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Frame-qualified dissipative entropy cancellation

Statement

Let $(M, g, N)$ , $u$ , $\nabla$ , and the fixed-composition state be as in Rest fluid state, frame and geometric current definitions and Local Gibbs, Euler, parcel first law and response identities, with $g (u, u) = - 1$ , $n > 0$ , and $T > 0$ . Define $Δ^{a b} = g^{a b} + u^{a} u^{b}$ , $Δ_{a b} = g_{a b} + u_{a} u_{b}$ , $Δ_{a}^{b} = δ_{a}^{b} + u_{a} u^{b}$ , and $D = u^{a} \nabla_{a}$ , $θ = \nabla_{a} u^{a}$ , $a^{b} = u^{a} \nabla_{a} u^{b}$ , and the spatial symmetric trace-free shear

$σ_{a b} = Δ_{a}^{c} Δ_{b}^{d} \nabla_{(c} u_{d)} - \frac{θ}{3} Δ_{a b},$

where parentheses denote the averaged symmetric part. Suppose smooth fields $Π$ , $Q^{a}$ , $V^{a}$ , and $π^{a b}$ obey $u_{a} Q^{a} = u_{a} V^{a} = 0$ , $u_{a} π^{a b} = 0$ , and $π^{a b}$ is symmetric with $g_{a b} π^{a b} = 0$ . Set

$T^{a b} = ϵ u^{a} u^{b} + (p + Π) Δ^{a b} + u^{a} Q^{b} + Q^{a} u^{b} + π^{a b}, J^{a} = n u^{a} + V^{a} .$

If $\nabla_{a} T^{a b} = f^{b}$ and $\nabla_{a} J^{a} = 0$ , the vector field

$S^{a} = s u^{a} + \frac{Q^{a}}{T} - \frac{μ V^{a}}{T}$

satisfies the exact identity

$\nabla_{a} S^{a} = - \frac{Π θ + π^{a b} σ_{a b}}{T} - \frac{Q^{a} (\nabla_{a} T + T a_{a})}{T^{2}} - V^{a} \nabla_{a} ⁣ (\frac{μ}{T}) - \frac{u_{b} f^{b}}{T} .$

This identity applies to the stated smooth fields and conventions; it does not select a constitutive law or assert that any such fields solve an evolution problem.

Facts & Assumptions

Given: A nonempty open set $O$ and the smooth Lorentzian data of Rest fluid state, frame and geometric current definitions, a $C^{2}$ state map into a nonempty open state domain, the thermodynamic identities of Local Gibbs, Euler, parcel first law and response identities, and the stress and particle balances displayed in the Statement.

[A1]

The spacetime and tensor construction assumes Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )); no stronger choice principle is used.

[F1]

Rest fluid state, frame and geometric current definitions (current batch-53 draft): on a nonempty open set $O$ , the observer field is smooth and future unit timelike; a local state has a nonempty open domain $X \subset (0, \infty) \times R$ , a specified $e \in C^{2} (X)$ , and a $C^{2}$ state map $(v, σ_{p}) : O \to X$ . It defines $n = v^{- 1} > 0$ , $s = n σ_{p}$ , $ϵ = n e$ , $T = e_{σ_{p}}$ , $p = - e_{v}$ , $H = e + p v$ , $h = ϵ + p = n H$ , and $μ = H - T σ_{p}$ .

[F2]

Local Gibbs, Euler, parcel first law and response identities (current batch-53 draft): under the stated nonempty open-domain and $C^{2}$ hypotheses, it gives $d ϵ = T d s + μ d n$ and $ϵ + p = T s + μ n$ . Applying the first identity to the state map along $D$ gives the directional identity used below. This use is conditional on the current draft carrier.

[F3]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: the induced tensor derivative satisfies the product rule, $\nabla g = 0$ , and contraction commutes with differentiation.

[F4]

In the SI convention of [F1], $u$ and $Δ$ are dimensionless, $D, θ, a, σ$ have inverse-length components, $ϵ, p, Π, Q, π$ have units Pa, $n, V$ have units $m a r k e r m^{- 3}$ , $s$ has units $J K^{- 1} m^{- 3}$ , $T$ has units K, and $μ$ has units J per marker. The force density $f$ has units $N m^{- 3}$ , the physical heat flux is $q = c Q$ , and $c$ has units $m s^{- 1}$ .

