Microstates, configurations and the Boltzmann distribution

A huge system is described by its most likely arrangement

Intuition

Flip four coins. There are 16 exact patterns, such as heads-tails-tails-heads. But you might only care how many heads came up. Many exact patterns give the same head count, and two heads can happen in more ways than four heads can. So two heads is the most likely result. Statistical mechanics runs the same idea on about $10^{23}$ molecules.

Nobody can write down where every one of $10^{23}$ molecules is. What we can know are bulk numbers: how many molecules, the temperature, the pressure, the volume. Statistical mechanics links the behaviour of single molecules to those bulk properties. It also asks what equilibrium is, what pushes a system there, what entropy is and what is special about temperature. Some quantities, such as temperature, mean nothing for one molecule and exist only for the whole system.

Two words matter here. A microstate is one exact arrangement of the individual molecules: which coin is heads, which molecule is in which quantum state. A configuration (also called a macrostate) is a group of microstates that give the same bulk property: how many heads. The fundamental postulate says all microstates are equally probable. So the most likely configuration is the one made of the most microstates.

Now keep tossing. The lecture shows the curve of “number of heads” for 10, 100 and 1000 tosses, and it gets narrower each time. The chance of getting 4 heads in 10 tosses is 0.205. The chance of 40 heads in 100 tosses is 0.0108. The chance of 400 heads in 1000 tosses is $4.6 \times 10^{-11}$. In each case that is 40% heads, and it becomes ever less likely. The most likely configuration takes over. In the thermodynamic limit, where $N \to \infty$, only the dominant configuration is ever seen.

Air in a room is the same story. Each molecule has a 50:50 chance of being in the left or right half. A 40 m³ room holds $10^{27}$ molecules. The chance that more than 51% of the air is on the left is very roughly

$\approx e^{-10^{23}}$

and 99% of the time the number in each half differs by less than 1 part in $10^{16}$. The air does not pile up in a corner because the even split is overwhelmingly the most likely.

Counting microstates gives the weight of a configuration

In chemistry we count quantum states. The lecture’s example is the harmonic oscillator, whose energy levels are $E_n = h\nu(n + \tfrac{1}{2})$. Ignore the constant zero-point energy and the levels are $E_n = h\nu n$: evenly spaced rungs on a ladder.

Picture three people (distinguishable oscillators) sharing out 3 coins (quanta of energy). A configuration is written by occupation numbers, $n_i$ = number of oscillators in state $i$. So $\{0, 3, 0, 0\}$ means no oscillator at level 0, three at level 1, none at 2 or 3. The weight $W$ of a configuration is the number of microstates that make it:

$W(n_0, n_1, n_2, \dots) = \frac{N!}{n_0!\, n_1!\, n_2! \cdots}$

Here $N$ is the total number of molecules and $n! = n(n-1)(n-2)\cdots 2 \cdot 1$ is the factorial.

Why that formula? Put $N$ balls into bins. There are $N!$ ways to order the balls as you hand them out. But inside bin $i$ the same $n_i$ balls could have come out in $n_i!$ different orders, and all of those give the same microstate. So divide the total by each $n_i!$ to stop double counting.

Try it: worked example

Find the weight of each configuration when 3 quanta are shared among 3 distinguishable oscillators, and the probability of {1, 1, 1, 0}.

Show answer and steps

AnswerOrange {0, 3, 0, 0}: W = 1. Purple {1, 1, 1, 0}: W = 6. Blue {2, 0, 0, 1}: W = 3. Total 10, so the purple probability is 6/10.

1. There are 3 oscillators, so $N = 3$ and $N! = 3! = 6$.
2. Orange $\{0,3,0,0\}$: $W = \dfrac{3!}{0!\,3!\,0!\,0!} = \dfrac{6}{1 \cdot 6 \cdot 1 \cdot 1} = 1$.
3. Purple $\{1,1,1,0\}$: $W = \dfrac{3!}{1!\,1!\,1!\,0!} = \dfrac{6}{1 \cdot 1 \cdot 1 \cdot 1} = 6$.
4. Blue $\{2,0,0,1\}$ (the slide’s own case): $W = \dfrac{3!}{2!\,0!\,0!\,1!} = \dfrac{6}{2 \cdot 1 \cdot 1 \cdot 1} = 3$.
5. The three groups hold $1 + 6 + 3 = 10$ microstates, matching the ten panels on the slide.
6. Every microstate is equally likely, so the probability of purple is $6/10$.

