Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics

I’ve discussed statistics, in the context of quantum mechanics, a couple of times already (see, for example, my post on amplitudes and statistics). However, I never took the time to properly explain those distribution functions which are referred to as the Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac distribution functions respectively. Let me try to do that now—without, hopefully, getting lost in too much math! It should be a nice piece, as it connects quantum mechanics with statistical mechanics, i.e. two topics I had nicely separated so far. 🙂

You know the Boltzmann Law now, which 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. The point is: we have some probability density function f, and it’s a function of the energy E, so we write:

f(E) = C·e−energy/kT = C/eenergy/kT

C is just a normalization constant (all probabilities have to add up to one, so the integral of this function over its domain must be one), and k and T are also usual suspects: T is the (absolute) temperature, and k is the Boltzmann constant, which relates the temperate to the kinetic energy of the particles involved. We also know the shape of this function. For example, when we applied it to the density of the atmosphere at various heights (which are related to the potential energy, as P.E. = m·g·h), assuming constant temperature, we got the following graph. The shape of this graph is that of an exponential decay function (we’ll encounter it again, so just take a mental note of it).

graph

A more interesting application is the quantum-mechanical approach to the theory of gases, which I introduced in my previous post. To explain the behavior of gases under various conditions, we assumed that gas molecules are like oscillators but that they can only take on discrete levels of energy. [That’s what quantum theory is about!] We denoted the various energy levels, i.e. the energies of the various molecular states, by E0, E1, E2,…, Ei,…, and if Boltzmann’s Law applies, then the probability of finding a molecule in the particular state Ei is proportional to e−Ei /kT. We can then calculate the relative probabilities, i.e. the probability of being in state Ei, relative to the probability of being in state E0, is:

Pi/P0 = e−Ei /kT/e−E0 /kT = e−(Ei–E0)/kT = 1/e(Ei–E0)/kT

Now, Pi obviously equals ni/N, so it is the ratio of the number of molecules in state Ei (ni) and the total number of molecules (N). Likewise, P0 = n0/N and, therefore, we can write:

ni/ne−(Ei−E0)/kT = 1/e(Ei–E0)/kT

This formulation is just another Boltzmann Law, but it’s nice in that it introduces the idea of a ground state, i.e. the state with the lowest energy level. We may or may not want to equate E0 with zero. It doesn’t matter really: we can always shift all energies by some arbitrary constant because we get to choose the reference point for the potential energy.

So that’s the so-called Maxwell-Boltzmann distribution. Now, in my post on amplitudes and statistics, I had jotted down the formulas for the other distributions, i.e. the distributions when we’re not talking classical particles but fermions and/or bosons. As you know, fermions are particles governed by the Fermi exclusion principle: indistinguishable particles cannot be together in the same state. For bosons, it’s the other way around: having one in some quantum state actually increases the chance of finding another one there, and we can actually have an infinite number of them in the same state.

We also know that fermions and bosons are the real world: fermions are the matter-particles, bosons are the force-carriers, and our ‘Boltzmann particles’ are nothing but a classical approximation of the real world. Hence, even if we can’t see them in the actual world, the Fermi-Dirac and Bose-Einstein distributions are the real-world distributions. 🙂 Let me jot down the equations once again:

Fermi-Dirac (for fermions): f(E) = 1/[Ae(E − EF)/kT + 1]

Bose-Einstein (for bosons):  f(E) = 1/[AeE/kT − 1]

We’ve got some other normalization constant here (A), which we shouldn’t be too worried about—for the time being, that is. Now, to see how these distributions are different from the Maxwell-Boltzmann distribution (which we should re-write as f(E) = C·e−E/kT = 1/[A·eE/kT] so as to make all formulas directly comparable), we should just make a graph. Please go online to find a graph tool (I found a new one recently—really easy to use), and just do it. You’ll see they are all like that exponential decay function. However, in order to make a proper comparison, we would actually need to calculate the normalization coefficients and, for the Fermi energy, we would also need the Fermi energy E(note that, for simplicity, we did equate E0 with zero). Now, we could give it a try, but it’s much easier to google and find an example online.

The HyperPhysics website of Georgia State University gives us one: the example assumes 6 particles and 9 energy levels, and the table and graph below compare the Maxwell-Boltzmann and Bose-Einstein distributions for the model.

Graph Table

Now that is an interesting example, isn’t it? In this example (but all depends on its assumptions, of course), the Maxwell-Boltzmann and Bose-Einstein distributions are almost identical. Having said that, we can clearly see that the lower energy states are, indeed, more probable with Bose-Einstein statistics than with the Maxwell-Boltzmann statistics. While the difference is not dramatic at all in this example, the difference does become very dramatic, in reality, with large numbers (i.e. high matter density) and, more importantly, at very low temperatures, at which bosons can condense into the lowest energy state. This phenomenon is referred to as Bose-Einstein condensation: it causes superfluidity and superconductivity, and it’s real indeed: it has been observed with supercooled He-4, which is not an everyday substance, but real nevertheless!

What about the Fermi-Dirac distribution for this example? The Fermi-Dirac distribution is given below: the lowest energy state is now less probable, the mid-range energies much more, and none of the six particles occupy any of the four highest energy levels. Again, while the difference is not dramatic in this example, it can become very dramatic, in reality, with large numbers (read: high matter density) and very low temperatures: at absolute zero, all of the possible energy states up to the Fermi energy level will be occupied, and all the levels above the Fermi energy will be vacant.

graph 2 Table 2

What can we make out of all of this? First, you may wonder why we actually have more than one particle in one state above: doesn’t that contradict the Fermi exclusion principle? No. We need to distinguish micro- and macro-states. In fact, the example assumes we’re talking electrons here, and so we can have two particles in each energy state—with opposite spin, however. At the same time, it’s true we cannot have three, or more, in any state. That results, in the example we’re looking at here, in five possible distributions only, as shown below.

Table 3

The diagram is an interesting one: if the particles were to be classical particles, or bosons, then 26 combinations are possible, including the five Fermi-Dirac combinations, as shown above. Note the little numbers above the 26 possible combinations (e.g. 6, 20, 30,… 180): they are proportional to the likelihood of occurring under the Maxwell-Boltzmann assumption (so if we assume the particles are ‘classical’ particles). Let me introduce you to the math behind the example by using the diagram below, which shows three possible distributions/combinations (I know the terminology is quite confusing—sorry for that!).

table 4

If we could distinguish the particles, then we’d have 2002 micro-states, which is the total of all those little numbers on top of the combinations that are shown (6+60+180+…). However, the assumption is that we cannot distinguish the particles. Therefore, the first combination in the diagram above, with five particles in the zero energy state and one particle in state 9, occurs 6 times into 2002 and, hence, it has a probability of 6/2002 ≈ 0.003 only. In contrast, the second combination is 10 times more likely, and the third one is 30 times more likely! In any case, the point is, in the classical situation (and in the Bose-Einstein hypothesis as well), we have 26 possible macro-states, as opposed to 5 only for fermions, and so that leads to a very different density function. Capito?

No? Well, this blog is not a textbook on physics and, therefore, I should refer you to the mentioned site once again, which references a 1992 textbook on physics (Frank Blatt, Modern Physics, 1992) as the source of this example. However, I won’t do that: you’ll find the details in the Post Scriptum to this post. 🙂

Let’s first focus on the fundamental stuff, however. The most burning question is: if the real world consists of fermions and bosons, why is that that we only see the Maxwell-Boltzmann distribution in our actual (non-real?) world? 🙂 The answer is that both the Fermi-Dirac and Bose-Einstein distribution approach the Maxwell–Boltzmann distribution if higher temperatures and lower particle densities are involved. In other words, we cannot see the Fermi-Dirac distributions (all matter is fermionic, except for weird stuff like superfluid helium-4 at 1 or 2 degrees Kelvin), but they are there!

Let’s approach it mathematically: the most general formula, encompassing both Fermi-Dirac and Bose-Einstein statistics, is:

Ni(Ei) ∝ 1/[e(Ei − μ)/kT ± 1]

If you’d google, you’d find a formula involving an additional coefficient, gi, which is the so-called degeneracy of the energy level Ei. I included it in the formula I used in the above-mentioned post of mine. However, I don’t want to make it any more complicated than it already is and, therefore, I omitted it this time. What you need to look at are the two terms in the denominator: e(Ei − μ)/kT and ± 1.

From a math point of view, it is obvious that the values of e(Ei − μ)/kT + 1 (Fermi-Dirac) and e(Ei − μ)/kT − 1 (Bose-Einstein) will approach each other if e(Ei − μ)/kT is much larger than ±1, so if e(Ei − μ)/kT >> 1. That’s the case, obviously, if the (Ei − μ)/kT ratio is large, so if (Ei − μ) >> kT. In fact, (Ei − μ) should, obviously, be much larger than kT for the lowest energy levels too! Now, the conditions under which that is the case are associated with the classical situation (such as a cylinder filled with gas, for example). Why?

Well… […] Again, I have to say that this blog can’t substitute for a proper textbook. Hence, I am afraid I have to leave it to you to do the necessary research to see why. 🙂 The non-mathematical approach is to simple note that quantum effects, i.e. the ±1 term, only apply if the concentration of particles is high enough. Indeed, quantum effects appear if the concentration of particles is higher than the so-called quantum concentration. Only when the quantum concentration is reached, particles will start interacting according to what they are, i.e. as bosons or as fermions. At higher temperature, that concentration will not be reached, except in massive objects such as a white dwarf (white dwarfs are stellar remnants with the mass like that of the Sun but a volume like that of the Earth). So, in general, we can say that at higher temperatures and at low concentration we will not have any quantum effects. That should settle the matter—as for now, at least.

You’ll have one last question: we derived Boltzmann’s Law from the kinetic theory of gases, but how do we derive that Ni(Ei) = 1/[Ae(Ei − μ)/kT ± 1] expression? Good question but, again, we’d need more than a few pages to explain that! The answer is: quantum mechanics, of course! Go check it out in Feynman’s third Volume of Lectures! 🙂

Post scriptum: combinations, permutations and multiplicity

The mentioned example from HyperPhysics is really interesting, if only because it shows you also need to master a bit of combinatorics to get into quantum mechanics. Let’s go through the basics. If we have n distinct objects, we can order hem in n! ways, with n! (read: n factorial) equal to n·(n–1)·(n–2)·…·3·2·1. Note that 0! is equal to 1, per definition. We’ll need that definition.

For example, a red, blue and green ball can be ordered in 3·2·1 = 6 ways. Each way is referred to as a permutation.

Besides permutations, we also have the concept of a k-permutation, which we can denote in a number of ways but let’s choose P(n, k). [The P stands for permutation here, not for probability.] P(n, k) is the number of ways to pick k objects out of a set of n objects. Again, the objects are supposed to be distinguishable. The formula is P(n, k) = n·(n–1)·(n–2)·…·(n–k+1) = n!/(n–k)!. That’s easy to understand intuitively: on your first pick you have n choices; on your second, n–1; on your third, n–2, etcetera. When n = k, we obviously get n! again.

There is a third concept: the k-combination (as opposed to the k-permutation), which we’ll denote by C(n, k). That’s when the order within our subset doesn’t matter: an ace, a queen and a jack taken out of some card deck are a queen, a jack, and an ace: we don’t care about the order. If we have k objects, there are k! ways of ordering them and, hence, we just have to divide P(n, k) by k! to get C(n, k). So we write: C(n, k) = P(n, k)/k! = n!/[(n–k)!k!]. You recognize C(n, k): it’s the binomial coeficient.

