Mean-field theory: replace the crowd with its average and a transition appears

When interactions are too hard, replace them with an average

Intuition

You are in a crowd at a concert and you want to know whether to sway left or right. Listening to every person around you is hopeless. So you look at the average mood of the crowd and follow that. Mean-field theory does the same for a particle: it stops tracking each neighbour and feels only their average.

Lecture 21 recaps where things stand. Phase transitions are sharp changes in a macroscopic parameter. They need interactions and the thermodynamic limit. Interacting systems are hard, because the configuration space is enormous and exact solutions are usually impossible. Simulations help, but it can be hard to get an intuitive explanation from them.

Mean-field theory is an approximate way of including interactions. It is sometimes crude, but it gives qualitative insight and, importantly, it lets us predict phase transitions. The lecture applies it to two systems: the Ising model of the ferromagnetic to paramagnetic transition (this page), and the van der Waals model of the real gas (next page).

The Ising model is the lattice of $\pm 1$ spins from the last lecture, with energy

$E = -B\sum_i \sigma_i - J\sum_{\langle ij\rangle}\sigma_i\sigma_j$

and net magnetisation $m = \sum_i \sigma_i$. The behaviour is $m \sim 0$ for $T \gg T_c$ and $|m| \sim 1$ for $T \ll T_c$. The lecture asks the question: how can we be sure that the transition is actually sharp?

First the easy case: no interactions at all

Drop the $J$ term. The energy becomes

$E = -B\sum_i\sigma_i$

Now each spin stands alone, so use one-spin quantities. A spin has two states, up (energy $-B$) and down (energy $+B$). The single-spin partition function and the probabilities are

$q = e^{\beta B} + e^{-\beta B} \qquad p_\uparrow = \frac{e^{\beta B}}{e^{\beta B} + e^{-\beta B}} \qquad p_\downarrow = \frac{e^{-\beta B}}{e^{\beta B} + e^{-\beta B}}$

Try it: worked example

Find the net magnetisation m of N non-interacting spins in a field B.

Show answer and steps

Answerm = N tanh(beta B).

1. The magnetisation is the number up minus the number down: $m = N_\uparrow - N_\downarrow$.
2. With $N$ spins, $N_\uparrow = Np_\uparrow$ and $N_\downarrow = Np_\downarrow$.
3. So $m = N\left(\dfrac{e^{\beta B}}{e^{\beta B}+e^{-\beta B}} - \dfrac{e^{-\beta B}}{e^{\beta B}+e^{-\beta B}}\right)$.
4. The two fractions share a denominator: $m = N\,\dfrac{e^{\beta B}-e^{-\beta B}}{e^{\beta B}+e^{-\beta B}}$.
5. That ratio is the definition of the hyperbolic tangent, so $m = N\tanh(\beta B)$.
6. Plain reading: the curve $m/N$ against $B$ is S-shaped. A bigger $\beta$ (colder) makes it steeper and it saturates at $\pm 1$. There is no sudden jump, so no transition.

Mean-field turns the neighbours into an extra field

Keep the interaction, but replace each neighbour’s spin with the average spin $\langle\sigma\rangle$. $d$ is the number of nearest neighbours of each spin.

$E = -B\sum_i \sigma_i - J\sum_{\langle ij\rangle}\sigma_i\sigma_j \quad\longrightarrow\quad E = -B\sum_i \sigma_i - J\sum_i \sigma_i\langle\sigma\rangle d$

Collect the terms:

$E = -\left(B + J\langle\sigma\rangle d\right)\sum_i \sigma_i$

This looks exactly like the non-interacting model. The neighbours act as an extra field $J\langle\sigma\rangle d$. Now set the applied field to zero, $B = 0$, and use $\langle\sigma\rangle = m/N$:

$E = -\frac{Jmd}{N}\sum_i \sigma_i$

This is the non-interacting model with “field” $Jmd/N$. Take the non-interacting answer $m = N\tanh(\beta B)$ and swap in that field:

$\frac{m}{N} = \tanh\left(\beta J d\,\frac{m}{N}\right)$

The self-consistency equation has one solution when hot and three when cold

This is an odd equation, with $m/N$ on both sides. The lecture solves it graphically. Let $x = \beta J d\, m/N$, so $m/N = x\,kT/(Jd)$. The equation becomes

$\frac{kT}{Jd}\,x = \tanh(x)$

The left side is a straight line through the origin with slope $kT/Jd$. The right side is the S-shaped tanh curve. The solutions are where they cross.

The sketch follows slide 10. Read it like this.

• High $T$. The line is steep (the $2x$ line). It meets the curve only at $x = 0$, so $m = 0$. This is the unmagnetised phase.
• Low $T$. The line is shallow (the $x/2$ line). It meets the curve at three points: $x = 0$ and two non-zero solutions near $x \approx \pm 1.9$ (read from the plot). The two non-zero solutions are the magnetic phases, spin-up and spin-down. The lecture notes that $x = 0$ is a free-energy maximum, so the system does not stay there.
• At $T_c$. The line of slope 1 meets the curve only at $0$.

The line has slope $kT/Jd$, so slope 1 means $kT = Jd$:

$T_c = \frac{Jd}{k}$

The slide wording is unclear here. It says the critical temperature “corresponds to $m = 1$”, but the figure shows the $x$ line touching the curve only at $0$ when $kT/Jd = 1$. This page follows the figure and the formula $T_c = Jd/k$. Check the wording against the lecture recording.

Try it: worked example

Which line in the slide 10 figure is the cold magnet, and what are its solutions?

Show answer and steps

AnswerThe x/2 line, slope kT/Jd = 1/2. Three solutions: x = 0 and about x = plus or minus 1.9.

1. The line has slope $kT/Jd$. A cold magnet has small $T$, so a small slope.
2. Slopes 1/2, 1 and 2 correspond to $kT/Jd = 1/2$, $1$ and $2$.
3. The shallow line (slope 1/2) cuts the S-curve near the origin and again once on each flat arm. That gives three crossings.
4. The crossings on the arms sit at about $x \approx \pm 1.9$ (read from the plot), meaning $|m|$ is close to $N$.
5. The middle crossing, $x = 0$, is a free-energy maximum. The system sits in one of the two outer ones.

Mean-field gets the shape right and the numbers wrong

Mean-field theory explains two important facts. There is a sudden transition from the magnetised to the unmagnetised phase as the Curie temperature $T_c$ is crossed. And spin-down and spin-up are equally likely solutions. The limitation is stated plainly: mean-field theory is an approximation, so its numerical predictions are incorrect.

Exam detail

Learning outcomes on slide 23: explain the approximations and assumptions in mean-field models, and understand what mean-field theory predicts for the Ising model and the van der Waals fluid. Know the chain: $E$, then mean-field $E$, then $m/N = \tanh(\beta Jd\,m/N)$, then $T_c = Jd/k$.

Recall

Why does the mean-field equation have three solutions only at low temperature?

At low $T$ the line $kT x/(Jd)$ is shallow enough to cut the tanh curve three times: at $0$ and on each arm. At high $T$ the line is steep and cuts it only at $0$.