Thermodynamic Potentials and the Fundamental Relation

Give a chemist the ideal-gas law $PV = Nk_BT$ and ask them for the entropy of the gas, or how much it warms when compressed adiabatically, and they cannot answer — the information simply is not in that equation. Give them instead a single function, the entropy $S(U,V,N)$ , and every equilibrium property falls out by differentiation: temperature, pressure, chemical potential, all the heat capacities, every response coefficient, the location of phase transitions. This astonishing economy is the central promise of thermodynamics, and it rests on one object: the fundamental relation. The fundamental relation exists in two equivalent forms — the energy representation $U(S,V,N)$ and the entropy representation $S(U,V,N)$ — related exactly as $y = f(x)$ relates to $x = f^{-1}(y)$ . Picture $U(S,V,N)$ as a smooth surface floating in the space of extensive coordinates. Every thermodynamic question is a question about the local geometry of that surface: its slopes are the intensive variables ( $T$ , $P$ , $\mu$ ), and its curvatures are the response functions and stability conditions. The first law guarantees $U$ is a state function; the second law guarantees $S$ is a state function maximized at equilibrium; the fundamental relation fuses both laws into one differentiable object. There is a hidden engine behind all this: extensivity. Doubling a system doubles $U$ , $S$ , $V$ , $N$ together, so the fundamental relation is a first-order homogeneous function. That single fact — not decoration — forces the Euler relation $U = TS - PV + \mu N$ , forces the Gibbs–Duhem constraint that the intensive variables are not independent, and forces the relation to be phrased among extensive quantities. An equation of state involving only intensive variables is merely a shadow of the full surface and throws information away. Finally, because the differential of a state function is exact, the second derivatives commute — instantly producing the Maxwell relations and the convexity conditions that govern stability. The fundamental relation is therefore not a formula to memorize but the organizing principle from which the entire structure of equilibrium thermodynamics unfolds.

Key takeaways

A single function — the fundamental relation $U(S,V,N)$ or equivalently $S(U,V,N)$ — contains all equilibrium thermodynamics of a simple system; differentiating it yields $T$ , $P$ , and $\mu$ .
Its differential $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V + \mu\,\mathrm{d}N$ packages the first and second laws into one exact differential.
Because $U$ is first-order homogeneous in its extensive arguments, the relation forces the Euler relation $U = TS - PV + \mu N$ and the Gibbs–Duhem constraint among intensive variables.
A single equation of state (like $PV = Nk_BT$ ) is only a projection of the fundamental relation and is thermodynamically incomplete.
Exactness of the differential makes mixed partials commute, giving the Maxwell relations; curvature gives stability.
Key idea: the fundamental relation $U(S,V,N)$ is a first-order homogeneous surface whose first derivatives are the intensive variables, whose Legendre transforms are the other potentials, and whose second-derivative structure is all response and stability information.

The core insight

A single first-order homogeneous function — the fundamental relation $U(S,V,N)$ or $S(U,V,N)$ — encodes all equilibrium thermodynamics: its first derivatives give the intensive variables $T$ , $P$ , $\mu$ ; homogeneity forces the Euler relation and Gibbs–Duhem constraint; and its second-derivative structure gives the Maxwell relations, response functions, and stability conditions.

Where this lives today

The fundamental-relation viewpoint is the working language of modern equilibrium physics, not a historical curiosity. In computational materials science and chemistry, databases such as CALPHAD store the Gibbs free energy $G(T,P,\{N_i\})$ — a fundamental relation in its natural variables — for thousands of phases; phase diagrams, solubility limits, and reaction equilibria are then computed by minimizing $G$ over competing phases, exactly the extremum principle Gibbs and Callen formalized. The same structure powers equation-of-state engines in process simulators and in planetary-interior models, where a single Helmholtz free energy $F(T,V)$ (as in the widely used IAPWS-95 formulation for water) reproduces every measured property of the substance by differentiation. Conceptually, the fundamental relation is also where thermodynamics hands off to statistical mechanics. The temperature $T = (\partial U/\partial S)^{-1}$ defined by the surface's slope is provably the same $T$ that appears in $1/T = (\partial S/\partial U)_{V,N}$ from Boltzmann's $S = k_B\ln\Omega$ and in the Boltzmann factor $e^{-E/k_BT}$ . Statistical mechanics computes the fundamental relation from a microscopic Hamiltonian; thermodynamics then reads off everything else. That two-level structure — micro-model produces the surface, macro-geometry produces the observables — is the organizing idea of the entire field.

The physical picture

Think of every equilibrium state of a simple system as a single point on a smooth surface $U(S,V,N)$ suspended in the space of its extensive coordinates. To ask "what is the temperature here?" is to ask for a slope of that surface; to ask "is this phase stable?" is to ask about its curvature. This is the whole geometric content of equilibrium thermodynamics: slopes are intensive variables, curvatures are stability. The two representations are the same surface viewed two ways. In the energy representation you fix $S, V, N$ and read off $U$ (the energy-minimum principle is transparent); in the entropy representation you fix $U, V, N$ and read off $S$ (the entropy-maximum principle is transparent). They carry identical information because $U$ is strictly increasing in $S$ , so the map $S \leftrightarrow U$ is invertible.

Figure

The fundamental surface U(S,V) at fixed N: its slopes are T and −P, its curvature is stability

Homogeneity is the hidden engine. Because scaling all extensive variables by $\lambda$ scales $U$ by $\lambda$ , the intensive variables can depend only on ratios — densities. This one geometric fact forces the Euler relation and the Gibbs–Duhem constraint, and it explains a common frustration: why a single equation of state is never enough. The plot below drives the punchline home — one intensive-variable equation of state pins only a single family of slopes; you need a second independent equation of state to reconstruct the full surface, and even then one integration constant (the absolute entropy) stays free until quantum mechanics fixes it.

Figure

One equation of state is a shadow: it fixes one slope family, not the surface

Legendre transforms are the last piece: they are a change of contact coordinates. Instead of labeling a point by where it sits ( $S$ ), you label it by the slope there ( $T$ ). No information is lost because a convex surface is fully reconstructible from its family of tangent planes — which is exactly why the Helmholtz, enthalpy, and Gibbs potentials are as complete as $U$ itself.

The core equations

The fundamental relation in the energy representation is the single exact differential that carries both laws. Read it as: the energy of the surface changes by a thermal piece, a mechanical piece, and a chemical piece as you move.

\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V + \mu\,\mathrm{d}N \qquad (1)

where: $U$ — internal energy (extensive); $S,V,N$ — entropy, volume, particle number (all extensive); $T = (\partial U/\partial S)_{V,N}$ — absolute temperature; $P = -(\partial U/\partial V)_{S,N}$ — pressure; $\mu = (\partial U/\partial N)_{S,V}$ — chemical potential. The three intensive variables are literally the three slopes of the surface. (Track convention: writing the first law as $\mathrm{d}U = \delta Q + \delta W$ with $W$ the work done on the system, the reversible pieces are $\delta Q_{\text{rev}} = T\,\mathrm{d}S$ and $\delta W_{\text{rev}} = -P\,\mathrm{d}V + \mu\,\mathrm{d}N$ .) Solving (1) for $\mathrm{d}S$ gives the equivalent entropy representation, the natural home of the entropy-maximum principle:

\mathrm{d}S = \frac{1}{T}\,\mathrm{d}U + \frac{P}{T}\,\mathrm{d}V - \frac{\mu}{T}\,\mathrm{d}N \qquad (2)

where: $1/T = (\partial S/\partial U)_{V,N}$ , $P/T = (\partial S/\partial V)_{U,N}$ , and $-\mu/T = (\partial S/\partial N)_{U,V}$ are the three entropy slopes. This is the form in which statistical mechanics defines temperature, via $1/T = (\partial S/\partial U)$ . Extensivity is the statement that the fundamental relation is a first-order homogeneous function of its extensive arguments:

U(\lambda S, \lambda V, \lambda N) = \lambda\, U(S,V,N) \qquad (3)

where: $\lambda > 0$ — an arbitrary scaling factor. Doubling the system doubles the energy: no surface tension, no long-range gravity, no boundary effects that would break additivity. Differentiating (3) with respect to $\lambda$ at $\lambda = 1$ (Euler's homogeneous-function theorem) collapses the whole surface into one algebraic identity:

