Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Perfect-gas entropy, isentropic law and temperature balance

Physical scope

Conditional smooth local-equilibrium calculation for an explicitly adopted calorically perfect gas with positive specific gas constant and specific heat. The temperature equation includes stress, heat flux and heat supply; entropy constancy and the isentropic pressure law are restricted to the stated adiabatic, inviscid, unsourced branch. No empirical gas data or universal gas model is asserted.

Statement

Assume the complete smooth material, state, contact and mass/momentum/angular-momentum/total-energy balance hypotheses of the mathematical Gibbs-entropy identity, including ${A C}_{ω}$ , as stated in Facts & Assumptions. Let $R > 0$ be a specific gas constant and $c_{v} > 0$ a constant specific heat at constant volume, both in $J k g^{- 1} K^{- 1}$ . Choose reference values $T_{r e f} > 0$ in $K$ , $ρ_{r e f} > 0$ in $k g m^{- 3}$ , and $s_{r e f}$ in $J k g^{- 1} K^{- 1}$ . On $v > 0, s \in R$ , adopt $T (v, s) = T_{r e f} \exp ⁣ (\frac{s - s_{r e f}}{c_{v}}) {(\frac{1}{ρ_{r e f} v})}^{R / c_{v}}, e (v, s) = c_{v} T (v, s), p (v, s) = \frac{R T (v, s)}{v} .$ These formulas give $T = e_{s} > 0$ , $p = - e_{v}$ , and $e \in C^{2} ((0, \infty) \times R)$ ; equivalently, with $ρ = 1 / v$ , $s = s_{r e f} + c_{v} \ln (T / T_{r e f}) - R \ln (ρ / ρ_{r e f})$ and $p = ρ R T$ . The logarithms have dimensionless arguments.

For a smooth solution, define $θ = div u$ , $σ = - p I_{3} + τ$ , and the local entropy residual $G = ρ D_{t} s + div (q / T) - ρ r / T$ . The temperature balance is $ρ c_{v} D_{t} T = - p θ + τ : D - div q + ρ r .$ On the additional adiabatic, inviscid, unsourced branch $τ = 0, q = 0, r = 0$ , one has $D_{t} s = 0$ . Along any trajectory on which the entropy value is $s_{0}$ , the pressure law is $p = K (s_{0}) ρ^{γ}, γ = 1 + \frac{R}{c_{v}}, K (s_{0}) = R T_{r e f} \exp ⁣ (\frac{s_{0} - s_{r e f}}{c_{v}}) ρ_{r e f}^{- R / c_{v}} > 0.$ Here $γ$ is dimensionless and $K (s_{0})$ has SI units $P a (k g m^{- 3})^{- γ}$ , so $K (s_{0}) ρ^{γ}$ has units of pressure. The factor $K (s_{0})$ can differ between trajectories with different entropy values. The gas closure is an adopted model, and the adiabatic conclusion does not apply when viscous work, heat conduction, or heat supply is present.

Statement references

David Tong, Fluid Mechanics, Part III lecture notes, §§4.4.2–4.4.3, printed pp. 146–149

An Introduction to Continuum Mechanics, Second Edition, Chapter 6, §§6.1.4 and 6.5.2, printed pp. 224–226 and 243–244

Facts & Assumptions

Given: The smooth positive-density material and balance hypotheses, the adopted local state relation, and the explicit perfect-gas constants and reference scales in the Statement.

[A1]

Single-component nonpolar continuum and material balances: smooth positive-density material fields obey no-source mass conservation and the separate contact, linear/angular-momentum, total-energy, and entropy assumptions on material subregions. These are physical premises, not conclusions. Its current batch-3 carrier and source-backed formulation were inspected, and step1Decision currently reports ready.

[A2]

Local simple-fluid equilibrium and Gibbs closure: $T = e_{s} (v, s) > 0$ , $p = - e_{v} (v, s)$ , and $ρ = 1 / v$ are adopted pointwise equilibrium values. The current batch-11 Postulate states this local relation and its positive-temperature scope.

[L1]

Gibbs cancellation and entropy-production identity: under ${A C}_{ω}$ , the stated $C^{2}$ orientation-preserving motion, positive $C^{1}$ state fields, local contact and stress regularity, continuity equation, and separate linear/angular-momentum and total-energy integral balances, one obtains $σ = σ^{T}$ and $ρ D_{t} e = σ : D - div q + ρ r$ . The exact Gibbs interface applies under all displayed assumptions; its proof step 3.1 supplies time recentering on any nonempty open interval.

[F1]

Classical frames, primitive quantities and SI units: $D_{t} = \partial_{t} + u \cdot \nabla$ , $θ = div u$ , and the SI types for $ρ, u, T, s, e, p$ . Its current batch-3 carrier was inspected; step1Decision currently reports ready. Continuity itself is the separate hypothesis in [L1].

[A3]

The positive constants, reference scales, and calorically perfect state relation are adopted assumptions. No equipartition argument, measured heat capacity, or empirical gas data is used.

Proof

1.1A1A2A3F1L1algebra

Differentiate the entropy formula along a material trajectory. The chain rule gives $D_{t} s = (c_{v} / T) D_{t} T - (R / ρ) D_{t} ρ$ . The continuity equation in [L1] is $D_{t} ρ + ρ θ = 0$ , so $D_{t} ρ = - ρ θ$ . Hence $ρ T D_{t} s = ρ c_{v} D_{t} T + ρ R T θ = ρ c_{v} D_{t} T + p θ$ .

2.1A1A2A3L1step 1.1algebra

The local internal-energy equation from [L1] and $σ = - p I_{3} + τ$ give $ρ D_{t} e = - p θ + τ : D - div q + ρ r$ , since $I_{3} : D = θ$ . With $e = c_{v} T$ , this is the stated full temperature balance.

3.1A3step 1.1step 2.1algebra

If $τ = q = r = 0$ , the temperature equation gives $D_{t} T = - (R / c_{v}) T θ$ . Substitution in the differentiated entropy formula of step 1.1 yields $D_{t} s = - (R θ) + (R θ) = 0$ .

4.1A3step 3.1algebra∎

On a trajectory with entropy value $s_{0}$ , solve the entropy formula for $T / T_{r e f} = \exp ((s_{0} - s_{r e f}) / c_{v}) (ρ / ρ_{r e f})^{R / c_{v}}$ . Using $p = ρ R T$ , collect powers of $ρ$ to obtain $p = K (s_{0}) ρ^{1 + R / c_{v}}$ with the displayed positive $K (s_{0})$ . Its dependence on $s_{0}$ shows that different entropy levels can have different coefficients.

Perfect-gas entropy, isentropic law and temperature balance

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources