Legendre transforms connecting them: Exercises — Physics deep dive

Everything here builds on the parent note. If a symbol feels unfamiliar, that note builds it from zero.

Level 1 — Recognition

These test whether you can read a differential and name the machinery. No heavy algebra.

Exercise 1.1 — Read the natural variables

Problem. You are handed the differential $d H = T d S + V d P + μ d N .$ (a) What are the natural (independent) variables of $H$ ? (b) Write down the three slope formulas that come straight out of this line.

Recall Solution

WHAT we do: in a differential $d Φ = (\dots) d x + (\dots) d y + \dots$ , the variables being differentiated ( $d x$ , $d y$ , …) are the natural variables, and the coefficient in front of each is the corresponding slope.

(a) The differentials appearing are $d S$ , $d P$ , $d N$ . So the natural variables are $H = H (S, P, N) .$ (b) Each coefficient is a partial derivative holding the other natural variables fixed: $T = (\frac{\partial H}{\partial S})_{P, N}, V = (\frac{\partial H}{\partial P})_{S, N}, μ = (\frac{\partial H}{\partial N})_{S, P} .$ Why this works: a differential is just a bookkeeping of "how much $Φ$ moves when each independent variable nudges." The coefficient of $d P$ is by definition the partial with respect to $P$ . (Recall $H = U + P V$ is enthalpy — energy plus the $P V$ room-making term.)

Exercise 1.2 — Spot the conjugate pairs

Problem. From $d U = T d S - P d V + μ d N$ , list the conjugate variable pairs and, for each, state which member is extensive (scales with system size) and which is intensive (independent of size).

Recall Solution

A conjugate pair is (variable being differentiated, its coefficient). Reading off $d U$ (the internal energy):

Pair	Extensive	Intensive
$(S, T)$	$S$	$T$
$(V, - P)$	$V$	$- P$
$(N, μ)$	$N$	$μ$

Why the split matters: you cannot dial an extensive knob directly (there is no "entropy dial"), but you can set an intensive one with a reservoir (a thermostat sets $T$ , a piston/atmosphere sets $P$ ). The Legendre transform exists precisely to swap the awkward extensive member for its convenient intensive partner. See Internal Energy and the First Law.

Exercise 1.3 — Which potential for which lab?

Problem. Match each experimental setup to the potential whose natural variables are the ones you control: (a) sealed rigid box in a thermostat; (b) open beaker on a bench at room temperature; (c) an adiabatic, insulated cylinder sealed by a movable but gas-tight piston, its outside face open to atmosphere.

Recall Solution

Identify the controlled variables, then pick the potential whose natural variables match.

(a) Fixed volume ( $V$ ) + thermostat ( $T$ ) + sealed ( $N$ ) $\Rightarrow$ controls $(T, V, N)$ $\Rightarrow$ Helmholtz $F = U - T S$ , $F (T, V, N)$ (usable work at fixed $T$ ).
(b) Atmosphere sets $P$ , bench sets $T$ , matter conserved $N$ $\Rightarrow$ controls $(T, P, N)$ $\Rightarrow$ Gibbs $G = U - T S + P V$ , $G (T, P, N)$ .
(c) Insulated means no heat exchange, so entropy $S$ is the conserved natural slot; the piston is open to atmosphere, so it transmits pressure and fixes $P$ . Why $N$ is still fixed: a piston is a solid, gas-tight wall — it slides to equalize pressure but no molecules cross it, so the gas count $N$ inside the cylinder cannot change. (Contrast this with a permeable membrane, which would let particles through and instead fix the chemical potential $μ$ — that is the grand-potential setup of Exercise 5.3.) So we control $(S, P, N)$ $\Rightarrow$ Enthalpy $H = U + P V$ , $H (S, P, N)$ .

Why: equilibrium is a minimum of the potential whose natural variables you hold fixed. Choosing the right potential makes the lab conditions the independent variables, so the minimization is clean.

