Temperature, absolute zero and the zeroth law

An average is each value weighted by how often it happens

Intuition

You rarely want the whole probability table. You want one number: the average energy, say. If marks in a class are 50, 60 and 90, you do not add them and divide by three if more students scored 60. You weight each mark by how many students got it. That is all an expectation value is.

The lecture calls an average of an observable $O$ its expectation value:

$\langle O \rangle = \sum_i p_i O_i$

Here $p_i$ is the probability of state $i$ and $O_i$ is the value of $O$ in that state. For the energy of one molecule, this becomes

$\langle \varepsilon \rangle = \sum_i p_i \varepsilon_i = \frac{\sum_i \varepsilon_i e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}$

For $N$ identical, independent molecules the total energy is $N$ times the average:

$E = N\langle\varepsilon\rangle = \sum_i n_i\varepsilon_i = N\,\frac{\sum_i \varepsilon_i e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}$

Temperature is the number beta, and beta comes from the energy

At equilibrium an isolated system with fixed $N$ and $E$ follows the Boltzmann distribution

$p_i = \frac{n_i}{N} = \frac{e^{-\beta\varepsilon_i}}{\sum_i e^{-\beta\varepsilon_i}}$

Where does $\beta$ come from? It is a Lagrange multiplier, a number that appears when you maximise something while holding the total energy fixed. Put the distribution into the energy equation and $\beta$ is whatever value makes the sum equal $E$. The lecture says this cannot be solved easily for $\beta$, but it makes clear that $\beta$ depends on $N$ and $E$.

Then comes the definition. Absolute temperature $T$ is defined so that

$\beta = \frac{1}{kT}$

$k$ is the Boltzmann constant, exactly $1.380649 \times 10^{-23}$ J K$^{-1}$. $\beta$ has units of inverse energy. The population ratio then reads

$\frac{n_i}{n_j} = e^{-\beta(\varepsilon_i-\varepsilon_j)}$

Plain version: hot means $\beta$ small, so the exponent is small and high levels are well filled. Cold means $\beta$ large, so high levels are nearly empty. The slide asks whether temperature defined this way behaves as expected. The lecture answers this later in the week, using the zeroth law and heat flow (covered in the thermodynamics topics).

Pitfall

Temperature is not a property of one molecule. Beta and T belong to an equilibrium configuration and describe the whole system. A non-equilibrium configuration has no temperature.

At absolute zero everything is in the ground state

Intuition

Think of a block of flats with floors at rising rent. When money is plentiful, people live on every floor. As money runs out everyone moves down. At zero money, everyone is on the ground floor, and there is no lower floor to go to.

The Boltzmann distribution shows what is special about $T = 0$. The energy gap in the exponent becomes infinite:

$\frac{\varepsilon_i - \varepsilon_0}{kT} = \infty$

So for any excited level, with $n_i \neq 0$ in the formula:

$n_i = n_0\, e^{-(\varepsilon_i-\varepsilon_0)/kT} = n_0\, e^{-\infty} = 0$

Every excited population is zero, which leaves

$n_0 = N$

All molecules sit in the ground state. There cannot be a lower temperature, because there are no lower energies.

The zeroth law is why a thermometer works

Imagine a cup of tea, a bowl of soup and a thermometer. If the thermometer settles with the tea, and then settles with the soup at the same reading, tea and soup are in equilibrium with each other. That is the zeroth law of thermodynamics: if two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.

The law defines thermal equilibrium. It also guarantees that one parameter, temperature, describes every system in equilibrium with each other. The previous topic showed that a single parameter $\beta$ sets the distribution, so all systems with the same temperature are in equilibrium with each other. A probe system that decides whether two systems will stay in equilibrium on contact is a thermometer. The absolute scale is usually defined with the ideal-gas thermometer.

Exam detail

Lecture 13 numbers and wording: $k = 1.380649 \times 10^{-23}$ J K$^{-1}$ exactly. At $T = 0$, $n_0 = N$. The lecture adds that later in the week the statistical temperature is shown to follow this rule, and that in an isolated system energy flows from higher to lower temperature.

Recall

Why can no system be colder than absolute zero in this picture?

At $T = 0$ all molecules are in the ground state, $n_0 = N$. There are no lower energies, so there is nothing lower to move to.