Statistical mechanics (intro)

The idea

Statistical mechanics rebuilds thermodynamics from counting. A macrostate is what you can measure (say, the total number of heads in a coin set, or a gas's pressure); a microstate is one complete microscopic arrangement consistent with it. The central postulate is that an isolated system is equally likely to be in any accessible microstate, so the probability of a macrostate is proportional to its multiplicity Ω — the number of microstates that realize it.

Entropy quantifies that count: S = k ln Ω, with Boltzmann's constant k = 1.38 × 10⁻²³ J/K. The logarithm makes entropy additive when multiplicities multiply, and the second law becomes almost a tautology: systems drift toward the macrostates with overwhelmingly more microstates simply because there are overwhelmingly more ways to be there. For everyday particle numbers (10²³), 'overwhelmingly' is so extreme that fluctuations away from equilibrium are unobservable.

The common misreading is that disorder is forced or that ordered states are forbidden. Nothing prevents all the air molecules from gathering in one corner of the room — every microstate remains equally likely. The point is purely statistical: the spread-out macrostates own astronomically more microstates, so that is where the system is found.

Worked example

Four distinguishable coins are tossed. Count the microstates, find the probability of the macrostate 'exactly two heads' and of 'all four heads', and compute the entropy S = k ln Ω of each macrostate using k = 1.38 × 10⁻²³ J/K.

1. Count all microstates: each coin has 2 outcomes, so Ω_total = 2⁴ = 16 equally likely sequences.
2. Count the microstates of 'exactly two heads': choosing which 2 of the 4 coins land heads gives C(4,2) = 6 microstates, so P = 6/16 = 0.375.
3. Count 'all four heads': only 1 sequence works, so P = 1/16 ≈ 0.063 — six times less likely than the even split, purely because fewer arrangements realize it.
4. Convert multiplicities to entropies: S(two heads) = k ln 6 = 1.38 × 10⁻²³ × 1.79 ≈ 2.5 × 10⁻²³ J/K, while S(all heads) = k ln 1 = 0.
5. Interpret the scaling: with 4 coins the even split is only modestly favored, but the binomial peak sharpens ferociously with N — for 10²³ molecules the analogous even spread is so dominant that deviations are never seen, which is the second law in miniature.

Answer. There are 16 microstates; 'two heads' has probability 0.375 and entropy about 2.5 × 10⁻²³ J/K, while 'all heads' has probability 1/16 and zero entropy.

Check your understanding

• Why does the macrostate with the most microstates dominate more and more completely as the particle number grows?
• What does the logarithm in S = k ln Ω accomplish when you combine two independent systems?
• How would you answer someone who claims the second law forbids ordered states from ever occurring?
• What changes in the counting if the coins (or particles) are indistinguishable, and why does that matter physically?

Build the foundations first

Statistical mechanics (intro) builds on these concepts. If any feel shaky, start there.

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