Level 2 — Application

Now you run the machine on concrete functions. In every step, when you see $g = f - p x$ , remember the picture: it is the height where the tangent (slope $p$ ) hits the vertical axis.

Exercise 2.1 — Legendre transform of a cubic

Problem. Let $f (x) = x^{3}$ for $x > 0$ . Compute its Legendre transform $g (p)$ and verify the slope relation $d g / d p = - x$ .

Recall Solution

Step 1 (WHAT & WHY): the slope becomes the new variable, so $p = f^{'} (x) = 3 x^{2} .$ Why this is legal: for $x > 0$ the slope $3 x^{2}$ is strictly increasing, so each $p$ names exactly one $x$ — the transform is well-defined (see the existence note in Exercise 3.2). Step 2 — invert (we need $g$ in terms of $p$ only; here $x > 0$ so we take the positive root): $x = (\frac{p}{3})^{1/2} .$ Step 3 — build the transform $g = f - p x$ (geometrically: drop the tangent line down to the axis): $g (p) = x^{3} - p x = x^{3} - 3 x^{2} \cdot x = - 2 x^{3} = - 2 (\frac{p}{3})^{3/2} .$ Step 4 — check the defining slope relation: $\frac{d g}{d p} = - 2 \cdot \frac{3}{2} (\frac{p}{3})^{1/2} \cdot \frac{1}{3} = - (\frac{p}{3})^{1/2} = - x . ✓$ The check confirms no information was lost — $g$ knows $x$ through its own slope.

Exercise 2.2 — Build $F$ and extract its slopes

Problem. A system has internal energy given implicitly through $d U = T d S - P d V$ (fixed $N$ ). Its Helmholtz free energy ( $F = U - T S$ , the usable work at fixed temperature) turns out to be $F (T, V) = - a T^{2} - N k_{B} T ln V,$ with $a > 0$ constant. Find the pressure $P (T, V)$ and the entropy $S (T, V)$ .

Recall Solution

First, WHY does $d F = - S d T - P d V$ ? We derive it, so nothing is assumed. Start from the definition $F = U - T S$ and take the differential of each side, using the product rule on the $T S$ term: $d F = d U - d (T S) = d U - (T d S + S d T) .$ Now substitute the first law $d U = T d S - P d V$ : $d F = (T d S - P d V) - T d S - S d T = - S d T - P d V .$ The two $T d S$ terms cancel — that cancellation is the whole point of the Legendre subtraction $- T S$ : it kills the awkward $d S$ and installs $d T$ as the driver. Which slopes? From $d F = - S d T - P d V$ we read off the two rules: $S = - (\frac{\partial F}{\partial T})_{V}, P = - (\frac{\partial F}{\partial V})_{T} .$ Entropy (differentiate in $T$ , holding $V$ ): $\frac{\partial F}{\partial T} = - 2 a T - N k_{B} ln V \Rightarrow S = 2 a T + N k_{B} ln V .$ Pressure (differentiate in $V$ , holding $T$ ): $\frac{\partial F}{\partial V} = - \frac{N k _{B} T}{V} \Rightarrow P = \frac{N k _{B} T}{V} .$ The ideal-gas law $P V = N k_{B} T$ appears — recovered as a slope of $F$ , exactly as the parent note promised. See Maxwell Relations for how these two derivatives are secretly linked.

Exercise 2.3 — Enthalpy sign, done right

Problem. Starting from $d U = T d S - P d V$ , construct the potential whose natural variable in the second slot is $P$ instead of $V$ . Show the sign carefully.

Recall Solution

Step 1 — read the slope's sign (this is the whole game): the coefficient of $d V$ in $d U$ is $- P$ , so $(\frac{\partial U}{\partial V})_{S} = - P .$ Step 2 — Legendre subtract (slope $\times$ variable): the transform subtracts $slope \times var = (- P) V$ (again: this is the tangent-line intercept, now in the $V$ direction): $H = U - (- P) V = U + P V .$ Step 3 — check the differential: $d H = d U + P d V + V d P = (T d S - P d V) + P d V + V d P = T d S + V d P .$ The $P d V$ terms cancel, leaving $d P$ as the new driver. Natural variables: $H = H (S, P)$ . ✓ ( $H = U + P V$ is the enthalpy.)

