Visual walkthrough — Free energy from partition function

We only assume you can add, multiply, and read a bar chart. Everything else — Boltzmann weights, the partition function, entropy, the logarithm trick — is built in front of you.

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Step 1 — A system is a list of energy levels

WHAT. Picture a tiny system. It cannot hold any energy it likes — it can only sit in one of a fixed set of microstates, each with its own energy. Label them $i=1,2,3,\dots$ and call their energies $E_1,E_2,E_3,\dots$ We draw them as rungs on a ladder: low rung = low energy, high rung = high energy.

WHY. Before we can count or weight anything, we need the thing being counted. The ladder of levels is the raw material of all of statistical mechanics.

PICTURE. Look at the ladder below. The height of each rung is its energy $E_i$. Nothing has been weighted yet — this is just the menu.

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Step 2 — Cold systems prefer low rungs: the Boltzmann weight

WHAT. The system touches a large heat bath at temperature $T$. The bath lets the system borrow or dump energy. The chance the system sits on rung $i$ is proportional to the Boltzmann weight $w_i=e^{-\beta E_i},\beta \equiv \frac{1}{k_{B}T}.$ Term by term: $e$ is the exponential (it turns adding in the exponent into multiplying — we'll cash that in later); $E_i$ is the rung's energy; $\beta =1/k_{B}T$ is "coldness" (big $\beta$ = cold, small $\beta$ = hot); $k_B$ is Boltzmann's constant, the tiny number that converts temperature into energy.

WHY use $e^{-\beta E}$ and not, say, $1/E$? Because we want a rule where energies add but weights multiply. If two independent pieces have energies $E_a$ and $E_b$, the joint energy is $E_a+E_b$, and $e^{-\beta (E_a+E_b)}=e^{-\beta E_a}{e}^{-\beta E_b}$ splits cleanly into a product. Only the exponential does this. That single property drives the whole derivation.

PICTURE. The weight $e^{-\beta E_i}$ is a falling curve: tall for low rungs, shrinking as you climb. Cold (large $\beta$) = steep fall, only the bottom rungs matter. Hot (small $\beta$) = gentle fall, all rungs roughly equal.

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Step 3 — Add up the weights: that sum IS $Z$

WHAT. Add every rung's weight into one number, the partition function: $Z={\sum }_i{e}^{-\beta E_i}.$ Here ${\sum }_i$ means "sum over all rungs." $Z$ is a single number that depends on $\beta$ (and on $V,N$ through the energies).

WHY. Two reasons. (1) To turn weights into honest probabilities we must divide by their total — that total is $Z$. (2) $Z$ secretly stores everything: once we have $Z$ we never touch individual rungs again. It is the "master number" the parent note promised.

PICTURE. Stack the weight-bars end to end. The full length of the stack is $Z$. The fraction of the stack owned by rung $i$ is its probability $p_i=\frac{e^{-\beta E_i}}{Z},{\sum }_i{p}_i=1.$ $p_i$ is "what share of the stack is rung $i$" — and shares always add to $1$.

This connects to Boltzmann distribution and microstates, where the $p_i$ shape is derived by maximising entropy.

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Step 4 — Average energy $U$ is the slope of $\ln Z$

WHAT. The internal energy $U$ is the average rung height, each rung weighted by its share: $U=⟨E⟩={\sum }_i{p}_i{E}_i=\frac{1}{Z}{\sum }_i{E}_i{e}^{-\beta E_i}.$

Now a small piece of calculus. A derivative $\partial /\partial \beta$ answers: "if I nudge the coldness $\beta$ a hair, how fast does this quantity change?" We use it because differentiating the exponential pulls the energy down in front: $\frac{\partial }{\partial \beta }{e}^{-\beta E_i}=-E_i{e}^{-\beta E_i}.$ That is exactly the $E_i$ we needed! Summing over $i$: ${\sum }_i{E}_i{e}^{-\beta E_i}=-\frac{\partial Z}{\partial \beta }.$ Divide by $Z$ and recognise $\frac{1}{Z}\frac{\partial Z}{\partial \beta }=\frac{\partial \ln Z}{\partial \beta }$ (the chain rule: the slope of $\ln Z$ is $\frac{1}{Z}$ times the slope of $Z$): $\boxed{U=-\frac{\partial \ln Z}{\partial \beta }}$

WHY. We replaced a messy weighted sum over all rungs with one derivative of one function. This is the entire payoff of statistical mechanics: differentiate $\ln Z$, never re-count microstates.

