Legendre transforms connecting them: Worked examples

Prerequisites we lean on: Internal Energy and the First Law, Maxwell Relations, Thermodynamic Stability and Convexity, and the mechanics cousin Lagrangian to Hamiltonian (Legendre in Mechanics). Chemistry payoff lands in Chemical Potential and Phase Equilibrium; the statistical origin of $F$ lives in Partition Functions.

The scenario matrix

Every Legendre problem is one (or a blend) of these cells:

Cell	Case class	What can go wrong / what to watch	Hit by
A	Convex power law $f = \frac{1}{2} a x^{2}$ , $a > 0$	the "textbook" case, sign of $g$	Ex 1
B	Sign trap: slope is negative ( $- P$ slot)	subtract a negative = add	Ex 2
C	Non-quadratic convex $f = e^{x}$	inverting $p = f^{'}$ needs $ln$ , domain $p > 0$	Ex 3
D	Degenerate: linear $f = c x$	slope constant → transform undefined/collapses	Ex 4
E	Limiting / boundary: convexity $a \to 0^{+}$	transform diverges, stability edge	Ex 5
F	Real-world word problem (lab: fixed $T, P$ )	choosing the right potential $G$	Ex 6
G	Full thermo swap on a real gas	recover $P$ , $S$ as slopes	Ex 7
H	Exam twist: double transform (involution)	prove you return home	Ex 8

We now hit each cell.

Example 1 — Cell A: the clean convex parabola

Figure — Legendre transforms connecting them

Example 2 — Cell B: the sign trap (the $- P$ slot)

Worked example Enthalpy from

$U$ : swap $V$ , whose slope is negative Suppose near equilibrium $U (V) = U_{0} + \frac{1}{2} k (V - V_{0})^{2}$ at fixed $S, N$ , with $k > 0$ . Build the transform on $V$ . Forecast: do we get $U - P V$ or $U + P V$ ? Guess first.

Step 1. Read the slope with its sign: $\frac{\partial U}{\partial V} = k (V - V_{0})$ . From the first law $d U = \dots - P d V$ , that slope is $- P$ . So $- P = k (V - V_{0})$ , i.e. $P = - k (V - V_{0})$ . Why this step? The whole trap is forgetting the minus in $- P d V$ . Reading the sign from the differential kills the trap.

Step 2. Legendre subtract = slope × variable = $(- P) V$ . So $H = U - (- P) V = U + P V .$ Why this step? "Subtract slope×variable" is literal; here slope is $- P$ , so we subtract a negative → add $P V$ .

Step 3. Differential check: $d H = d U + P d V + V d P = (T d S - P d V) + P d V + V d P = T d S + V d P$ . Why this step? Confirms the $- P d V$ cancelled and $H$ now rides on $d P$ — pressure became the driver.

Verify (numbers): let $k = 4, V_{0} = 1, U_{0} = 0, V = 3$ . Then $U = \frac{1}{2} (4) (2)^{2} = 8$ , $P = - 4 (2) = - 8$ , so $H = U + P V = 8 + (- 8) (3) = 8 - 24 = - 16$ . And the wrong formula $U - P V = 8 - (- 24) = 32$ — clearly different, showing the sign matters. ✓

Example 3 — Cell C: non-quadratic, inversion needs a log

Example 4 — Cell D: the degenerate straight line

Worked example What happens for

$f (x) = c x$ (a line)? Forecast: guess whether the transform exists at all.

Step 1. Slope: $p = f^{'} (x) = c$ — a constant, independent of $x$ . Why this step? We always start with the slope; here it never changes.

Step 2. Try to invert $p = c$ for $x$ . Impossible: for $p = c$ every $x$ qualifies; for $p \neq = c$ no $x$ does. Why this step? The transform requires a one-to-one slope map. A straight line has one slope everywhere → the map collapses.

Step 3. Interpret. Along $p = c$ : $g = f - p x = c x - c x = 0$ . The "curve" degenerates to a single point $(c, 0)$ — all information about which $x$ is lost. Why this step? Shows the failure geometrically: a line has no curvature, so "tangent lines" don't fan out — they're all the same line, carrying no positional info.

Verify: Second derivative $f^{''} = 0$ . The Legendre transform is well-defined exactly when $f^{''} \neq = 0$ (strict convexity/concavity). Here $f^{''} = 0 \Rightarrow$ degenerate. ✓ This is why stability ( $\partial^{2} U / \partial S^{2} = T / C_{V} > 0$ ) is what guarantees the thermo transforms work.

Example 5 — Cell E: limiting behaviour, convexity vanishing

Worked example Watch

$g (p) = - \frac{p ^{2}}{2 a}$ as $a \to 0^{+}$ Forecast: guess what happens to $g$ at fixed $p$ as the parabola flattens.

Step 1. Fix $p \neq = 0$ , let $a \to 0^{+}$ : $g = - \frac{p ^{2}}{2 a} \to - \infty$ . Why this step? $a$ is the curvature $f^{''}$ . Sending $a \to 0$ approaches the degenerate line of Cell D — the transform must break, and it does so by diverging.

Step 2. Meanwhile the $x$ that produces slope $p$ is $x = p / a \to \infty$ . Why this step? To keep slope $p$ on an ever-flatter parabola you must go ever further out — the tangent point runs to infinity, dragging the intercept to $- \infty$ .

