Constrained entropy criticality, Tolman–Klein equivalence and TOV consequence

For a smooth compactly supported variation $(\dot{ϵ},\dot{n})$ in $(0,R)$, differentiating the mass definition and $b=1-2GM\mathrm{/}(c^{2}r)$ gives $\dot{E}=4\pi {\int }_0^R\dot{ϵ}{r}^{2}dr,\dot{b}(r)=-\frac{2K}{r}{\int }_0^r{t}^2\dot{ϵ}(t)dt,\frac{d}{d\lambda }{\mid }_0{b}_{\lambda }(r{)}^{-1\mathrm{/}2}=\frac{K}{r}b(r{)}^{-3\mathrm{/}2}{\int }_0^r{t}^2\dot{ϵ}(t)dt\mathrm{.}$ The compact-range and positive-$b$ hypotheses give uniform differentiability under the integrals by [F6]. Small multiples of these variations remain admissible because the profile range is compactly contained in the open chart and $b\ge b_{\ast }>0$.

These first variations also apply to every admissible sup-norm $C^1$ profile curve, with the same tangent at zero. Indeed the linear operator $Lf(r):=-2K{r}^{-1}{\int }_0^r{t}^{2}f(t)dt$ for $r>0$, with $Lf(0)=0$, maps $C^0([0,R])$ to itself and satisfies $\mathrm{\parallel }Lf{\mathrm{\parallel }}_{\mathrm{\infty }}\le (2K{R}^2\mathrm{/}3)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Applying this bound to difference quotients shows that $b_{\lambda }=1+L{ϵ}_{\lambda }$ is $C^1$ in the sup norm with derivative $L{ϵ}_{\lambda }^{\mathrm{\prime }}$. Evaluation of a sup-norm $C^1$ curve and of its derivative gives jointly continuous functions of $(\lambda ,r)$: continuity follows by splitting the change into a sup-norm parameter change and a spatial change at a fixed parameter. On a sufficiently small closed parameter interval the profiles stay in a common compact subset of $\mathcal{U}$ and $b_{\lambda }\ge b_{\ast }\mathrm{/}2$, by sup-norm continuity. Consequently the integrands defining $S,N_b,E$ and their parameter derivatives are jointly continuous, using the $C^1$ state functions and the positive-base power rule. The compact-rectangle differentiation theorem in [F6] therefore applies directly to these curves, so the first variations depend only on the tangent at zero, with no assumption of affine dependence on $\lambda$.

Using $s_{ϵ}=1\mathrm{/}\mathrm{\Theta }$ and $s_n=-\mu \mathrm{/}\mathrm{\Theta }$, the first variations are $\dot{S}=4\pi {\int }_0^R{r}^2[b^{-1\mathrm{/}2}(\frac{\dot{ϵ}}{\mathrm{\Theta }}-\frac{\mu }{\mathrm{\Theta }}\dot{n})+s\frac{d}{d\lambda }{\mid }_0{b}_{\lambda }^{-1\mathrm{/}2}]dr,{\dot{N}}_b=4\pi {\int }_0^R{r}^2[b^{-1\mathrm{/}2}\dot{n}+n\frac{d}{d\lambda }{\mid }_0{b}_{\lambda }^{-1\mathrm{/}2}]dr\mathrm{.}$ For a nonnegative bump ${\eta }_n\equiv ̸0$, the continuous integrand ${\eta }_n{r}^2{b}^{-1\mathrm{/}2}$ is nonnegative and nonzero on $(0,R)$, so [F10] gives ${\dot{N}}_b=4\pi \int {\eta }_n{r}^2{b}^{-1\mathrm{/}2}dr>0$ for the variation $(0,{\eta }_n)$, whose $\dot{E}=0$. For another nonnegative bump ${\eta }_{ϵ}\equiv ̸0$, [F10] similarly gives $\dot{E}=4\pi \int {\eta }_{ϵ}{r}^{2}dr>0$. Hence the derivative map $(\dot{ϵ},\dot{n})\mapsto (\dot{E},{\dot{N}}_b)$ is onto ${\mathbb{R}}^2$.

Fix any tangent variation $(\dot{ϵ},\dot{n})$ with $\dot{E}={\dot{N}}_b=0$, and choose the two bump directions from step 2.1 whose constraint derivative columns are linearly independent. Define $G(\lambda ,a_{ϵ},a_n)$ to be the pair of constraint changes $(E-E[ϵ,n],N_b-N_b[ϵ,n])$ along the profiles $(ϵ+\lambda \dot{ϵ}+a_{ϵ}{\eta }_{ϵ},n+\lambda \dot{n}+a_n{\eta }_n)$. The compact-in-chart and positive-$b$ hypotheses give an open parameter neighbourhood where these profiles remain admissible; [F6] makes $G$ a $C^1$ map there. Its derivative in $(a_{ϵ},a_n)$ at $(0,0,0)$ has the two independent columns from step 2.1 and is invertible. Apply the $C^1$ implicit function theorem [F4] with path parameter $\lambda$ and correction variables $(a_{ϵ},a_n)$. It supplies $C^1$ corrections $a_{ϵ}(\lambda ),a_n(\lambda )$ with $a_{ϵ}(0)=a_n(0)=0$ that keep both constraints fixed. Differentiating the constraint equations at zero and using $\dot{E}={\dot{N}}_b=0$ gives the invertible correction matrix times $(a_{ϵ}^{\mathrm{\prime }}(0),a_n^{\mathrm{\prime }}(0))=0$, so both correction derivatives vanish. The correction path remains admissible for small $\lambda$, and its derivative at zero is the given smooth compactly supported tangent. Therefore constrained stationarity implies $\dot{S}=0$ for every tangent variation in the kernel of $(\dot{E},{\dot{N}}_b)$.

Put $q=b^{-1\mathrm{/}2}$ and $\mathcal{L}=S+\alpha N_b-\beta E$. Substituting the variations from steps 1.1 and 2.1 and integrating by parts in the single radial variable gives $\dot{\mathcal{L}}=4\pi {\int }_0^R{r}^2[q(\alpha -\frac{\mu }{\mathrm{\Theta }})\dot{n}+\{\frac{q}{\mathrm{\Theta }}+K{\int }_r^R\frac{t(s+\alpha n)}{b(t{)}^{3\mathrm{/}2}}dt-\beta \}\dot{ϵ}]dr\mathrm{.}$ For the nonlocal term, set $Q(r)={\int }_0^r{t}^2\dot{ϵ}(t)dt$ and $H(r)=K{\int }_r^{R}t(s+\alpha n)b^{-3\mathrm{/}2}dt$. By [F8]–[F9], $Q,H$ are differentiable with $Q^{\mathrm{\prime }}(r)=r^2\dot{ϵ}(r)$ and $H^{\mathrm{\prime }}(r)=-Kr(s+\alpha n)b^{-3\mathrm{/}2}$. Also $H(R)=Q(0)=0$, so integration by parts gives $K{\int }_0^{R}r(s+\alpha n)b^{-3\mathrm{/}2}Q(r)dr={\int }_0^R{r}^{2}H(r)\dot{ϵ}(r)dr$. Thus the displayed formula has no omitted boundary term.

Because the constraint derivative is onto and $\dot{S}$ vanishes on its kernel, finite-dimensional linear algebra gives constants $\alpha ,\beta$ such that $\dot{S}+\alpha {\dot{N}}_b-\beta \dot{E}=0$ for every compactly supported variation. Conversely, if such multipliers exist, then $\dot{S}=0$ for every variation with $\dot{E}={\dot{N}}_b=0$. The multiplier argument uses only the two bump directions and finite-dimensional algebra.

The two bracket coefficients in step 3.2 are continuous. Each can be tested against every smooth compactly supported variation; if a coefficient were positive or negative at an interior point, continuity and [F5] would give a nonnegative bump supported where it keeps that sign. The resulting continuous integrand, or its negative in the negative case, is nonnegative and not identically zero, so [F10] makes its integral strictly positive; this contradicts the vanishing first variation. Since $q>0$, this gives $\mu \mathrm{/}\mathrm{\Theta }=\alpha$ and $\frac{q}{\mathrm{\Theta }}+K{\int }_r^R\frac{t(s+\alpha n)}{b(t{)}^{3\mathrm{/}2}}dt=\beta \mathrm{.}$ By the Euler identity in [F2], $s+\alpha n=(ϵ+p)\mathrm{/}\mathrm{\Theta }$.

Differentiate the last identity on $(0,R)$; [F9] differentiates its variable-endpoint integral, and [F7] differentiates $q=b^{-1\mathrm{/}2}$, giving $\frac{{\mathrm{\Theta }}^{\mathrm{\prime }}}{\mathrm{\Theta }}=-\frac{b^{\mathrm{\prime }}}{2b}-\frac{Kr(ϵ+p)}{b}\mathrm{.}$ The mass and $b^{\mathrm{\prime }}$ identities in [F3] yield $b^{\mathrm{\prime }}=2GM\mathrm{/}(c^2{r}^2)-2Krϵ$, so $\frac{{\mathrm{\Theta }}^{\mathrm{\prime }}}{\mathrm{\Theta }}=-\frac{GM\mathrm{/}(c^2{r}^2)+Krp}{b}=-{\varphi }^{\mathrm{\prime }}\mathrm{.}$ Thus $\mathrm{\Theta }{e}^{\varphi }$ is constant. Since $\mu \mathrm{/}\mathrm{\Theta }=\alpha$, $\mu e^{\varphi }=\alpha \mathrm{\Theta }{e}^{\varphi }$ is constant as well. At $r=R$, the boundary normalization makes $e^{\varphi (R)}=\sqrt{b(R)}$, and the energy bracket has integral zero there; hence $\beta =1\mathrm{/}(\mathrm{\Theta }(R)\sqrt{b(R)})=1\mathrm{/}{\mathrm{\Theta }}_K$.

Differentiating $ϵ+p=\mathrm{\Theta }s+\mu n$ using the calculus rules [F7] and substituting $dϵ=\mathrm{\Theta }ds+\mu dn$ gives $dp=sd\mathrm{\Theta }+nd\mu$. Since $\mu \mathrm{/}\mathrm{\Theta }=\alpha$, ${\mu }^{\mathrm{\prime }}=\alpha {\mathrm{\Theta }}^{\mathrm{\prime }}$, so $p^{\mathrm{\prime }}=s{\mathrm{\Theta }}^{\mathrm{\prime }}+n{\mu }^{\mathrm{\prime }}=(s+\alpha n){\mathrm{\Theta }}^{\mathrm{\prime }}=\frac{ϵ+p}{\mathrm{\Theta }}{\mathrm{\Theta }}^{\mathrm{\prime }}=-(ϵ+p){\varphi }^{\mathrm{\prime }}\mathrm{.}$ Substituting the lapse equation from [F3] gives the TOV pressure equation.

Conversely, suppose the stated Tolman and Klein products are constant with the boundary normalization. Then $\mu \mathrm{/}\mathrm{\Theta }$ is constant. The Euler and differential identities in [F2] imply $p^{\mathrm{\prime }}=-(ϵ+p){\varphi }^{\mathrm{\prime }}$, so the TOV pressure equation follows from [F3]. Reversing the algebra in step 5.1 shows that the energy bracket in step 4.2 has derivative zero; its boundary value at $R$ is $\beta =1\mathrm{/}{\mathrm{\Theta }}_K$, so it equals $\beta$ throughout. The number bracket vanishes because $\mu \mathrm{/}\mathrm{\Theta }=\alpha$. Therefore step 3.2 gives $\dot{\mathcal{L}}=0$ for every variation, and any constraint-preserving variation has $\dot{S}=0$. This proves constrained stationarity and both directions.