[A2]

Define $Δ^{a b} = g^{a b} + u^{a} u^{b}$ , $D = u^{a} \nabla_{a}$ , $θ = \nabla_{a} u^{a}$ , and $a^{b} = u^{a} \nabla_{a} u^{b}$ . The stress decomposition, spatial symmetries, and balance equations are exactly those in the Statement.

Proof

1.1A1A2F1F3algebra

From metric compatibility, $\nabla_{a} Δ^{a b} = θ u^{b} + a^{b}$ and $u_{b} \nabla_{a} u^{b} = 0$ . The perfect energy term contracts as $u_{b} \nabla_{a} (ϵ u^{a} u^{b}) = - D ϵ - ϵ θ$ ; the isotropic pressure term contributes $- (p + Π) θ$ . Orthogonality and the product rule give $u_{b} \nabla_{a} (u^{a} Q^{b}) = - Q^{b} a_{b}$ and $u_{b} \nabla_{a} (Q^{a} u^{b}) = - \nabla_{a} Q^{a}$ . Finally $u_{b} \nabla_{a} π^{a b} = - π^{a b} \nabla_{a} u_{b} = - π^{a b} σ_{a b}$ : the spatial symmetric trace-free tensor $π$ annihilates the antisymmetric part of $\nabla u$ and its isotropic spatial trace. Thus the stress balance is $D ϵ + h θ + Π θ + \nabla_{a} Q^{a} + Q^{a} a_{a} + π^{a b} σ_{a b} = - u_{b} f^{b} .$

2.1A2F2step 1.1algebra

The particle balance gives $D n + n θ = - \nabla_{a} V^{a}$ . By [F2], $D ϵ + h θ = T (D s + s θ) + μ (D n + n θ) = T \nabla_{a} (s u^{a}) - μ \nabla_{a} V^{a} .$ Substitution in step 1.1 yields $T \nabla_{a} (s u^{a}) = - Π θ - π^{a b} σ_{a b} - \nabla_{a} Q^{a} - Q^{a} a_{a} + μ \nabla_{a} V^{a} - u_{b} f^{b} .$

3.1step 2.1A2algebra∎

By the definition $S^{a} = s u^{a} + Q^{a} / T - μ V^{a} / T$ , its divergence is $\nabla_{a} (s u^{a}) + \nabla_{a} (Q^{a} / T) - \nabla_{a} (μ V^{a} / T)$ : the last derivative is subtracted. The product rule gives $\nabla_{a} ⁣ (\frac{Q^{a}}{T}) = \frac{\nabla_{a} Q^{a}}{T} - \frac{Q^{a} \nabla_{a} T}{T^{2}}, \nabla_{a} ⁣ (\frac{μ V^{a}}{T}) = \frac{μ}{T} \nabla_{a} V^{a} + V^{a} \nabla_{a} ⁣ (\frac{μ}{T}) .$ Substituting these expressions and step 2.1 gives $\nabla_{a} S^{a} = \frac{- Π θ - π^{a b} σ_{a b} - \nabla_{a} Q^{a} - Q^{a} a_{a} + μ \nabla_{a} V^{a} - u_{b} f^{b}}{T} + \frac{\nabla_{a} Q^{a}}{T} - \frac{Q^{a} \nabla_{a} T}{T^{2}} - \frac{μ \nabla_{a} V^{a}}{T} - V^{a} \nabla_{a} ⁣ (\frac{μ}{T}) .$ The $\nabla_{a} Q^{a} / T$ and $μ \nabla_{a} V^{a} / T$ terms cancel with the displayed opposite-sign terms. The remaining heat terms combine as $- Q^{a} \nabla_{a} T / T^{2} - Q^{a} a_{a} / T = - Q^{a} (\nabla_{a} T + T a_{a}) / T^{2}$ , proving the stated identity. Since $Q, V$ are spatial, only the rest-space projections of the corresponding gradients contribute.

Frame-qualified dissipative entropy cancellation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources