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Static isotropic Einstein-fluid and supplied EOS model — Examples

1 · Prerequisites

2 · Summary

The examples specialize the A-page results to half compactness and to a half-capacity finite bath, and exhibit the divergence of central pressure as the constant-density compactness approaches its family endpoint. The star calculations remain within the specified constant-density model; the bath calculation remains within the adopted scalar exchange law. They report no measurement, generic compactness bound, causal fluid dynamics, or stability property of a relativistic star.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Half-compactness constant-density Einstein star

Physical scope

The explicit constant-density family at u=1/2, with its adopted vacuum exterior; not an observed star, realistic EOS, or generic compactness result.

Example

Specialize the constant-density family of Verified constant-density static star and vacuum exterior to $u = 1 / 2$ . Then $a = \frac{1}{\sqrt{2}}, \frac{P (0)}{ε_{0}} = \frac{\sqrt{2} - 1}{3 - \sqrt{2}} \in (0, 1), e^{ϕ (0)} = \frac{3 / \sqrt{2} - 1}{2} > 0.$ At the surface, $b (R) = 1 / 2$ and $e^{ϕ (R)} = 1 / \sqrt{2}$ . The prescribed vacuum exterior has $M = M_{R}$ , $b (r) = 1 - R / (2 r)$ and $e^{ϕ (r)} = \sqrt{b (r)}$ for $r \geq R$ .

Facts & Assumptions

Given: The exact static star family and its equations from Verified constant-density static star and vacuum exterior, with the specialization $u = 1 / 2$ .

[F1]

Verified constant-density static star and vacuum exterior gives $a = \sqrt{1 - u}$ , $B (r) = \sqrt{1 - u r^{2} / R^{2}}$ , $P (0) / ε_{0} = (1 - a) / (3 a - 1)$ , $e^{ϕ (0)} = (3 a - 1) / 2$ , and the vacuum exterior $b (r) = 1 - u R / r$ , $e^{ϕ} = \sqrt{b}$ .

Verification

1.1givenF1algebra

Substitution of $u = 1 / 2$ gives $a = 1 / \sqrt{2}$ and hence $P (0) / ε_{0} = (\sqrt{2} - 1) / (3 - \sqrt{2})$ . Both numerator and denominator are positive, and $\sqrt{2} - 1 < 3 - \sqrt{2}$ because $\sqrt{2} < 2$ ; thus the ratio lies strictly between zero and one. Also $e^{ϕ (0)} = (3 / \sqrt{2} - 1) / 2 > 0$ , $b (R) = a^{2} = 1 / 2$ , and $e^{ϕ (R)} = a = 1 / \sqrt{2}$ .

2.1F1step 1.1∎

With $u = 1 / 2$ , the exterior formula in [F1] becomes $b (r) = 1 - R / (2 r)$ ; it is positive for $r \geq R$ and has $e^{ϕ} = \sqrt{b}$ . The theorem supplies the matching and field-equation verification for this family, so this specialization is a finite, positive-pressure example within the adopted model.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Central pressure is unbounded at the constant-density branch limit

Physical scope

A counterexample within the explicit constant-density branch as u approaches 8/9 from below; no general Buchdahl theorem or claim about realistic matter.

Statement refuted

For fixed energy density $ε_{0} > 0$ , there is a finite constant $K_{0}$ such that every radius $R > 0$ with $u = 8 π G ε_{0} R^{2} / (3 c^{4}) \in (0, 8 / 9)$ in the constant-density family satisfies $P (0) \leq K_{0} ε_{0}$ .

Facts & Assumptions

Given: The constant-density family from Verified constant-density static star and vacuum exterior, including its strict range $0 < u < 8 / 9$ and central-pressure formula.

[F1]

Verified constant-density static star and vacuum exterior states $P (0) / ε_{0} = (1 - a) / (3 a - 1)$ with $a = \sqrt{1 - u}$ for every $0 < u < 8 / 9$ ; this is a family-specific formula, not a general compactness theorem.

Counterexample

Given: Fixed $ε_{0} > 0$ , varying radii with $u = 8 π G ε_{0} R^{2} / (3 c^{4}) \in (0, 8 / 9)$ , the uniform-bound statement above, and the family-specific formula [F1].

1.1givenF1constructalgebra

For each integer $n \geq 2$ , set $u_{n} = 8 / 9 - 1 / n$ and $R_{n} = c^{2} \sqrt{3 u_{n} / (8 π G ε_{0})} > 0$ . Keep $ρ_{0} = ε_{0} / c^{2}$ fixed and set $M_{n} = 4 π ρ_{0} R_{n}^{3} / 3$ . Then $2 G M_{n} / (c^{2} R_{n}) = 8 π G ε_{0} R_{n}^{2} / (3 c^{4}) = u_{n} \in (0, 8 / 9)$ , so every $(ρ_{0}, R_{n})$ is an admissible member of [F1] at the fixed density. Moreover $a_{n} = \sqrt{1 - u_{n}} = \sqrt{1 / 9 + 1 / n} > 1 / 3$ and $a_{n} ↓ 1 / 3$ .

2.1F1step 1.1algebra∎

By [F1], $P_{n} (0) / ε_{0} = (1 - a_{n}) / (3 a_{n} - 1)$ . The numerator tends to $2 / 3 > 0$ and the denominator tends to $0$ through positive values, so the ratios tend to $+ \infty$ . Therefore no finite $K_{0}$ can bound all ratios, refuting the displayed statement. At the limiting value $u = 8 / 9$ , the denominator is zero and the endpoint is not an admissible regular finite-pressure member.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Half-capacity bath attracts the selected exchange state

Physical scope

A single common-temperature point of the adopted finite-bath scalar exchange model with C_b=|C_G|/2; no causal transport or generic star-stability claim.

Example

At a selected subsystem state with $T_{*} > 0$ and $C_{G} < 0$ , choose the finite bath capacity $C_{b} = ∣ C_{G} ∣ / 2$ and bath energy $E_{b} = E_{0} + C_{b} T_{*}$ . In the adopted scalar exchange model, the common-temperature state is a strict local entropy maximum and is locally exponentially attracting; its entropy curvature is $S_{t o t}^{''} (x_{*}) = - \frac{1}{T_{*}^{2} ∣ C_{G} ∣} < 0.$

Facts & Assumptions

Given: The finite-bath exchange theorem Finite bath stabilizes the selected negative-capacity exchange model and its adopted matched-temperature point.

[F1]

Finite bath stabilizes the selected negative-capacity exchange model gives $S_{t o t}^{''} (x_{*}) = - T_{*}^{- 2} (1 / C_{G} + 1 / C_{b})$ and the strict local maximum, forward-invariant local basin, exponential attraction, and entropy production $L (S_{t o t}^{'})^{2} \geq 0$ whenever $C_{G} < 0$ and $0 < C_{b} < ∣ C_{G} ∣$ .

Verification

1.1givenF1algebra

With $C_{b} = ∣ C_{G} ∣ / 2$ , the bath capacity is positive and satisfies $C_{b} < ∣ C_{G} ∣$ . Setting $E_{b} = E_{0} + C_{b} T_{*}$ gives $T_{b} = (E_{b} - E_{0}) / C_{b} = T_{*}$ , so the selected state is common-temperature as required in [F1].

2.1F1step 1.1algebra∎

Since $C_{G} < 0$ , $1 / C_{G} = - 1 / ∣ C_{G} ∣$ and $1 / C_{b} = 2 / ∣ C_{G} ∣$ . Hence [F1] gives $S_{t o t}^{''} (x_{*}) = - 1 / (T_{*}^{2} ∣ C_{G} ∣) < 0$ and its stated local exponential attraction and nonnegative entropy production. This is the specified exchange ODE, not a stellar evolution law.

Static isotropic Einstein-fluid and supplied EOS model — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Half-compactness constant-density Einstein star

Example

Facts & Assumptions

Verification

Central pressure is unbounded at the constant-density branch limit

Statement refuted

Facts & Assumptions

Counterexample

Half-capacity bath attracts the selected exchange state

Example

Facts & Assumptions

Verification

Sources