The second law and entropy

Systems drift towards the arrangement with the most microstates

Intuition

Drop a drop of ink in water and it spreads. You never see the spread-out ink gather itself back into a drop. It is not forbidden. It is just wildly unlikely, because there are far more ways to be spread out than to be bunched up. That is the whole second law in everyday words.

The first law says which processes are possible. The second law says which ones actually happen by themselves. The lecture gives three versions:

1. The entropy of an isolated system always increases, $\Delta S \ge 0$.
2. Heat can never pass from a colder body to a hotter one without some other change occurring.
3. An engine cannot extract heat from a reservoir and convert it completely into work.

Systems spontaneously move towards configurations with more microstates. The end point is equilibrium, the state with the maximum number of microstates. The lecture says this one-way tendency fixes the forward direction of time, and that it is strictly true only in the thermodynamic limit. For small systems there can be fluctuations away from equilibrium.

The slide shows two chambers split by a membrane. At first the particles are crowded on the left. Later they are spread evenly. A crowded lift empties into a hall and spreads out. Nobody pushes people into an even spread. It just has the most ways to happen.

The dog-flea model shows equilibrium arriving

Rover and Rex are two dogs. At the start, all $2R$ fleas are on Rover. Every second, one randomly chosen flea jumps to the other dog. With $N_{\text{Rover}}$ fleas on Rover, the chance the jump is from Rover to Rex is

$P_+(N_{\text{Rover}}) = \frac{N_{\text{Rover}}}{2R}, \qquad P_- = 1 - P_+$

In plain words, the more fleas on Rover, the more likely the next flea chosen lives there. At the start every jump goes off Rover. As fleas build up on Rex, jumps back become likely. The lecture describes this as a system of $2R$ particles and two degenerate energy levels, Rover and Rex.

At equilibrium the system no longer changes measurably, but jumps still happen. Jumps off and on balance:

$N_{\text{Rover}} = N_{\text{Rex}} = R$

The slide’s simulations (R = 20, 80, 200, four experiments each) show three things. The average decays to $R$. Each experiment has its own random wobble. The more fleas, the smaller the wobble.

Entropy is a count of microstates, turned into a logarithm

At constant $N$ and $E$, systems move to configurations with more microstates, $\Delta W \ge 0$. Logs are easier, so the lecture writes $d\ln W \ge 0$, with $d\ln W = 0$ at equilibrium. Then it defines

$S = k\ln W$

and the first form of the second law becomes, for an isolated system,

$dS \ge 0$

Why a logarithm? Because of additivity. Take two independent, isolated systems $x$ and $y$. Every microstate of one can pair with every microstate of the other, so the weights multiply, $W_{\text{total}} = W_x W_y$. The log turns a product into a sum, $\ln W_{\text{total}} = \ln W_x + \ln W_y$, so

$S_{\text{total}} = S_x + S_y$

Entropy is additive, also called extensive: double the system and you double the entropy.

Counting microstates leads to dS = dQ_rev / T

Intuition

Heat is molecules moving up to higher levels. More ways to arrange them means more entropy. So heat going in should raise entropy, and hotter systems should gain less entropy per bit of heat. The algebra gives exactly that.

The steps, as on the slides. Start with the weight and use Stirling’s approximation from the Boltzmann topic:

$W = \frac{N!}{n_0!\,n_1!\,n_2!\cdots}, \qquad \ln W = N\ln N - \sum_i n_i\ln n_i$

With $N$ constant ($dN = 0$), the total differential is

$d\ln W = -\sum_i \ln n_i\,dn_i$

Now use the equilibrium populations $n_i = n_0 e^{-\beta\varepsilon_i}$, so $\ln n_i = \ln n_0 - \beta\varepsilon_i$:

$d\ln W = -\sum_i (\ln n_0 - \beta\varepsilon_i)\,dn_i = -\ln n_0\sum_i dn_i + \beta\sum_i \varepsilon_i\,dn_i$

The first term is struck out on the slide because $\sum_i dn_i = 0$: the total number of molecules does not change. What is left, $\sum_i \varepsilon_i\,dn_i$, is the heat term from the first-law topic, populations moving between levels. The slide writes

$d\ln W = \beta\,dQ_{\text{rev}}$

Multiply by $k$ and use $\beta = 1/kT$:

$dS = \frac{dQ_{\text{rev}}}{T}$

This is the original thermodynamic definition of entropy, now explained by counting microstates. A reversible process is one that is always at equilibrium, which means infinitesimally slow.

Pitfall

Entropy is a state function, so it depends only on the configuration, not the route taken. For a reversible and an irreversible route between the same states 1 and 2, $\Delta S_{\text{rev}} = S_2 - S_1 = \Delta S_{\text{irrev}}$. Only the reversible route is calculated with $\int dQ_{\text{rev}}/T$. Do not plug the real heat of an irreversible route into that integral.

Optional aside

Slide 13 writes $\Delta S = \int dQ_{\text{rev}}/T \ge 0$ for reversible processes. The slide does not explain the inequality, and it sits oddly beside $dS = dQ_{\text{rev}}/T$ as a definition. Treat $dS = dQ_{\text{rev}}/T$ as the lecture’s definition and check the inequality with the lecturer. The same lecture also gives the total entropy from the partition function, $S = U/T + Nk\ln q$, and says there is no need to memorise it. Its point is that derivatives of $q$ carry macroscopic information.

Exam detail

Lecture 15 summary slide: entropy measures the statistical weight of the configuration and always increases in an isolated system. Slide numbers: dog-flea simulations at $R = 20, 80, 200$. Three versions of the second law, as listed above.

Recall

Why does taking the logarithm of W make entropy additive?

Weights of independent systems multiply, $W_{\text{total}} = W_x W_y$. A log turns the product into a sum, so $S = k\ln W$ adds: $S_{\text{total}} = S_x + S_y$.