Circular timelike orbit and reduced radial stability

By [F1], the reduced vector field is $F_{\mathrm{\ell }}(r,w)=(w,-W_{\mathrm{\ell }}^{\mathrm{\prime }}(r)\mathrm{/}2)$ on the open set $(s,\mathrm{\infty })\times \mathbb{R}$. It is smooth there and [F4] gives a unique local solution through every state; uniqueness lets one glue all solutions through the same initial state on the union of their intervals to obtain its maximal solution. The change of variables $(q,v,T)=(r\mathrm{/}s,w\mathrm{/}c,c\tau \mathrm{/}s)$ gives $dq\mathrm{/}dT=v$ and $dv\mathrm{/}dT=-s{W}_{\mathrm{\ell }}^{\mathrm{\prime }}(sq)\mathrm{/}(2c^2)$ on $(1,\mathrm{\infty })\times \mathbb{R}$. This invertible scaling preserves maximality, compact containment and existence for all nonnegative time. Differentiating $W_{\mathrm{\ell }}=f(c^2+{\mathrm{\ell }}^2\mathrm{/}{r}^2)$ gives $W_{\mathrm{\ell }}^{\mathrm{\prime }}=c^{2}s\mathrm{/}{r}^2-2{\mathrm{\ell }}^2\mathrm{/}{r}^3+3s{\mathrm{\ell }}^2\mathrm{/}{r}^4$; substituting the circular value of ${\mathrm{\ell }}^2$ from [F2] yields $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r_c)=0$, so $F_{\mathrm{\ell }}(r_c,0)=0$. Along every solution, $H_{\mathrm{\ell }}(r,w):=\frac{1}{2}(w^2+W_{\mathrm{\ell }}(r))$ satisfies $d{H}_{\mathrm{\ell }}\mathrm{/}d\tau =w(-W_{\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{/}2)+(W_{\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{/}2)w=0$. Thus $H_{\mathrm{\ell }}$ is constant on each solution interval.

Suppose $r_c>3s$. By [F2], $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}(r_c)>0$, so continuity gives an interval about $r_c$ on which $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}>0$. Since $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r_c)=0$, the derivative is negative to the left and positive to the right there; hence $W_{\mathrm{\ell }}$ has a strict local minimum. Fix any dimensionless phase-space radius $\epsilon >0$ whose closed ball about $(r_c\mathrm{/}s,0)$ lies in $(1,\mathrm{\infty })\times \mathbb{R}$. Choose a length $0<\eta <s\epsilon \mathrm{/}2$ so $[r_c-\eta ,r_c+\eta ]\subset (s,\mathrm{\infty })$ and both boundary values of $W_{\mathrm{\ell }}$ exceed $W_{\mathrm{\ell }}(r_c)$. By continuity of $H_{\mathrm{\ell }}$, choose $0<\delta \le \epsilon$ so every state at distance $d<\delta$ from $z_c$ has $\mathrm{\mid }r-r_c\mathrm{\mid }<\eta \mathrm{/}2$, energy below both boundary potential halves, and $H_{\mathrm{\ell }}-W_{\mathrm{\ell }}(r_c)\mathrm{/}2<c^2{\epsilon }^2\mathrm{/}8$. Conservation in 1.1 prevents such a solution from reaching either radial boundary, since there $H_{\mathrm{\ell }}\ge W_{\mathrm{\ell }}(r_c\pm \eta )\mathrm{/}2$. Inside the resulting radial interval $W_{\mathrm{\ell }}(r)\ge W_{\mathrm{\ell }}(r_c)$, so $w^2\mathrm{/}2=H_{\mathrm{\ell }}-W_{\mathrm{\ell }}(r)\mathrm{/}2<c^2{\epsilon }^2\mathrm{/}8$ and $\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c<\epsilon \mathrm{/}2$. The state stays in a compact subset of the open domain; [F5] rules out a finite forward maximal endpoint. With $\mathrm{\mid }r-r_c\mathrm{\mid }\mathrm{/}s<\eta \mathrm{/}s<\epsilon \mathrm{/}2$ and $\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c<\epsilon \mathrm{/}2$, it remains inside the prescribed phase-space ball for every $\tau \ge 0$. This is Lyapunov stability by [F3].

Suppose $3s\mathrm{/}2<r_c<3s$. By [F2], $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}(r_c)<0$; shrink $\eta >0$ so $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r)<0$ on $(r_c,r_c+\eta ]$. For any proposed initial-state radius $\delta >0$, choose $r_0\in (r_c,r_c+\min \{\eta \mathrm{/}2,s\delta \mathrm{/}2\})$ and initial data $(r_0,0)$. [F4] supplies a local solution. Its acceleration is $-W_{\mathrm{\ell }}^{\mathrm{\prime }}(r_0)\mathrm{/}2>0$, and conservation of $H_{\mathrm{\ell }}$ gives, on the outgoing branch, $w=\sqrt{W_{\mathrm{\ell }}(r_0)-W_{\mathrm{\ell }}(r)}$. The travel-time integral to $r_c+\eta$ is ${\int }_{r_0}^{r_c+\eta }[W_{\mathrm{\ell }}(r_0)-W_{\mathrm{\ell }}(r){]}^{-1\mathrm{/}2}dr$, whose only endpoint singularity is bounded by a constant times $(r-r_0{)}^{-1\mathrm{/}2}$ because $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r_0)<0$. Before reaching that radius, the energy identity bounds $w$ on the compact radial interval $[r_0,r_c+\eta ]$, so the state stays in a compact subset of $(s,\mathrm{\infty })\times \mathbb{R}$ near any finite maximal endpoint. The compact-continuation result [F5] therefore prevents finite-time termination before the finite travel time; the solution reaches $r_c+\eta$ and leaves the fixed $d$-ball of dimensionless radius $\eta \mathrm{/}s$. Since $d((r_0,0),z_c)=\mathrm{\mid }{r}_0-r_c\mathrm{\mid }\mathrm{/}s$ can be made smaller than every proposed $\delta$, this contradicts [F3].

At $r_c=3s$, the circular relation gives ${\mathrm{\ell }}^2=3c^2{s}^2$. Differentiating $W_{\mathrm{\ell }}(r)=c^2(1-s\mathrm{/}r)+{\mathrm{\ell }}^2(r^{-2}-s{r}^{-3})$ at this fixed $\mathrm{\ell }$ gives $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r)=c^{2}s(r-3s{)}^2\mathrm{/}{r}^4$ and $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}(r)=-2c^{2}s(r-3s)(r-6s)\mathrm{/}{r}^5$. Thus $W_{\mathrm{\ell }}^{\mathrm{\prime }}$ is positive except at $3s$ and $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}$ changes sign there, so the equilibrium is a stationary inflection. Fix $0<\eta <s$ and, for any $\delta >0$, choose $r_0\in (3s-\min \{\eta \mathrm{/}2,s\delta \mathrm{/}2\},3s)$ with initial data $(r_0,0)$. [F4] supplies a local solution. The acceleration is negative. On the inward branch energy conservation yields $w=-\sqrt{W_{\mathrm{\ell }}(r_0)-W_{\mathrm{\ell }}(r)}$ for $3s-\eta \le r<r_0$, since $W_{\mathrm{\ell }}$ is strictly increasing on that interval. The travel-time integral to $3s-\eta$ is finite; near $r_0$ its integrand is bounded by a constant times $(r_0-r{)}^{-1\mathrm{/}2}$ because $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r_0)>0$. The energy identity bounds $w$ while $r$ stays in the compact interval $[3s-\eta ,r_0]\subset (s,\mathrm{\infty })$, so [F5] extends the solution past any finite maximal endpoint before its finite travel time. It therefore reaches $3s-\eta$, leaving the fixed $d$-ball of dimensionless radius $\eta \mathrm{/}s$. Since $d((r_0,0),z_c)=\mathrm{\mid }{r}_0-r_c\mathrm{\mid }\mathrm{/}s$ can be made smaller than every proposed $\delta$, [F3] says the equilibrium is not Lyapunov stable.

