Free Energy Competition

A system in contact with a heat bath at temperature T doesn't minimize its energy U — it minimizes F = U − TS. Compare two candidate states of the same N-dipole paramagnet: the fully-ordered ground state A (all spins aligned, zero entropy) against a more disordered alternative B. At T=0 the lower-energy state always wins. Turn up T and watch entropy start to matter — past a crossover temperature, the messier state actually has lower free energy and becomes favored.

N (dipoles)

State B: N_↑ (up-spins)

State A is fixed at N_↑=N (fully ordered, S=0). Drag B anywhere — 50/50 is maximum entropy.

Temperature T (μB₀/k_B)

State A — Ordered

U_A / μB₀: -

S_A / k_B: -

F_A / μB₀: -

State B — Alternative

U_B / μB₀: -

S_B / k_B: -

F_B / μB₀: -

Crossover T*: -

Why F, Not U?

$$F = U - TS$$

Held at fixed T, a system trades energy with the bath freely, so U alone doesn't decide what's favored — the combined entropy of system + bath does, and the Second Law says that's what climbs. It turns out maximizing that combined entropy is exactly the same as minimizing F for the system by itself: F is what the Second Law looks like when you only have to watch the system, not the whole bath.

Since Ω_A=1 (only one way to have every dipole aligned), S_A=0 always, so F_A=U_A is flat — a horizontal line in the chart on the right. F_B is a straight line with slope −S_B: the more disordered B is, the faster its free energy falls as T rises. They cross at

$$T^* = \frac{U_B - U_A}{S_B}$$

Below T*, A's lower energy wins. Above T*, B's larger entropy wins — not because B suddenly has less energy, but because the TS term it earns outgrows A's energy advantage.

Isn't this just a phase transition?

Yes — this is the same logic behind melting, boiling, and magnetic ordering. A crystal (ordered, low S) competes against a liquid (disordered, higher S) at the same U roughly; below the melting point the crystal's lower F wins, above it the liquid's higher S wins. Real phase transitions compare a huge number of candidate states, not just two, but the mechanism — a low-S state losing to a high-S state as T climbs past U_gap/ΔS — is exactly this.

Try dragging State B's N_↑ away from 50/50 toward the extremes — S_B shrinks, and T* shoots up: a state that's only slightly disordered needs a much higher temperature before its small entropy advantage is worth anything.

Where A and B Sit on the Entropy Landscape

S(N_↑) for every possible state

State A

State B

A and B are just two points out of every N_↑ the system could have — F only compares the two you've picked, not the whole landscape.

Free Energy Competition

State A — Ordered

State B — Alternative

Why F, Not U?

Free Energy vs. Temperature

F = U − TS, Broken Down at Current T

Where A and B Sit on the Entropy Landscape