Phase transitions need interactions, and only sharpen with many particles

A phase transition is an abrupt jump, and it needs particles that push and pull on each other

Intuition

Think of a crowd walking through a station. If everyone ignores everyone else, the crowd just thins out or thickens smoothly as you add people. Add pushing and pulling between people and something new can happen: the crowd suddenly locks into a jam. That sudden change is a phase transition, and it only exists because the people interact.

Lecture 19 starts from what has been done so far. Only single phases and non-interacting particles, such as the ideal gas and the Einstein solid, have been treated. For those, the partition function of the whole system is built from one particle:

$Q = q^N \ \text{(distinguishable)} \qquad Q = \frac{q^N}{N!} \ \text{(indistinguishable)}$

and totals are just multiples of the single-particle values: $E_n = nE_1$, $S_n = nS_1$, $G_n = nG_1$. That is why you could mostly work with $q$, the molecular partition function.

Two definitions. A phase is a state whose properties change continuously, for example ice, liquid water, water vapour. A phase transition is an abrupt change in a physical property. Water is liquid below 100 °C and gas above it, and its density jumps at 100 °C.

Phase transitions occur because of interactions. Compressed real gases liquefy once the molecules cannot avoid each other. At low $T$ the atoms in a solid vibrate together, and the Einstein model, which treats them as independent, fails.

With interactions you must count states of the whole system, not one molecule

For non-interacting molecules the molecular partition function is

$q = \sum_{i=1}^{N} e^{-\beta\varepsilon_i}$

summed over the states $i$ of one molecule, with $\beta = 1/kT$. This works because many molecules sit in each state, so the average for one molecule matches the ensemble average. That reasoning breaks once molecules interact, because you can no longer look at each molecule alone.

The fix is to imagine many copies of the entire system. This is the canonical ensemble: every copy has the same $N$, $V$ and $T$. The same argument as before gives the canonical partition function:

$Q = \sum_{j=1}^{\tilde N} e^{-\beta E_j}$

The sum is over the $\tilde N$ states of the whole system, indexed $j$, each with total energy $E_j$. $E_j$ depends on all the degrees of freedom of all the molecules. The slide text writes $E_i$ here, but the equation uses $E_j$, so this page uses $E_j$.

The DNA zipper is the simplest exactly solvable phase transition

Intuition

A zip on a jacket. You can only open it from one end, and a tooth can only come apart if every tooth before it has already come apart. The teeth are not independent. That dependence is the interaction. Heat the zip enough and it runs open all at once.

At high temperature DNA unzips. The lecture’s model has two rules.

• Each open link has energy $\varepsilon$ and orientational degeneracy $g$ (an open link can point in $g$ ways).
• The zipper unzips only from the left: link $s$ can open only if links $1, 2, \dots, s-1$ are already open.

So the state of the zipper is just $s$, the number of open links. It has energy $s\varepsilon$ and $g^s$ ways to arrange itself. The partition function is

$Q = \sum_{s=0}^{N-1} g(s)\,e^{-\beta E(s)} = \sum_{s=0}^{N-1} g^s e^{-\beta s\varepsilon} = \sum_{s=0}^{N-1} x^s = \frac{1-x^N}{1-x}, \qquad x = g e^{-\beta\varepsilon}$

$x$ packs the two competing effects into one number. $g$ rewards opening (more ways), and $e^{-\beta\varepsilon}$ penalises it (energy). When $x$ passes 1, opening wins.

To predict the transition we need the average number of open links, $\langle s\rangle$.

$\langle s\rangle = \sum_{s=0}^{N-1} s\,p_s = \sum_{s=0}^{N-1}\frac{s\,x^s}{Q}$

The lecture uses a trick: $s x^s = x\,\dfrac{d}{dx}x^s$. Then $x\,d/dx$ can be pulled outside the sum, which turns the sum back into $Q$.