The two subcritical cases in 2.2–2.3 are not Lyapunov stable, while the supercritical case in 2.1 is Lyapunov stable. These are deductions under the adopted Schwarzschild free-fall model [A1] and concern only its fixed-angular-momentum radial subsystem.

In each equatorial spherical chart the components in [F1] are independent of $x^0$ and $\phi$, and the coordinate components of ${\mathrm{\partial }}_0$ and the axial generator ${\mathrm{\partial }}_{\phi }$ are constant. The flow pullback derivative in [F2] therefore gives ${\mathcal{L}}_{{\mathrm{\partial }}_0}g={\mathcal{L}}_{{\mathrm{\partial }}_{\phi }}g=0$, so both fields are locally Killing. By the conservation part of [F2] and the affine geodesic equation [F3], $-g({\mathrm{\partial }}_0,{\gamma }^{\mathrm{\prime }})=fd{x}^0\mathrm{/}d\lambda =E$ and $g({\mathrm{\partial }}_{\phi },{\gamma }^{\mathrm{\prime }})=r^{2}d\phi \mathrm{/}d\lambda =L$ are constant on each chart interval; longitude charts differ by additive constants, so these values agree on overlaps and are constant along $\gamma$.

At $\vartheta =\pi \mathrm{/}2$, the null condition and [F1] give $0=-f{\left(\frac{d{x}^0}{d\lambda }\right)}^2+f^{-1}{\left(\frac{dr}{d\lambda }\right)}^2+r^2{\left(\frac{d\phi }{d\lambda }\right)}^2\mathrm{.}$ A nonzero null tangent cannot have $d{x}^0\mathrm{/}d\lambda =0$, since the remaining quadratic form is positive definite. Its time component has constant sign on $J$ by continuity; the future orientation relative to $N={\mathrm{\partial }}_0$ gives $-g(N,{\gamma }^{\mathrm{\prime }})=E>0$. Parameter reversal flips both $E$ and $L$, leaving their ratio invariant. Substitute $d{x}^0\mathrm{/}d\lambda =E\mathrm{/}f$ and $d\phi \mathrm{/}d\lambda =L\mathrm{/}{r}^2$, then multiply by $f>0$. The result is $(dr\mathrm{/}d\lambda {)}^2=E^2[1-(L\mathrm{/}E{)}^{2}f\mathrm{/}{r}^2]=E^2[1-b^{2}H(r)]$, valid for either sign of $L$.

Differentiating $H(r)=r^{-2}-s{r}^{-3}$ yields $H^{\mathrm{\prime }}(r)=-2r^{-3}+3s{r}^{-4}=(3s-2r)\mathrm{/}{r}^4$. Since $r^4>0$, this derivative is positive on $(s,3s\mathrm{/}2)$ and negative on $(3s\mathrm{/}2,\mathrm{\infty })$; the mean value theorem in [F4] gives strict increase and decrease on these respective intervals, hence the unique maximum is at $3s\mathrm{/}2$. Substitution gives $H(3s\mathrm{/}2)=4\mathrm{/}(27s^2)$, and solving $b^{2}H(3s\mathrm{/}2)=1$ gives $b_{\mathrm{c}}=3\sqrt{3}s\mathrm{/}2$.

The right side of the radial first integral vanishes at a radius precisely when $b^{2}H(r)=1$. Since $H$ is positive on $r>s$, tends to zero at both ends, and has the maximum computed in the preceding step, no root exists for $0\le b<b_{\mathrm{c}}$, whereas $b=b_{\mathrm{c}}$ gives the root $3s\mathrm{/}2$. For $b>b_{\mathrm{c}}$, choose $R>3s\mathrm{/}2$ so large that $H(R)<1\mathrm{/}{b}^2<H(3s\mathrm{/}2)$; continuity and the intermediate value theorem in [F4] on $[3s\mathrm{/}2,R]$ give an exterior root. Thus the threshold for the existence of an exterior root is exactly $b_{\mathrm{c}}$. This is a first-integral statement only; it asserts no global scattering continuation.

