TNK LAB

Two Paramagnets: Counting Spin Microstates

Paramagnet A (NA dipoles) and Paramagnet B (NB dipoles) sit in the same fixed external field B0. Every dipole is a simple two-state compass needle — it points either along the field (↑, energy −μB0) or against it (↓, energy +μB0). The two paramagnets exchange energy with each other (a ↓→↑ flip in one, paired with an ↑→↓ flip in the other), which keeps the total number of up-spins across both systems fixed. For every way to split that fixed total between A and B, count the multiplicity Ω — the split with the most microstates is the one you'll observe.

Capped at NA+NB — can't have more up-spins than dipoles.

Most Likely Split

NA↑* (peak): -
Ωtotal at peak: -
P(NA↑*): -
UA*, UB* (μB0): -

Sanity Check

Σ Ωtotal: -
C(NA+NB, m): -

Multiplicity of One Two-State Paramagnet

$$\Omega(N,N_\uparrow) = \binom{N}{N_\uparrow} = \frac{N!}{N_\uparrow!\,(N-N_\uparrow)!}$$
The number of ways to choose which N↑ of the N dipoles point along the field. Energy in units of μB0,
$$U(N,N_\uparrow) = N - 2N_\uparrow$$
For the combined system with a fixed total up-spin count m = NA↑+NB↑,
$$\Omega_{\text{total}}(N_{A\uparrow}) = \Omega(N_A,\,N_{A\uparrow})\cdot\Omega(N_B,\,m-N_{A\uparrow})$$
Every microstate of the combined system is equally likely, so the split with the largest Ωtotal is overwhelmingly the one you'll observe — and that margin grows sharper the larger NA and NB get.
$$S = k_B \ln\Omega, \qquad S_{\text{total}} = S_A + S_B$$
Stotal peaks at exactly the split with the largest Ωtotal, but stays a normal, human-sized number no matter how large NA, NB get.
Why does the sum equal a single binomial coefficient?
$$\sum_{N_{A\uparrow}=0}^{N_A}\binom{N_A}{N_{A\uparrow}}\binom{N_B}{m-N_{A\uparrow}} = \binom{N_A+N_B}{m}$$
This is Vandermonde's identity: summing the split-by-split counts is the same as directly counting the ways to choose m up-spins out of all NA+NB dipoles at once, ignoring which paramagnet each one belongs to. The Sanity Check panel verifies it numerically for whatever NA, NB, m you've dialed in.
Bonus: the textbook example of negative temperature
$$\frac{1}{T} = \frac{\partial S}{\partial U}$$
Look at the entropy chart below: SA (red) peaks when exactly half of A's dipoles point along the field (NA↑ = NA/2, UA = 0) and falls off toward either extreme. Since 1/T = ∂S/∂U, T stays positive on the everyday branch (NA↑ > NA/2, UA < 0) — but swings negative once fewer than half the dipoles align with the field (NA↑ < NA/2, UA > 0): a "population-inverted" state that holds more energy than the maximum-entropy NA↑=NA/2 configuration. The two-state paramagnet is the standard system used to introduce negative absolute temperature.

System

Paramagnet A
NA = 4
Energy
Paramagnet B
NB = 4

m = 4 total up-spins, shared between the two paramagnets in every possible way — each row in the table below is one such split.

Multiplicity Table

NA↑ ΩA UA/μB0 NB↑ ΩB UB/μB0 Ωtotal = ΩAΩB SA/kB SB/kB Stotal/kB
Σ Ωtotal = C(NA+NB, m) - — — —

Ωtotal vs. NA↑

Ωtotal(NA↑), relative to its peak
Most likely split

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

Entropy vs. NA↑

SA/kB = ln ΩA
SB/kB = ln ΩB
Stotal/kB = SA+SB

Stotal peaks at exactly the same NA↑ as Ωtotal does above. Note SA and SB individually are not monotonic — see the negative-temperature note in the sidebar.