Geometry inside the event horizon

Question

I'm trying to understand intuitively the geometry as it would look to an observer entering the event horizon of a Schwarszchild black hole. I would appreciate any insights or corrections to the above.

Immediately after your enter the event horizon, if you look back and try to reach again for the horizon, it will seem to be expanding away quicker than the speed of light. Near this region, the apparent shape of the horizon is a sphere expanding away, and we are inside the sphere

Near the singularity, we really don't know what happens. I've heard that spaghettification is not a necessarily occurrence, since the metric field diagonal components are shrinking as the curvature grows, so it could very well be the case that a infinite length hyper-cylinder $S^3 \times R^+$ of constant physical radius is being conformally mapped to the $S^3 - \{0\}$ region around the singularity, or that in general a region around the singularity can be mapped to anything in the other end, which is basically because the degrees of freedom of curvature and stress-energy in our end of the spacetime cannot really predict what sort of topology endpoint will connect to the matter in the other end. Since the metric components are tending to zero at the singularity, this argument sounds pretty interesting, since it would seem to imply that observers will "shrink" relative to kruskal coordinates, because the local physics would always be that physical observers will stay fixed relative to their local metric, since the metric is covariantly constant!.

However, i'm not expert on how to describe the asymptotic physics in the neighbourhood of the schwarszchild singularity. (which is why i'm asking on this site, after all!). Question: does this argument hold any water?

Answer 1

The geometry in your picture is too classical. Once you pass the event horizon, it doesn't look like a sphere surrounding you anymore, and you don't see it as a special surface anyway. If you look back along a radial direction, you will see the same horizon point ahead of you (in the past) and behind you (also in the past), at different affine parameter along the horizon (this is clear in a Penrose diagram). But you won't see the horizon as a sphere.

When you approach a Schwarzschild singularity, there is no way to avoid getting compressed to oblivion, because all the volume you carry is compressed to a tiny volume near r=0. The radial area is r, and the area of a sphere is $4\pi r^2$ always, and r is time inside the horizon, and you are necessarily drawn to r=0, which is the singularity. You can't save yourself by conformal mapping, because the actual physical distances are shrunk--- even if you were to conformally get shrunk to zero size, your matter is not conformally invariant, the atoms set a scale.

The dr component of the metric doesn't vanish at the singularity, it's limiting value is ${1\over 2m}$. This means that you are losing a certain unit of r per unit time as you fall in, which means your radial volume is shrinking to zero quadratically with time. The time part of the metric (which is spatial now) goes to ${2m\over r}$, and so you gain a linearly diverging space in exchange, but the quadratic compression doesn't make up in volume for the quadratic sphere shrinking. Further, this is not a conformal transformation in any reasonable sense, it's spaghettification.

The real caveat about black holes is that this whole story assumes the black hole is neutral and nonspinning. For spinning or charged black holes, the interior structure is altered in radical ways, and there is nothing classically wrong with going in and coming out, except for some dubious arguments about what happens when you hit the Cauchy horizon in the interior.