Choose $R>r_{\mathrm{p}\mathrm{h}}$ so $b^{2}H(R)<1$. At $r=R$, set $\vartheta =\pi \mathrm{/}2$, ${\dot{x}}^0=E\mathrm{/}f(R)$, $\dot{r}=-E\sqrt{1-b^{2}H(R)}$, $\dot{\vartheta }=0$, and $\dot{\phi }=L\mathrm{/}{R}^2$, with $L=bE$. Substitution into the diagonal Schwarzschild metric gives $g(\dot{\eta },\dot{\eta })=-\frac{E^2}{f(R)}+\frac{E^2}{f(R)}(1-b^{2}H(R))+\frac{L^2}{R^2}=0.$ Also $g(\dot{\eta },{\mathrm{\partial }}_0)=-E<0$, so the initial tangent is future directed, and [F2] gives a unique local geodesic. It remains equatorial: define the sphere reflection $S(y_1,y_2,y_3)=(y_1,y_2,-y_3)$ and $\mathrm{\Psi }(x^0,r,\omega )=(x^0,r,S\omega )$. The reflection is orthogonal, so it preserves the induced round metric; [F4] therefore makes $\mathrm{\Psi }$ an isometry of the exterior metric. Pulling back the metric-selected connection by $\mathrm{\Psi }$ gives another torsion-free, metric-compatible connection, hence the same connection by uniqueness in [F4]; thus $\mathrm{\Psi }\circ \eta$ is an affinely parametrized geodesic. At the initial event $y_3=0$, and $\dot{\vartheta }=0$ makes the initial tangent lie in the fixed tangent plane of $S$, so $\mathrm{\Psi }$ fixes both initial data. Geodesic uniqueness [F2] gives $\mathrm{\Psi }\circ \eta =\eta$ on the connected maximal interval, and its image lies in the fixed set $y_3=0$, the equatorial plane. Finally [F5] preserves the initial null norm along the geodesic, so it is null throughout.

By [F1], along the equatorial null geodesic ${\dot{r}}^2=E^2(1-b^{2}H(r))$. For $r\ge R$, $H$ is decreasing and $b^{2}H(r)<1$, so the past-directed parameter branch has $\dot{r}<0$ and extends through every finite radial interval $[R,R^{\mathrm{\prime }}]$: the first integral bounds $\dot{r}$, while $f(r)\ge f(R)>0$ bounds ${\dot{x}}^0=E\mathrm{/}f(r)$ and $\dot{\phi }=L\mathrm{/}{r}^2$; compact continuation [F2] excludes a finite-parameter endpoint there. As $r\to \mathrm{\infty }$, $H(r)\to 0$ and $\dot{r}\to -E$, so the affine distance to infinity is ${\int }_R^{\mathrm{\infty }}dr\mathrm{/}[E\sqrt{1-b^{2}H(r)}]=\mathrm{\infty }$. Hence the constructed solution supplies an incoming ray from infinity for every $b\ge 0$.

By [F1], $H$ increases on $(s,r_{\mathrm{p}\mathrm{h}})$, decreases on $(r_{\mathrm{p}\mathrm{h}},\mathrm{\infty })$, and vanishes at both exterior ends. If $b>b_{\mathrm{c}}$, then $b^{2}H(r_{\mathrm{p}\mathrm{h}})>1$, so there is one root $r_{-}\in (s,r_{\mathrm{p}\mathrm{h}})$ and one simple outer root $r_{\ast }>r_{\mathrm{p}\mathrm{h}}$. The affine time from any finite large radius to $r_{\ast }$ is finite because near the simple root $[E\sqrt{1-b^{2}H(r)}{]}^{-1}=O((r-r_{\ast }{)}^{-1\mathrm{/}2})$.

If $b=b_{\mathrm{c}}$, direct factorization gives $1-b_{\mathrm{c}}^{2}H(r)=(2r-3s{)}^2(r+3s)\mathrm{/}(4r^3)$. The incoming radial speed vanishes linearly as $r\downarrow r_{\mathrm{p}\mathrm{h}}$; hence the affine integral diverges like $\int dr\mathrm{/}(r-r_{\mathrm{p}\mathrm{h}})$. The angular integrand has the same simple-pole behavior, so the ray approaches the photon sphere only at infinite affine parameter and accumulates logarithmically divergent azimuth.

If $0\le b<b_{\mathrm{c}}$, then $b^{2}H(r)<1$ for every $r>s$ by [F1]. The inward branch has no turning point and reaches the exterior boundary $r=s$ in finite affine parameter because $1-b^{2}H(r)\to 1$ there. The exterior calculation ends at that boundary. If a future continuation is placed in the Kruskal model and enters $U\ge 0$, [F3] prevents it from returning to the right exterior.

For $b>b_{\mathrm{c}}$, [F2] extends the incoming ray through the outer turn. Away from $r_{\ast }$, differentiating the radial first integral and dividing by $\dot{r}\mathrm{\ne }0$ gives $\ddot{r}=-E^2{b}^2{H}^{\mathrm{\prime }}(r)\mathrm{/}2$; smoothness of the geodesic makes $\ddot{r}$ continuous at the simple root, so this identity holds there as a limit. Since $H^{\mathrm{\prime }}(r_{\ast })<0$, the acceleration is positive and the ray leaves the turn with $\dot{r}>0$. On every finite segment with $r\ge r_{\ast }$, the conserved quantities bound $\mathrm{\mid }\dot{r}\mathrm{\mid }$, $\mathrm{\mid }{\dot{x}}^0\mathrm{\mid }=E\mathrm{/}f(r)$ and $\mathrm{\mid }\dot{\phi }\mathrm{\mid }=L\mathrm{/}{r}^2$; the state stays in a compact subset of the smooth exterior tangent domain and cannot end at a finite affine parameter by [F2]. On the outgoing branch $\dot{r}\to E>0$, while $\mathrm{\mid }\dot{r}\mathrm{\mid }\le E$, so it reaches arbitrarily large radii only at infinite affine parameter. Also $\mathrm{\mid }d\phi \mathrm{/}dr\mathrm{\mid }=b\mathrm{/}[r^2\sqrt{1-b^{2}H(r)}]$ is $O((r-r_{\ast }{)}^{-1\mathrm{/}2})$ at the turn and $O(r^{-2})$ at infinity; integrating both halves gives the convergent formula in clause 1.

The simple outer root, critical double root and no-turn regimes in 3.1–3.3 give all three clauses. Only the last clause uses the separate Kruskal continuation, and no observation or apparatus model is used.

By [F1], $U=cn$ is the static observer four-velocity because $g(U,U)=-c^2$. Since $c$ is constant, ${\mathrm{\nabla }}_{U}U=c^2{\mathrm{\nabla }}_{n}n=c^{2}s{\mathrm{\partial }}_r\mathrm{/}(2r^2)$. With $g({\mathrm{\partial }}_r,{\mathrm{\partial }}_r)=f^{-1}$ from [F3], its squared norm is $c^4{s}^2\mathrm{/}(4r^{4}f)$, so its positive magnitude is $c^{2}s\mathrm{/}(2r^2\sqrt{f})$. Substituting $s=2G_N{m}_{\ast }\mathrm{/}{c}^2$ gives the displayed Newton-constant form.

