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Radiation Gibbs and perfect linear sound deduction

Statement

Fix $a_{R} > 0$ and the maps of Separate entropy-only radiation state maps. First, their temperature and entropy coordinates are inverse, $d ε_{T} = T d s_{T}$ , $h_{T} = T s_{T}$ , and $d p_{T} / d ε_{T} = 1 / 3$ on $T > 0$ .

Next, assume countable choice ${A C}_{ω}$ . Let $M$ be a smooth four-manifold without boundary and let $g$ be a smooth symmetric nondegenerate covariant $2$ -tensor field of signature $(-, +, +, +)$ . Write $g^{a b}$ for the inverse-metric components and $u_{b} : = g_{b c} u^{c}$ . Use the Levi-Civita connection supplied by Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields. Let $c > 0$ , let $u$ be a smooth vector field with $g_{a b} u^{a} u^{b} = - c^{2}$ , and let $T : M \to (0, \infty)$ be smooth. Put $ε = ε_{T} (T)$ , $p = p_{T} (T)$ , $s = s_{T} (T)$ , and $h = ε + p$ . Define the symmetric tensor field

$T^{a b} : = \frac{h}{c^{2}} u^{a} u^{b} + p g^{a b} .$

If $\nabla_{a} T^{a b} = 0$ , then $\nabla_{a} (s u^{a}) = 0$ . This is a conditional mathematical identity for the specified tensor and state maps; it assumes no particle-number current.

Finally, on $R^{4}$ with Minkowski coordinates $(x^{0}, x^{1}, x^{2}, x^{3}) = (c t, x)$ and metric $η = diag (- 1, 1, 1, 1)$ , specify the linearization as follows. For some $ϵ > 0$ , let $λ \mapsto (T_{λ}, u_{λ})$ be jointly smooth on $(- ϵ, ϵ) \times R^{4}$ , with $T_{λ} (x) > 0$ , $T_{0} (x) = T_{*}$ , $u_{0} = (c, 0, 0, 0)$ , and $η_{a b} u_{λ}^{a} u_{λ}^{b} = - c^{2}$ . Form $T_{λ}^{a b}$ by the same stress formula above with $(T, u, g)$ replaced by $(T_{λ}, u_{λ}, η)$ , using $ε_{λ} : = ε_{T} (T_{λ})$ and $p_{λ} : = p_{T} (T_{λ})$ . Write $δ f : = {\frac{d}{d λ} f_{λ} ∣}_{λ = 0}$ ; the linearized equation means $δ (\partial_{a} T_{λ}^{a b}) = 0$ . Put $h_{*} : = ε_{T} (T_{*}) + p_{T} (T_{*})$ , $r : = δ ε$ , and $w^{i} : = δ u^{i} / c$ . Its resulting constant-coefficient linear system is

$r_{t} + c h_{*} div w = 0, w_{t} + \frac{c}{3 h_{*}} \nabla r = 0,$

where $\nabla$ and $div$ are the Euclidean spatial gradient and divergence. With $U = (r, w)$ , write this system as $U_{t} + \sum_{j = 1}^{3} A_{j} \partial_{j} U = 0$ , where

$A_{j} : = c (\begin{array}{cc} 0 & h_{*} e_{j}^{T} \\ (1 / (3 h_{*})) e_{j} & 0_{3 \times 3} \end{array}), e_{j} is the j th standard basis vector of R^{3} .$

The matrix $M_{*} : = diag (1 / (3 h_{*}), h_{*} I_{3})$ is a positive symmetrizer in the explicit sense that it is symmetric positive definite and $M_{*} A_{j}$ is symmetric for each $j$ . For each unit spatial vector $ν \in R^{3}$ , define the directional coefficient matrix $A (ν) : = \sum_{j = 1}^{3} ν_{j} A_{j}$ ; its eigenvalues are $0$ (multiplicity two) and $\pm c / \sqrt{3}$ . For Schwartz initial perturbations on $R^{3}$ or smooth periodic initial perturbations on $T_{L}^{3}$ with $L > 0$ , where Schwartz means each component belongs to $S (R^{3})$ as defined in Schwartz space and its seminorms, the global smooth evolution, weighted Sobolev contraction, and finite-propagation conclusions are conditional on the exact symmetrized-system interface of Complete symmetrized constant-coefficient evolution stated in [A1]. That current carrier is draft, so this item does not record those conclusions as independently accepted.

Facts & Assumptions

Given: $a_{R} > 0$ , $T > 0$ , and the smooth fields and tensor equation in the Statement. Assume ${A C}_{ω}$ where indicated; no stronger choice principle is used.

