Smooth static spherical geometry and state domain — Examples

The map $ϵ(p)={ϵ}_0+\lambda p$ is $C^{\mathrm{\infty }}$ and positive on $I_p=(0,\mathrm{\infty })$. Since $p_1>0$, any $0<\delta <p_1$ gives the required closed pressure margin around $P_c$ inside $I_p$.

Apply [F1] under [A1] with this $ϵ$ and $P_c$. There is one radius $R_0>0$ valid for every $p_c\in [p_1,p_2]$, and the corresponding solutions depend continuously on $p_c$ in the uniform norm; they satisfy $v(0)=K({ϵ}_0+\lambda p_c)\mathrm{/}3$, $p^{\mathrm{\prime }}(0)=v^{\mathrm{\prime }}(0)=0$, and have positive lapse and $b\ge 1\mathrm{/}2$.

Substitution of the chosen ${\rho }_0$ gives $M_R=c^{2}R\mathrm{/}(4G)$ and hence $u=1\mathrm{/}2\in (0,8\mathrm{/}9)$. Therefore [F1] applies to the displayed functions.

The central pressure ratio is $\frac{p(0)}{{ϵ}_0}=\frac{1-1\mathrm{/}\sqrt{2}}{3\mathrm{/}\sqrt{2}-1}=\frac{2\sqrt{2}-1}{7}>0,$ and the central lapse is $(3\mathrm{/}\sqrt{2}-1)\mathrm{/}2>0$; the surface pressure is $p(R)=0$ and surface lapse is $1\mathrm{/}\sqrt{2}$. The theorem gives the regular Cartesian center and no-layer matching to the vacuum exterior.

Fix ${\rho }_0={ϵ}_0\mathrm{/}{c}^2>0$. For integers $n\ge 2$, set $u_n=8\mathrm{/}9-1\mathrm{/}n$ and $R_n=c^2\sqrt{3u_n\mathrm{/}(8\pi G{ϵ}_0)}>0$. With $M_n=4\pi {\rho }_0{R}_n^3\mathrm{/}3$, direct substitution gives $2G{M}_n\mathrm{/}(c^2{R}_n)=8\pi G{ϵ}_0{R}_n^2\mathrm{/}(3c^4)=u_n\in (0,8\mathrm{/}9)$. Thus each $({\rho }_0,R_n)$ supplies an admissible regular star of [F1], rather than varying $u$ at fixed radius. By [F2], after reindexing $k=n-1$, $1\mathrm{/}n=1\mathrm{/}(k+1)\to 0$. Set $a_n=\sqrt{1-u_n}=\sqrt{1\mathrm{/}9+1\mathrm{/}n}>1\mathrm{/}3$; then $0<a_n-\frac{1}{3}=\frac{1\mathrm{/}n}{a_n+1\mathrm{/}3}\le \frac{3}{2n},$ For any real $\eta >0$, [F3] gives rational $q$ with $0<q<2\eta \mathrm{/}3$. By [F2], eventually $1\mathrm{/}n<q$, so $\mathrm{\mid }{a}_n-1\mathrm{/}3\mathrm{\mid }\le 3\mathrm{/}(2n)<3q\mathrm{/}2<\eta$. This proves $a_n\to 1\mathrm{/}3$ directly from the convergence definition, including all rational positive error thresholds.

For $n\ge 9$, step 1.1 gives $a_n\le 1\mathrm{/}2$, so ${ϵ}_0(1-a_n)\ge {ϵ}_0\mathrm{/}2$ and $0<3a_n-1\le 9\mathrm{/}(2n)$. Hence $p_c(u_n)\ge {ϵ}_{0}n\mathrm{/}9$. Given $M>0$, apply [F3] to $(0,{ϵ}_0\mathrm{/}(9M))$ to choose a rational $q>0$ with $q<{ϵ}_0\mathrm{/}(9M)$; by [F2], for all sufficiently large $n$, $1\mathrm{/}n<q$, so $p_c(u_n)>M$. Also $0<e^{\varphi (0)}(u_n)=(3a_n-1)\mathrm{/}2\le 9\mathrm{/}(4n)$; for any $\eta >0$, apply [F3] to $(0,4\eta \mathrm{/}9)$ to choose rational $q>0$ with $q<4\eta \mathrm{/}9$ and apply [F2] to get $e^{\varphi (0)}(u_n)<\eta$ eventually. Thus the central pressure diverges to $+\mathrm{\infty }$ and the central lapse tends to zero. At $u=8\mathrm{/}9$ the denominator is exactly zero with positive numerator, so no finite central pressure or positive central lapse is obtained by extending these formulas.

Choose $u=1\mathrm{/}2$. The center and the interior point $r=R\mathrm{/}2$ both have energy density ${ϵ}_0$, while their pressures differ. By [F1], $B(0)=1>B(R\mathrm{/}2)=\sqrt{7\mathrm{/}8}>1\mathrm{/}\sqrt{2}=a$; the denominators $3a-B(0)$ and $3a-B(R\mathrm{/}2)$ are positive. Subtracting the exact pressure formulas gives $\frac{p(0)-p(R\mathrm{/}2)}{{ϵ}_0}=\frac{2a(B(0)-B(R\mathrm{/}2))}{(3a-B(0))(3a-B(R\mathrm{/}2))}>0,$ so $p(0)>p(R\mathrm{/}2)>0$.

If a single-valued relation $p=P(ϵ)$ described this full static configuration, it would require both $P({ϵ}_0)=p(0)$ and $P({ϵ}_0)=p(R\mathrm{/}2)$, contradicting $p(0)>p(R\mathrm{/}2)$. Thus the static solution does not supply a single-valued map $P$ satisfying the stated relation, so no value of $P^{\mathrm{\prime }}({ϵ}_0)$, finite sound-speed parameter, or causal bound can be inferred from this solution alone.