Level 3 — Analysis

Take the structure apart and see why it holds.

Exercise 3.1 — Derive a Maxwell relation from $G$

Problem. From $d G = - S d T + V d P$ derive the Maxwell relation connecting $(\partial S / \partial P)_{T}$ to a volume derivative. State which quantity is measurable and which is not.

Recall Solution

Key fact: $d G$ is an exact differential, so its mixed second partials are equal regardless of the order of differentiation: $\frac{\partial}{\partial P} (\frac{\partial G}{\partial T})_{P} = \frac{\partial}{\partial T} (\frac{\partial G}{\partial P})_{T} .$ Now substitute the first-order slopes read from $d G$ ( $G = U - T S + P V$ is the Gibbs energy): $(\frac{\partial G}{\partial T})_{P} = - S, (\frac{\partial G}{\partial P})_{T} = V .$ Differentiating each: $(\frac{\partial ( - S )}{\partial P})_{T} = (\frac{\partial V}{\partial T})_{P} \Rightarrow - (\frac{\partial S}{\partial P})_{T} = (\frac{\partial V}{\partial T})_{P} .$ Measurability: $(\partial V / \partial T)_{P}$ is directly measurable (thermal expansion — heat a fluid at constant pressure and watch it expand). $(\partial S / \partial P)_{T}$ is not directly measurable (no entropy meter). Maxwell trades the unmeasurable for the measurable — for free. See Maxwell Relations.

Exercise 3.2 — Convexity flips to concavity (and why the transform exists at all)

Problem. The transform of the convex parabola $f (x) = \frac{1}{2} a x^{2}$ (with $a > 0$ ) is $g (p) = - \frac{p ^{2}}{2 a}$ , which is concave. (a) Confirm the concavity by computing $g^{''} (p)$ . (b) Explain geometrically why a Legendre transform of a convex function is concave. (c) State the condition on $f$ that guarantees the transform even exists.

Recall Solution

(a) $g (p) = - \frac{p ^{2}}{2 a}$ , so $g^{'} (p) = - \frac{p}{a}, g^{''} (p) = - \frac{1}{a} < 0.$ Since $g^{''} < 0$ , $g$ is concave. Compare $f^{''} (x) = a > 0$ (convex). In fact $g^{''} (p) \cdot f^{''} (x) = (- 1/ a) (a) = - 1$ : the curvatures are reciprocal-and-negated. (b) Geometric reason: the transform variable is the slope $p = f^{'} (x)$ . For a convex $f$ , as $x$ increases the slope $p$ increases (that's what convex means). The transform's own slope is $d g / d p = - x$ , which decreases as $p$ (hence $x$ ) increases. A quantity whose slope decreases is concave. So convex $\leftrightarrow$ concave under the transform. This is why $U$ (convex in $S$ ) becomes $F$ (concave in $T$ ) — see Thermodynamic Stability and Convexity. (c) Existence condition (the point often skipped). To build $g (p)$ we must invert $p = f^{'} (x)$ to get a single $x$ for each $p$ . That inversion is guaranteed exactly when $f^{'}$ is strictly monotonic, i.e. when $f$ is strictly convex ( $f^{''} > 0$ ) or strictly concave ( $f^{''} < 0$ ) — never flat, never wiggling. If $f^{'}$ turned around (some $p$ hit by two different $x$ 's), "which $x$ ?" would be ambiguous and $g (p)$ ill-defined. This is why thermodynamics insists potentials be convex/concave: stability ( $C_{V} > 0$ , etc.) is exactly the condition that makes every Legendre transform between $U, H, F, G$ well-defined. The figure below shows the two curvatures side by side.

