How interior of Schwarzschild horizon represents black hole?

Question

Consider a Schwarzschild metric:

$$g=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta \, d \phi^2)$$

It is static, which means there is a timelike Killing vector which becomes spacelike when you cross the horizon.

Is this feature alone enough to say that the interior of Schwarzschild really represents a black hole?

Edit: If not then what is the way to determine Interior of Schwarzschild represent Black Hole by looking at the metric tensor. When Karl Schwarzschild gave his solution, then how the interior represent this trapped region ??

You can calculate the Kretschmann scalar invariant as well which blows up at $r=0$, but can we just rely on the sign change of the Killing vector?

If so, can we say this for any arbitrary metric tensor?

How some metric tensors represent Black Hole?

Answer 1

Is this feature alone enough to say that the interior of Schwarzschild really represents a black hole?

No. A clear counter example is Minkowski spacetime. It has a Killing vector field defined by the Rindler congruence. That Killing vector field has the behavior you described; it changes from timelike to spacelike at the Rindler horizon. But Minkowski spacetime is not considered to be a black hole.

Answer 2

Based on your definitions, what you are asking is if the Schwarzschild horizon is a trapped surface and how to see this from the metric. And more specifically you are asking if the change of the radial coordinate from spacelike to timelike is sufficient.

Dale’s answer is correct that the change of the nature of the radial coordinate alone is insufficient. The Rindler example he gives shows that the Rindler horizon is not a trapped surface, because the null infinity in the opposite direction is still accessible from behind this horizon.

Therefore you need an additional condition. Not only that every point on the horizon must be trapped (as defined by the change of the radial coordinate from spacelike to timelike or equivalently by the horizon being lightlike), but also that there must be no path anywhere around from behind the horizon to the null infinity (like in the Rindler case). In other words, you need to show that the horizon is also fully enclosing and thus represents a trapped surface.

Luckily in the Schwarzschild case this is given by the spherical symmetry. The horizon is a sphere and thus fully enclosing. In conjunction with its lightlike nature this makes the horizon a trapped surface thus making the Schwarzschild solution a black hole, as is immediately evident from the metric.