# Is it simply an assumption that spacetime fills the region under the event horizon?

For the black-hole solution of the Schwarzschild metric to be at all possible would require spacetime to fill the region under the event horizon. Is this requirement just an unspoken or unexamined assumption or is there some mathematical or theoretical reason that it must be true?

Given the AMPS paradox with its postulation of a firewall at the horizon (space and time seem to "somehow" end there), recent LIGO data tentatively indicating signs of firewalls or other exotic physics at the horizon and some versions of string theory predicting a structure there, I'm left wondering if such an assumption (if that's all it is) is justified.

• It is a general assumption in general relativity that realistic spacetimes are maximally extended, that is, there is no regular boundary point (no boundary point without a singularity where something goes bad, like the curvature diverging). If you have an event horizon with nothing behind it, from the point of view of an infalling observer, he just disappears when crossing the horizon for no adequate reason. Commented Jan 5, 2018 at 9:07
• @Slereah: If spacetime ends at the event horizon, an infalling observer would never get past the horizon. That seems like the whole point of the AMPS paradox and its firewall and Raphael Bousso's (and others') comments about how spacetime "somehow" seems to end at the horizon. The event horizon itself would be a spherical singularity. This is why I'm questioning the assumption (if that's all it is) that allows for the internal portion of the two solutions to the field equations in the Schwarzschild metric. Commented Jan 5, 2018 at 19:54
• @dcgeorge "The event horizon itself would be a spherical singularity." - It is: indiana.edu/~fluid/… - Accordingly spacetime inside is not Schwarzschild and nothing can enter it. Commented Nov 2, 2019 at 7:39
• @safesphere Thanks for the interesting link. The statement "Nothing gets inside a black hole, and nothing gets out of the black hole either" seems correct. If the manifold ends at the event horizon, there is no inside as a place for things to get to and, consequently, there is nothing there to come out. Commented Nov 4, 2019 at 16:49
• @dcgeorge The solution known today as "Schwarzschild" was in fact given by Hilbert. In the actual solution by Schwarzschild, there is no event horizon, because it's radius is zero (read the foreword in the PDF): arxiv.org/abs/physics/9905030 Commented Nov 4, 2019 at 17:21

You can take Minkowski space and remove all points with $t\ge 0$ (for some Minkowski coordinate time $t$). Then what you have left is a perfectly well-behaved manifold that is a solution to the Einstein field equations. I suppose it could be considered an "assumption" that time will not just end at some arbitrarily chosen time $t=0$, but there is no clear reason to worry about the assumption, because $t=0$ isn't special.

The same logic applies to the event horizon of a black hole. There is nothing special about the event horizon. It doesn't have unusually high curvature or any other unusual properties. The only reason we single it out is because it has certain light-cone relationships with respect to distant regions of spacetime (such as the singularity and null infinity).

It's true that semiclassical gravity tends to predict that crazy stuff happens at the event horizon of a black hole. This is a reason to be very skeptical about all predictions of semiclassical gravity. Note that no prediction of semiclassical gravity has ever been verified, although some of its predictions have been falsified. When it makes obviously false predictions, its practitioners try to fix up the theory by doing renormalizations.

• I would be interested in some examples of falsified predictions. Commented Jan 6, 2018 at 17:56
• @ErikJörgenfelt: Before renormalization, they get all kinds of unphysical results, such as divergences of various quantities at the event horizon of a black hole.
– user4552
Commented Jan 22, 2018 at 19:57
• I'm sorry Ben but I don't understand your answer or how it's related to my question. My question isn't about time ending "... at some arbitrarily chosen time ..." (whatever that means) but about whether or not the entire spacetime manifold ends at the event horizon. Commented Jan 27, 2018 at 15:52
• "There is nothing special about the event horizon. It doesn't have unusually high curvature or any other unusual properties." - In any frame of reference (defined as the coordinate system of a physical observer), time stops and the spatial metric diverges at the horizon. (Note that a falling observer does not have a frame at the horizon.) Consequently the horizon has a lot of special properties, even if spacetime is asymptotically flat there. For example, assuming the Schwarzschild interior, the singularity is not a cause, but a result of the horizon (based on the time direction inside). Commented Nov 2, 2019 at 8:37

We don't yet understand what goes on behind the horizon; it is a topic of active research. The reason to expect spacetime to exist behind the horizon is classical: we can solve Einstein's equations and find the Schwarzschild solution, which extends smoothly across the horizon. This is the classical picture, and we can ask whether quantum corrections can change the answer drastically. One can argue that large quantum effects will show up when the curvature is large, and that for large black holes the curvature at the horizon is small. According to this line of reasoning one expects the classical picture to hold behind the horizon (as long as we don't come too near to the singularity, where the curvature blows up). This is the textbook argument for why there should be spacetime behind the horizon.

However, this argument may be too naive, and perhaps quantum effects can be large even if the curvature is small. This was suggested for example by AMPS. We do not yet know whether AMPS implies there is a firewall, or whether some loop hole (like complementarity) avoids this conclusion. (There are also other reasons besides AMPS to be skeptical of the classical answer.) It may even be that the answer depends on the precise quantum state of the black hole, namely that for some states (like the eternal black hole / thermofield double) there is a smooth interior, while for other states there is a firewall.

• Thanks for your thoughts. You say the reason to expect spacetime to exist behind the horizon is that we can solve the field equations for the region. But the very fact that the equations can be solved there depends on the existence of the manifold there. So, it looks to me like a bit of circular reasoning. It's always seemed significant to me that there had to be a separate solution (with its assumption of a manifold) for the interior that was then mathematically welded to the exterior solution. I've never seen an argument for the assumption. Hence the reason for my question. Commented Jan 6, 2018 at 17:29

From the naive effective field theory perspective, one expects the classical spacetime to be the scenario where a bunch of particles live (photons, gravitons, electrons, ...). This is the semiclassical gravity picture, and it is assumed to be some low energy limit of quantum gravity. Note that these particles can backreact in the geometry, as long as the backreaction is small, through their interaction with gravitons. Thus, one expects the Equivalence Principle to hold, and this means that the horizon is not a special place locally, implying smoothness of the geometry.

Although semiclassical gravity (+ the Equivalence Principle) encodes Hawking's calculation of a black hole radiating thermodynamically, people believe that this effective description cannot capture some very important effects of the whole theory, such as very tiny corrections to the density matrix that make information to be conserved or even the counting of microstates of a stable black hole.

Nonetheless, and even though it is true that the authors you mention argue against the smoothness of the horizon (it is still an open theoretical problem), there are some very reasonable theoretical arguments in the context of holography that go against the AMPS and other proposals, at least within the context of typical black hole states that reach equilibrium. One of the most famous ones is the Papadodimas-Raju proposal, which shows how one can define the black hole interior with a smooth horizon, using some new state-dependent operators.

It would be very nice if we had any experimental hint regarding this problem, and although I doubt we can see if there is a firewall or not using LIGO data, it would be quite amazing if someone could actually extract some information from it.

Q: "For the black-hole solution of the Schwarzschild metric to be at all possible would require spacetime to fill the region under the event horizon. Is this requirement ...".

It would not "fill" the region, it must be slightly smaller to be a black hole.

For example: The Earth's Schwarzschild radius is 8.87 mm and the Sun's is ~2.95 km; because those objects are larger they are not black holes, if they were almost exactly equal they would be a black hole until they increased their diameter beyond the $$r_s$$, if they were smaller they would be black holes (the escape velocity is greater than $$c$$). Small black holes are therefore much more dense than large ones.

