# Does Birkhoff's theorem hold inside the event horizon?

Can Birkhoff's theorem be used to say that the blackhole exterior and interior sections of Kruskal-Szekeres's solution (or coordinate transformations of it like Gullstrand–Painlevé coordinates, etc.) are unique all the way down to the singularity? Or are there different options for how to extend the exterior Schwarzschild solution beyond the event horizon?

The wikipedia entry states that the exterior solution is unique, but doesn't comment on the interior.

There is more than one way of stating Birkhoff's theorem, but one statement is that spherically symmetric vacuum solutions are Schwarzschild. Staticity is not an assumption, nor does staticity come out as a result, since the Schwarzschild metric is not static inside the horizon. However, it is asymptotically static. For a proof that I think handles this correctly, see section 7.4.6 of my GR book.

Ben Crowell's answer is right. I hope I can clarify a little. If one starts from the following two assumptions:

1. $$R_{ab} = 0\;$$ (i.e. field equation in vacuum)
2. The metric can be written in the form $${\rm d}s^2 = g_{uu}(u,v) {\rm d}u^2 + 2 g_{uv}(u,v) {\rm d}u {\rm d}v + g_{vv}(u,v) {\rm d}v^2 + f^2(u,v) \left( {\rm d}\theta^2 + \sin^2(\theta) \, {\rm d}\phi^2 \right)$$

then it can be shown (Birkhoff) that there exist coordinates in which the metric takes the Schwarzschild form. Adopting the usual coordinate names $$t,r,\theta,\phi$$ it is not necessary to assume that $$r$$ is a spacelike coordinate; the result is valid both inside and outside the horizon.

Item (2) above is, of course, an attempt to capture the notion of "spherical symmetry" without imposing any further assumptions. One might provide further arguments to show why this form must hold if the situation has rotational symmetry about two different axes.

The solution is not unique if you can have sources. And Birkhoff can still work with sources.

In fact Birkhoff covers the case of a spherically symmetric star fairly easily. You can paste together a solution of parameter $M$ and a solution of parameter $M+m$ just cut out the inside of the parameter $M+m$ solution and cut out he the outside of the parameter $M$ solution and sew them together on a surface of common surface area and place $mc^2$ of energy on the interface.

That's what a real (spherically symmetric) star would look like. Note that when you encounter a layer of energy, the solution below that layer changes. So knowing where the energy is located is important.

So lets look at a black hole. If we expected it to form from layers of the star contracting we notice that it's only the layers history before they reach the event horizon that affect us. A very inner layer is quite time dilated and we are still seeing it from before the horizon formed. A medium layer is fairly time dilated and we are still seeing it from before the event horizon got up to that layer. And for an outer layer it is a bit time dilated and we see it from before the event horizon go to that outer layer.

We could model how the layers fell in and thus get solutions for the inside, but they will depend on exactly when each layer crossed.

One simple to compute example is a single layer of dust that collapsed under free fall from some initial radius r. It looks like Minkowski on the inside. Schwarzschild of parameter $M$ on the outside and the boundary between inside and outside moves inwards. You could (mathematically) trace the shell all the way until it hits the singularity and at that point there is no more Minkowski spacetime.

But you could also have an eternal black hole where it was always Schwarzschild outside for every event outside. But that doesn't tell you what shells are inside.

One solution is that there is a shell of zero rest mass energy on the event horizon moving away at the speed of light. This eternal solution has Schwarzschild spacetime (outside of the event horizon) of parameter $M$ and has Minkowski spacetime (inside the event horizon). And always has an always had that solution. There is just energy on the interface. The energy happens to have no rest mass so moves at the speed of light. And it happens to be moving outwards. So the solution is just static forever since the energy stays at the event horizon.

Another solution has no enegy at all on the event horizon or anywhere else, its all Schwarzschild of parameter $M$ all the way (you can call it Kruskal-Szekeres I'm just trying to say that without energy or sources it can be the same parameter all the way).

" after you got inside you'd still see all that mass there inside with you" ... I think this is incorrect. The material that fell in is causally disconnected from you, regardless of your motions after that.

You don't have to think about it, you can just look at a Penrose-Carter diagram. A simple one is just Kruskal-Szekeres coordinates instead of $r$ and $t.$ Depending on how much earlier the matter entered compared to when you went in you might not be to rendezvous with it, and you might not be able to see its entire presingularity future, but when you cross the horizon you see everything event where something crossed earlier. And after you get through you start to see parts after they went through.

Draw a Kruskal-Szekeres coordinate system. Pick an event on the event horizon where you cross. Take the forward cone with the part rushing in at the speed of light (that's simply the continuation of the inward going light cone from the outside if inward confuses you inside) then at the event where it intersects the singularity, take the past light cone of that event. Then take the past light cone of all the events on the singularity closer to your outside than than event. Their union is all the events you could see.

But the forward light cone of the infalling matter that went before you includes your worldline and there is not getting around it (at least for the parts that fell in similar enough to where you went, and assuming it isn't destroyed which it would if your star was made of layers of matter and antimatter, but you'd still see the radiation from the annihilation of those layers). Again, you don't have to think, just draw the picture.

If you insist on considering something formed from matter, then consider a finite spherical distribution of stress-energy within an apparent horizon. Let it evolve and an event horizon will form and all matter will cross it in finite proper time, and will hit a singularity in finite proper time. In finite proper time we will have a vacuum solution.

That's so vague as to be impossible to read. If you had a shell of source infalling then by definition at the event horizon light from those events could move outwards and stay on the event horizon, so you'd literally see it when you cross. Your crossing would be an event where you see the light from it crossing reach you as you cross. And if it is infalling until it hits the singularity then the outwards going light from that infall is going to fill the interior events of the event horizon.

Hence you keep seeing the infalling matter. Sure if it was a shell falling slowly and you came in after you might be able to rush in faster and get to a region of Minkowski spacetime. But the point still stands, the matter is inside and you see it when you cross (as is clear from a diagram) so real black holes aren't vacuum even to events that are outside the past light cone of the crossing of the outermost layer. Everyone always sees the matter than formed it.

• I'm not sure why you brought up all the issues with mass. This is discussing vacuum solutions. The restriction that the vacuum solution is spherical, Birkhoff's theorem shows us the solution must be asymptotically flat and is unique down to an event horizon. Is the vacuum solution also unique beyond the event horizon? In Schwarzschild coordinates, beyond the event horizon the radial coordinate is time like. So I'm not sure how to consider spherical symmetry here. Feb 15, 2016 at 0:22
• @BuddyJohn From the outside we don't know if it is vacuum on the inside. I wanted to discuss realistic black holes before addressing eternal ones. For a realistic one, even if you entered after all the other mass then after you got inside you'd still see all that mass there inside with you. And a maximal eternal solution has two asymptomatically flat ends outside the horizons. Whether you want to identify them is an issue too. I've even seen people discuss about whether a future pointing curve in one should be identified with a past pointing curve in the other. Feb 15, 2016 at 0:30
• " after you got inside you'd still see all that mass there inside with you" ... I think this is incorrect. The material that fell in is causally disconnected from you, regardless of your motions after that. If you insist on considering something formed from matter, then consider a finite spherical distribution of stress-energy within an apparent horizon. Let it evolve and an event horizon will form and all matter will cross it in finite proper time, and will hit a singularity in finite proper time. In finite proper time we will have a vacuum solution. Feb 15, 2016 at 0:39
• @BuddyJohn regarding "I am not sure how to consider spherical symmetry here" (i.e. beyond horizon when $r$ is timelike) one can proceed as in the answer I added just now. Jul 2, 2019 at 14:09