Figure (above). Blue is the convex parent $f (x) = \frac{1}{2} a x^{2}$ ("curves up", $f^{''} > 0$ ); red is its Legendre transform $g (p) = - \frac{p ^{2}}{2 a}$ ("curves down", $g^{''} < 0$ ). Read it as the visual statement of parts (a)–(b): the transform reflects a bowl into a dome. The reason it can be drawn at all is part (c) — the convex bowl has a strictly increasing slope, so every slope value $p$ names one unique point, and the red dome is unambiguous.

Exercise 3.3 — The involution, one clean convention

Problem. Show that transforming $g (p) = - \frac{p ^{2}}{2 a}$ a second time returns the original $f (x) = \frac{1}{2} a x^{2}$ . Use one convention throughout: the Legendre transform of a function $ϕ (u)$ is $ψ (w) = ϕ - w u$ with $w \equiv ϕ^{'} (u)$ .

Recall Solution

We apply the same rule we always use — no new sign convention — treating $g (p)$ as the function to be transformed. Its own slope variable, call it $w$ , is $w \equiv g^{'} (p)$ .

Step 1 — new slope. $w = \frac{d g}{d p} = - \frac{p}{a} .$ Step 2 — invert (solve for $p$ in terms of $w$ ): $p = - a w .$ Step 3 — apply the transform $ψ (w) = g (p) - w p$ : $ψ (w) = - \frac{p ^{2}}{2 a} - w p .$ Substitute $p = - a w$ term by term: $- \frac{p ^{2}}{2 a} = - \frac{( - a w ) ^{2}}{2 a} = - \frac{a w ^{2}}{2}, - w p = - w (- a w) = + a w^{2} .$ Step 4 — add the two pieces: $ψ (w) = - \frac{a w ^{2}}{2} + a w^{2} = + \frac{a w ^{2}}{2} .$ Step 5 — read the result and match variables. We have $ψ (w) = \frac{1}{2} a w^{2}$ , which is exactly the original $f$ with $w$ in the role of $x$ . To confirm the variable really is $x$ : the first transform obeyed $d g / d p = - x$ , and here $w = d g / d p$ , so $w = - x$ . Substituting $w = - x$ into $ψ$ : $ψ = \frac{1}{2} a (- x)^{2} = \frac{1}{2} a x^{2} = f (x) . ✓$ Takeaway: with a single consistent rule, the double transform lands back on $f$ ; the only subtlety is that the intermediate slope variable equals $- x$ , which is precisely what the relation $d g / d p = - x$ has been tracking all along. Two applications = identity: the Legendre transform is an involution.

Level 4 — Synthesis

Combine transform + physics into a single argument.

Exercise 4.1 — Gibbs energy of an ideal gas from $F$

Problem. For a gas with $F (T, V, N)$ satisfying $d F = - S d T - P d V$ and $P = N k_{B} T / V$ , construct $G (T, P, N)$ by Legendre transforming the $V$ -slot, and confirm $G$ depends on $P$ (not $V$ ).

Recall Solution

Recall $F = U - T S$ (Helmholtz) and we are heading to $G = U - T S + P V$ (Gibbs). Step 1 — which slope to swap. We want to trade $V$ for its conjugate. The slope of $F$ in $V$ is $(\frac{\partial F}{\partial V})_{T} = - P .$ Step 2 — Legendre transform on the $V$ -slot (subtract slope $\times$ var $= (- P) V$ , i.e. the tangent intercept in $V$ ): $G = F - (- P) V = F + P V .$ Step 3 — check the differential: $d G = d F + P d V + V d P = (- S d T - P d V) + P d V + V d P = - S d T + V d P .$ The $d V$ cancels; the driver is now $d P$ . So $G = G (T, P, N)$ . ✓ Step 4 — sanity with the ideal gas: $P V = N k_{B} T$ , so the added term $P V = N k_{B} T$ is finite and well-defined; substituting $V = N k_{B} T / P$ removes every trace of $V$ from $G$ . The transform did exactly its job: convert $F (T, V)$ into $G (T, P)$ losslessly.

