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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.

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  • $\begingroup$ The linked "duplicate" has a computation outside the event horizon and is using Schwarzschild coordinates. Thus it is answering a different question; and Schwarzschild coordinates used there give an imaginary line element for coordinates inside the event horizon. $\endgroup$
    – emacs drives me nuts
    Jan 24, 2020 at 16:12
  • $\begingroup$ It's the same equation - just change the limits $\endgroup$
    – John Rennie
    Jan 24, 2020 at 16:13
  • $\begingroup$ So one can integrate over the complex numbers, and the quotient of the imaginary part of the integral and the Schwarzschild radius is the solution? $\endgroup$
    – emacs drives me nuts
    Jan 24, 2020 at 16:18
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    $\begingroup$ @emacsdrivesmenuts Your questions is bit like asking "What is the distance from where I am sitting right now to midnight?" Surely I will reach midnight in finite time, so what is the distance from here to then? $\endgroup$
    – MBN
    Jan 27, 2020 at 10:45
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    $\begingroup$ @emacsdrivesmenuts No, it is not. That is the whole point. The analogy of a sphere in Euclidean space is misleading. Black holes are nothing like that. $\endgroup$
    – MBN
    Jan 27, 2020 at 14:30

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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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  • $\begingroup$ Of course it is static, or do any of the terms in the metric tensor depend on t or τ? I would say they do not, even in Raindrop or Finkelstein coordinates, see en.wikipedia.org/wiki/Static_spacetime - The question does have a meaningful answer, it is the one you downvoted at physics.stackexchange.com/questions/524731/… $\endgroup$
    – Yukterez
    Jan 25, 2020 at 2:13
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    $\begingroup$ @Yukterez: No, that's incorrect. See, for example, Misner, Thorne, and Wheeler, p. 838. Staticity isn't defined in a coordinate-dependent way, it's defined in terms of a timelike Killing vector. The Killing vector is spacelike inside the horizon. $\endgroup$
    – user4552
    Jan 25, 2020 at 2:59
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    $\begingroup$ Why is this answer down voted!!! $\endgroup$
    – MBN
    Jan 25, 2020 at 13:13
  • $\begingroup$ @safesphere: The Schwarzschild singularity is not a surface, but a coordinate Euclidean line removed from the manifold. We've discussed this previously in comments. As I explained earlier, there is a set of specialized definitions that allows us to use terms like "spacelike" and "surface" to discuss singularities. This answer describes these terms in more detail, and gives a reference to a paper by Penrose on this topic: physics.stackexchange.com/a/60903/4552 $\endgroup$
    – user4552
    Jan 25, 2020 at 18:21
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    $\begingroup$ As safesphere says, there's no need for the spacetime to be static (although I agree with you that the first part of their comment is incorrect). $\endgroup$
    – tparker
    Jan 27, 2020 at 13:40
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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.

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  • $\begingroup$ If the singularity cannot be used directly, then one could use a point close to it and take the limit? $\endgroup$
    – emacs drives me nuts
    Jan 27, 2020 at 16:39
  • $\begingroup$ @emacsdrivesmenuts I don't see what that would accomplish - you already have spacelike curves with every possible length; what more information could taking a limit give you? $\endgroup$
    – tparker
    Jan 27, 2020 at 16:54
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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.

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    $\begingroup$ This line however is infinitely long in the spacelike t coordinate. This is a statement without any physical meaning, because the metric breaks down at the singularity. See physics.stackexchange.com/questions/144447/… $\endgroup$
    – user4552
    Jan 25, 2020 at 1:01
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    $\begingroup$ Thus you can say that geometrically the Schwarzschild singularity is a singular line (r=0,−∞<t<+∞) removed from the spacetime manifold. This sounds wrong to me. The topology of the Schwarzschild spacetime is $R^2\times S^2$, which you can see based on the Penrose diagram (which omits the $S^2$). See, e.g., arxiv.org/abs/1111.5790 or MTW, p. 837, fig 31.5a. I don't think this is the same as $R^4$ with a line removed, is it? I could be wrong, but I think that's not simply connected, whereas $R^2\times S^2$ is simply connected. $\endgroup$
    – user4552
    Jan 25, 2020 at 1:52
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    $\begingroup$ 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 timelike path of a free fall This doesn't make sense. The metric integrated along a timelike path isn't zero, it's either real or imaginary, depending on your choice of signature. $\endgroup$
    – user4552
    Jan 25, 2020 at 1:53
  • $\begingroup$ @BenCrowell Thanks for catching my typo. I’ve edited to change timelike to lightlike for null dust (e.g. neutrinos, approximately, of course). The proper distance cannot be imaginary, because it is measured along spacelike intervals (and marginally lightlike, as for neutrinos). The equivalent of the proper distance measured along timelike intervals is proper time. You other comments are wrong, of course. Your errors in visualizing the Schwarzschild geometry and singularity have been pointed out time and again by Lubos, myself, and others. Please get up to speed before criticizing others :) $\endgroup$
    – safesphere
    Jan 25, 2020 at 4:53
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    $\begingroup$ I wouldn't call it a line. For $r=0$ and $t, \varphi, \theta$ arbitrary is a three dimensional hypersurface. Of course it isn't part of the manifold so all this is very imprecise to begin with, but it is definitely not a line. $\endgroup$
    – MBN
    Jan 25, 2020 at 13:11

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