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How far away is the event horizon of a (Schwarzschild) black hole away from the central singularity for a radially infalling observer starting with $v=0$ somewhere outside the black hole? After crossing the event horizon, such an observer hits the singularity in a finite time, hence such an observer would also assign a finite distance from the horizon to the singularity.

"Crossing the horizon" shall mean that the observer moves from outside the black hole (there are future world-lines, including non-radial and non-freefalling ones, that do /not/ hit the singularity) to inside the black hole (all future world lines hit the singularity).

The radius of a black hole is defined as follows: Take a ball $B$ in flat (Euclidean) space that has the same surface area like the event horizon of the black hole. Then the Schwarzschild radius of the black hole is defined to be the radius of $B$.

I'd guess that the so defined Schwarzschild radius is not the same (smaller?) like the proper distance from the event horizon to the center, but what is the ratio of these two values exactly, for example in terms of the mass $M$ of the black hole?

[EDIT]: Clarified that it's for a free falling observer.

emacs drives me nuts
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3 Answers3

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You refer to the "central singularity," but the singularity of a Schwarzschild black hole is not a point at the center of the event horizon. It's a spacelike surface that is in the future of all observers. It's also not a point. See Is a black hole singularity a single point? .

The question you ask doesn't have a meaningful answer. From a point on the horizon, you can draw a null geodesic that intersects the singularity, and its metric length is zero. You can also draw a timelike geodesic, in which case the metric length will be (for +--- signature), a positive real number of order M in geometrized units. You can also draw a spacelike curve whose length in this metric is an imaginary number.

You refer to "proper distance," but that doesn't succeed in resolving this ambiguity. Proper distance is distance defined by a ruler at rest relative to the thing being measured. Inside the horizon, we can't have a ruler at rest. The spacetime inside the horizon is not static.

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In GR, the proper distance is a property of curves connecting two points, not of the points by themselves. If two points are causally disconnected, then you can define a "distance" between them as the minimum proper distance over all the spacelike curves that connect them (which will necessarily be attained by a spacelike geodesic).

But this doesn't really work for a black hole singularity. As Ben Crowell says, a (curvature) singularity is not actually part of the spacetime manifold, so it doesn't really have a well-defined topology, dimension, etc., but in some situations (including this one) it's best thought of as being "like" a spacelike hypersurface. There are timelike, lightlike, and spacelike curves connecting any point on the horizon to different "points" "in" the event horizon hypersurface, and the spacelike curves have every positive proper distance, no matter how large or small. Since the proper distances get arbitrarily small, I suppose you could say that in some sense the "distance" between the event horizon and the singularity is zero, but this isn't really a particular useful way to think about it.

tparker
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The proper distance is defined along a spacelike path between two events in spacetime:

$$ L = c \int_P \sqrt{-g_{\mu\nu} dx^\mu dx^\nu} $$

However, the Schwarzschild singularity is not an event. It is a moment in time $r=0$ ($r$ is timelike inside the horizon) that happens everywhere in space $-\infty<t<+\infty$ ($t$ is spacelike inside the hirizon). Thus you can say that geometrically the Schwarzschild singularity is a singular line $(r=0,-\infty<t<+\infty)$ removed from the spacetime manifold. See: Is the schwarzschild singularity stretched in space as a straight line

This line however is infinitely long in the spacelike $t$ coordinate. Therefore you can pick an event asymptotically close to the singularity in such a way that it would be arbitrarily far away in proper distance from any event you pick asymptotically close to the horizon.

Accordingly, the answer to your question is that the proper distance between the horizon and Schwarzschild singularity is not uniquely defined. It can be anything from zero along a lightlike path of a null dust to arbitrary large, because the future timelike eternity of the universe translates to a spacelike infinity inside a Schwarzschild black hole.

safesphere
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