In every system so far the particles exerted no force on one another. Here we put in an interaction for the first time. On each site of a lattice we place an arrow pointing up or down, and we grant one rule only: when neighbours agree, the energy goes down. The expression is E = -J \sum s_i s_j, the sum running over neighbouring pairs.
The lattice in the figure is that. It alternates between an ordered and a disordered arrangement. The ordered one has the lower energy, the disordered one the larger count of states. Which of them wins is settled, as always, by the temperature.
So it is a contest between E and TS in F = E - TS. When it is cold E wins and the arrows agree; when it is hot TS wins and they scatter. The changeover happens suddenly, at one temperature. That is a phase transition.
While neighbours are tied to one another, settling one spin means knowing all the spins around it, and those are tied to their own neighbours in turn. As it stands there is nothing to solve. So we stop looking at the surrounding spins one by one and replace them by their average m.
The left of the figure is the actual surroundings, the right what is left after the replacement. All the central spin now feels is a single field h = Jzm, where z is the number of neighbours, which on a square lattice is 4.
A many-body problem has turned into a one-body problem, of exactly the shape of the two-level system we solved earlier. Only one thing is different: the field h has m inside it. To get the answer we need the answer.
Taking the two-level result as it stands gives m = \tanh(\beta J z m). Measuring temperature on a scale where kT_c = Jz is 1, this reads m = \tanh(m/T). The m on the left and the m on the right are the same m.
The figure lays the straight line of the left side over the curve of the right side. The crossings are the answers, and moving the temperature changes only how steeply the curve rises.
When it is hot the curve lies flat and the origin is the only crossing. Lower the temperature and the rise at the origin steepens, until at some point two crossings appear away from it. The number of answers goes from one to three, and that boundary is the transition.
The curve overtakes the line at the origin when its slope there reaches 1. The slope of \tanh(m/T) at the origin is 1/T, so the boundary is T = 1, that is kT_c = Jz. The temperature of the transition has come out of nothing but the strength of the interaction and the number of neighbours.
The figure is the spontaneous magnetisation that follows. Above T_c it stays 0; below it rises, and approaches 1 as the temperature falls.
Just below T_c it rises as m \propto (T_c - T)^{1/2}. Being a square root, its slope at T_c is infinite: drop the temperature by the smallest amount and a magnetisation is already there.
Two crossings appear away from the origin, at +m and -m, with exactly the same energy and exactly the same entropy. The two branches in the figure are those.
Nothing in E = -J \sum s_i s_j tells up from down. Turn every spin over and the expression is unchanged. And yet below T_c the system picks one of the two. Which one it picks is not settled; that it picks one is.
The answer fails to hold a symmetry that the expression holds. This is called broken symmetry. Out of counting alone, and the contest in F = E - TS, we have the temperature at which a magnet becomes a magnet. The rest of this chapter uses this same shape again and again.