Pitfall

Do not confuse the two words. A microstate is one exact arrangement. A configuration is a bundle of microstates with the same occupation numbers. Probabilities are worked out for configurations by counting the microstates inside them.

The winner is found by maximising W under two rules

Intuition

Equilibrium is where the system spends nearly all its time. Left alone, jostling and swapping energy carry it from an odd starting arrangement towards the configuration with the biggest weight. Once there, it keeps changing microstate but stays overwhelmingly close to the same configuration.

For a given $N$ and total energy $E$ (an isolated system), equilibrium is the one dominant configuration. To find it we want the largest $W$ as a function of all the $n_i$. Factorials of huge numbers are awkward, so the lecture uses Stirling’s approximation:

$\ln n! \approx n \ln n - n$

Take the log of $W$ and apply it to each factorial. The $-N$ and the $+\sum n_i$ cancel because $\sum_i n_i = N$, which leaves

$\ln W = N \ln N - \sum_i n_i \ln n_i$

Maximising $\ln W$ is the same as maximising $W$, and it is easier to handle. The $n_i$ cannot be chosen freely. Two things are held fixed:

1. The number of molecules: $\sum_i n_i = N$.
2. The total energy: $\sum_i \varepsilon_i n_i = E$, where $\varepsilon_i$ is the energy of level $i$.

Think of sharing out a fixed pile of coins among people, where each person also has a price tag and the total spend is fixed. You cannot give everybody everything. Solving this problem uses Lagrange multipliers, a standard way to maximise something under fixed rules. The lecture says to see the textbook for the working, so no steps are shown on the slides.

The result is the Boltzmann distribution

The answer to the maximising problem gives the probability $p_i$ that a molecule sits in level $i$:

$p_i = \frac{n_i}{N} = \frac{e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}$

The bottom line is so central that it has its own name, the partition function:

$q = \sum_i e^{-\beta\varepsilon_i}$

In plain words: give every level a score $e^{-\beta\varepsilon_i}$ that shrinks as the energy rises. Add the scores up to get $q$. A level’s share of the molecules is its score divided by $q$.

The lecture makes three points. First, the single parameter $\beta$ fixes the most probable populations of all the states. Second, as the energy of a state rises, its occupation drops exponentially. Third, the ratio of two populations is

$\frac{n_i}{n_j} = e^{-\beta(\varepsilon_i - \varepsilon_j)}$

So only the energy gap matters in a ratio. The slide ends by saying that 1/$\beta$ will turn out to behave like temperature, and temperature is defined so that

$\beta = \frac{1}{kT}$

where $k$ is the Boltzmann constant and $T$ the temperature. Lecture 13 explains why.

Exam detail

Lecture 12 summary slide: microstates are arrangements of individual molecules. Configurations are collections of microstates giving the same bulk properties. All microstates are assumed equally likely, so you find the probability of a configuration by counting microstates. In the thermodynamic limit the most probable configuration dominates and defines equilibrium. Occupations and probabilities at equilibrium follow the Boltzmann distribution. Numbers on the slides: 0.205, 0.0108 and $4.6 \times 10^{-11}$ for the coin tosses; $10^{27}$ molecules in a 40 m³ room; $W(2,0,0,1) = 3$.

Optional aside

The figure on slide 16 plots $W/W_{\max}$ against $N_3$ and shows a single sharp peak near $N_3 \approx 1200$. The slide does not say what system it is for, so only the shape (one narrow peak) is a lecture point.

Recall

What two things are held fixed when finding the most probable configuration, and what distribution comes out?

The number of molecules ($\sum n_i = N$) and the total energy ($\sum \varepsilon_i n_i = E$). The result is the Boltzmann distribution, $p_i = e^{-\beta\varepsilon_i}/q$ with $q = \sum_i e^{-\beta\varepsilon_i}$.