In many standard textbooks about general relativity, we write the Schwarzschild solution in Schwarzschild coordinates for $r>2GM$: $$ \mathrm{d}s^2 = -\left(1-\frac{2GM}{r}\right)\mathrm{d}t^2+\left(1-\frac{2GM}{r}\right)^{-1}\mathrm{d}r^2+r^2 \mathrm{d}\theta^2+r^2\sin^2 \theta \mathrm{d}\phi^2 \tag{1} $$ and because the Schwarzschild coordinates are not good enough for the entire spacetime - a coordinate singularity at $r=2GM$ - we change to the Kruskal coordinates: $$ \mathrm{d}s^2 = \frac{32G^3M^3}{r}\mathrm{e}^{-r/2GM}(-\mathrm{d}T^2 + \mathrm{d}R^2) +r^2 \mathrm{d}\theta^2+r^2\sin^2 \theta \mathrm{d}\phi^2, r=r(T,R) $$ which cover what we should think of as the entire manifold, the horizon and the so-called interior included.

To make it more clear, I could post a diagram from Sean Carroll's Spacetime and Geometry: An Introduction to General Relativity: The Kruskal diagram

The extension to Kruskal coordinates is mathematically OK to me, but my question arises physically: Why can we describe the interior of the black hole, from which any signal can't reach us?

Or in another situation, given the metric of the form $(1)$ and told it is only appropriate for $r>2GM$, I, who was outside the black hole, transferred to the so-called "Kruskal coordinates" and claimed that I successfully worked out the interior geometry. But the interior is undetectable to outside observers. For example, we can calculate the path that a test particle follows when given a metric, but we can never know whether a test particle inside really follows the predicted path. So, how to validate our interior solution?

The question is not specific in this Schwarzschild case. I don't know how to believe those "extensions" since they describe a spacetime region that doesn't have a causal influence on us, who are standing on the "outside" and waiting to test the extended solution.

  • $\begingroup$ It is a misconception that the Kruskal and other “regular” coordinates “remove” the horizon singularity. They don’t. They simply hide it by moving it from the metric equation to the transformation equation. This way the metric becomes “well behaved”, but the transformation from the new coordinated to the coordinates of any external observer becomes singular and mathematically invalid. The horizon and its singularity are a physical reality with distinct propertied that cannot be made “unreal” by a mathematical trick. This is especially evident in semiclassical gravity (e.g. Hawking radiation). $\endgroup$
    – safesphere
    Commented Apr 25, 2022 at 16:35

2 Answers 2


Well, the answer is that you are right: in practice, you cannot know this unless you are an infalling observer starting from Schwarzschild exterior and "enter" the black hole and see if the interior is what it is. The maximal extension is the mathematical extension of the geometry that has exactly the same metric given by the metric function $f(r)$.

To drive this point, see the paper here by Hsu and Reeb on the so-called "bag of gold" spacetime: you essentially can sew into the Schwarzschild interior an entire Friedmann-Robertson-Walker (FRW) spacetime with compact spatial topology ($k=1$), here called Kruskal-FRW gluing. Einstein field equations do not forbid this construction, even if there may be other reasons to exclude this geometry from being physical.

  • $\begingroup$ So, is it better to see the extension as purely mathematcal rather than a physical, or in other words, we don't insist that's what really is inside the black hole, right? But then what's the physical meaning of extension and the necessity to explore that? $\endgroup$ Commented Apr 26, 2022 at 8:05
  • $\begingroup$ I like to think of this as being the natural assumption until you have other reasons otherwise. I mean, even outside, if you are never near a static spherically symmetric mass, ALL of them have Schwarzschild metric at its exterior so you cannot tell a black hole from a star far away without further information (e.g., collecting radiation). $\endgroup$ Commented Apr 27, 2022 at 4:07

You can't! That's the trick: it's not actually a prediction we can test, unless you both a) had interstellar travel and b) a deathwish, and even then only 1 person could, for that last little bit of time, know the validity of said prediction.

To get a sense of this, note that string theory - or at least one proposal for it - replaces pretty much exactly the whole interior of the black hole with a "fuzzball", i.e. a very exotic kind of matter! Yet it makes essentially no difference to the externally-accessible data! This is a big part of why trying to test any theories of that sort is so difficult.


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