LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Local Gibbs, Euler, parcel first law and response identities

Statement

Let $X \subset (0, \infty) \times R$ be nonempty and open, let $e \in C^{2} (X, R)$ , and define $Y : = {(n, s) \in (0, \infty) \times R : (1 / n, s / n) \in X}, ε (n, s) : = n e (1 / n, s / n) .$ For $(n, s) \in Y$ , put $v = 1 / n$ , $σ_{p} = s / n$ , $T = e_{σ_{p}}$ , $p = - e_{v}$ , $H = e + p v$ , $h = ε + p = n H$ , and $μ = H - T σ_{p}$ , as in Rest fluid state, frame and geometric current definitions. Then the following identities hold as equalities of differentials on $Y$ : $d ε = T d s + μ d n, ε + p = T s + μ n, d p = s d T + n d μ .$

For parcel totals $S \in R$ (entropy), $V > 0$ (proper volume), and $N > 0$ (marker count) with $(V / N, S / N) \in X$ , define $E (S, V, N) : = N e (V / N, S / N) .$ Then $E$ is $C^{2}$ , is positively homogeneous of degree one, and satisfies $d E = T d S - p d V + μ d N, E = T S - p V + μ N .$ If $E$ is viewed on its open domain in $(S, V, N)$ , its Hessian annihilates the radial vector $(S, V, N)$ .

The local response statements are conditional. At a state with $T > 0$ and $e_{σ_{p} σ_{p}} > 0$ , temperature is a valid local coordinate at fixed $v$ , and the per-marker heat capacity $c_{v} : = {(\frac{\partial e}{\partial T})}_{v} = \frac{T}{e_{σ_{p} σ_{p}}}$ is positive. At a state with $e_{v v} > 0$ , the isentropic compressibility $κ_{σ_{p}} : = - \frac{1}{v} {(\frac{\partial v}{\partial p})}_{σ_{p}} = \frac{1}{v e_{v v}}$ is positive. At a state with $e_{v v} > 0$ and $h > 0$ , the local pressure-energy response $α_{σ_{p}} : = {(\frac{\partial p}{\partial ε})}_{σ_{p}} = \frac{e_{v v}}{n h}$ is positive. If $X$ is convex and $e$ is convex on $X$ , then $D^{2} e$ is positive semidefinite. Positivity of the displayed diagonal responses is not asserted to imply convexity, a sound cone, or an evolution theorem.

Facts & Assumptions

Given: $X \subset (0, \infty) \times R$ is nonempty and open, $e \in C^{2} (X, R)$ , and $(n, s) \in Y$ .

[F1]

Rest fluid state, frame and geometric current definitions: $v = 1 / n$ , $σ_{p} = s / n$ , $ε = n e$ , $T = e_{σ_{p}}$ , $p = - e_{v}$ , $H = e + p v$ , $h = ε + p = n H$ , and $μ = H - T σ_{p}$ .

[F2]

Local graph form at a nonsingular derivative minor: a $C^{1}$ map from an open Euclidean set to $R^{q}$ whose derivative has an invertible $q \times q$ minor has a local $C^{1}$ graph for its zero set, unique in suitable neighborhoods, with the derivative formula stated there.

[F3]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : if $f$ and $g$ are totally differentiable at the relevant points, then $D (g \circ f) = D g (f (\cdot)) \circ D f$ .

[F4]

Concave supporting bound, affine maxima, and Hessian sign: for a $C^{2}$ concave function, the Hessian is negative semidefinite at each point of a convex open region.

[A1]

At states where the heat-capacity clause is applied, $T > 0$ and $e_{σ_{p} σ_{p}} > 0$ .

[A2]

At states where the compressibility clause is applied, $e_{v v} > 0$ .

[A3]

At states where the pressure-energy response is applied, $e_{v v} > 0$ and $h > 0$ .

[A4]

For the last clause, $X$ is convex and $e$ is convex on $X$ .

Proof

1.1givenF1F3algebra

The map $(n, s) \mapsto (v, σ_{p}) = (1 / n, s / n)$ is smooth on $(0, \infty) \times R$ , so $Y$ is open as the inverse image of $X$ . Direct differentiation gives $d v = - d n / n^{2}$ and $d σ_{p} = d s / n - s d n / n^{2}$ .

1.2F1algebra

From $h = ε + p = n (e + p v)$ and $μ = H - T σ_{p}$ , we obtain $T s + μ n = T (n σ_{p}) + (H - T σ_{p}) n = n H = h = ε + p$ .

