An exact turning ray and the critical Schwarzschild impact parameter

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.