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I've been looking for a satisfying proof of this, and can't quite find it. I read the brief proof of the black hole area theorem in Wald, which is similar, but doesn't quite come down to the actual statement $\frac{dA}{dt} \geq 0$. I had pictured something fairly succinct relating $T_{ab}k^ak^b \geq 0$ (weak energy condition) directly to $\frac{dA}{dt}$, and I tried but I couldn't get anything to quite work out. Anyway- can anyone point me to a good source, or even just give the proof?

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If you want the full proof I suggest you consult "The large scale structure of spacetime" by S. Hawking and G.R.F. Ellis. A slightly incomplete proof is contained in P. Townsend's lecture notes – Daniel Grumiller Aug 18 '11 at 5:46
The proof I presented is presented in Hawking/Ellis of course, but I think their presentation is suboptimal. The important physical idea is drowned in a soup of arcane symbols. – Ron Maimon Aug 19 '11 at 15:54

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up vote 11 down vote accepted

The proof is very simple, Hawking saw it in a flash. While writing this, I found the proof presented in the Wikipedia page for the Raychoudhuri equation. To understand it, there are three background results you need to be comfortable with.

Horizon area makes physical sense

Given three nearly parallel infinitesimally separated light rays in Minkowski space moving in the direction perpendicular to their plane of separation, you can define the area they span by cutting them with a space time plane, and asking what is the area contained in the triangle they form. Unlike in Euclidean geometry, where this is never true for any area, in Minkowski geometry the area of this triangle doesn't depend on the orientation of the plane. In other words, you can slide each of the three intersection points up and down on the light ray without changing the area of the triangle.

To see this, first note that if you have two parallel light rays l,l' whose separation is perpendicular to their line of motion, and you have a line between two points on l and l' respectively, the length of this line does not depend on which two points it connects. The length s in Minkowski space is defined by

$s^2= A\cdot A$

and if you add a null-vector N to A, the result is unchanged because $A\cdot N$ and $N \cdot N$ are both zero. This implies that if you have three light rays which are all moving in the same direction and are separated perpendicular to the direction of motion, you connect any three points these rays and the three side-lengths of the resulting triangle are the same. Side-side-side congruence is true in Minkowski space too (when the sidelengths are nonzero).

So the area of a black hole is well defined--- cut up the surface into infinitesimal triangles which are close to some space-like surface cutting the horizon and no matter what orientation these triangles have with the light rays, you get the same answer. This is an important point which is not explained well in usual treatments.

Deviation equation

If two parallel geodesics are moving along a shared direction with a separation $\Delta$, you can think of their relative dynamics as being determined by a Newton's law. Parametrizing the space-time in a neighborhood of one of them with local coordinates, the deviation of the other is determined by expanding the arc-length formula to second order. The result is motion with a linear restoring/expelling force.

The spring-constant of the restoring/expelling force is one of the equivalent defining properties of the Riemann curvature:

$ {d^2 \Delta^\mu\over d\tau^2} = R^\mu_{\nu\sigma\lambda} \Delta^\nu \dot{x}^\sigma\dot{x}^\lambda$

Consider a simplex made out of nearby points, lying on a certain infinitesimal space-like plane, and set these points moving along the timelike direction perpendicular to the plane. The rate at which the log of their volume increases as a function of their shared proper time s is called the expansion $\theta$. The derivative of $\theta$ is interesting because it takes the Ricci trace of the Riemann tensor when you evaluate it in terms of the geodesic deviation above:

${d\theta \over ds} = - {1\over 2}\theta^2 - \sigma^2 - R_{\mu\nu} \dot{x}^\mu \dot{x^\nu}$

Where $\sigma$ is not important because it always makes a negative contribution. This is the Raychoudhuri equation (for vanishing vorticity--- the vorticity vanishes for these parallel infinitesimally separated geodesics for all times, because the curvature forces are Hooke springs, directed in and out). For null paths, the volume becomes an area (two of the perpendicular spanning directions of the simplex become parallel), and the proper time becomes the affine parameter.

Note that on a null vector like $\dot{x}$, $T_{\mu\nu}$ is equal to $R_{\mu\nu}$, so if the weak energy condition holds, this quantity is always positive. This tells you that once the divergence is negative, it must crash to minus infinity in a finite proper time. If the area of a little triangle of parallel light-rays is decreasing along their affine parameter at any instant, if the weak energy condition is satisfied, then this area will crash to zero area after a finite affine parameter. This means that some two neighboring geodesics in the congruence have collided, or "focused".

Focusing theorem

When two nearby geodesics collide, they cannot be "shortest" paths between their endpoints, for a simple reason: first, the deviations between them are always to first infinitesimal order, so that they have the same arclength (this is the same reason that particles which are moving slowly relative to each other agree on their shared Newtonian time, and this notion extends to affine parameter by limits). If geodesics 1 and 2 started at point P, became parallel for a while, then intersected, you can follow 1 from P to the intersection point, then follow 2 until the end, and this is the same length as following 2 all the way from P to the end. But the first path bends a corner. If light bends a corner, it is possible to catch up to it with a time-like path, so past the intersection point, a massive particle could have taken the path from P to the extension.

This holds for two geodesics that start off going in a direction perpendicular to a given space-like plane just as well as those from a common point, replacing "distance from the point" with "distance from the plane".

The Area theorem

The event horizon of a black hole is defined as those light-paths which just barely do not escape. If the light rays are pushed just a little out, then they escape to infinity, and if they are pushed just a little in, then they are sucked into the black hole. This means that any massive object which hits these light rays must never be able to catch up with these light rays, it must fall into the black hole.

Hawking's insight was that the if the area is decreasing at any one moment in any little triangle on this horizon, it must crash to zero in a finite affine parameter. This means that two nearby null geodesics on the extension of this simplex forward in affine parameter will focus. But this means that their extension can be linked to the original simplex by a time-like path, so that their extension must be inside the black hole, which means they could not be on the boundary.

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wow. i'm speechless. +1 – lurscher Aug 19 '11 at 14:14
Could you write where precise statement of first theorem (on invariance of area) can be found. I've been searching in Hawking Ellis and can't find it. – Blazej Aug 6 at 17:30
@Blazej: It isn't there! I was annoyed by this. You have to work it out for yourself, and I did that years ago when studying this stuff, and put the main unstated theorem here, it doesn't appear anywhere else. – Ron Maimon Aug 7 at 0:17
It is easy how this works out in specific examples. For example in Schwarzschild metric on horizon takes form $ds^2=r^2 d \Omega ^2$ but there is one more parameter, one can choose t. If we choose any $t=f(\theta,\phi)$ slice it has the same metric, so the same volume. Same is true for light cone in Minkowski. What bothers me is that I don't know how to convince myself that nonvanishing part of metric doesn't depend on t ("null coordinate") in general case. If we had surface parametrized by (t,x,y) with metric $ds^2=f(t)dx^2+g(t)dy^2$ then area would depend on choosing $t$-slice - wouldn't it? – Blazej Aug 7 at 11:36
@Blazej: The examples you give are too simple, the boosted Schwarzschild black hole is better, or using a slice which is wiggling like t(r)=t_0 + Acos(r). The area is independent of the wiggles. The area depends on the slice only if there is a divergence (positive expansion) of the null geodesics making up the horizon between the two slices you are comparing. Otherwise, you are free to slide the points up and down and the area doesn't depend on the slice. The principle of it is the Minkowsi triangle thing I said. – Ron Maimon Aug 7 at 19:07

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