Now, the HyperPhysics example illustrating the three mentioned distributions (Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac) is a bit more complicated: we need to associate q energy levels with N particles. Every possible configuration is referred to as a micro-state, and the total number of possible micro-states is referred to as the multiplicity of the system, denoted by Ω(N, q). The formula for Ω(N, q) is another binomial coefficient: Ω(N, q) = (q+N–1)!/[q!(N–1)!]. Ω(N, q) = Ω(6, 9) = (9+6–1)!/[9!(6–1)!] = 2002.

In our example, however, we do not have distinct particles and, therefore, we only have 26 macro-states (as opposed to 2002 micro-states), which are also referred to, confusingly, as distributions or combinations.

Now, the number of micro-states associated with the same macro-state is given by yet another formula: it is equal to N!/[n1!·n2!·n3!·…·nq!], with ni! the number of particles in level i. [See why we need the 0! = 1 definition? It ensures unoccupied states do not affect the calculation.] So that’s how we get those numbers 6, 60 and 180 for those three macro-states.

But how do we calculate those average numbers of particles for each energy level? In other words, how do we calculate the probability densities under the Maxwell-Boltzmann, Fermi-Dirac and Bose-Einstein hypothesis respectively?

For the Maxwell-Boltzmann distribution, we proceed as follows: for each energy level j (or Ej, I should say), we calculate n= ∑nij·Pi over all macro-states i. In this summation, we have nij, which is the number of particles in energy level j in micro-state i, while Pi is the probability of macro-state i as calculated by the ratio of (i) the number of micro-states associated with macro-state i and (ii) the total number of micro-states. For Pi, we gave the example of 3/2002 ≈ 0.3%. For 60 and 180, we get 60/2002 ≈ 3% and 180/2002 ≈ 9%. Calculating all the nj‘s for j ranging from 1 to 9 should yield the numbers and the graph below indeed.

M-B graphOK. That’s how it works for Maxwell-Boltzmann. Now, it is obvious that the Fermi-Dirac and the Bose-Einstein distribution should not be calculated in the same way because, if they were, they would not be different from the Maxwell-Boltzmann distribution! The trick is as follows.

For the Bose-Einstein distribution, we give all macro-states equal weight—so that’s a weight of one, as shown below. Hence, the probability Pi  is, quite simply, 1/26 ≈ 3.85% for all 26 macro-states. So we use the same n= ∑nij·Pformula but with Pi = 1/26.

BE

Finally, I already explained how we get the Fermi-Dirac distribution: we can only have (i) one, (ii) two, or (iii) zero fermions for each energy level—not more than two! Hence, out of the 26 macro-states, only five are actually possible under the Fermi-Dirac hypothesis, as illustrated below once more. So it’s a very different distribution indeed!

Table 3

Now, you’ll probably still have questions. For example, why does the assumption, for the Bose-Einstein analysis, that macro-states have equal probability favor the lower energy states? The answer is that the model also integrates other constraints: first, when associating a particle with an energy level, we do not favor one energy level over another, so all energy levels have equal probability. However, at the same time, the whole system has some fixed energy level, and so we cannot put the particles in the higher energy levels only! At the same time, we know that, if we have q particles, and the probability of a particle having some energy level j is the same for all j, then they are likely not to be all at the same energy level: they’ll be distributed, effectively, as evidenced by the very low chance (0.3% only) of having 5 particles in the ground state and 1 particle at a higher level, as opposed to the 3% and 9% chance of the other two combinations shown in that diagram with three possible Maxwell-Boltzmann (MB) combinations.

So what happens when assigning an equal probability to all 26 possible combinations (with value 1/26) is that the combinations that were previously rather unlikely – because they did have a rather heavy concentration of particles in the ground state only – are now much more likely. So that’s why the Bose-Einstein distribution, in this example at least, is skewed towards the lowest energy level—as compared to the Maxwell-Boltzmann distribution, that is.