[F1]

Separate entropy-only radiation state maps defines $ε_{T} = a_{R} T^{4}$ , $p_{T} = ε_{T} / 3$ , $s_{T} = 4 a_{R} T^{3} / 3$ , and the candidate entropy-coordinate maps $τ (s) = (3 s / (4 a_{R}))^{1 / 3}$ , $ε_{s} = a_{R} τ^{4}$ , $p_{s} = ε_{s} / 3$ .

[F2]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: for a smooth four-manifold without boundary, smooth symmetric nondegenerate metric, and ${A C}_{ω}$ , there is a unique torsion-free metric-compatible connection; its induced tensor derivative obeys the Leibniz and contraction rules and annihilates $g$ and $g^{- 1}$ . We use metric compatibility and the Leibniz rule only.

[F3]

Schwartz space and its seminorms: $S (R^{3})$ is the class of smooth complex-valued functions for which every seminorm $\sup_{x} ∣ x^{α} \partial^{β} f (x) ∣$ is finite; real components are viewed as complex-valued functions with zero imaginary part, and the condition is imposed componentwise on vector-valued data.

[A1]

Conditional evolution supplier Complete symmetrized constant-coefficient evolution: assume ${A C}_{ω}$ , $m \geq 1$ , constant real matrices $A_{1}, A_{2}, A_{3}, B, M \in R^{m \times m}$ , symmetric positive-definite $M$ , symmetric $M A_{j}$ , and positive-semidefinite $S : = (M B + B^{T} M) / 2$ . For $U_{t} + \sum_{j} A_{j} \partial_{j} U + B U = 0$ , the supplier gives a unique global smooth solution for Schwartz initial data on $R^{3}$ and for smooth periodic data on $T_{L}^{3}$ . With its negative-sign, $2 π$ -normalized Fourier transform $\hat{U} (ξ) : = \int_{R^{3}} e^{- 2 π i x \cdot ξ} U (x) d x$ , its Euclidean weighted norms are $∥ U ∥_{H_{M}^{s}}^{2} : = \int_{R^{3}} (1 + ∣ ξ ∣^{2})^{s} \hat{U} (ξ)^{*} M \hat{U} (ξ) d ξ$ for $s \geq 0$ ; they contract for $t \geq 0$ . On the torus, ${\hat{U}}_{k} : = L^{- 3} \int_{[0, L]^{3}} U (x) e^{- 2 π i k \cdot x / L} d x$ and $∥ U ∥_{H_{M}^{s} (T_{L}^{3})}^{2} : = \sum_{k \in Z^{3}} (1 + ∣ k / L ∣^{2})^{s} {\hat{U}}_{k}^{*} M {\hat{U}}_{k}$ ; these norms contract. For a unit $ν \in R^{3}$ , put $A (ν) = \sum_{j} ν_{j} A_{j}$ and $v_{*} : = \sup_{∣ ν ∣ = 1} \max {∣ λ ∣ : λ \in spec (M^{1 / 2} A (ν) M^{- 1 / 2})}$ . If $v_{*} > 0$ , $R > v_{*} t$ , and $U_{0}$ vanishes on $B_{R} (x_{0})$ , its solution vanishes on $B_{R - v_{*} t} (x_{0})$ for $0 \leq t < R / v_{*}$ , with the analogous torus statement using geodesic distance; if $v_{*} = 0$ , each $A_{j} = 0$ . Here $m = 4$ , $U = (r, w)$ , $B = 0$ , and $M = M_{*}$ . The current supplier carrier and proof are draft; this item uses its interface only conditionally.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) defines ${A C}_{ω}$ as a choice function for each countable family of nonempty sets. This is the only choice strength carried here, through [F2] and [A1].

Proof

1.1givenF1algebra

Differentiating the maps in [F1] gives $ε_{T}^{'} (T) = 4 a_{R} T^{3} = T s_{T}^{'} (T)$ and $s_{T}^{'} (T) = 4 a_{R} T^{2} > 0$ . For $T > 0$ and $s > 0$ , direct substitution gives $τ (s_{T} (T)) = T$ and $s_{T} (τ (s)) = s$ , since $3 s_{T} (T) / (4 a_{R}) = T^{3}$ and $τ (s)^{3} = 3 s / (4 a_{R})$ . Hence the explicitly defined $ε_{s}, p_{s}$ are the energy and pressure expressed in the entropy coordinate, and $d ε_{T} = T d s_{T}$ .

