Circular timelike orbit and reduced radial stability — Examples

Use $s>0$ and the reduced potential $W_{\mathrm{\ell }}(r)=(1-s\mathrm{/}r)(c^2+{\mathrm{\ell }}^2\mathrm{/}{r}^2)$ of Lyapunov stability of Schwarzschild circular orbits in the fixed-angular-momentum radial system. At a circular orbit, ${\mathrm{\ell }}^2=c^{2}s{r}_0^2\mathrm{/}(2r_0-3s)$.

1. Stable radial equilibrium at $r_0=4s$. The circular value is ${\mathrm{\ell }}^2=16c^2{s}^2\mathrm{/}5$, and

   $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}(4s)=\frac{2c^{2}s(4s-3s)}{(4s{)}^3(8s-3s)}=\frac{c^2}{160s^2}>0.$

   The fixed-$\mathrm{\ell }$ radial equilibrium is Lyapunov stable.

2. Unstable radial equilibrium at $r_0=2s$. Here ${\mathrm{\ell }}^2=4c^2{s}^2$ and

   $W_{\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}(2s)=\frac{2c^{2}s(2s-3s)}{(2s{)}^3(4s-3s)}=-\frac{c^2}{4s^2}<0.$

   The fixed-$\mathrm{\ell }$ equilibrium is unstable.

3. Degenerate endpoint at $r_0=3s$. Here ${\mathrm{\ell }}^2=3c^2{s}^2$ and $W_{\mathrm{\ell }}^{\mathrm{\prime }}(r)=c^{2}s(r-3s{)}^2\mathrm{/}{r}^4$. It has a stationary inflection rather than a strict extremum, and the radial equilibrium is not Lyapunov stable by the one-sided perturbation argument in the theorem.

Use the SI Schwarzschild convention of the scattering model: $s=2G_N{m}_{\ast }\mathrm{/}{c}^2>0$ and the areal radius $r>s$ are lengths in metres, with $m_{\ast }$ in kilograms, $G_N$ in cubic metres per kilogram per squared second, and $c$ in metres per second. The impact parameters $b,b_c$ are also lengths in metres; $H(r)=(1-s\mathrm{/}r)\mathrm{/}{r}^2$ has units inverse square metres, so $b^{2}H(r)$ is dimensionless. No numerical mass or measured length is stipulated. Take the incoming equatorial future-null rays with $E>0$ and $L\ge 0$ of the scattering supplier, so $b=L\mathrm{/}E=\mathrm{\mid }L\mathrm{\mid }\mathrm{/}E$. The critical impact parameter is $b_c=3\sqrt{3}s\mathrm{/}2$. Choose $b=2\sqrt{2}s>b_c$. The radial factor is $1-b^2(1-s\mathrm{/}r)\mathrm{/}{r}^2$.

1. At $r=2s$, this factor is $1-(8s^2)(1\mathrm{/}2)\mathrm{/}(4s^2)=0$. Because $2s>3s\mathrm{/}2$ and $H(r)=(1-s\mathrm{/}r)\mathrm{/}{r}^2$ is strictly decreasing for $r>3s\mathrm{/}2$, $r_{\ast }=2s$ is the unique outer turning point. The ideal incoming ray turns there and returns outward.

2. At the critical value, direct factorization gives

   $1-b_c^{2}H(r)=\frac{(2r-3s{)}^2(r+3s)}{4r^3}\mathrm{.}$

   Thus the critical inward ray from infinity approaches $r=3s\mathrm{/}2$ asymptotically, and the azimuthal integral diverges logarithmically; it does not complete a finite number of windings before reaching the circular orbit in affine parameter.

Let $(M,g,N)$ be the adopted Schwarzschild exterior datum with $s>0$ and $f(r)=1-s\mathrm{/}r$. Compare ideal static clocks at radii $r_E=2s$ and $r_R=4s$ over a common Schwarzschild coordinate-time interval of duration $\mathrm{\Delta }t>0$. For the frequency ratio, choose a specified smooth, affinely parametrized future-directed null geodesic $\eta :J\to M$ in the exterior with endpoint parameters ${\lambda }_E<{\lambda }_R$ on those static worldlines, and use the ideal geometric-optics frequencies assigned by the endpoint static observers. Here $f(2s)=1\mathrm{/}2$ and $f(4s)=3\mathrm{/}4$.

1. The exact model ratios are

   $\frac{{\omega }_R}{{\omega }_E}=\sqrt{\frac{f(2s)}{f(4s)}}=\sqrt{\frac{2}{3}},\frac{\mathrm{\Delta }{\tau }_R}{\mathrm{\Delta }{\tau }_E}=\sqrt{\frac{f(4s)}{f(2s)}}=\sqrt{\frac{3}{2}}\mathrm{.}$

   The static proper acceleration at $2s$ is $a(2s)=c^2\mathrm{/}(4\sqrt{2}s)$.

2. Write $r=s(1+ϵ)$ for $ϵ>0$. The exact static proper acceleration is

   $a(r)=\frac{c^2}{2s\sqrt{ϵ}(1+ϵ{)}^{3\mathrm{/}2}},$

   which tends to $+\mathrm{\infty }$ as $ϵ\downarrow 0$. In contrast, the curvature scalar has limit $K\to 12\mathrm{/}{s}^4$ by Orthonormal-frame curvature components and the Kretschmann scalar 12s^2/r^6 of the Schwarzschild exterior metric, and the advanced extension chart is smooth through $r=s$ by Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart. The acceleration divergence is therefore a property of maintaining a static worldline near the boundary, not evidence of a curvature blow-up there.

Fix $m,\kappa ,K,b>0$ and let $c\to \mathrm{\infty }$ with $W_c=m{c}^2+K$ and $J_c=b\sqrt{A_c}$, where $A_c=(W_c^2-m^2{c}^4)\mathrm{/}{c}^2$. For sufficiently large $c$, $J_c>\kappa \mathrm{/}c$ because $J_c^2-{\kappa }^2\mathrm{/}{c}^2\to 2mK{b}^2>0$.

1. The relativistic scattering angle tends to the nonrelativistic repulsive Coulomb angle:

   ${\theta }_c⟶2\arctan \left(\frac{\kappa }{2Kb}\right)\mathrm{.}$

   This agrees with Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $\mathrm{\Delta }=\pi \mathrm{/}2-\arctan (\kappa \mathrm{/}(2Eb))$ at kinetic energy $K$ and impact parameter $b$.

2. The limit compares ideal mathematical model formulas while holding the indicated parameters fixed; it supplies no experimental evidence for either model.