PICTURE. Plot $\ln Z$ against $\beta$. It slopes downward (more coldness $\to$ fewer accessible rungs $\to$ smaller $Z$). The steepness of that downhill slope, made positive by the minus sign, is $U$.

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Step 5 — Entropy $S$ is "how spread out" the shares are

WHAT. Entropy measures how spread across many rungs the probability is. The Gibbs definition: $S=-k_B{\sum }_i{p}_i\ln p_i.$ Term by term: $p_i$ is the share of rung $i$; $\ln p_i$ is a large negative number for a tiny share; the overall minus and $k_B$ make $S$ a positive energy-per-temperature.

Substitute what we already know: from Step 3, $p_i=e^{-\beta E_i}/Z$, so $\ln p_i=-\beta E_i-\ln Z.$ Plug in: $S=-k_B{\sum }_i{p}_i(-\beta E_i-\ln Z)=k_B\beta \underset{=U}{\underbrace{\sum _i{p}_i{E}_i}}+k_B\ln Z\underset{=1}{\underbrace{\sum _i{p}_i}}.$ The first bracket is $U$ (Step 4), the second is $1$ (shares sum to one). Using $k_B\beta =1/T$: $\boxed{S=\frac{U}{T}+k_B\ln Z}$

WHY. We deliberately used $\ln p_i$ so that the $\ln Z$ term would break loose — it is that loose $k_B\ln Z$ piece that will become the free energy in the next step.

PICTURE. One share is drawn as tall spike (ordered, low $S$); many equal shares as a flat plateau (spread out, high $S$). Below each we show the size of $-\sum p_i\ln p_i$.

See Entropy — Gibbs and Boltzmann definitions for why $-k_B\sum p_i\ln p_i$ collapses to $k_B\ln \Omega$ when all shares are equal.

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Step 6 — Assemble $F=U-TS$ and watch $U$ vanish

WHAT. The Helmholtz free energy is defined as $F=U-TS$ (its natural variables are $T,V,N$; see Helmholtz vs Gibbs free energy). Substitute the Step 5 result for $S$: $F=U-T\left(\frac{U}{T}+k_B\ln Z\right)=U-U-k_{B}T\ln Z.$ The $U$ terms are equal and opposite — they annihilate: $\boxed{F=-k_{B}T\ln Z}$

WHY it must cancel. $F$ is supposed to depend on $T,V,N$ through $Z$ alone; the leftover average energy $U$ has no business being there. The cancellation is the theory telling us the bookkeeping is consistent.

PICTURE. Two stacked bars, $+U$ from internal energy and $-U$ hidden inside $-TS$, cancel to zero; the surviving strip is the lavender $-k_{B}T\ln Z$.

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Step 7 — Why a logarithm: joining systems

WHAT. Take two independent systems, $a$ and $b$. A joint microstate is "$a$ on rung $\alpha$ AND $b$ on rung $\gamma$", with energy $E_{\alpha }+E_{\gamma }$. Then $Z_{ab}={\sum }_{\alpha ,\gamma }{e}^{-\beta (E_{\alpha }+E_{\gamma })}=\left({\sum }_{\alpha }{e}^{-\beta E_{\alpha }}\right)\left({\sum }_{\gamma }{e}^{-\beta E_{\gamma }}\right)=Z_a{Z}_b.$ Partition functions multiply. But real thermodynamic quantities like energy and free energy must be extensive — join two identical tanks of gas and $F$ should double, not square. The cure is the logarithm: $F_{ab}=-k_{B}T\ln (Z_a{Z}_b)=\underset{F_a}{\underbrace{-k_{B}T\ln Z_a}}+\underset{F_b}{\underbrace{-k_{B}T\ln Z_b}}.$ $\ln$ turns the product into a sum, so $F$ adds. That is the entire reason $F$ is $-k_{B}T\ln Z$ and not $-k_{B}TZ$.