$\langle s\rangle = \frac{1}{Q}\,x\frac{d}{dx}\sum_{s=0}^{N-1}x^s = \frac{1}{Q}\,x\frac{d}{dx}Q$

“Standard algebra” then gives

$\langle s\rangle = \frac{Nx^N}{x^N-1} - \frac{x}{x-1}, \qquad \frac{\langle s\rangle}{N} = \frac{x^N}{x^N-1} - \frac{x}{N(x-1)}$

Try it: worked example

Show how the closed form for the average number of open links follows from Q = (1 - x^N)/(1 - x). The lecture skips this algebra.

Show answer and steps

AnswerDifferentiate Q, multiply by x, divide by Q. The result is N x^N/(x^N - 1) - x/(x - 1).

1. This algebra is not on the slides, so it is written out here. The slide only says it is standard.
2. Start with $Q = \dfrac{1-x^N}{1-x}$.
3. Differentiate with the quotient rule: $\dfrac{dQ}{dx} = \dfrac{-Nx^{N-1}(1-x) + (1-x^N)}{(1-x)^2}$.
4. Form $\langle s\rangle = \dfrac{x}{Q}\dfrac{dQ}{dx}$. The $(1-x)$ in $Q$ cancels one factor of $(1-x)$ in the denominator.
5. That leaves $\langle s\rangle = \dfrac{x\left[-Nx^{N-1}(1-x) + (1-x^N)\right]}{(1-x)(1-x^N)}$.
6. Split into two fractions: $\dfrac{-Nx^N}{1-x^N} + \dfrac{x}{1-x}$.
7. Flip the signs in both denominators: $\dfrac{Nx^N}{x^N-1} - \dfrac{x}{x-1}$. This matches slide 11.

The transition is abrupt from fully zipped to fully unzipped at $x = 1$, that is, at

$T_c = \frac{\varepsilon}{k\ln g}$

Try it: worked example

Show that x = 1 gives T_c = epsilon/(k ln g).

Show answer and steps

AnswerSetting g exp(-epsilon/kT) = 1 and taking logs gives epsilon/kT = ln g, so T_c = epsilon/(k ln g).

1. Start with $x = g\,e^{-\varepsilon/kT} = 1$.
2. Take the natural log of both sides: $\ln g - \dfrac{\varepsilon}{kT} = 0$.
3. Move the energy term across: $\dfrac{\varepsilon}{kT} = \ln g$.
4. Solve for $T$: $T_c = \dfrac{\varepsilon}{k\ln g}$.
5. At higher $T$, $x$ grows towards $g$, which is above 1, so the zipper is open. Below $T_c$, $x$ is under 1 and it stays shut.

The jump only becomes sharp when N is huge

The partition function looks smooth. It is a finite sum of smooth terms.

$Q = \sum_{s=0}^{N-1} g^s e^{-s\varepsilon/kT}$

A finite sum of smooth functions cannot jump. Discontinuities can appear only as $N \to \infty$. This is the thermodynamic limit. The lecture’s plot of $\langle s\rangle/N$ against $x$ shows this: for $N = 20$ the curve is gradual and passes about 0.47 at $x = 1$ (read from the plot), while for $N = 10^5$ it is a step from 0 to 1 at $x = 1$. The curves for $N = 20$, $100$, $500$ and $10^5$ steepen with $N$.

The general statement on the slide: phase transitions can only occur in the thermodynamic limit. You can even define a phase as a state whose properties vary smoothly. Even everyday transitions stop being sharp when there are few particles. A gold nanoparticle a few tens of atoms across has no sharp melting temperature: is it solid or liquid?

Exam detail

Learning outcomes on slide 16: understand the meaning of the canonical partition function, average over the canonical ensemble to get an expectation value, state the cause and conditions of a phase change, and predict phase behaviour near a transition including how it depends on system size.

Pitfall

The molecular partition function $q$ and the canonical partition function $Q$ are different. $q$ sums over states of one molecule and only works without interactions. $Q$ sums over states of the whole system and works with interactions.

Recall

Why can a system of 20 zipper links not show a perfectly sharp unzipping?

The partition function is a finite sum of smooth terms, so every thermodynamic function is continuous. Only as $N \to \infty$ can a discontinuity appear.