# Family of nested spheres in Schwarzschild spacetime

Mentor

http://en.wikipedia.org/wiki/Schwarzschild_coordinates

...it talks about a "family of nested spheres": each surface of constant t and r is a 2-sphere (i.e., setting dt = dr = 0 and r = constant in the metric results in a Euclidean 2-sphere). At the top of the page, however, it says:

We should also note that the extension of the exterior region of the Schwarzschild vacuum solution inside the event horizon of a spherically symmetric black hole is not static inside the horizon, and the family of (spacelike) nested spheres cannot be extended inside the horizon, so the Schwarzschild chart for this solution necessarily breaks down at the horizon.

I've bolded the sentence that I'm wondering about. I understand that the chart as a whole is singular at the horizon, so the interior and exterior charts are disconnected. But it seems to me that there are some subtleties about the nested 2-spheres that are worth mentioning:

(1) In the Schwarzschild *interior* chart, the t coordinate is spacelike and the r coordinate is timelike. But the angular part of the metric in these coordinates is the same inside the horizon as outside, so setting dt = dr = 0 and r = constant should work the same. Yes, r is a timelike coordinate, but that only means $\partial / \partial r$ is a timelike vector instead of a spacelike one; it doesn't affect the physical meaning of a constant value of r relative to the area of a 2-sphere at r, correct?

(2) Another way of expressing #1 would be to point out that in Painleve coordinates, for example, the physical definition of r is the same: a 2-sphere at radial coordinate r has physical area $4 \pi r^{2}$. The only difference is that in these coordinates, $\partial / \partial r$ is a spacelike vector all the way down to r = 0. So since a given physical 2-sphere is labeled by the same r in both coordinate charts, its physical area must be $4 \pi r^{2}$ regardless of which chart we are regarding r as a part of; i.e., the family of nested 2-spheres, physically, must run all the way in to r = 0.

(3) Bringing up Painleve coordinates also raises another issue: at the horizon, Schwarzschild coordinates are singular, but physically, there is still a 2-sphere there, with physical area $4 \pi r^{2} = 16 \pi M^{2}$. Painleve coordinates are not singular at r = 2M so this can be seen directly in those coordinates by setting r = 2M, dt = dr = 0.

(4) Finally, the bit about Schwarzschild spacetime not being static inside the horizon: that means that a curve of constant r, theta, phi, which is timelike outside the horizon, is spacelike inside the horizon (and null *on* the horizon). But that doesn't affect the fact that a surface of constant t and r (or constant Painleve time T and r) is a spatial 2-sphere. It just means that, on the horizon, a curve that stays on that 2-sphere for all time is null (the path of a light ray), and inside the horizon, a curve that stays on that 2-sphere is spacelike (i.e., no object can move on it, not even light).

Have I got all the above correct?