Step 3. Thermodynamic reading: if $C_{V} \to \infty$ (curvature of $U$ in $S$ vanishing), the system loses its ability to set a unique temperature-entropy relation — a phase-transition-like edge. Why this step? Connects the pure-math limit to real physics: diverging response = onset of instability.

Verify: at $a = 0.1, p = 2$ : $g = - \frac{4}{0.2} = - 20$ . At $a = 0.01$ : $g = - \frac{4}{0.02} = - 200$ . Monotonically to $- \infty$ . ✓

Example 6 — Cell F: the lab word problem

Worked example "A reaction runs in an open beaker on the bench. Which potential do I minimize?"

The beaker is open to the atmosphere (pressure $P$ fixed by the room) and sits in the lab (temperature $T$ fixed by the room). Particle number $N$ can change as reactants convert. Forecast: guess the potential before computing.

Step 1. List the controlled variables: $T$ (room) and $P$ (atmosphere). Why this step? The right potential is the one whose natural variables are exactly what you hold fixed — then equilibrium becomes a clean minimum.

Step 2. Match to the fundamental relations. Only $G$ has $d G = - S d T + V d P + μ d N$ , natural in $(T, P, N)$ . Why this step? $U (S, V, N)$ , $H (S, P, N)$ , $F (T, V, N)$ all still carry an uncontrolled natural variable ( $S$ or $V$ ). $G$ alone is native to the bench.

Step 3. At fixed $T, P$ : $d G = μ d N$ . Equilibrium $\Leftrightarrow d G = 0 \Leftrightarrow$ chemical potentials balance. Why this step? This is the whole reason chemists live in $G$ ; see Chemical Potential and Phase Equilibrium.

Verify (units): $[μ d N] = (J/mol) (mol) = J$ , matching $[G] = J$ . And $d G = - S d T + V d P + μ d N$ with $d T = d P = 0$ leaves only $μ d N$ . ✓

Example 7 — Cell G: real gas, recover $P$ and $S$ as slopes

Worked example From

$F (T, V, N)$ of an ideal gas, read off pressure and entropy Take the Helmholtz free energy piece $F = - N k_{B} T ln V + (terms not depending on V)$ at fixed $T, N$ . Forecast: guess whether $P = + \partial F / \partial V$ or $- \partial F / \partial V$ .

Step 1. From $d F = - S d T - P d V + μ d N$ , the $V$ -slope is $(\frac{\partial F}{\partial V})_{T, N} = - P$ . Why this step? Read the sign straight from the differential (Cell B lesson).

Step 2. Compute: $\frac{\partial}{\partial V} (- N k_{B} T ln V) = - \frac{N k _{B} T}{V}$ . Set equal to $- P$ : $- P = - \frac{N k _{B} T}{V} \Rightarrow P = \frac{N k _{B} T}{V} .$ Why this step? The ideal gas law drops out as a slope of a potential — no separate postulate needed.

Step 3. Entropy from the other slot: $S = - (\frac{\partial F}{\partial T})_{V, N}$ . Why this step? Same machine, different variable — a Maxwell/slope pairing (see Maxwell Relations).

Verify (numbers): $N = 2, k_{B} = 1$ (natural units), $T = 300, V = 10$ : $P = \frac{2 \cdot 1 \cdot 300}{10} = 60$ . And $\partial F / \partial V = - \frac{2 \cdot 300}{10} = - 60 = - P$ . ✓ Units of $N k_{B} T / V$ : $\frac{( J/K ) ( K )}{m ^{3}} = J/m^{3} = Pa$ . ✓

Example 8 — Cell H: exam twist, prove the involution

Worked example Transform

$f (x) = \frac{1}{2} a x^{2}$ twice and show you return home Forecast: guess what the second transform gives — the original, or something scaled?

Step 1. First transform (from Ex 1): $g (p) = - \frac{p ^{2}}{2 a}$ , with slope $\frac{d g}{d p} = - \frac{p}{a}$ . Why this step? We need $g$ and its slope as the input to the second transform.

Step 2. New conjugate variable $q = \frac{d g}{d p} = - \frac{p}{a}$ , so $p = - a q$ . Why this step? The involution proof requires treating $g$ as a fresh $f$ and repeating the recipe.

Step 3. Second transform $h (q) = g - q p = - \frac{p ^{2}}{2 a} - q (- a q) = - \frac{( - a q ) ^{2}}{2 a} + a q^{2} = - \frac{a q ^{2}}{2} + a q^{2} = \frac{a q ^{2}}{2}$ . Why this step? This is $\frac{1}{2} a q^{2}$ — identical in form to the original $f$ . The transform is its own inverse.

Verify (numbers): $a = 2, q = 3$ : $h = \frac{1}{2} (2) (9) = 9 = f (3) = \frac{1}{2} (2) (9) = 9$ . ✓ Involution confirmed — this is the mathematical guarantee that no information was ever lost across all four potentials.

Recall Which cell fails, and why?

Only Cell D (linear $f$ , $f^{''} = 0$ ) has no well-defined transform ::: because the slope map $p = f^{'} (x)$ is not invertible — one slope for all $x$ . What guarantees thermo transforms always work ::: stability/convexity, e.g. $\partial^{2} U / \partial S^{2} = T / C_{V} > 0$ .

Worked examples — Legendre transforms connecting them