Exercise 4.2 — Equilibrium criterion at fixed $T, P$

Problem. A beaker holds a reaction at fixed $T$ and $P$ . Using $d G = - S d T + V d P + \sum_{i} μ_{i} d N_{i}$ (the Gibbs differential with several species $i$ , each with chemical potential $μ_{i}$ ), explain why equilibrium is $\sum_{i} μ_{i} d N_{i} = 0$ , and why no other potential gives this as cleanly.

Recall Solution

Recall $G = U - T S + P V$ is the Gibbs energy, and $μ_{i}$ is the energy cost of adding one particle of species $i$ . Step 1 — impose the fixed conditions. Fixed $T$ and $P$ mean $d T = 0$ and $d P = 0$ . Then $d G = - S 0 d T + V 0 d P + \sum_{i} μ_{i} d N_{i} = \sum_{i} μ_{i} d N_{i} .$ Step 2 — spontaneity and equilibrium. The second law, specialized to fixed $(T, P)$ , says a spontaneous change can only lower $G$ , i.e. $d G \leq 0$ . The system slides downhill in $G$ until it can go no lower — that resting point is the minimum, where every allowed change gives $d G = 0$ . Combining with Step 1: $i \sum μ_{i} d N_{i} = 0.$ In words: at equilibrium the chemical potentials are balanced so that shuffling particles among species (allowed $d N_{i}$ 's tied by the reaction's stoichiometry) changes $G$ not at all. Step 3 — why $G$ specifically. Only for $G$ are the controlled lab variables $(T, P)$ the natural variables. For $U$ , $H$ , or $F$ , at least one held-fixed lab variable is not natural, so the differential does not collapse to the clean $\sum_{i} μ_{i} d N_{i}$ form — you'd carry leftover $S d T$ or $P d V$ terms that are not zero under these lab conditions, muddying the criterion. That is precisely why the constant- $T, P$ chemist reaches for $G$ . See Chemical Potential and Phase Equilibrium.

Level 5 — Mastery

Prove or create.

Exercise 5.1 — Legendre in mechanics matches thermodynamics

Problem. In mechanics the Hamiltonian is $H = p \overset{q}{˙} - L$ , where $L (\overset{q}{˙})$ is the Lagrangian and $p = \partial L / \partial \overset{q}{˙}$ . This is a Legendre transform swapping velocity $\overset{q}{˙}$ for momentum $p$ . (a) Show it has the same structure as $F = U - T S$ . (b) Verify the analogue of the slope relation.

Recall Solution

(a) Match the structures. The thermodynamic transform swapping $S \to T$ is $F = U - T S, T = \frac{\partial U}{\partial S} .$ The mechanical transform is $H = p \overset{q}{˙} - L = - (L - p \overset{q}{˙}), p = \frac{\partial L}{\partial q ˙} .$ Identify $L \leftrightarrow U$ , $\overset{q}{˙} \leftrightarrow S$ (the variable being traded away), $p \leftrightarrow T$ (its conjugate slope). Apart from an overall sign convention (mechanics writes $p \overset{q}{˙} - L$ ; thermo writes $U - T S$ ), it is the same operation: subtract slope $\times$ variable to change independent variable from $\overset{q}{˙}$ to $p$ . See Lagrangian to Hamiltonian (Legendre in Mechanics). (b) Slope relation. Differentiate $H = p \overset{q}{˙} - L$ : $d H = \overset{q}{˙} d p + p d \overset{q}{˙} - \frac{\partial L}{\partial q ˙} d \overset{q}{˙} = \overset{q}{˙} d p + = 0 (p - \frac{\partial L}{\partial q ˙}) d \overset{q}{˙} = \overset{q}{˙} d p .$ So $(\frac{\partial H}{\partial p}) = \overset{q}{˙}$ — the exact analogue of $d g / d p = - x$ (here the sign convention gives $+ \overset{q}{˙}$ ). Just as in thermodynamics, the awkward variable ( $\overset{q}{˙}$ ) is eliminated and its conjugate ( $p$ ) becomes the driver.

Exercise 5.2 — Non-convex $f$ breaks the transform

Problem. Consider $f (x) = - \frac{1}{2} a x^{2}$ with $a > 0$ over $x \in R$ . (a) Compute $p = f^{'} (x)$ and show that the naive substitution "eliminate $x$ " produces $g (p) = + \frac{p ^{2}}{2 a}$ . (b) Explain what physical pathology (in thermodynamic language) a concave-where-it-should-be-convex function corresponds to, and why real stable systems avoid it.

Recall Solution

(a) $p = f^{'} (x) = - a x$ , invert $x = - p / a$ . Then $g (p) = f - p x = - \frac{1}{2} a (\frac{p}{a})^{2} - p (- \frac{p}{a}) = - \frac{p ^{2}}{2 a} + \frac{p ^{2}}{a} = + \frac{p ^{2}}{2 a} .$ The algebra runs, and the inverse $x (p) = - p / a$ is single-valued here because $f^{'} = - a x$ is strictly monotonic (this $f$ is strictly concave). The danger appears only when $f^{'}$ is non-monotonic (a bump): then some $p$ is hit by two different $x$ 's and the transform is ambiguous — exactly the existence condition from Exercise 3.2(c). (b) Physical meaning. If $U (S)$ were concave, then $\partial^{2} U / \partial S^{2} = T / C_{V} < 0$ would force negative heat capacity $C_{V} < 0$ — a system that gets colder as you add heat. That's a runaway instability: any hot spot grows hotter, any cold spot colder, until the phase separates. Real stable equilibria require $U$ convex in $S$ (and in $V$ ), which is precisely why the Legendre transform is always well-defined for them. See Thermodynamic Stability and Convexity.

Figure (above). Green is a stable $U (S)$ : it curves up (convex), so its slope $T = \partial U / \partial S$ steadily increases and $C_{V} > 0$ . Red-dashed is an unstable concave stretch, where adding entropy would lower the slope — negative heat capacity, the runaway of part (b). The yellow line is one tangent, whose slope is the temperature at that point: convexity is exactly "every such tangent lies below the curve," which is what keeps the Legendre transform single-valued.

Exercise 5.3 — Design a potential

Problem. Invent a "grand potential" $Ω$ for a system that exchanges particles with a reservoir (so $μ$ is controlled, not $N$ ). Starting from $F (T, V, N)$ with $d F = - S d T - P d V + μ d N$ , Legendre-transform the $N$ -slot and give $Ω$ , its differential, and its natural variables.

Recall Solution

Recall $F = U - T S$ is Helmholtz and $μ$ is the chemical potential (energy to add one particle). Step 1 — the slope to swap. We want $μ$ (intensive, reservoir-controlled) as the independent variable instead of $N$ . The slope of $F$ in $N$ is $(\frac{\partial F}{\partial N})_{T, V} = μ .$ Step 2 — transform on the $N$ -slot (subtract slope $\times$ var $= μ N$ ): $Ω = F - μ N .$ Step 3 — differential: $d Ω = d F - μ d N - N d μ = (- S d T - P d V + μ d N) - μ d N - N d μ = - S d T - P d V - N d μ .$ Step 4 — natural variables: the drivers are $d T$ , $d V$ , $d μ$ , so $Ω = Ω (T, V, μ),$ the grand potential — the natural object for the grand canonical ensemble. This is exactly the potential that pairs with the grand partition function via $Ω = - k_{B} T ln Z$ .

Quick self-check

Recall Which potential is minimized at fixed

$(T, P, N)$ ? The Gibbs free energy $G = U - T S + P V$ , because $(T, P, N)$ are its natural variables — see Exercise 4.2.

Recall Sign of the Legendre term for enthalpy vs grand potential?

Enthalp