1.3givenF1F3algebra

The map $(S, V, N) \mapsto (V / N, S / N)$ is smooth on the open set $R \times (0, \infty)^{2}$ , so the parcel domain $D = {(S, V, N) : V > 0, N > 0, (V / N, S / N) \in X}$ is open. The definition of $E$ is a $C^{2}$ composition. Differentiating it gives $\partial_{S} E = e_{σ_{p}} = T, \partial_{V} E = e_{v} = - p, \partial_{N} E = e - v e_{v} - σ_{p} e_{σ_{p}} = e + p v - T σ_{p} = μ .$ Thus $d E = T d S - p d V + μ d N$ .

1.4F1F2F3A1constructalgebra

For the heat-capacity clause, define $F (v, σ_{p}, T) = e_{σ_{p}} (v, σ_{p}) - T$ . This is $C^{1}$ , vanishes at the given state with $T = e_{σ_{p}}$ , and its derivative in $σ_{p}$ is $e_{σ_{p} σ_{p}} > 0$ . By [F2] its zero set is locally a unique $C^{1}$ graph $σ_{p} = φ (v, T)$ . Differentiating at fixed $v$ gives $\partial_{T} σ_{p} = 1 / e_{σ_{p} σ_{p}}$ , whence $c_{v} = e_{σ_{p}} \partial_{T} σ_{p} = T / e_{σ_{p} σ_{p}} > 0$ .

1.5F1F2A2constructalgebra

For the compressibility clause, define $G (v, σ_{p}, p) = p (v, σ_{p}) - p = - e_{v} (v, σ_{p}) - p$ . It is $C^{1}$ and its derivative in $v$ is $- e_{v v} < 0$ . By [F2], $v$ is a local $C^{1}$ function of $(σ_{p}, p)$ , with $(\partial v / \partial p)_{σ_{p}} = - 1 / e_{v v}$ . Therefore $κ_{σ_{p}} = 1 / (v e_{v v}) > 0$ .

1.6F1F2F3A3algebra

At fixed $σ_{p}$ , $ε (n, σ_{p}) = n e (1 / n, σ_{p})$ and $p (n, σ_{p}) = - e_{v} (1 / n, σ_{p})$ . Differentiating gives $\partial_{n} ε = e - e_{v} / n = e + p / n = h / n$ and $\partial_{n} p = e_{v v} / n^{2}$ . Since $h / n > 0$ by [A3], the local implicit-rank theorem makes $ε$ a valid local coordinate in place of $n$ . Taking the ratio of the two derivatives yields $α_{σ_{p}} = e_{v v} / (n h) > 0$ .

2.1step 1.1F1F3algebra

By the chain rule, $d e = e_{v} d v + e_{σ_{p}} d σ_{p} = - p d v + T d σ_{p}$ . Since $ε = n e$ , the product rule and step 1.1 give $d ε = e d n + n d e = T d s + (e + \frac{p}{n} - T σ_{p}) d n = T d s + μ d n .$ Here $v = 1 / n$ , so $e + p / n - T σ_{p} = H - T σ_{p} = μ$ .

2.2step 1.3F3algebra

For $λ > 0$ , $E (λ S, λ V, λ N) = λ E (S, V, N)$ because the intensive ratios are unchanged. Differentiating this identity in $λ$ at $1$ yields $E = T S - p V + μ N$ . At $z_{0} = (S, V, N)$ , differentiate Euler's identity $D E (z_{0}) [z_{0}] = E (z_{0})$ in an arbitrary direction $w$ . The two first-derivative terms cancel, leaving $D^{2} E (z_{0}) [w, z_{0}] = 0$ for every $w$ ; symmetry of the Hessian gives $D^{2} E (z_{0}) [z_{0}, w] = 0$ .

3.1step 2.1step 1.2F3algebra

Differentiating $ε + p = T s + μ n$ and substituting step 2.1 cancels $T d s$ and $μ d n$ , leaving $d p = s d T + n d μ$ .

4.1F4A4algebra∎

Apply [F4] to $- e$ , which is concave when $e$ is convex on the convex open domain $X$ . Its Hessian is $- D^{2} e$ and is negative semidefinite, so $D^{2} e$ is positive semidefinite. This conditional curvature statement does not promote the response inequalities to convexity without [A4].

Local Gibbs, Euler, parcel first law and response identities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources