# Why isn't the Schwarzschild radius defined as the distance from the event horizon to the “place” where all mass is concentrated?

The Schwarzschild radius is defined as the radius of the sphere in 3-dimensional space surrounding a black hole. Let's switch to 2-dimensional space to make things easier to visualize. Why is the Schwarzschild radius defined as the radius of the circle, which has a circumference of $2\pi R$, where R is the radius of a circle in the flat plane. Isn't it more logical to define the radius as the distance from any point on the circle to the distribution of mass in the center? You can argue that this radius will have infinite length, but it's easy to imagine that the center isn't a point, but an extended structure which has a certain distance to the circle (or the sphere, in 3-dimensional space), which we can call the Schwarzschild radius.

• – Qmechanic Jun 14 '17 at 11:31
• Are you aware that (1) the Schwarzschild coordinate $r$ is defined such that the area of the surface of constant $r$ is $4\pi r^2$, (2) that the surface area of the event horizon is coordinate independent and (3) that the radial coordinate within the horizon is timelike? – Alfred Centauri Jun 14 '17 at 12:28
• So if we look in 2-dimensional space the circumference of a circle on a sphere is still $2\pi R$, if we define R (or t, in which the radial coordinate changes while passing the event horizon, according to the Schwarzwald metric) in such a way that it gives the desired result of $2\pi R$, i.e. R is not the distance from the center of the circle to the circle itself? It's obvious that a circle circumference (which on a random variety is defined as the set of points with equal distance to a certain point on the variety) has nothing to do with the coordinates. – Deschele Schilder Jun 14 '17 at 22:41
• But the relation between the circumference and the radius is not $2\pi R$, a sign that the variety is curved. So how is R defined on the Schwarzschild sphere, looking at the matter in 4-d again, to make the surface area of the event horizon $4\pi R^2$? Just as the radius of a sphere in flat 3-d space? – Deschele Schilder Jun 14 '17 at 22:42

In differential geometry, defining 3-surfaces is pretty easy. In particular, if you're looking for the horizon, you can define a null three-surface that meets these criteria (let $q_{ab}$ be the (degenerate) 3-metric of this 3-surface, with geometry induced by the enveloping 4-space, and let $\ell^{a}$ be the null tangent to this surface, and $k^{a}$ be the null normal to this surface, and normalize them so that $k^{a}\ell_{a} = -1$):

1. Every spacelike section of the surface is closed
2. it satisfies $q^{ab}\nabla_{a}\ell_{b} = 0$
3. it satisfies $q^{ab}\nabla_{a}k_{b} < 0$
4. For some timelike vector $t^{a}$, $£_{t}q_{ab} = 0$

Then, you can prove that there is a section of this surface that has maximum circumference, and that this circumference is necessarily spacelike. Once you've done this, you just need to divide this by $2\pi$ to get a radius.

The key point is that I had to define very little. Most of the above conditions just define whether or not it is a horizon. Nothing depends on what is inside the horizon. If I were to engage in your program, I would first have to define the horizon, and after I had done that, then I would have to find a spacelike slicing of the interior of the horizon that intersects with the central singularity. Doing this, I would run into complications such as:

1. The Schwarzschild singularity is spacelike, while the Kerr and Nørdstrom singularities are timelike
2. The Kerr singularity isn't even a point, it is actually a ring, so what does "intersecting" it mean? The center, the edge? Do I need a surface that hits the center without intersecting the ring?
3. Handling a distance integral that deals with the infinite curvature singularity
4. While the geometry at the horizon is invariant under boosts, there's no guarantee that this is true of the interior black hole geometry
5. While these spacelike paths are cleanly definable irrespective of geometry, the Kerr horizon is not spherically symmetric, and for $a> \frac{\sqrt{3}}{2}$, the Kerr horizon does not even embed in 3-dimensional Euclidean space due to negative curvature at the poles. A max and min circumference always make sense, but where do I start measuring my "radius" from? The pole or the equator?

For these reasons, while it may be possible to come up with a consistent definition based on "black hole radius", in practice, it is much more theoretically clean to work with circumference, and infer radius.

TLDR: radius is, in general, ill-defined, but a max and min circumference is not.