Over the same coordinate-time interval, [F1] gives $\mathrm{\Delta }{\tau }_R\mathrm{/}\mathrm{\Delta }{\tau }_E=\sqrt{f(r_R)\mathrm{/}f(r_E)}$. [F2] gives the inverse ratio for received to emitted ideal-ray frequency, so the two displayed exact ratios follow. Each ratio is a conditional model result for the specified static observers and ideal ray.

By [F4], $L^{\mathrm{\prime }}(y)=1\mathrm{/}y$ for $y>0$ and $L(1)=0$. For fixed $q>0$, the function $a\mapsto L(a\mathrm{/}q)-L(a)$ has derivative $1\mathrm{/}a-1\mathrm{/}a=0$ on $(0,\mathrm{\infty })$ by the chain rule. It is constant by [F6], and its value at $a=q$ is $-L(q)$, so $L(a\mathrm{/}q)=L(a)-L(q)$. Likewise $y\mapsto L(y^2)-2L(y)$ has derivative $2\mathrm{/}y-2\mathrm{/}y=0$ for $y>0$ and value zero at $1$, hence vanishes by [F6]. Setting $y=\sqrt{a\mathrm{/}q}>0$ yields $L(\sqrt{a\mathrm{/}q})=\frac{1}{2}[L(a)-L(q)]$. Taking $a=1-x_E$ and $q=1-x_R$ in the positive frequency ratio of [F2] establishes the logarithmic ratio identity used below.

For $0<x\le \rho <1$, the chain rule and [F4] show that $d\log (1-x)\mathrm{/}dx=-1\mathrm{/}(1-x)=-1-x\mathrm{/}(1-x)$. By [F4] the function $\log (1-x)+x+{\int }_0^{x}t\mathrm{/}(1-t)dt$ has derivative zero for $0<x<1$ and is continuous at $0$, where it vanishes. Applying [F6] on $[0,x]$ gives $\log (1-x)=-x-{\int }_0^x\frac{t}{1-t}dt,0\le {\int }_0^x\frac{t}{1-t}dt\le \frac{1}{1-\rho }{\int }_0^{x}tdt=\frac{x^2}{2(1-\rho )}\mathrm{.}$ For $t\in [0,x]$, $0\le t\mathrm{/}(1-t)\le t\mathrm{/}(1-\rho )$. Thus [F5] and ${\int }_0^{x}tdt=x^2\mathrm{/}2$ give the displayed bound. By [F2] and step 1.3, $\log ({\omega }_R\mathrm{/}{\omega }_E)=\frac{1}{2}[\log (1-x_E)-\log (1-x_R)]$. Inserting the integral identity separates the linear term $(x_R-x_E)\mathrm{/}2=s(1\mathrm{/}{r}_R-1\mathrm{/}{r}_E)\mathrm{/}2$. The remaining terms have absolute value at most half the sum of their nonnegative integral bounds, namely $(x_E^2+x_R^2)\mathrm{/}(4(1-\rho ))$.

The exact ratios, acceleration magnitude and uniform remainder bound are all consequences of the stated model identities; no empirical input is used.

Fix $x\mathrm{\ne }0$. By [F1] and [F2], sine is continuous on the closed interval with endpoints $0,x$ and differentiable in its interior. By [F3] there is ${\xi }_x$ strictly between $0$ and $x$ such that $\sin x\mathrm{/}x=(\sin x-\sin 0)\mathrm{/}(x-0)=\cos {\xi }_x$.

Given $\epsilon >0$, [F1] and [F2] make cosine continuous at $0$, so choose $\delta >0$ such that $\mathrm{\mid }t\mathrm{\mid }<\delta$ implies $\mathrm{\mid }\cos t-1\mathrm{\mid }<\epsilon$. If $0<\mathrm{\mid }x\mathrm{\mid }<\delta$, then $\mathrm{\mid }{\xi }_x\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }<\delta$ and step 1.1 yields $\mathrm{\mid }\sin x\mathrm{/}x-1\mathrm{\mid }=\mathrm{\mid }\cos {\xi }_x-1\mathrm{\mid }<\epsilon$. This is the stated limit.