns out that, if we’d have, for example, a diatomic molecule, consisting of an A and B atom, the internal rotational and vibratory motion would, indeed, also absorb energy, and we’d have a total energy equal to (3/2)kT + (3/2)kT = 2×(3/2)kT = 3kT. Now, that amount (3kT) can be split over (i) the energy related to the CM motion, which must still be equal to (3/2)kT, and (ii) the average kinetic energy of the internal motions of the diatomic molecule excluding the bodily motion of the CM. Hence, the latter part must be equal to 3kT – (3/2)kT = (3/2)kT. So, for the diatomic molecule, the total energy happens to consist of two equal parts.

Now, there is a more general theorem here, for which I have to introduce the notion of the degrees of freedom of a system. Each atom can rotate or vibrate or oscillate or whatever in three independent directions–namely the three spatial coordinates x, y and z. These spatial dimensions are referred to as the degrees of freedom of the atom (in the kinetic theory of gases, that is), and if we have two atoms, we have 2×3 = 6 degrees of freedom. More in general, the number of degrees of freedom of a molecule composed of r atoms is equal to 3rNow, it can be shown that the total energy of an r-atom molecule, including all internal energy as well as the CM motion, will be 3r×kT/2 = 3rkT/2 joules. Hence, for every independent direction of motion that there is, the average kinetic energy for that direction will be kT/2. [Note that ‘independent direction of motion’ is used, somewhat confusingly, as a synonym for degree of freedom, so we don’t have three but six ‘independent directions of motion’ for the diatomic molecule. I just wanted to note that because I do think it causes confusion when reading a textbook like Feynman’s.] Now, that total amount of energy, i.e.  3r(kT/2), will be split as follows according to the “theorem concerning the average energy of the CM motion”, as Feynman terms it:

  1. The kinetic energy for the CM motion of each molecule is, and will always be, (3/2)kT.
  2. The remainder, i.e. r(3/2)kT – (3/2)kT = (3/2)(r–1)kt, is internal vibrational and rotational kinetic energy, i.e. the sum of all vibratory and rotational kinetic energy but excluding the energy of the CM motion of the molecule.

Phew! That’s quite something. And we’re not quite there yet.

The analysis for photon gas

Photon gas? What’s that? Well… Imagine our box is the gas in a very hot star, hotter than the sun. As Feynman writes it: “The sun is not hot enough; there are still too many atoms, but at still higher temperatures in certain very hot stars, we may neglect the atoms and suppose that the only objects that we have in the box are photons.” Well… Let’s just go along with it. We know that photons have no mass but they do have some very tiny momentum, which we related to the magnetic field vector, as opposed to the electric field. It’s tiny indeed. Most of the energy of light goes into the electric field. However, we noted that we can write p as p = E/c, with c the speed of light (3×108). Now, we had that = (2/3)n〈mv2/2〉 formula for gas, and we know that the momentum p is defined as p = mv. So we can substitute mvby (mv)v = pv. So we get = (2/3)n〈pv/2〉 = (1/3)n〈pv〉.

Now, the energy of photons is not quite the same as the kinetic energy of an atom or an molecule, i.e. mv2/2. In fact, we know that, for photons, the speed v is equal to c, and pc = E. Hence, multiplying by the volume V, we get

PV = U/3

So that’s a formula that’s very similar to the one we had for gas, for which we wrote: PV = NkT = 2U/3. The only thing is that we don’t have a factor 2 in the equation but so that’s because of the different energy concepts involved. Indeed, the concept of the energy of a photon (E = pc) is different than the concept of kinetic energy. But so the result is very nice: we have a similar formula for the compressibility of gas and radiation. In fact, both PV = 2U/3 and PV = U/3 will usually be written, more generally, as:

PV = (γ – 1)U 

Hence, this γ would be γ = 5/3 ≈ 1.667 for gas and 4/3 ≈ 1.333 for photon gas. Now, I’ll skip the detail (it involves a differential analysis) but it can be shown that this general formula, PV = (γ – 1)U, implies that PVγ (i.e. the pressure times the volume raised to the power γ) must equal some constant, so we write:

PVγ = C

So far so good. Back to our problem: blackbody radiation. What you should take away from this introduction is the following:

  1. Temperature is a measure of the average kinetic energy of the atoms or molecules in a gas. More specifically, it’s related to the mean kinetic energy of the CM motion of the atoms or molecules, which is equal to (3/2)kT, with k the Boltzmann constant and T the temperature expressed in Kelvin (i.e. the absolute temperature).
  2. If gas atoms or molecules have additional ‘degrees of freedom’, aka ‘independent directions of motion’, then each of these will absorb additional energy, namely kT/2.

Energy and radiation

The atoms in the box are atomic oscillators, and we’ve analyzed them before. What the analysis above added was that average kinetic energy of the atoms going around is (3/2)kT and that, if we’re talking molecules consisting of r atoms, we have a formula for their internal kinetic energy as well. However, as an oscillator, they also have energy separate from that kinetic energy we’ve been talking about alrady. How much? That’s a tricky analysis. Let me first remind you of the following:

  1. Oscillators have a natural frequency, usually denoted by the (angular) frequency ω0.
  2. The sum of the potential and kinetic energy stored in an oscillator is a constant, unless there’s some damping constant. In that case, the oscillation dies out. Here, you’ll remember the concept of the Q of an oscillator. If there’s some damping constant, the oscillation will die out and the relevant formula is 1/Q = (dW/dt)/(ω0W) = γ0, with γ the damping constant (not to be confused with the γ we used in that PVγ = C formula).

Now, for gases, we said that, for every independent direction of motion there is, the average kinetic energy for that direction will be kT/2. I admit it’s a bit of a stretch of the imagination but so that’s how the blackbody radiation analysis starts really: our atomic oscillators will have an average kinetic energy equal to kT/2 and, hence, their total energy (kinetic and potential) should be twice that amount, according to the second remark I made above. So that’s kT. We’ll note the total energy as W below, so we can write:

W = kT

Just to make sure we know what we’re talking about (one would forget, wouldn’t one?), kT is the product of the Boltzmann constant (1.38×10–23 J/K) and the temperature of the gas (so note that the product is expressed in joule indeed). Hence, that product is the average energy of our atomic oscillators in the gas in our furnace.

Now, I am not going to repeat all of the detail we presented on atomic oscillators (I’ll refer you, once again, to Feynman) but you may or may not remember that atomic oscillators do have a Q indeed and, hence, some damping constant γ. So we can use and re-write that formula above as

dW/dt = (1/Q)(ω0W) = (ω0W)(γ/ω0) = γW, which implies γ = (dW/dt)/W

What’s γ? Well, we’ve calculated the Q of an atomic oscillator already: Q = 3λ/4πr0. Now, λ = 2πc/ω(we just convert the wavelength into (angular) frequency using λν = c) and γ = ω0/Q, so we get γ = 4πr0ω0/[3(2πc/ω0)] = (2/3)r0ω02/c. Now, plugging that result back into the equation above, we get

dW/dt = γW = (2/3)(r0ω02kT)/c

Just in case you’d have difficulty following – I admit I did 🙂 – dW/dt is the average rate of radiation of light of (or near) frequency ω02. I’ll let Feynman take over here:

Next we ask how much light must be shining on the oscillator. It must be enough that the energy absorbed from the light (and thereupon scattered) is just exactly this much. In other words, the emitted light is accounted for as scattered light from the light that is shining on the oscillator in the cavity. So we must now calculate how much light is scattered from the oscillator if there is a certain amount—unknown—of radiation incident on it. Let I(ω)dω be the amount of light energy there is at the frequency ω, within a certain range dω (because there is no light at exactly a certain frequency; it is spread all over the spectrum). So I(ω) is a certain spectral distribution which we are now going to find—it is the color of a furnace at temperature T that we see when we open the door and look in the hole. Now how much light is absorbed? We worked out the amount of radiation absorbed from a given incident light beam, and we calculated it in terms of a cross section. It is just as though we said that all of the light that falls on a certain cross section is absorbed. So the total amount that is re-radiated (scattered) is the incident intensity I(ω)dω multiplied by the cross section σ.