U = TS - PV + \mu N \qquad (4)

where: $TS$ — thermal, $-PV$ — mechanical, $\mu N$ — chemical contributions. This Euler relation is a finite (not differential) statement — a remarkable consequence of extensivity alone. Taking the differential of the Euler relation (4) and subtracting the fundamental relation (1) yields a constraint linking the changes of the three intensive variables:

S\,\mathrm{d}T - V\,\mathrm{d}P + N\,\mathrm{d}\mu = 0 \qquad (5)

where: the three intensive differentials $\mathrm{d}T, \mathrm{d}P, \mathrm{d}\mu$ are not independent. This Gibbs–Duhem equation is why a one-component system has only two independent intensive variables, not three. Because $\mathrm{d}U$ is exact, mixed second derivatives commute, giving relations like

\left(\frac{\partial T}{\partial V}\right)_{S,N} = -\left(\frac{\partial P}{\partial S}\right)_{V,N} \qquad (6)

where: this is the Maxwell relation belonging to $U$ ; the other potentials give three more. Curvature gives stability: $\left(\partial^2 U/\partial S^2\right)_{V,N} = T/C_V \ge 0$ (thermal) and $\left(\partial^2 U/\partial V^2\right)_{S,N} = 1/(V\kappa_S) \ge 0$ (mechanical).

Derivation AEuler relation and the two equations of state from the fundamental surface

The goal here is to prove the two structural facts that make the fundamental relation special, using nothing but homogeneity and differentiation. First: extensivity (first-order homogeneity) forces the Euler relation $U = TS - PV + \mu N$ — a finite algebraic identity, not a differential one — via Euler's theorem on homogeneous functions. Second: a genuine fundamental relation like the Sackur–Tetrode entropy $S(U,V,N)$ regenerates both equations of state ( $PV = Nk_BT$ and $U = \tfrac32 Nk_BT$ ) by differentiation, whereas either equation of state alone cannot recover the entropy. We do both explicitly and check them symbolically.

Step 1. Differentiate the homogeneity identity $U(\lambda S, \lambda V, \lambda N) = \lambda U(S,V,N)$ with respect to $\lambda$ and set $\lambda = 1$ . The left side, by the chain rule, is $S\,U_S + V\,U_V + N\,U_N$ ; the right side is $U$ . Why: this is Euler's theorem — the scaling derivative of a degree-1 function equals the function itself.

Step 2. Insert the definitions $U_S = T$ , $U_V = -P$ , $U_N = \mu$ to obtain $TS - PV + \mu N = U$ , the Euler relation (4). Why: the slopes of the surface are exactly the intensive variables, so the abstract theorem becomes physics.

Step 3. Now take a concrete fundamental relation, the Sackur–Tetrode entropy $S = Nk_B\!\left[\ln\!\big(\tfrac{V}{N}(\tfrac{U}{N})^{3/2}C\big) + \tfrac52\right]$ ( $C$ lumps the constants $m, h$ ). Compute $(\partial S/\partial U)_{V,N} = \tfrac32 Nk_B/U$ , which is $1/T$ , giving immediately $U = \tfrac32 Nk_B T$ — the caloric equation of state. Why: the energy slope of the entropy surface is the inverse temperature, so one differentiation delivers the internal energy.

Step 4. Compute $(\partial S/\partial V)_{U,N} = Nk_B/V$ , which is $P/T$ ; multiplying by $T$ gives $PV = Nk_BT$ , the mechanical equation of state. Thus the single function $S(U,V,N)$ regenerates both equations of state simultaneously — the hallmark of a fundamental relation, which a lone equation of state can never do. Why: the volume slope is $P/T$ , so the same surface that gave $U(T)$ also gives $P(V,T)$ ; only a complete relation carries both. *Verified: homogeneity of the Sackur–Tetrode $U(S,V,N)$ , the Euler identity $S U_S + V U_V + N U_N = U$ , and the two equations of state ( $U = \tfrac32 NkT$ , $PV = NkT$ ) recovered by differentiation are all checked symbolically in verify_aq-sm-thermodynamic-potentials-fundamental-relation.py.*

Go deeper · Derivation B · 1 minMaxwell relation and internal pressure from exactness of dU

Derivation BMaxwell relation and internal pressure from exactness of dU

The goal is to show that the Maxwell relations and the "energy equation" $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ are automatic consequences of $\mathrm{d}U$ being exact — no new physics, just the commutativity of mixed partial derivatives of a state function. We derive one Maxwell relation from $\mathrm{d}U$ directly and then use the entropy representation to extract the internal pressure, the quantity that distinguishes a real gas from an ideal one.

Step 1. From $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ (fixed $N$ ), read off $T = (\partial U/\partial S)_V$ and $-P = (\partial U/\partial V)_S$ . Because $U \in C^2$ , $\partial^2 U/\partial V\,\partial S = \partial^2 U/\partial S\,\partial V$ . Why: mixed partials of a smooth state function are equal (Clairaut/Schwarz).

Step 2. Differentiate the two first derivatives crosswise: $(\partial T/\partial V)_S = -(\partial P/\partial S)_V$ , Equation (6). Why: applying $\partial/\partial V$ to $T=U_S$ and $\partial/\partial S$ to $-P=U_V$ gives the same mixed partial, so the two are equal.

Step 3. For the internal pressure, start from $\mathrm{d}S = (\partial S/\partial T)_V\,\mathrm{d}T + (\partial S/\partial V)_T\,\mathrm{d}V$ and use the Helmholtz Maxwell relation $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ . Then $(\partial U/\partial V)_T = T(\partial S/\partial V)_T - P = T(\partial P/\partial T)_V - P$ . Why: entropy is unmeasurable directly, but $(\partial P/\partial T)_V$ is an ordinary equation-of-state derivative, so exactness converts one into the other.

Step 4. For an ideal gas $P = Nk_BT/V$ gives $(\partial P/\partial T)_V = Nk_B/V = P/T$ , so $(\partial U/\partial V)_T = 0$ — energy depends only on $T$ . For a van der Waals gas $P = Nk_BT/(V-Nb) - aN^2/V^2$ gives $(\partial U/\partial V)_T = aN^2/V^2 > 0$ : attractive interactions store energy in volume. Why: this single Maxwell-relation consequence is the operational difference between an ideal and a real gas, and it follows purely from exactness. *Verified: the mixed-partial Maxwell identity and the van der Waals internal pressure $aN^2/V^2$ are checked symbolically in the verify script; the ideal-gas internal pressure is zero as required.*

Limiting cases

How the fundamental relation specializes in important regimes:

Case	Result	Meaning
Ideal monatomic gas (Sackur–Tetrode)	$S = Nk_B\!\left[\ln\!\big(\tfrac{V}{N}(\tfrac{4\pi m U}{3Nh^2})^{3/2}\big) + \tfrac52\right]$	A genuine fundamental relation: differentiating gives both $PV = Nk_BT$ and $U = \tfrac32 Nk_BT$ .
Single equation of state	$PV = Nk_BT$ alone	A projection: cannot recover $S$ , $U$ , or $\mu$ without a second equation of state.
Two-phase coexistence	$\mathrm{d}P/\mathrm{d}T = \Delta s/\Delta v$ (Clausius–Clapeyron)	Gibbs–Duhem per phase + $\mu_\alpha = \mu_\beta$ ; the two fundamental surfaces intersect.
Third-law boundary	$\lim_{T\to 0} S = S_0$ (independent of $V,N$ )	The surface flattens in entropy as $T\to 0$ ; heat capacities and expansivities vanish.
Broken extensivity (gravity)	$U(\lambda\cdots) \ne \lambda U$	Long-range forces violate homogeneity; Euler and Gibbs–Duhem fail (self-gravitating systems).

Common trap“

PV = Nk_BT

is the fundamental relation of an ideal gas”

It is not. $PV = Nk_BT$ involves only intensive ratios and fixes just one slope family of the entropy surface, $(\partial S/\partial V)_U = P/T = Nk_B/V$ . It cannot give you $S(U,V,N)$ or the energy dependence — you also need $U = \tfrac32 Nk_BT$ , and even then the absolute entropy carries an integration constant that only quantum statistics ( $h$ ) can fix (the Sackur–Tetrode $\tfrac52$ and the $h^3$ inside the log). A fundamental relation is always a relation among extensive variables; any relation among intensive variables alone (like $P = P(T)$ along coexistence) carries strictly less information.