So that’s what’s behind, and that should also answer the other question you surely have when looking at those five acceptable Fermi-Dirac configurations: why don’t we have the same five configurations starting from the top down, rather than from the bottom up? Now you know: such configuration would have much higher energy overall, and so that’s not allowed under this particular model.

There’s also this other question: we said the particles were indistinguishable, but so then we suddenly say there can be two at any energy level, because their spin is opposite. It’s obvious this is rather ad hoc as well. However, if we’d allow only one particle at any energy level, we’d have no allowable combinations and, hence, we’d have no Fermi-Dirac distribution at all in this example.

In short, the example is rather intuitive, which is actually why I like it so much: it shows how bosonic and fermionic behavior appear rather gradually, as a consequence of variables that are defined at the system level, such as density, or temperature. So, yes, you’re right if you think the HyperPhysics example lacks rigor. That’s why I think it’s such wonderful pedagogic device. 🙂

The Quantum-Mechanical Gas Law

In my previous posts, it was mentioned repeatedly that the kinetic theory of gases is not quite correct: the experimentally measured values of the so-called specific heat ratio (γ) vary with temperature and, more importantly, their values differ, in general, from what classical theory would predict. It works, more or less, for noble gases, which do behave as ideal gases and for which γ is what the kinetic theory of gases would want it to be: γ = 5/3—but we get in trouble immediately, even for simple diatomic gases like oxygen or hydrogen, as illustrated below: the theoretical value is 9/7 (so that’s 1.286, more or less), but the measured value is very different.

Heat ratioLet me quickly remind you how we get the theoretical number. According to classical theory, a diatomic molecule like oxygen can be represented as two atoms connected by a spring. Each of the atoms absorbs kinetic energy, and for each direction of motion (x, y and z), that energy is equal to kT/2, so the kinetic energy of both atoms – added together – is 2·3·kT/2 = 3kT. However, I should immediately add that not all of that energy is to be associated with the center-of-mass motion of the whole molecule, which determines the temperature of the gas: that energy is and remains equal to the 3kT/2, always. We also have rotational and vibratory motion. The molecule can rotate in two independent directions (and any combination of these directions, of course) and, hence, rotational motion is to absorb an amount of energy equal to 2·kT/2 = kT. Finally, the vibratory motion is to be analyzed as any other oscillation, so like a spring really. There is only one dimension involved and, hence, the kinetic energy here is just kT/2. However, we know that the total energy in an oscillator is the sum of the kinetic and potential energy, which adds another kT/2 term. Putting it all together, we find that the average energy for each diatomic particle is (or should be) equal to 7·kT/2 = (7/2)kT. Now, as mentioned above, the temperature of the gas (T) is proportional to the mean molecular energy of the center-of-mass motion only (in fact, that’s how temperature is defined), with the constant of proportionality equal to 3k/2. Hence, for monatomic ideal gases, we can write: U = N·(3k/2)T and, therefore, PV = NkT = (2/3)·U. Now, γ appears as follows in the ideal gas law: PV = (γ–1)U. Therefore, γ = 2/3 + 1 = 5/3, but so that’s for monatomic ideal gases only! The total kinetic energy of our diatomic molecule is U = N·(7k/2)T and, therefore, PV = (2/7)·U. So γ must be γ = 2/7 + 1 = 9/7 ≈ 1.286 for diatomic gases, like oxygen and hydrogen.

Phew! So that’s the theory. However, as we can see from the diagram, γ approaches that value only when we heat the gas to a few thousand degrees! So what’s wrong? One assumption is that certain kinds of motions “freeze out” as the temperature falls—although it’s kinda weird to think of something ‘freezing out’ at a thousand degrees Kelvin! In any case, at the end of the 19th century, that was the assumption that was advanced, very reluctantly, by scientists such as James Jeans. However, the mystery was about to be solved then, as Max Planck, even more reluctantly, presented his quantum theory of energy at the turn of the century itself.

But the quantum theory was confirmed and so we should now see how we can apply it to the behavior of gas. In my humble view, it’s a really interesting analysis, because we’re applying quantum theory here to a phenomenon that’s usually being analyzed as a classical problem only.

