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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 ε0>0, there is a finite constant K0 such that every radius R>0 with u=8πGε0R2/(3c4)∈(0,8/9) in the constant-density family satisfies P(0)≤K0ε0.

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)/ε0=(1−a)/(3a−1) with a=1−u for every 0<u<8/9; this is a family-specific formula, not a general compactness theorem.

Counterexample

Given: Fixed ε0>0, varying radii with u=8πGε0R2/(3c4)∈(0,8/9), the uniform-bound statement above, and the family-specific formula [F1].

1.1givenF1constructalgebra

For each integer n≥2, set un=8/9−1/n and Rn=c23un/(8πGε0)>0. Keep ρ0=ε0/c2 fixed and set Mn=4πρ0Rn3/3. Then 2GMn/(c2Rn)=8πGε0Rn2/(3c4)=un∈(0,8/9), so every (ρ0,Rn) is an admissible member of [F1] at the fixed density. Moreover an=1−un=1/9+1/n>1/3 and an↓1/3.

2.1F1step 1.1algebra∎

By [F1], Pn(0)/ε0=(1−an)/(3an−1). The numerator tends to 2/3>0 and the denominator tends to 0 through positive values, so the ratios tend to +∞. Therefore no finite K0 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.

Depends on

Dependency tree · 2 levels

Sources