Worked examples

1 · Regenerating both equations of state from one entropy surface (confidence builder)

Take the Sackur–Tetrode entropy $S(U,V,N) = Nk_B[\ln(\tfrac{V}{N}(\tfrac{U}{N})^{3/2}C) + \tfrac52]$ . Show by differentiation that it contains both the mechanical and caloric equations of state, then evaluate the temperature and pressure for $1\ \mathrm{mol}$ of argon with $U = 3740\ \mathrm{J}$ and $V = 22.4\ \mathrm{L}$ .

Step 1. Energy slope: $\tfrac{1}{T} = (\partial S/\partial U)_{V,N} = \tfrac32 Nk_B/U$ . Rearranged, $U = \tfrac32 Nk_BT$ — the caloric equation of state, obtained by one differentiation.

Step 2. Volume slope: $\tfrac{P}{T} = (\partial S/\partial V)_{U,N} = Nk_B/V$ . Multiplying by $T$ : $PV = Nk_BT$ — the mechanical equation of state, from the same surface. A single equation of state could never yield both.

Step 3. Temperature for argon: $T = \tfrac{2U}{3Nk_B} = \tfrac{2(3740)}{3(6.022\times10^{23})(1.381\times10^{-23})} = \tfrac{7480}{24.94} = \mathbf{300\ K}$ . (Here $Nk_B = R = 8.314\ \mathrm{J/K}$ for one mole, so $T = 2U/3R = 7480/24.94$ .)

Step 4. Pressure: $P = Nk_BT/V = (8.314)(300)/(22.4\times10^{-3}) = 2494/0.0224 = \mathbf{1.11\times10^{5}\ Pa}$ ≈ 1 atm — exactly the standard-conditions consistency check. The one entropy surface delivered a self-consistent $(T,P)$ .

2 · Internal pressure: ideal gas versus van der Waals (exam level)

Using the Maxwell relation $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ , compute the internal pressure for (a) an ideal gas and (b) a van der Waals gas $P = \tfrac{Nk_BT}{V-Nb} - \tfrac{aN^2}{V^2}$ . Then estimate the energy stored per mole of CO $_2$ ( $a = 0.364\ \mathrm{Pa\,m^6/mol^2}$ ) when compressed from $V = \infty$ to $V = 1\ \mathrm{L/mol}$ .

Step 1. Ideal gas: $(\partial P/\partial T)_V = Nk_B/V = P/T$ , so $(\partial U/\partial V)_T = T\cdot P/T - P = 0$ . Internal energy depends only on temperature — the classic ideal-gas result, here derived from exactness.

Step 2. Van der Waals: $(\partial P/\partial T)_V = \tfrac{Nk_B}{V-Nb}$ , so $(\partial U/\partial V)_T = \tfrac{Nk_BT}{V-Nb} - P = \tfrac{Nk_BT}{V-Nb} - \big(\tfrac{Nk_BT}{V-Nb} - \tfrac{aN^2}{V^2}\big) = \mathbf{\tfrac{aN^2}{V^2}}$ . Attractive interactions store energy as the gas expands.

Step 3. Energy to pull the molecules apart (per mole, so $N^2 \to 1$ in molar units with molar $a$ ): $\Delta U = \int_{V_0}^{\infty} \tfrac{a}{V^2}\,\mathrm{d}V = \tfrac{a}{V_0}$ . For CO $_2$ at $V_0 = 1\times10^{-3}\ \mathrm{m^3/mol}$ : $\Delta U = 0.364/10^{-3} = \mathbf{364\ J/mol}$ .

Step 4. Sanity: $364\ \mathrm{J/mol}$ is about $0.15\,RT$ at room temperature — a modest but real correction, exactly the scale of intermolecular attraction that makes CO $_2$ cool on throttled expansion (the Joule–Thomson effect). Every number here came from one Maxwell relation, itself a consequence of $\mathrm{d}U$ being exact.

Origins · 1 minthermodynamic potential and how this was discovered

thermodynamic potential

concept · energy/entropy representations

The idea that a whole science can be compressed into a single function of the right variables is due above all to J. Willard Gibbs, whose 1873–78 memoirs On the Equilibrium of Heterogeneous Substances introduced the fundamental relation, the potentials, and the geometric picture of thermodynamic surfaces. A thermodynamic potential is any of the family $U, F, H, G$ (plus the grand potential and Massieu functions) obtained from the fundamental relation by Legendre transformation; each is a complete encoding of the equilibrium thermodynamics, minimized under the constraints matching its natural variables. Callen's twentieth-century axiomatic reformulation made the "fundamental relation + extremum principle" structure the modern backbone of the subject, elevating the potential from a computational convenience to the logical starting point of thermodynamics itself.

History

1755

— Euler proves the homogeneous-function theorem that later forces the Euler relation from extensivity.

1865

— Clausius names entropy and states dU = T dS − P dV for reversible changes.

1869

— Massieu introduces the first "characteristic functions" (Massieu functions), precursors of the free energies.

1873

— Gibbs publishes graphical/geometric thermodynamics, the surface U(S,V) with slopes T and −P.

1875

— Gibbs begins On the Equilibrium of Heterogeneous Substances, defining the fundamental relation, μ, and the potentials.

1960

— Callen's Thermodynamics recasts the subject on the fundamental relation plus the entropy-maximum postulate.

Going further

References. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., chs. 1–5 · Pathria & Beale, Statistical Mechanics, 3rd ed., ch. 1 · Kardar, Statistical Physics of Particles, ch. 1 · Reif, Fundamentals of Statistical and Thermal Physics, ch. 5 · Landau & Lifshitz, Statistical Physics Part 1, ch. 2 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, chs. 3 & 6.

Provenance. All equations are standard textbook thermodynamics. A verifier should: (1) confirm $\mathrm{d}U = T\mathrm{d}S - P\mathrm{d}V + \mu\mathrm{d}N$ and its entropy-representation rearrangement (Callen chs. 2–3); (2) check the Euler relation via Euler's homogeneous-function theorem and Gibbs–Duhem as its differential (Callen ch. 3); (3) verify the Maxwell relation from $\mathrm{d}U$ and the internal-pressure formula $T(\partial P/\partial T)_V - P$ (Reif §5.6); (4) reproduce both ideal-gas equations of state by differentiating the Sackur–Tetrode entropy (Pathria & Beale ch. 1). Numerical instances — argon $T = 2U/3R = 300\ \mathrm{K}$ , $P = 1.11\times10^{5}\ \mathrm{Pa}$ ; van der Waals internal pressure $aN^2/V^2$ ; CO $_2$ $\Delta U = a/V_0 = 364\ \mathrm{J/mol}$ ( $a = 0.364\ \mathrm{Pa\,m^6/mol^2}$ ) — are recomputed in verify_aq-sm-thermodynamic-potentials-fundamental-relation.py. $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ (CODATA).

Builds on: First Law of Thermodynamics and Internal Energy · Second Law of Thermodynamics and Entropy · Work, Heat, and Inexact Differentials · Temperature, Pressure, and Chemical Potential as Entropy Derivatives. Leads to: Legendre Transforms · Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy · Maxwell Relations · Euler Relation and Gibbs–Duhem Equation.

What exactly is the difference between a "fundamental relation" and an "equation of state"?

A fundamental relation is a single function among extensive variables — $U(S,V,N)$ or $S(U,V,N)$ — that contains all equilibrium information: differentiate it and every equation of state, every response function, and every stability condition falls out. An equation of state is a relation among some of the variables — usually intensive ones, like $PV = Nk_BT$ or $U = \tfrac32 Nk_BT$ — that captures only one facet of the surface. You need at least two independent equations of state (plus an integration constant) to reconstruct the fundamental relation. The slogan: the fundamental relation is the whole surface; an equation of state is a single family of slopes.

Why must the fundamental relation be written among extensive variables and not intensive ones?