1.2givenF2algebra

Write $D : = u^{a} \nabla_{a}$ and $θ : = \nabla_{a} u^{a}$ . Metric compatibility and $u_{b} u^{b} = - c^{2}$ give $u^{a} u_{b} \nabla_{a} u^{b} = \frac{1}{2} u^{a} \nabla_{a} (u_{b} u^{b}) = 0$ . Expanding the divergence by the Leibniz rule and contracting with $u_{b}$ yields $u_{b} \nabla_{a} T^{a b} = \frac{u_{b} u^{b}}{c^{2}} \nabla_{a} (h u^{a}) + \frac{h}{c^{2}} u^{a} u_{b} \nabla_{a} u^{b} + u^{a} \nabla_{a} p = - D h - h θ + D p = - D ε - h θ$ . The last pressure term uses $\nabla_{a} g^{a b} = 0$ and $u_{b} g^{a b} = u^{a}$ . Thus the assumed tensor equation implies $D ε + h θ = 0$ .

2.1givenF1step 1.1algebra

From [F1], $h_{T} = ε_{T} + p_{T} = \frac{4}{3} a_{R} T^{4} = T s_{T}$ , and $d p_{T} / d ε_{T} = p_{T}^{'} (T) / ε_{T}^{'} (T) = 1 / 3$ . In particular $h_{T} > 0$ for all $T > 0$ .

3.1step 1.1step 2.1step 1.2algebra

By step 1.1 and the chain rule, $D ε = T D s$ ; by step 2.1, $h = T s$ . Substitution in step 1.2 gives $0 = T (D s + s θ) = T \nabla_{a} (s u^{a})$ . Since $T > 0$ , division gives $\nabla_{a} (s u^{a}) = 0$ .

3.2givenstep 2.1algebra

By the linearization definition in the Statement, differentiating $η_{a b} u_{λ}^{a} u_{λ}^{b} = - c^{2}$ at $λ = 0$ gives $- 2 c δ u^{0} = 0$ , so $δ u^{0} = 0$ ; also $δ u^{i} = c w^{i}$ . Since $r = δ ε$ and step 2.1 gives $d p_{T} / d ε_{T} = 1 / 3$ , the state variations satisfy $δ p = r / 3$ and $δ h = r + δ p$ . The first variations of the stress components are therefore $δ T^{00} = δ h - δ p = r$ , $δ T^{0 i} = h_{*} w^{i}$ , and $δ T^{i j} = δ p δ^{i j}$ . Differentiating $\partial_{a} T_{λ}^{a b} = 0$ at $λ = 0$ , with $\partial_{0} = c^{- 1} \partial_{t}$ , yields the displayed equations for $b = 0$ and $b = i$ .

4.1givenstep 3.2A1A2algebra

Write the system of step 3.2 as $U_{t} + \sum_{j} A_{j} \partial_{j} U = 0$ with $U = (r, w)$ , $A_{j} = c (\begin{array}{cc} 0 & h_{*} e_{j}^{T} \\ (1 / (3 h_{*})) e_{j} & 0_{3 \times 3} \end{array})$ , and $M_{*} = diag (1 / (3 h_{*}), h_{*} I_{3})$ . Because $h_{*} > 0$ , $M_{*}$ is positive definite, and $M_{*} A_{j} = c (\begin{array}{cc} 0 & \frac{1}{3} e_{j}^{T} \\ \frac{1}{3} e_{j} & 0_{3 \times 3} \end{array})$ is symmetric. For a unit $ν$ , $A (ν)$ has the two transverse zero eigenvalues and the longitudinal eigenvalues $λ = \pm c / \sqrt{3}$ , obtained from $λ^{2} = c^{2} / 3$ on the span of $(1, 0)$ and $(0, ν)$ . The supplier matrix $M_{*}^{1 / 2} A (ν) M_{*}^{- 1 / 2}$ is similar to $A (ν)$ because $M_{*}$ is positive definite, so it has the same spectrum and the supplier speed is $v_{*} = c / \sqrt{3}$ . Thus the exact matrices meet the algebraic hypotheses in [A1] with $B = 0$ and $S = 0$ .

5.1A1F3step 3.2step 4.1∎

Conditional on [A1], apply its stated global-evolution and finite-propagation conclusions to the matrices of steps 3.2 and 4.1; the resulting speed is $v_{*} = c / \sqrt{3}$ . The Schwartz data class is the componentwise class in [F3]. The linearization and characteristic calculation in steps 3.2 and 4.1 are unconditional, while these PDE conclusions remain conditional on the current draft supplier interface.

Radiation Gibbs and perfect linear sound deduction

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources