Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.
share|improve this question
2  
can you give a link for this claim? –  anna v Jan 27 '13 at 15:11
    
Absolutely! See above. –  user12345 Jan 27 '13 at 15:42
1  
There is some background to it here (but still not a full calculation of how that number is obtained). Maybe a trawl though Chris Doran's publications would find it somewhere... –  twistor59 Jan 27 '13 at 16:32
    
This is actually left as an exercise (25.25) for the reader in MTW. I just spent an hour trying to solve it, but MTW is hopelessly confused in this section, with numerous inconsistent equations. I have no doubt the answer is right, but perhaps there is a text that gets at it in a better way. –  Chris White Jan 27 '13 at 22:09

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer

The cross section referred to in the papers is the overall cross section for light shining on the black hole. You might think that the cross section of the black hole is just the area of the event horizon i.e. $\pi r_s^2$. However any light approaching more closely than $r = 3r_s/2$ (the last stable orbit) will spiral into the black hole, so this increases the cross section. And finally, originally parallel light rays travelling from infinity get focused in towards the black hole so even a light ray further away than $r = 3r_s/2$ will be focused in and hit the last stable orbit. The end result is that for parallel light from infinity the black hole appears to have a cross section of $27\pi r_s^2/4$.

I say all this with great confidence, but I have absolutely no idea how to calculate this figure. I'm just quoting from this book. The book provides references is you want to dig in to how the calculation is done, but I would guess the calculations are messy and complicated.

Any particle moving fast enough will see a similar cross section to light.

share|improve this answer
    
shouldn't there be a dependence somewhere on the momentum of the incoming photon/particle? –  lurscher Jan 27 '13 at 20:59
    
@lurscher Massless particles all follow the same null geodesics, so being captured is independent of energy/frequency/momentum/wavelength/whatever. This finds the area of the section of an incoming wall of initially parallel photons that eventually hit the event horizon. –  Chris White Jan 27 '13 at 22:12
    
@ChrisWhite right. The momentum of the incoming particle will matter only when the wavelength becomes of the same magnitude as the schwarzschild radius, but for this classical analysis it is a valid simplification. –  lurscher Jan 27 '13 at 22:29
    
@lurscher Yes, there is a dependence on incoming speed, see page 31 of the link given in twistor59's comment. The formula above is for the limit of $v \rightarrow c$. –  user12345 Jan 28 '13 at 11:57
    
@John Your intuition with $3r_s/2$ was right (see my answer), but I'm still not entirely sure why that value and not some other. Also, that is the radius of the innermost circular orbit, which, at the maximum of the effective radial potential, is unstable. $3r_s$ is the RISCO, but for some reason the right answer requires $3r_s/2$. –  Chris White Jan 28 '13 at 23:14

1) Let us work in units where the speed-of-light $c=1$ is one.

In Ref. 1 is derived the radial geodesic equation for a particle in the equatorial plane

$$\tag{7.47} (\frac{dr}{d\lambda})^2+2V(r)~=~E^2, $$

with potential

$$ \tag{7.48} 2V(r)~:=~(1-\frac{r_s}{r})((\frac{L}{r})^2+\epsilon). $$

Here $\epsilon=0$ for a massless particle and $\epsilon=1$ for a massive particle. The energy $E$ and angular momentum $L$ are constants of motion (which reflect Killing-symmetries of the Schwarzschild metric); $\lambda$ is the affine parameter of the geodesic; and $r_s\equiv\frac{2GM}{c^2}$ is the Schwarzschild-radius. (More precisely, in the massive case $\epsilon=1$, the quantities $E$ and $L$ are specific quantities, i.e. quantities per unit rest mass; and $\lambda$ is proper time.)

2) By differentiating eq. (7.47) wrt. $\lambda$, we find that the condition for a circular orbit

$$r(\lambda)~\equiv~ r_{*} \qquad\Rightarrow\qquad \frac{dr}{d\lambda}~\equiv~0$$

is

$$\tag{1}V'(r_{*})~=~0\qquad\Leftrightarrow\qquad \frac{2r_{*}}{r_s}~=~3+\epsilon(\frac{r_{*}}{L})^2.$$

3) Let us next investigate an incoming particle, which has non-constant radial coordinate $\lambda\mapsto r(\lambda)$, and that is precisely on the critical border between being captured and not being captured by the black hole. It would have a radial turning point $\frac{dr}{d\lambda}=0$ precisely at the radius $r=r_{*}$, so that

$$\tag{2} 2V(r_{*})~=~E^2\qquad\Leftrightarrow\qquad (1-\frac{r_s}{r_{*}})((\frac{L}{r_{*}})^2+\epsilon)~=~E^2.$$

4) The massless case $\epsilon=0$. Eq. (1) yields

$$\tag{3}r_{*}~=~\frac{3}{2}r_s.$$

Plugging eq. (3) into eq. (2) then yields the ratio

$$\tag{4} \frac{L}{E}~=~\frac{3}{2}\sqrt{3}r_s. $$

We next use that $L$ and $E$ are constants of motion, so that we can easily identify them at spacial infinity $r=\infty$, where special relativistic formulas apply. The critical impact parameter $b$ is precisely this ratio

$$\tag{5} b~=~\frac{L}{p}~=~\frac{L}{E}~\stackrel{(4)}{=}~\underline{\underline{\frac{3}{2}\sqrt{3}r_s}}. $$

5) The non-relativistic case $v_{\infty}\ll 1$. The specific energy $E\approx 1$ consists mostly of rest energy. Solving eqs. (1) and (2) then leads to a unique solution

$\tag{6}r_{*}~\approx~ 2r_s~\approx~ L.$

The critical impact parameter $b$ becomes

$$\tag{5} b~=~\frac{L}{v_{\infty}}~\approx~\underline{\underline{2r_s\frac{c}{v_{\infty}}}}, $$

cf. Ref. 2. The cross section is $\sigma=\pi b^2$.

References:

  1. S. Carroll, Lecture Notes on General Relativity, Chapter 7, p.172-179. The pdf file is available from his website.

  2. V.P. Frolov and I.D. Novikov, Black Hole Physics: Basic Concepts and New Developments, p.48.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.