Two Einstein Solids — Counting Microstates

Two Einstein Solids: Counting Microstates

Solid A (N_A oscillators) and Solid B (N_B oscillators) are in thermal contact, sharing a fixed pool of q indistinguishable energy quanta. For every way to split q between the two solids, count how many microscopic arrangements — the multiplicity Ω — each split allows. The split with the most arrangements is the one you'll actually find the system in.

N_A (oscillators in A)

N_B (oscillators in B)

q (total quanta shared)

Most Likely Split

q_A* (peak): -

Ω_total at peak: -

P(q_A*): -

Sanity Check

Σ Ω_total: -

C(q+N_A+N_B−1, q): -

Multiplicity of One Einstein Solid

$$\Omega(N,q) = \binom{q+N-1}{q} = \frac{(q+N-1)!}{q!\,(N-1)!}$$

The number of ways to deal q indistinguishable energy quanta among N distinguishable oscillators (the standard "stars and bars" count). For the combined system with a fixed total q,

$$\Omega_{\text{total}}(q_A) = \Omega(N_A,\,q_A)\cdot\Omega(N_B,\,q-q_A)$$

Every microstate of the combined system is equally likely (the fundamental assumption of statistical mechanics), so the split with the largest Ω_total is overwhelmingly the one you'll observe — and that margin grows sharper the larger N_A and N_B get. Try cranking N_A, N_B up and watch the bar chart collapse to a single spike.

$$S = k_B \ln\Omega, \qquad S_{\text{total}} = S_A + S_B$$

Entropy is just the log of multiplicity, so it inherits the same maximization: S_total peaks at exactly the split with the largest Ω_total. But being a log, S stays a normal, human-sized number even where Ω itself has grown too large to write out.

Why does the sum equal a single binomial coefficient?

$$\sum_{q_A=0}^{q}\Omega(N_A,q_A)\,\Omega(N_B,q-q_A) = \binom{q+N_A+N_B-1}{q}$$

Summing the split-by-split counts is the same as directly counting the ways to deal q quanta among all N_A+N_B oscillators at once, ignoring which solid each one belongs to — a combinatorial identity (Vandermonde's convolution for "stars and bars"), not a coincidence. The Sanity Check panel verifies it numerically for whatever N_A, N_B, q you've dialed in.

System

Solid A

N_A = 3

Energy

Solid B

N_B = 3

q = 6 energy quanta, shared between the two solids in every possible way — each row in the table below is one such split.

Multiplicity Table

q_A	Ω_A	q_B	Ω_B	Ω_total = Ω_AΩ_B	S_A/k_B	S_B/k_B	S_total/k_B
Σ Ω_total = C(q+N_A+N_B−1, q)				-	—	—	—

Ω_total vs. q_A

Ω_total(q_A), relative to its peak

Most likely split

Bar height is Ω_total normalized to its own peak (always safe to plot, however large N_A, N_B, q get) — hover a bar for the actual Ω_total value. The exact numbers live in the table above.

Entropy vs. q_A

S_A/k_B = ln Ω_A

S_B/k_B = ln Ω_B

S_total/k_B = S_A+S_B

Unlike Ω_total, entropy is a log, so it stays a normal, human-sized number no matter how large N_A, N_B, q get — and S_total peaks at exactly the same q_A as Ω_total does above.