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Conditional redshift heat relaxation

Physical scope

Conditional deduction for the fixed-background supported parabolic heat model of post-tgr-supported-static-heat-model, with compatible positive data; no claim of empirical realization, full Einstein-fluid dynamics, causality, or finite propagation.

Statement

Use the fixed static box, positive bounded lapse $N$ , constants, material state function, observer field, heat flux, support, initial-data class, and SI conventions of Supported static heat model. Let ${\tilde{ϑ}}_{0}$ be its positive $C^{2}$ even periodic extension, and define $∣ Ω ∣ = L_{1} L_{2} L_{3}$ , $\overline{ϑ} = ∣ Ω ∣^{- 1} \int_{Ω} ϑ_{0} d x$ , and $C_{Ω} = \frac{3}{2} (\max_{i} L_{i})^{2}$ . Use Euclidean Lebesgue volume $d x$ and the associated $L^{2} (Ω)$ norm. Then the model has a unique solution

$T \in C ([0, \infty) \times \overline{Ω}) \cap C^{2, 1} ((0, \infty) \times \overline{Ω}),$

where $C^{2, 1}$ means two continuous spatial derivatives and one continuous time derivative up to the box faces at positive times. With $ϑ = N T$ and $Γ (z, τ) = (4 π τ)^{- 3 / 2} \exp (- ∣ z ∣^{2} / (4 τ))$ for $z \in R^{3}$ and $τ > 0$ , the solution is

$ϑ (t_{K}, x) = \int_{R^{3}} Γ ⁣ (x - y, \frac{A_{0}}{C_{V}} t_{K}) {\tilde{ϑ}}_{0} (y) d y (t_{K} > 0), ϑ (0, x) = ϑ_{0} (x), T = \frac{ϑ}{N} .$

It satisfies $C_{V} \partial_{t_{K}} ϑ = A_{0} Δ_{x} ϑ$ , the six insulated-face conditions $\partial_{ν} ϑ = 0$ , and $m_{0} \leq ϑ \leq M_{0}$ . The lapse-weighted material energy

$E_{K} (t_{K}) = \int_{Ω} N (x) ϵ (t_{K}, x) d x$

is constant. For the entropy current $S^{a} = s u^{a} + Q^{a} / T$ and total entropy $S (t_{K}) = \int_{Ω} s (t_{K}, x) d x$ , the exact model rate is

$\frac{d S}{d t_{K}} = A_{0} \int_{Ω} \frac{∣ \nabla_{x} ϑ ∣^{2}}{ϑ^{2}} d x \geq 0.$

Moreover,

${∥ ϑ (t_{K}, \cdot) - \overline{ϑ} ∥}_{L^{2} (Ω)} \leq \exp ⁣ (- \frac{A_{0} t_{K}}{C_{V} C_{Ω}}) {∥ ϑ_{0} - \overline{ϑ} ∥}_{L^{2} (Ω)} .$

Thus $T$ converges in $L^{2} (Ω)$ to the positive proper-temperature profile $T_{e q} (x) = \overline{ϑ} / N (x)$ , while the redshifted temperature $N T$ converges to the constant $\overline{ϑ}$ . These are conditional predictions of the adopted model. They do not establish that a real material obeys the postulates, and they do not assert a full Einstein-fluid solution, causal or stable relativistic-fluid evolution, or finite propagation speed. No empirical data are used.

Facts & Assumptions

Given: The static box model, positive lapse bounds, material branch, constants, SI units, compatible initial profile, heat equation, support condition, and entropy current specified in Supported static heat model. The same current is required to satisfy the selected material's local admissibility condition in Local entropy-current admissibility in a separately defined model.

[A1]

Supported static heat model: on $O = (0, \infty) \times Ω$ , $g = - N^{2} (d x^{0})^{2} + \sum_{i} (d x^{i})^{2}$ , $ξ = \partial_{x^{0}}$ , $u = N^{- 1} ξ$ , and $ϑ = N T$ ; $N$ is smooth and bounded strictly above zero on the box. The material EOS gives $p = 0$ and $ϵ = C_{V} T + n_{0} e_{0}$ . The postulate also declares the open first-jet regime $R$ and requires a positive $C^{\infty} (O)$ temperature with the stated initial trace. The adopted flux, coefficient, energy balance and boundary data are $q = - (A_{0} / N) (\nabla_{x} T + T \nabla_{x} \log N)$ , $N C_{V} \partial_{t_{K}} T + {div}_{x} (N^{2} q) = 0$ , and $N^{2} q \cdot ν = 0$ .

[A2]

Local entropy-current admissibility in a separately defined model: for the separately specified smooth current $S^{a}$ of this material model, adopt $\nabla_{a} S^{a} \geq 0$ throughout its declared regime. This is an applicability postulate; it is not used to replace the calculation of the entropy rate below.

[F1]

Constructed supported static heat evolution: for the positive bounded lapse, rectangular box, fixed material state, coefficient $A_{0} / N$ , insulated faces and positive even-reflected periodic $C^{2}$ data stated there, the exact heat model has a unique all-time solution with $m_{0} \leq N T \leq M_{0}$ , conserved lapse-weighted energy, the displayed nonnegative entropy rate, and exponential $L^{2}$ relaxation of $N T$ to its initial mean. Its complete proof is conditional on its exact listed mathematical suppliers and uses exactly Countable Choice ${A C}_{ω}$ through that dependency chain, not unrestricted Axiom of Choice. The consumer uses this actual statement and its hypotheses, not its draft label as an acceptance stamp.

Proof

1.1givenA1F1algebra

In the static metric of [A1], $ξ$ is independent of $x^{0}$ and the lapse is positive, so the selected rest field is $u = N^{- 1} ξ$ . The specialization of the first-order closure in [A1] has $\nabla_{x} T + T \nabla_{x} \log N = N^{- 1} \nabla_{x} (N T) = N^{- 1} \nabla_{x} ϑ$ . Therefore $N^{2} q = - A_{0} \nabla_{x} ϑ$ , and since $N$ is time independent, $N \partial_{t_{K}} T = \partial_{t_{K}} ϑ$ . Substitution into the adopted balance gives $C_{V} \partial_{t_{K}} ϑ = A_{0} Δ_{x} ϑ$ ; the insulated condition is exactly $\partial_{ν} ϑ = 0$ . The EOS and support/current conventions in [A1] are the same as those required by [F1].

2.1step 1.1A1F1algebra

The extension in [A1] is bounded and uniformly continuous because it is continuous and periodic, and it has the positive even-reflected boundary compatibility required by [F1]. Every box length is positive, $A_{0}, C_{V} > 0$ , and $N_{\min} > 0$ ; thus all positivity and regularity hypotheses of [F1] hold. Apply [F1] to obtain its unique solution, Gaussian representation, positive bounds, conserved $E_{K}$ , entropy-rate identity, and exponential estimate. The heat convolution used in the proof of [F1] is smooth for positive times; since $N$ is smooth and strictly positive, $T = ϑ / N$ and its state maps, including $μ / T$ , are smooth on $O$ , with smooth extensions to the box faces on compact positive-time intervals. The constitutive law then makes $q$ , $Q$ , $J$ , $T$ , $f^{b} = \nabla_{a} T^{a b}$ , and $S^{a}$ smooth on $O$ , so the local balances and entropy-current divergence required by the postulate are classical. The initial slice has the stated continuous trace; no smoothness across $t_{K} = 0$ is used. By [A1], this is the solution of the selected physical model with its stated support and units.

3.1step 2.1A1A2F1algebra

The entropy current in [F1] is $S^{a} = s u^{a} + Q^{a} / T$ with $Q = q / c$ , exactly as in [A1]. The adopted local admissibility condition [A2] is an explicit physical premise for this current. Independently, [F1] gives the total entropy rate as the displayed integral, whose integrand is nonnegative because $A_{0} > 0$ and $ϑ \geq m_{0} > 0$ ; hence total entropy is nondecreasing. The same supplier gives conservation of the lapse-weighted energy.

4.1step 2.1F1algebra∎

Since $N (x) \geq N_{\min} > 0$ , $∣ T (t_{K}, x) - T_{e q} (x) ∣ = ∣ ϑ (t_{K}, x) - \overline{ϑ} ∣ / N (x) \leq N_{\min}^{} ∣ ϑ (t_{K}, x) - \overline{ϑ} ∣$ . Taking the $L^{2} (Ω)$ norm and using the estimate from [F1] proves $T \to \overline{ϑ} / N$ in that norm. The limit is positive because $\overline{ϑ} \geq m_{0} > 0$ , and $N T_{e q} = \overline{ϑ}$ is constant. These are conditional model deductions, not empirical findings.

Conditional redshift heat relaxation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources