general relativity - What is the capture cross-section of a black hole region for ultra-relativistic particles?

What is the capture cross-section of a black hole region for ultra-relativistic particles? I have read that it is

$$\sigma ~=~ \frac{27}{4}\pi R^{2}_{s}$$

for a Schwarzschild BH in the geometric optics limit. Where does the coefficient come from?

Edit - Sources:

1. Absorption and emission spectra of a Schwarzschild black hole.


2. Fermion absorption cross section of a Schwarzschild black hole.

Answer

What follows is a very rough adaption of chapter 25 in Gravitation by Misner, Thorne, and Wheeler.

Begin with the Schwarzschild metric with polar angle $\theta$ fixed at $\pi/2$: $$ ds^2 = -\left(1 - \frac{R_\mathrm{S}}{r}\right) \mathrm{d}t^2 + \frac{1}{1-R_\mathrm{S}/r} \mathrm{d}r^2 + r^2 \mathrm{d}\phi^2. $$ For a test particle of rest mass $m$,¹ we know by definition $$ g_{\mu\nu} p^\mu p^\nu + m^2 = 0, $$ where $\vec{p}$ is the 4-momentum of the particle. For an affine parameter $\lambda$ parametrizing the worldline of the particle, these two equations can be combined to give $$ -\left(1 - \frac{R_\mathrm{S}}{r}\right) \left(\frac{\mathrm{d}t}{\mathrm{d}\lambda}\right)^2 + \frac{1}{1-R_\mathrm{S}/r} \left(\frac{\mathrm{d}r}{\mathrm{d}\lambda}\right)^2 + r^2 \left(\frac{\mathrm{d}\phi}{\mathrm{d}\lambda}\right)^2 + m^2 = 0 $$

Now the derivative in the first term is simply the energy $E$, which is related to the conserved energy at infinity $E_\infty$ by $E = E_\infty/(1-R_\mathrm{S}/r)$. Furthermore, the definition of angular momentum is $L = r^2 (\mathrm{d}\phi/\mathrm{d}\lambda)$, which is also conserved. Inserting these definitions gives $$ -\frac{E_\infty^2}{1-R_\mathrm{S}/r} + \frac{1}{1-R_\mathrm{S}/r} \left(\frac{\mathrm{d}r}{\mathrm{d}\lambda}\right)^2 + \frac{L^2}{r^2} + m^2 = 0, $$ or $$ \left(\frac{\mathrm{d}r}{\mathrm{d}\lambda}\right)^2 = E_\infty^2 - \left(1 - \frac{R_\mathrm{S}}{r}\right) \left(\frac{L^2}{r^2} + m^2\right). $$ Using again the definition of $L$, we can write $$ \left(\frac{\mathrm{d}r}{\mathrm{d}\phi}\right)^2 = \frac{r^4}{L^2} \left(E_\infty^2 - \left(1 - \frac{R_\mathrm{S}}{r}\right) \left(\frac{L^2}{r^2} + m^2\right)\right), $$ which we can rewrite to be \begin{align} \left(\frac{\mathrm{d}r}{\mathrm{d}\phi}\right)^2 & = r^4 \frac{E_\infty^2-m^2}{L^2} \left(\frac{E_\infty^2}{E_\infty^2-m^2} - \left(1 - \frac{R_\mathrm{S}}{r}\right) \left(\frac{1}{r^2} \left(\frac{L^2}{E_\infty^2-m^2}\right) + \frac{m^2}{E_\infty^2-m^2}\right)\right) \\ & = \frac{r^4}{b^2} \left(\frac{E_\infty^2}{E_\infty^2-m^2} - \left(1 - \frac{R_\mathrm{S}}{r}\right) \left(\frac{b^2}{r^2} + \frac{m^2}{E_\infty^2-m^2}\right)\right), \end{align} where $$ b = \frac{L}{\sqrt{E_\infty^2-m^2}} $$ is the impact parameter, defined to be the ratio of angular to linear momentum.

At this point, we have a nice general result, but to apply it to photons we take the limit $m \to 0$, which gives $$ \left(\frac{\mathrm{d}r}{\mathrm{d}\phi}\right)^2 = \frac{r^4}{b^2} \left(1 - \frac{b^2}{r^2} \left(1 - \frac{R_\mathrm{S}}{r}\right)\right). $$ The radius of closest approach will be the value $r = r_\text{min}$ for which $\mathrm{d}r/\mathrm{d}\phi$ vanishes: \begin{gather} \frac{r_\text{min}^4}{b^2} \left(1 - \frac{b^2}{r_\text{min}^2} \left(1 - \frac{R_\mathrm{S}}{r_\text{min}}\right)\right) = 0; \\ \frac{b^2}{r_\text{min}^2} \left(1 - \frac{R_\mathrm{S}}{r_\text{min}}\right) = 1; \\ b^2 (r_\text{min} - R_\mathrm{S}) = r_\text{min}^3. \end{gather}

The only thing left is to decide what $r_\text{min}$ is allowed to be if the light is to escape. Since at this point in the trajectory the photon's velocity is entirely in the tangential direction by construction, the question becomes how close in can a photon have a circular orbit? The answer is $r_\text{min} = 3R_\mathrm{S}/2$. Thus $b_\text{max}$, the maximum impact parameter for a photon to be captured, obeys $$ b_\text{max}^2 \left(\frac{3}{2} R_\mathrm{S} - R_\mathrm{S}\right) = \left(\frac{3}{2} R_\mathrm{S}\right)^3, $$ or $$ b_\text{max}^2 = \frac{27}{4} R_\mathrm{S}^2. $$ This is exactly what we sought, since it immediately tells us $$ \sigma = \pi b_\text{max}^2 = \frac{27}{4} \pi R_\mathrm{S}^2. $$

---

¹ We could set $m = 0$ here, but by holding off we get slightly more general intermediate results.