Singularities and black holes

The Schwarzschild solution appears to have singularities at $$r = 0$$ and $$r = r_s$$; some of the metric components "blow up" at these radii. Since the Schwarzschild metric is only expected to be valid for radii larger than the radius $$R$$ of the gravitating body, there is no problem as long as $$R > r_s$$. For ordinary stars and planets this is always the case. For example, the radius of the Sun is approximately 700000 km, while its Schwarzschild radius is only 3 km.

The singularity at $$r = r_s$$ divides the Schwarzschild coordinates in two disconnected patches.

The exterior Schwarzschild solution with $$r > r_s$$ is the one that is related to the gravitational fields of stars and planets. The interior Schwarzschild solution with $$0 ≤ r < rs$$, which contains the singularity at $$r = 0$$, is completely separated from the outer patch by the singularity at $$r = r_s$$. The Schwarzschild coordinates therefore give no physical connection between the two patches, which may be viewed as separate solutions.

The singularity at $$r = r_s$$ is an illusion however; it is an instance of what is called a coordinate singularity. As the name implies, the singularity arises from a bad choice of coordinates or coordinate conditions.

When changing to a different coordinate system (for example Lemaitre coordinates, Eddington–Finkelstein coordinates, Kruskal–Szekeres coordinates, Novikov coordinates, or Gullstrand–Painlevé coordinates) the metric becomes regular at $$r = r_s$$ and can extend the external patch to values of $$r$$ smaller than $$r_s$$. Using a different coordinate transformation one can then relate the extended external patch to the inner patch.

When choosing coordinate conditions, it is important to beware of illusions or artifacts that can be created by that choice. For example, the Schwarzschild metric may include an apparent singularity at a surface that is separate from the point-source, but that singularity is merely an artifact of the choice of coordinate conditions, rather than arising from actual physical reality.

A coordinate singularity occurs when an apparent singularity or discontinuity occurs in one coordinate frame, which can be removed by choosing a different frame.

Both the Schwarzschild metric and the AMPS firewall fail to consider the ergosphere, the horizon caused by the rotation of the black hole. The Kerr metric and the Kerr–Newman metric are believed to be representative of all rotating black hole solutions, in the exterior region.

The ergosphere touches the event horizon at the poles of a rotating black hole and extends to a greater radius at the equator. With a low spin of the central mass the shape of the ergosphere can be approximated by an oblate spheroid, while with higher spins it resembles a pumpkin-shape. The equatorial (maximum) radius of an ergosphere corresponds to the Schwarzschild radius of a non-rotating black hole; the polar (minimum) radius can be as little as half the Schwarzschild radius (the radius of a non-rotating black hole) in the case that the black hole is rotating maximally (at higher rotation rates the black hole could not have formed).

As a black hole rotates, it twists spacetime in the direction of the rotation at a speed that decreases with distance from the event horizon. This process is known as the Lense-Thirring effect or frame-dragging. Because of this dragging effect, an object within the ergosphere cannot appear stationary with respect to an outside observer at a great distance unless that object was to move at faster than the speed of light (an impossibility) with respect to the local spacetime.

There's a chance of a non-rotating black hole occurring, and one where the region under the event horizon is filled, it's also reasonable that it is an unusual case. Usually the object would be smaller than $$r_s$$ and rotate due to in-falling matter.

• "A coordinate singularity occurs when an apparent singularity or discontinuity occurs in one coordinate frame, which can be removed by choosing a different frame" - This is an urban legend in cosmology. Any coordinate transformation pretending to remove the so labeled "coordinate singularity" is necessarily singular and therefore mathematically forbidden. Schwarzschild coordinates are spherical coordinates with no singularity anywhere. Thus the horizon singularity must be embedded in the field equations and therefore is physical and non-removable. Commented Nov 2, 2019 at 7:56
• @safesp, indeed that statement has some popularity, as a search for that sentence returns 53 results; one mine and one Wikipedia's (quoting Stephen Hawking), along with other non-results. I believe Mitra covers this in 1 (page 19), and 2 (still reading, to determine if it's supportive).
– Rob
Commented Nov 2, 2019 at 15:51