How to prove Hawking area theorem using Raychaudhuri equation?

Question

I recently started studying black holes and I am trying to understand the Hawking area theorem and its mathematical basis (at least at a simplified level). Some resources say that the Raychaudhuri equation can be used for this purpose (https://en.wikipedia.org/wiki/Raychaudhuri_equation). I have read the derivation of the Raychaudhuri equation and the math is clear to me. As I understand it, this equation describes the change in the 4-divergence of the velocity (denoted as θ) along the geodesic. I also understand that with the strong energy condition, the right-hand side of this equation will not be positive, so the derivative of the 4-divergence will be negative at the event horizon. But I cannot go further because I do not understand the physical meaning of the Raychaudhuri equation very well. I have read that θ describes the change in the volume of the beam, or its divergence/compression, but I do not really understand what exactly this means and how to imagine it. Perhaps that is why I also do not understand how to connect this equation with the event horizon, which can be represented as a 3-dimensional surface (a sphere in the case of a Schwarzschild black hole). Can someone help me build a meaningful chain from the Raychaudhuri equation to the law (if it can be called a law) of non-decreasing area of ​​the event horizon? I will also be grateful for any links to educational literature.

Answer 1

In this response I limit myself only to some remarks from Light Rays, Singularities, and All That by E. Witten. In my opinion it is an excellent reference where you can find the information you are looking for.

Here is the logic chain:

1. To prove Hawking's theorem you need the null Raychaudhuri equation. This is nothing but Einstein's first equation \begin{equation} R_{00} = 8\pi G \left( T_{00} -\frac{1}{2} g_{00}T^\alpha_\alpha \right) \end{equation} written in a special coordinate system. In your question you referred to the horizon as a 3-dimensional surface. That is incorrect; the surface is 2-dimensional. The coordinate system is constructed as follows: on the horizon consider the set of orthogonal outgoing null geodesics spanning a 1-manifold $Y$. You can then consider the spatial coordinates $\lbrace x^1,x^2 \rbrace$ on the horizon and the affine parameter $u$ of the null outgoing geodesics as coordinates on $Y$.
2. The coordinate system can be extended to a neighborhood of $Y$ by also considering ingoing null geodesics $\lbrace u,v,x^1,x^2 \rbrace$. The metric in these coordinates takes the general form \begin{equation} ds^2 = -g_{vu}dv du + g_{AB}dx^A dx^B + 2g_{vA}dv dx^A + g_{vv}dv^2 \end{equation} where $u$ is the affine coordinate for outgoing null geodesics and $v$ for the ingoing null geodesics and $A = 1,2$.
3. Define \begin{equation} A = \sqrt{\mathrm{det}g_{AB}} \end{equation} that define the transverse area of a little bundle of orthogonal null geodesics. The expansion is then \begin{equation} \theta = \frac{\dot{A}}{A} \end{equation} where the dot is the derivative respect to $u$. Then $\theta$ measure the relative change of the transverse area.
4. Assuming the null energy condition from the null Raychadhury equation it follows that \begin{equation} \partial_u \left(\frac{1}{\theta} \right) \ge \frac{1}{2} \end{equation} from which one want to prove that $\theta \ge 0$ for the BH horizon.
5. We want to prove that $\theta \ge 0$ for null generators of the horizon. Now, if there were some point for which $\theta < 0$ we would have that there exists a value of the affine parameter such that the null generators of the horizon cannot be continued, given the presence of a focal point. However, it is possible to show that the null generators are complete, thus containing no focal points from which we get that necessarily $\theta \ge 0$.

This is just the logical chain for the proof of the theorem and I have skipped a huge number of details. However, these can be found in the reference I have given.