# Why is this the probability that an incoming wavepacket is absorbed by the black hole?

I'm reading Parker's book "Quantum Field Theory in Curved Spacetime: Quantized Fields and Gravity" and when talking about Hawking radiation, there's a claim that I've not been able to understand.

He says (page 169):

Recall that the quantity $$\Gamma(\omega)$$ is the fraction of a purely outgoing wavepacket that when propagated from $$\mathcal{I}^+$$ backward in time would enter the collapsing body just before it had formed a black hole. At sufficiently late times this fraction is the same as the fraction of the wave packet that would enter the black hole past event horizon if the collapsing body were replaced in the spacetime by the analytic extension of the black hole spacetime. This means that $$\Gamma_{lm}(\omega)$$ is also the probability that a purely incoming wavepacket that starts from $$\mathcal{I}^-$$ at late times will enter the black hole event horizon, that is, will be absorbed by the black hole.

I simply can't get what is going on here.

What is the point with this argument of replacing the gravitational collapse spacetime with the analytic extension of the black hole? How does it leads to the author's conclusions?

I also don't get why to compute the probability that one incoming wavepacket is absorbed we need to propagate back one outgoing wavepacket. Shouldn't we be studying the behavior exactly of one incoming wavepacket?

Why is $$\Gamma(\omega)$$ the probability that a purely incoming wavepacket is absorbed by the black hole?

This argument involving $$\Gamma(\omega)$$ is made since in a naive approach, in order to obtain the Hawking radiation, one generally drops the effective potential. Let me make this statement clearer:

In a curved spacetime, the Lagrangian of a free massless scalar field is: $$$$\mathcal{L}=\frac{1}{2}g^{\mu\nu} \partial_\mu\phi\,\partial_\nu \phi$$$$ where the metric $$g^{\mu\nu}$$ is the solution of the Einstein equations: $$G_{\mu\nu}=8\pi G\,T_{\mu\nu}$$ in a spacetime with a source $$T_{\mu\nu}$$.

Now, by making use of the Euler-Lagrange equations, you can write down the equations of motion, which reads: $$\hat{\square}\phi=0$$ where $$\hat{\square}$$ is the Laplace-Beltrami operator: $$\frac{1}{\sqrt{g}}\partial_\mu (\sqrt{g}\,g^{\mu\nu}\partial_\nu)$$ with $$g=det(g_{\mu\nu})$$.

If you now consider the Schwarzchild metric given by $$ds^2=(1-\frac{2m}{r})dt^2-\frac{dr^2}{1-\frac{2m}{r}}-r^2d\Omega^2$$ Since we are in a spherically symmetric spacetime, we can expand the field in spherical harmonics: $$\phi=\sum_{l,m}\frac{F(r,t)}{r}Y_l^m(\theta,\phi)$$ And now, by putting this expansion in the equation of motion, and considering the radial part, you will obtain (after some algebraic steps): $$\left(\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial {r^*}^2}+V_l\right)F_l(r,t)=0$$ with the effective potential $$V_l=(1-\frac{2m}{r})(\frac{2m}{r^3}+\frac{l(l+1)}{r^2})$$.

In a first approximation, you can drop the $$V_l$$ term, by saying that your solution will be valid in the $$r\to\infty$$ regions (the asympotical regions). What you will find is a mean particle value given by the planckian distribution at late time: $$\langle in|N^R_\omega|in\rangle=\frac{1}{e^{8\pi m\omega}-1}$$ (where $$N^R_\omega={a^R_\omega}^\dagger a^R_\omega$$, with this creation and annhilation operators referring to the modes $$u^R_\omega\sim e^{-i\omega u}$$ and with $$u=t-r^*$$, one of the null coordinates in the double-null extension of the Schwarzchild metric).

This value is clearly divergent for $$\omega\to 0$$, and this is clear since we have dropped the effective potential, which would shield the modes allowed to go to the infinity. If you consider the effective potential, the equation of motion will obviously change, and you can say that there will be some Transmitted modes (T-modes) and some Reflected modes (R-modes) by the potential. Since we are considering modes $$\sim e^{-i\omega u}$$, which are directed through the future null infinity $$\mathcal{I^+}$$ (in a penrose diagram), only the T-modes will arrive at this $$\mathcal{I}^+$$, since the R-modes will be reflected by the potential into the Black-Hole. Thus, by an asymptotic analysis, you can calculate this $$T$$ and $$R$$ probability, and you will find that: $$|T|^2\simeq 16m^2\omega^2\sim A_H\omega^2\;\;\;\; \text{with }A_H\text{ the horizon surface area}$$ (and $$|R^2|=1-|T^2|$$).

Your Planckian distribution will then become: $$\langle in|N^R_\omega|in\rangle=\frac{|T|^2}{e^{8\pi m\omega}-1}=\frac{\Gamma(\omega)}{e^{8\pi m\omega}-1}$$ and this is correct since the low modes are shielded by this $$\Gamma$$.

However, in your reference, Parker is considering the modes $$\sim e^{-i\omega v}$$, with $$v=t+r^*$$, the other null-coordinate of the double-null extension of the Schwarzchild metric, then you have that the modes which will go to infinity, will be those that are reflected by the potential, and by doing the same analysis I have done, you will arrive at the same conclusion.

• Thanks for the help! There's one probably extremelly basic point, which I confess I'm having a very hard time to grasp. Hawking and others state that these $T,R$ are the probabilities that particles fall into the hole or escape to infinity. But what one is doing here is to consider the scattering of a classical scalar wave isn't it? I mean, I don't see why these are quantum mechanical probabilities for quantum incoming particles. As far as I know in QFT the KG equation is a field equation, not one state evolution equation as in non-relativistic QM. So what allows this interpretation? – user1620696 Feb 23 at 14:23
• Your Welcome :) and well, remember that the modes $u^I_\omega \sim e^{-i\omega v}$ and $u^R_\omega \sim e^{-i\omega u}$ are modes corresponding to the quantum field $\hat{\phi}=\sum{\hat{a}^R_\omega u_\omega^R+\hat{a}^I_\omega u_\omega^I+h.c.}$. – Kevin De Notariis Feb 23 at 17:00