Boltzmann’s Law

We derived Boltzmann’s Law in our post on the First Principles of Statistical Mechanics. To be precise, we gave Boltzmann’s Law for the density of a gas (which we denoted by n = N/V)  in a force field, like a gravitational field, or in an electromagnetic field (assuming our gas particles are electrically charged, of course). We noted, however, Boltzmann’s Law was also applicable to much more complicated situations, like the one below, which shows a potential energy function for two molecules that is quite characteristic of the way molecules actually behave: when they come very close together, they repel each other but, at larger distances, there’s a force of attraction. We don’t really know the forces behind but we don’t need to: as long as these forces are conservative, they can combine in whatever way they want to combine, and Boltzmann’s Law will be applicable. [It should be obvious why. If you hesitate, just think of the definition of work and how it affects potential energy and all that. Work is force times distance, but when doing work, we’re also changing potential energy indeed! So if we’ve got a potential energy function, we can get all the rest.]

randomBoltzmann’s Law itself is illustrated by the graph below, which also gives the formula for it: n = n0·e−P.E/kT.

graph

It’s a graph starting at n = n0 for P.E. = 0, and it then decreases exponentially. [Funny expression, isn’t it? So as to respect mathematical terminology, I should say that it decays exponentially.] In any case, if anything, Boltzmann’s Law shows the natural exponential function is quite ‘natural’ indeed, because Boltzmann’s Law pops up in Nature everywhere! Indeed, Boltzmann’s Law is not limited to functions of potential energy only. For example, Feynman derives another Boltzmann Law for the distribution of molecular speeds or, so as to ensure the formula is also valid in relativity, the distribution of molecular momenta. In case you forgot, momentum (p) is the product of mass (m) and velocity (u), and the relevant Boltzmann Law is:

f(p)·dp = C·e−K.E/kT·dp

The argument is not terribly complicated but somewhat lengthy, and so I’ll refer you to the link for more details. As for the f(p) function (and the dp factor on both sides of the equation), that’s because we’re not talking exact values of p but some range equal to dp and some probability of finding particles that have a momentum within that range. The principle is illustrated below for molecular speeds (denoted by u = p/m), so we have a velocity distribution below. The illustration for p would look the same: just substitute u for p.

Distribution

Boltzmann’s Law can be stated, much more generally, as follows:

The probability of different conditions of energy (E), potential or kinetic, is proportional to e−E/kT

As Feynman notes, “This is a rather beautiful proposition, and a very easy thing to remember too!” It is, and we’ll need it for the next bit.

The quantum-mechanical theory of gases

According to quantum theory, energy comes in discrete packets, quanta, and any system, like an oscillator, will only have a discrete set of energy levels, i.e. states of different energy. An energy state is, obviously, a condition of energy and, hence, Boltzmann’s Law applies. More specifically, if we denote the various energy levels, i.e. the energies of the various molecular states, by E0, E1, E2,…, Ei,…, and if Boltzmann’s Law applies, then the probability of finding a molecule in the particular state Ei will be proportional to e−Ei /kT.

Now, we know we’ve got some constant there, but we can get rid of that by calculating relative probabilities. For example, the probability of being in state E1, relative to the probability of being in state E0, is:

P1/P0 = e−E1 /kT/e−E0 /kT = e−(E1–E0)/kT

But the relative probability Pshould, obviously, also be equal to the ratio n1/N, i.e. the ratio of the number of molecules in state E1 and the total number of molecules. Likewise, P= n0/N. Hence, P1/P0 = n1/nand, therefore, we can write:

n = n0e−(E1–E0)/kT

What can we do with that? Remember we want to explain the behavior of non-monatomic gas—like diatomic gas, for example. Now we need some other assumption, obviously. As it turns out, the assumption that we can represent a system as some kind of oscillation still makes sense! In fact, the assumption that our diatomic molecule is like a spring is equally crucial to our quantum-theoretical analysis of gases as it is to our classical kinetic theory of gases. To be precise, in both theories, we look at it as a harmonic oscillator.