WHY here. This is the missing "why the log" from the boxed formula — the property that makes $F$, not $Z$, the natural thermodynamic potential.

PICTURE. Left: two boxes side by side, weights multiplying ($Z_a{Z}_b$). Right: after $\ln$, the two contributions lie end to end, adding ($F_a+F_b$).

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Edge & limiting cases — the two-level system checks every corner

WHAT. Test the machinery on the simplest system: two rungs at $0$ and $\epsilon$. $Z=e^0+e^{-\beta \epsilon }=1+e^{-\beta \epsilon },U=-\frac{\partial \ln Z}{\partial \beta }=\frac{\epsilon }{e^{\beta \epsilon }+1}.$

WHY. A formula you trust must survive the extremes. Two limits:

• Cold, $T\to 0$ ($\beta \to \infty$): $e^{\beta \epsilon }\to \infty$, so $U\to 0$. The system freezes into the ground rung. ✓
• Hot, $T\to \infty$ ($\beta \to 0$): $e^{\beta \epsilon }\to 1$, so $U\to \epsilon /2$ — both rungs equally likely, average height is the midpoint. ✓
• Degenerate $\epsilon \to 0$: the two rungs merge, $Z\to 2$, $F\to -k_{B}T\ln 2$ — pure entropy of a two-way coin flip, zero energy. ✓

PICTURE. The energy curve $U(T)$ rises smoothly from $0$ to the plateau $\epsilon /2$; the two dashed asymptotes are the cold and hot limits.

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The one-picture summary

WHAT. The whole derivation compressed into a single flow: ladder of levels $\to$ Boltzmann weights $\to$ sum them into $Z$ $\to$ from $\ln Z$ pull out $U$ (slope) and $S$ (spread) $\to$ assemble $F=U-TS$ $\to$ $U$ cancels $\to$ $F=-k_{B}T\ln Z$. The logarithm on the side note explains extensivity.

Recall Feynman retelling of the whole walkthrough

Picture a ladder where each rung is an energy the system can have. A warm bath outside makes low rungs comfy and high rungs unlikely — the "comfiness" of rung $i$ is the falling number $e^{-\beta E_i}$. Add up all the comfinesses and you get one grand total, $Z$. Divide any rung's comfiness by that total and you get its honest probability — its share of the ladder. Two things live inside $Z$. First, if you ask "how fast does $\ln Z$ drop as I cool the system?", the answer is the average energy $U$ — no need to re-add rung by rung. Second, "how spread across many rungs are the shares?" is the entropy $S$, and when you work it out you find $S=U/T+k_B\ln Z$. Now build the free energy $F=U-TS$: the $U$ from the energy and the $U$ hiding inside $TS$ are equal and opposite, so they wipe each other out, and all that survives is $F=-k_{B}T\ln Z$. Why the log? Glue two systems together and their comfiness-totals multiply; but the free energy of two tanks should add. The logarithm is the machine that turns multiplying into adding — that's the whole reason nature hands us $\ln Z$ instead of $Z$.

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Active recall

• What single number stores all thermodynamics once you have the energy ladder? ::: The partition function $Z={\sum }_i{e}^{-\beta E_i}$.
• How do you get $U$ from $Z$ without summing rung by rung? ::: $U=-\partial (\ln Z)/\partial \beta$ (the downhill slope of $\ln Z$ vs $\beta$).
• In $F=U-TS$, what happens to $U$? ::: It cancels exactly against the $U/T$ piece inside $S$, leaving $F=-k_{B}T\ln Z$.
• Why a logarithm and not just $Z$? ::: Independent systems multiply $Z$; $\ln$ turns the product into a sum so $F$ is extensive (additive).
• Two-level system as $T\to \infty$: what is $U$? ::: $\epsilon /2$ — both rungs equally likely, average is the midpoint.
• Two-level system as $T\to 0$: what is $U$? ::: $0$ — the system freezes into the ground rung.