OK. That makes sense. I’ll not copy the rest of his story though, because this is a post in a blog, not a textbook. What we need to find is that I(ω). So I’ll refer you to Feynman for the details (these ‘details’ involve fairly complicated calculations, which are less important than the basic assumptions behind the model, which I presented above) and just write down the result:

blackbody radiation formula

This formula is Rayleigh’s law. [And, yes, it’s the same Rayleigh – Lord Rayleigh, I should say respectfully – as the one who invented that criterion I introduced in my previous post, but so this law and that criterion have nothing to do with each other.] This ‘law’ gives the intensity, or the distribution, of light in a furnace. Feynman says it’s referred to as blackbody radiation because “the hole in the furnace that we look at is black when the temperature is zero.” […] OK. Whatever. What we call it doesn’t matter. The point is that this function tells us that the intensity goes as the square of the frequency, which means that if we have a box at any temperature at all, and if we look at it, the X- and gamma rays will be burning out eyes out ! The graph below shows both the theoretical curve for two temperatures (Tand 2T0), as derived above (see the solid lines), and then the actual curves for those two temperatures (see the dotted lines).

Blackbody radation graph

This is the so-called UV catastrophe: according to classical physics, an ideal black body at thermal equilibrium should emit radiation with infinite power. In reality, of course, it doesn’t: Rayleigh’s law is false. Utterly false. And so that’s where Planck came to the rescue, and he did so by assuming radiation is being emitted and/or absorbed in finite quanta: multiples of h, in fact.

Indeed, Planck studied the actual curve and fitted it with another function. That function assumed the average energy of a harmonic oscillator was not just proportional with the temperature (T), but that it was also a function of the (natural) frequency of the oscillators. By fiddling around, he found a simple derivation for it which involved a very peculiar assumption. That assumption was that the harmonic oscillator can take up energies only ħω at the time, as shown below.

Equally space energy levels

Hence, the assumption is that the harmonic oscillators cannot take whatever (continous) energy level. No. The allowable energy levels of the harmonic oscillators are equally spaced: E= nħω. Now, the actual derivation is at least as complex as the derivation of Rayleigh’s law, so I won’t do it here. Let me just give you the key assumptions:

  1. The gas consists of a large number of atomic oscillators, each with their own natural frequency ω0.
  2. The permitted energy levels of these harmonic oscillator are equally spaced and ħωapart.
  3. The probability of occupying a level of energy E is P(Eαe–E/kT.

All the rest is tedious calculation, including the calculation of the parameters of the model, which include ħ (and, hence, h, because h = 2πħ) and are found by matching the theoretical curves to the actual curves as measured in experiments. I’ll just mention one result, and that’s the average energy of these oscillators:

energy

As you can see, the average energy does not only depend on the temperature T, but also on their (natural) frequency. So… Now you know where h comes from. As I relied so heavily on Feynman’s presentation here, I’ll include the link. As Feynman puts it: “This, then, was the first quantum-mechanical formula ever known, or ever discussed, and it was the beautiful culmination of decades of puzzlement. Maxwell knew that there was something wrong, and the problem was, what was right? Here is the quantitative answer of what is right instead of kT.”

So there you go. Now you know. 🙂 Oh… And in case you’d wonder: why the h? Well… Not sure. It’s said the h stands for Hilfsgrösse, so that’s some constant which was just supposed to help him out with the calculation. At that time, Planck did not suspect it would turn out to be one of the most fundamental physical constants. 🙂

Post scriptum: I went quite far in my presentation of the basics of the kinetic theory of gases. You may wonder now. I didn’t use that theoretical PVγ = C relation, did I? And why all the fuss about photon gas? Well… That was just to introduce that PVγ = C relation, so I could note, here, in this post scriptum, that it has a similar problem. The γ exponent is referred to as the specific heat ratio of a gas, and it can be calculated theoretically as well, as we did–well… Sort of, because we skipped the actual derivation. However, their theoretical value also differs substantially from actually measured values, and the problem is the same: one should not assume that a continuous value for 〈E〉. Agreement between theory and experiment can only be reached when the same assumptions as those of Planck are used: discrete energy levels, multiples of ħ and ω: E= nħω. Also, the specific functional form which Planck used to resolve the blackbody radiation problem is also to be used here. For more details, I’ll refer to Feynman too. I can’t say this is easy to digest, but then who said it would be easy? 🙂

The point to note is that the blackbody radiation problem wasn’t the only problem in the 19th century. As Feynman puts it: “One often hears it said that physicists at the latter part of the nineteenth century thought they knew all the significant physical laws and that all they had to do was to calculate more decimal places. Someone may have said that once, and others copied it. But a thorough reading of the literature of the time shows they were all worrying about something.” They were, and so Planck came up with something new. And then Einstein took it to the next level and then… Well… The rest is history. 🙂