Microstates, Configurations and Weight

Introduction

Suppose a fixed amount of energy is shared among a group of molecules. There are many ways to do this: one molecule might hold all the energy, or it might be spread thinly across many. Statistical thermodynamics rests on a simple principle — every individual arrangement consistent with the total energy is equally likely. Counting those arrangements carefully reveals why systems settle into one particular distribution and stay there.

Core explanation

Microstates. Imagine N distinguishable molecules, each able to occupy energy levels ε₀, ε₁, ε₂, … A microstate specifies exactly which molecule is in which level: "molecule A in ε₁, molecule B in ε₀, …". The principle of equal a priori probabilities states that, for an isolated system of fixed total energy, all microstates are equally probable.

Configurations. Usually we do not care which molecule is where, only how many are in each level. A configuration lists these occupation numbers, {n₀, n₁, n₂, …}. Any configuration must satisfy two constraints:

- the total number is fixed: Σ nᵢ = N; - the total energy is fixed: Σ nᵢεᵢ = E.

Weight of a configuration. Many microstates share the same configuration. The number of them is the weight :

W = N! / (n₀! n₁! n₂! …)

This is the number of distinct ways of dividing N objects into groups of sizes n₀, n₁, … (remember 0! = 1). The configuration {N, 0, 0, …}, with every molecule in the ground state, has W = 1. A spread-out configuration has a much larger weight.

Probability of a configuration. Because every microstate is equally likely, the probability of a configuration is proportional to its weight. The configuration with the greatest W is the dominant configuration .

Stirling's approximation. For large numbers, factorials are unmanageable, so we work with ln W:

ln W = ln N! − Σ ln nᵢ!

and use Stirling's approximation, ln x! ≈ x ln x − x. This leads to

ln W ≈ N ln N − Σ nᵢ ln nᵢ

which is a smooth function that can be maximised using calculus.

Why one configuration dominates. As N grows, the weight of the most probable configuration grows so much faster than that of its neighbours that the peak in W becomes extremely sharp. Its relative width shrinks as about 1/√N. For a mole of molecules, configurations that differ noticeably from the dominant one have utterly negligible probability. The system is, in effect, always found in its dominant configuration, and this configuration defines thermal equilibrium.

Formulae

W = N!/(n₀! n₁! n₂! …); ln W ≈ N ln N − Σ nᵢ ln nᵢ (Stirling); constraints Σ nᵢ = N and Σ nᵢεᵢ = E.

Step-by-step reasoning

To find the weights for a small system:

1. List every set of occupation numbers {n₀, n₁, …} that satisfies both the number and energy constraints. 2. For each configuration, calculate W = N!/(n₀! n₁! …). 3. Add the weights to obtain the total number of microstates. 4. Divide each weight by the total to obtain the probability of each configuration. 5. Identify the configuration with the largest weight.

Visual explanation

Plot W against a variable describing the spread of the configuration. For four molecules the plot is a broad, lumpy hump. For a thousand molecules it narrows sharply, and for 10²³ molecules it becomes a spike so thin that, on any practical scale, only one configuration is visible.

Real-world analogy

Toss 20 coins. Exactly one outcome gives 20 heads, but 184 756 different sequences give exactly 10 heads and 10 tails. The "evenly mixed" result is not favoured by any force — it simply corresponds to vastly more arrangements. Toss 10²³ coins and a noticeable departure from half heads essentially never happens.

Real-world example

A gas released into an empty flask spreads to fill the whole flask. No force pushes the molecules apart; rather, the configurations with molecules spread uniformly have overwhelmingly greater weight than those with all molecules in one corner. The same counting argument explains why heat flows from hot to cold bodies and why mixing is spontaneous.

Why?

Why use ln W rather than W itself? Weights for macroscopic systems are numbers such as 10^(10²³), far too large to handle directly. The logarithm turns products into sums, makes Stirling's approximation usable, and, as later pages show, ln W is directly proportional to entropy.

Common misconception

"The equilibrium state is the only allowed configuration." Every configuration that satisfies the constraints is allowed, and the system constantly moves between microstates. Equilibrium corresponds to the dominant configuration because it, together with its near neighbours, contains almost all the microstates.

Worked example

Question: Four distinguishable molecules share two quanta of energy. Each molecule can occupy levels of energy 0, ε or 2ε. Find all configurations and their weights.

Reasoning: The allowed configurations {n₀, n₁, n₂} are {3, 0, 1} (one molecule holds both quanta) and {2, 2, 0} (two molecules hold one quantum each). Their weights are W = 4!/(3! 0! 1!) = 4 and W = 4!/(2! 2! 0!) = 24/4 = 6. The total is 10 microstates.

Answer: {3, 0, 1} with W = 4 (probability 0.4) and {2, 2, 0} with W = 6 (probability 0.6); the more spread-out configuration is dominant.

Quick check

1. What is the weight of the configuration in which all N molecules are in the ground state? Answer: W = N!/N! = 1, because there is only one way to put every molecule in the same level.

Exam focus

Be ready to enumerate configurations for small systems, apply the weight formula correctly (including 0! = 1), and explain in words why a sharply peaked weight function means a macroscopic system is always found in its dominant configuration.

Advanced insight

Finding the dominant configuration means maximising ln W subject to the two constraints. This is done with Lagrange multipliers, α for the number constraint and β for the energy constraint. The result is the Boltzmann distribution, and β turns out to be 1/kT. Temperature therefore enters statistical thermodynamics as the multiplier that enforces conservation of energy.

Summary

A microstate specifies the state of every molecule; a configuration specifies only the occupation numbers of each level. The weight W = N!/(n₀!n₁!…) counts microstates per configuration. Since all microstates are equally probable, the configuration of greatest weight is the most probable, and for macroscopic N it dominates overwhelmingly. Stirling's approximation makes ln W tractable for large N.

Practice questions

1. Calculate the weight of the configuration {4, 2, 1} for seven molecules. Answer: W = 7!/(4! 2! 1!) = 5040/48 = 105. 2. Use Stirling's approximation to estimate ln 100!. Answer: ln 100! ≈ 100 ln 100 − 100 = 460.5 − 100 = 360.5 (the exact value is about 363.7). 3. Five molecules share one quantum of energy ε. How many microstates are there, and what is the only configuration? Answer: The configuration is {4, 1}; its weight is 5!/(4! 1!) = 5, so there are five microstates. 4. Explain why a gas never spontaneously collects in one half of its container. Answer: Configurations with molecules spread through the whole volume have enormously greater weight, so the chance of all molecules being in one half, about (1/2)^N, is negligible for large N.