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The extrinsic Schwarzschild solution is a vacuum solution, meaning that it is valid for regions of spacetime where there is no matter or energy. This seems to imply that the Schwarzschild solution is only valid outside the spherical body. For inside the spherical object, the internal Schwarzschild solution is used, where the stress energy tensor is no longer 0. However, when we use kruskal coordinates, the coordinate system covers the inside of the spherical object (usually taken to be a black hole). How can the kruskal coordinates accurately describe interior region of black hole when it is only derived for when there’s a vacuum?

The question linked in the comments does not have an answer to this question, because the answers say that there can be spacetime curvature in a vacuum solution. My question is NOT “how can there be spacetime curvature in a vacuum?”. The question is why do we talk about the spacetime INSIDE a black hole (which supposedly has a mass) as Schwarzschild?

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  • $\begingroup$ Does this answer your question? If black holes are just empty vacuum of space inside, then what causes the curvature? $\endgroup$ Commented Jan 21, 2023 at 22:13
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    $\begingroup$ The Kruskal Szekeres is for the exterior solution where there is no homogenous sphere of rg>9rs/8 but all the mass is inside the singularity at r=0. You also spell the name Schwarzschild wrong: you have a black child, but it is supposed to be the german translation of a black shield. $\endgroup$
    – Yukterez
    Commented Jan 21, 2023 at 22:26
  • $\begingroup$ @Yukterez at least he Used The Schwartz $\endgroup$
    – JEB
    Commented Jan 22, 2023 at 0:45
  • $\begingroup$ @ConnorBehan I am not asking how there is curvature in a vacuum solution. I am asking why we talk about schwartzchild spacetime inside the spherical body if the vacuum solution is only valid outside the mass. $\endgroup$
    – user310742
    Commented Jan 22, 2023 at 1:27
  • $\begingroup$ Because inside the spherical "body" there is still no mass as explained in the linked question. $\endgroup$ Commented Jan 22, 2023 at 3:39

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However, when we use kruskal coordinates, the coordinate system covers the inside of the spherical object (usually taken to be a black hole).

The interior (by which I mean, $r<r_S$) of a Schwarzschild black hole is also a vacuum, insofar as the stress-energy tensor vanishes there.


But it’s a black hole..doesn’t it have mass/energy?

The Schwarzschild metric is a vacuum solution to the Einstein equations. It describes the metric in a vacuum ($T_{\mu\nu}=0$) under the assumptions that the metric is spherically symmetric and stationary. If you model a star as a spherical fluid with a sharp boundary with pure vacuum outside it, then the exterior region would be described by the Schwarzschild metric while the interior of the star (where $T_{\mu\nu}\neq 0$) would be described by the interior Schwarzschild metric. In this case, the parameter $M$ which appears in the exterior Schwarzschild solution is the mass of the star.

One could ask, however, what happens when you assume that $T_{\mu\nu}=0$ everywhere. The most general spherically symmetric, stationary vacuum solution to Einstein's equations is

$$\mathrm ds^2 = -\left(1-\frac{\alpha}{r}\right)c^2\mathrm dt^2 + \left(1-\frac{\alpha}{r}\right)^{-1} \mathrm dr^2 + r^2 \mathrm d\Omega^2$$

where $\alpha$ is an unknown constant. If $\alpha = 0$, then this is just the standard Minkowski spacetime from special relativity. However, it's entirely possible that $\alpha \neq 0$, provided that we remove the set of points with $r=0$ from the spacetime manifold (otherwise, the metric would be undefined there).

One might object immediately by noting that for $r=\alpha$, the metric component $g_{rr}$ is undefined. However, this is what is called a coordinate singularity; it's not actually a problem with the spacetime, just with this choice of coordinates. One can switch to the Kruskal coordinate system to demonstrate this fact. In contrast, the singularity at $r=0$ is a genuine one, and so that set of points must be removed.

So if $\alpha \neq 0$, then what is it? Comparison with the exterior Schwarzschild solution outside of a star provides a physical interpretation - this is what you'd expect to see if a total mass $M\equiv \alpha c^2/2G$ was crushed down to a single point in space (which, to avoid infinities, must be excised from the spacetime manifold). However, it's important to note that at all points (other than the singularity, which again has been removed), $T_{\mu\nu}=0$. This is true both for $r>\alpha$ and $0<r<\alpha$.

[...] from observations black holes are not a single point. Supermassive black holes have a radius of ~15 million miles.

The 15 million miles you refer to is the event horizon of the black hole - the $\alpha$ which appears in my explanation above. Again, for a Schwarzschild black hole, $T_{\mu\nu}=0$ both outside the event horizon and inside the event horizon.

[...] For these black holes there is surely mass and energy, so it seems like we cannot speak of geodesics “inside” the black hole if we are still sticking with schwartzchild space time

One of the reasons I keep using the qualifier Schwarzschild to refer to a black hole is because this metric is the simplest possible model for understanding the phenomenon. It is not necessarily a realistic model for an actual black hole, which would generically be e.g. rotating and surrounded by an accretion disk which is slowly falling below the event horizon. Rather, you should think of it as a toy model for understanding the nuances of the event horizon and other highly non-intuitive phenomena. Adding a steady influx of mass or some other more complex features would only obscure these issues.

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  • $\begingroup$ But it’s a black hole..doesn’t it have mass/energy? $\endgroup$
    – user310742
    Commented Jan 21, 2023 at 23:13
  • $\begingroup$ @Obama2020 Nope. $\endgroup$
    – TimRias
    Commented Jan 22, 2023 at 0:09
  • $\begingroup$ @TimRias Then what does the “m” in the schwartzchild metric stand for? Mass of what? Why do we say that black hole mass decreases due to Hawking radiation if it has no mass? $\endgroup$
    – user310742
    Commented Jan 22, 2023 at 1:21
  • $\begingroup$ @PeterShor from observations black holes are not a single point. Supermassive black holes have a radius of ~15 million miles. For these black holes there is surely mass and energy, so it seems like we cannot speak of geodesics “inside” the black hole if we are still sticking with schwartzchild space time $\endgroup$
    – user310742
    Commented Jan 22, 2023 at 1:39
  • $\begingroup$ @Obama2020 I have added a substantial addendum to my answer to address your questions. $\endgroup$
    – J. Murray
    Commented Jan 22, 2023 at 2:34
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Schwarzschild solution is valid only outside of matter's sphere and it is static - there is no time dependence. In case of black hole it ends on the even horizon. Behind it, time and space change their place and the resulting metric ("extended" Schwarzschild) is no more static - it depends explicitly on time. Hence, the black hole spacetime have two regions, separated by the event horizon, that are described by two different metrics. The use of the name Schwarzschild for both of them is somewhat misleading but due to history. Schwarzschild entitled his paper "On the gravitational field of a mass point according to Einstein's theory". He was not aware of the immense meaning of his result and the fact that in general relativity there is no something like a "mass point".

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  • $\begingroup$ Your answer is correct +1, but you statement on Schwarzschild being "unaware" of the meaning of his solution is not. In his paper, he states that the radius of the horizon defined as $r=\sqrt{x^2+y^2+z^2}$ equals zero, which is completely true, and therefore a black hole is indeed a mass point of a zero spatial radius. In the modern definition, the radius of the horizon is not zero, but its value is timelike, not spacelike. This means that the "distance" from the horizon to the origin is a not a distance in space, but a period of time. Thus spatially a black hole still is a mass point. $\endgroup$
    – safesphere
    Commented Jan 23, 2023 at 6:44
  • $\begingroup$ @safesphere I have to admit your last sentence I find difficult to accept. Please look into physics.stackexchange.com/a/682496/281096 where similar question and explanations have been given. Is there something there you would agree? $\endgroup$
    – JanG
    Commented Jan 23, 2023 at 8:58
  • $\begingroup$ The spatial Schwarzschild radial distance from the horizon to the origin is zero. As we observe from outside, the volume inside the horizon is zero as well: arxiv.org/abs/0801.1734 - It may seem hard to accept, because the popular propaganda has been painting a wrong picture for decades, but mathematics proves Schwarzschild right. There is nothing in the quote from t’Hooft that contradicts Schwarzschild. To get from the horizon to the singularity, you don’t need to move, just wait. You don’t need to change your location - the spatial location of the horizon and singularity is the same. $\endgroup$
    – safesphere
    Commented Jan 23, 2023 at 15:43
  • $\begingroup$ @safesphere I am with you. The spatial location of the event horizon and the "singularity", where pressure diverges by constant energy density, is the same. I believe to have proved it for the case of perfect fluid spheres. $\endgroup$
    – JanG
    Commented Jan 23, 2023 at 15:54
  • $\begingroup$ @safesphere "In his paper Schwarzschild states that the radius of the horizon, defined as $r=\sqrt(x2+y2+z2)$, equals zero". Would you mind giving me a link to this paper and a reference to where he states this? $\endgroup$
    – JanG
    Commented Jun 13 at 11:00

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