Don’t panic. A harmonic oscillator is, quite simply, a system that, when displaced from its equilibrium position, experiences some kind of restoring force. Now, for it to be harmonic, the force needs to be linear. For example, when talking springs, the restoring force F will be proportional to the displacement x). It basically means we can use a linear differential equation to analyze the system, like m·(d2x/dt2) = –kx. […] I hope you recognize this equation, because you should! It’s Newton’s Law: F = m·a with F = –k·x. If you remember the equation, you’ll also remember that harmonic oscillations were sinusoidal oscillations with a constant amplitude and a constant frequency. That frequency did not depend on the amplitude: because of the sinusoidal function involved, it was easier to write that frequency as an angular frequency, which we denoted by ω0 and which, in the case of our spring, was equal to ω0 = (k/m)1/2. So it’s a property of the system. Indeed, ωis the square root of the ratio of (1) k, which characterizes the spring (it’s its stiffness), and (2) m, i.e. the mass on the spring. Solving the differential equation yielded x = A·cos(ω0t + Δ) as a general solution, with A the (maximum) amplitude, and Δ some phase shift determined by our t = 0 point. Let me quickly jot down too more formulas: the potential energy in the spring is kx2/2, while its kinetic energy is mv2/2, as usual (so the kinetic energy depends on the mass and its velocity, while the potential energy only depends on the displacement and the spring’s stiffness). Of course, kinetic and potential energy add up to the total energy of the system, which is constant and proportional to the square of the (maximum) amplitude: K.E. + P.E. = E ∝ A2. To be precise, E = kA2/2.

That’s simple enough. Let’s get back to our molecular oscillator. While the total energy of an oscillator in classical theory can take on any value, Planck challenged that assumption: according to quantum theory, it can only take up energies equal to ħω at a time. [Note that we use the so-called reduced Planck constant here (i.e. h-bar), because we’re dealing with angular frequencies.] Hence, according to quantum theory, we have an oscillator with equally spaced energy levels, and the difference between them is ħω. Now, ħω is terribly tiny—but it’s there. Let me visualize what I just wrote:

Equipartition-3

So our expression for P1/P0 becomes P1/P0 = e−ħω/kT/e−0/kT = e−ħω/kT. More generally, we have Pi/P0 = e−i·ħω/kT. So what? Well… We’ve got a function here which gives the chance of finding a molecule in state Pi relative to that of finding it in state E0, and it’s a function of temperature. Now, the graph below illustrates the general shape of that function. It’s a bit peculiar, but you can see that the relative probability goes up and down with temperature. The graph makes it clear that, at extremely low temperatures, most particles will be in state E0 and, of course, the internal energy of our body of gas will be close to nil.

Capture-2

Now, we can look at the oscillators in the bottom state (i.e. particles in the molecular energy state E0) as being effectively ‘frozen’: they don’t contribute to the specific heat. However, as we increase the temperature, our molecules gradually begin to have an appreciable probability to be in the second state, and then in the next state, and so on, and so the internal energy of the gas increases effectively. Now, when the probability is appreciable for many states, the quantized states become nearly indistinguishable and, hence, the situation is like classical physics: it is nearly indistinguishable from a continuum of energies.

Now, while you can imagine such analysis should explain why the specific heat ratio for oxygen and hydrogen varies as it does in the very first graph of this post, you can also imagine the details of that analysis fill quite a few pages! In fact, even Feynman doesn’t include it in his Lectures. What he does include is the analysis of the blackbody radiation problem, which is remarkably similar. So… Well… For more details on that, I’ll refer you to Feynman indeed. 🙂

I hope you appreciated this little ‘lecture’, as it sort of wraps up my ‘series’ of posts on statistical mechanics, thermodynamics and, central to both, the classical theory of gases. Have fun with it all!