Because of extensivity and its consequence, the Gibbs–Duhem equation. The intensive variables $T, P, \mu$ are not independent — Gibbs–Duhem $S\mathrm{d}T - V\mathrm{d}P + N\mathrm{d}\mu = 0$ ties them together, so you cannot specify all three freely. A "fundamental relation" phrased purely in intensive variables would therefore be over-determined and could not encode the system's size or absolute scale. The extensive variables $S, V, N$ can all be specified independently, so a function of them is a complete, non-redundant encoding. This is why the natural variables of every thermodynamic potential always include the right mix of extensive quantities.

Where does the Euler relation U = TS − PV + μN actually come from? It looks like it should need a derivation.

It comes from extensivity alone, via Euler's theorem on homogeneous functions. Extensivity says $U(\lambda S, \lambda V, \lambda N) = \lambda U$ — the function is first-order homogeneous. Euler's theorem states that for such a function, $S\,U_S + V\,U_V + N\,U_N = U$ . Since $U_S = T$ , $U_V = -P$ , $U_N = \mu$ , this is exactly $TS - PV + \mu N = U$ . No thermodynamic law beyond additivity is needed — which is why the relation fails for self-gravitating systems, where energy is not extensive. It is a purely mathematical consequence of the surface being homogeneous of degree one.

If the entropy and energy representations are equivalent, why bother with both?

Because different equilibrium principles are transparent in different representations. In the energy representation $U(S,V,N)$ , equilibrium at fixed $S,V,N$ is the minimum of energy — natural for mechanically isolated systems. In the entropy representation $S(U,V,N)$ , equilibrium at fixed $U,V,N$ is the maximum of entropy — natural for thermally isolated systems and the starting point of statistical mechanics. They contain identical information (each is the inverse function of the other), but choosing the representation whose constraints match your physical setup saves enormous effort. The same reason motivates the Legendre transforms to $F$ , $H$ , $G$ : pick the potential whose natural variables you actually control.

Why do the second derivatives of the fundamental relation control stability?

Because equilibrium is not just an extremum but a stable one. Entropy is maximized at equilibrium, so $S(U,V,N)$ must be concave — its second derivatives must have the right sign, or a fluctuation could raise the entropy and the state would spontaneously fall apart. Concavity of $S$ translates to convexity of $U$ : $(\partial^2 U/\partial S^2)_V = T/C_V \ge 0$ forces the heat capacity positive, and $(\partial^2 U/\partial V^2)_S = 1/(V\kappa_S) \ge 0$ forces the compressibility positive. Where these curvature conditions fail — for instance inside the spinodal of a van der Waals fluid — the homogeneous phase is unstable and the system phase-separates. So the first derivatives give you the intensive variables; the second derivatives tell you whether the state can actually exist.

How is the temperature defined here the same as the one in the Boltzmann factor?

The fundamental relation defines $T$ geometrically: $1/T = (\partial S/\partial U)_{V,N}$ , the energy slope of the entropy surface. Statistical mechanics computes that surface from a microscopic model: $S = k_B\ln\Omega(U)$ . Differentiating gives $1/T = k_B\,\partial\ln\Omega/\partial U$ , and the canonical ensemble derived from it produces the Boltzmann factor $e^{-E/k_BT}$ with this same $T$ . So the temperature an engineer reads from the slope of a fundamental relation and the temperature a physicist writes in a partition function are provably the same object — a consistency that ties thermodynamics and statistical mechanics into one theory.

Does the fundamental relation still work when interactions are long-range, like gravity?

No — and this is one of its most instructive failure modes. The entire structure rests on extensivity: $U(\lambda S, \lambda V, \lambda N) = \lambda U$ . Long-range forces such as gravity break this, because the potential energy of a self-gravitating system scales roughly as $N^2$ , not $N$ . When homogeneity fails, so do all its consequences: the Euler relation, the Gibbs–Duhem equation, and even the equivalence of ensembles. Self-gravitating systems can have negative heat capacity and are not described by a convex fundamental relation. This is why star clusters and black holes require a more careful, non-extensive thermodynamics — a reminder that the beautiful economy of the fundamental relation is bought with the assumption of short-range interactions.