# Hypermassive (?) black holes

1. May 25, 2014

### Staff: Mentor

Bear with me, I am (not even) a chemist.

This is something that bothers me for some time.

The larger the black hole, the smaller the density and the smaller the tidal forces. Supermassive black holes (expected to exist in centers of galaxies) have density comparable to that of water, and I was told (in some other thread at PF) that tidal forces on the event horizon are "bearable" - that is, it can be in theory possible to cross the event horizon without being changed into a pretzel or rotini.

For any given mass we can easily calculate Schwarzschild radius, and the the density of the matter inside.

Here comes the part that can get me banned. A hypermassive black hole can have density comparable to that of the Milky Way, or a local group of galaxies. If so, is there anything that would make those living inside aware of the fact they are inside of a black hole? Or is there some other problem I am not aware of?

Instant ban: if the situation I described above is possible, what stops black holes from being nested?

2. May 25, 2014

### Bill_K

The inside of a black hole is distinctly anisotropic. Once you cross the event horizon, the coordinate "r" becomes timelike and the coordinate "t" becomes spacelike. Which means the translation symmetry t → t + C becomes a space translation.

The world inside a hole has more or less the symmetry of a paper towel tube : spherical cross-section combined with translational symmetry in the "t" direction. What's worse, while the spherical cross-section is contracting, and will lead you inescapably to the singularity at r = 0, at the same time the translational direction is lengthening. So an insider watching other insiders would see blueshifted ones in some directions and redshifted ones in the other.

3. May 25, 2014

### phinds

Are you sure about that? I don't know how to do the calculations but it seems REALLY unrealistic. There's a lot of mass in such objects but the size is stupendous by comparison. I mean we're talking about a radius of tens or even hundreds of thousands of light years !!!

I would be surprised to hear that a galaxy could have a density that was anything greater than than several orders of magnitude less than water, which is, as you say, the approximate density of some supermassive black holes.

I too have heard, and have no trouble believing, that supermassive black holes have tidal forces which are negligible at the EH.

I also remember a recent thread which said that even if dark matter were as dense in our solar system as it is in the densest parts of galaxies, the effect would be about that of a small asteroid. Since dark matter is most of what galaxies are made of, I conclude from all that that the density of galaxies differs from zero by a rounding error in the 10th decimal place or so. AS I said, I can't do any acutal calcs since I dont' have the data, so I could be wrong, but I'll be surprised if I am

4. May 25, 2014

### Staff: Mentor

Actually, what I think you are referring to here as the "density" is not really a physically valid calculation. I think by "density" you mean taking the mass of the hole and dividing it by the volume of a sphere whose radius is the hole's horizon radius, i.e.,

$$\rho = \frac{3 M}{4 \pi r_s^3}$$

Substituting $r_s = 2M$ then gives

$$\rho = \frac{3}{32 \pi M^2}$$

But this is not the actual density of matter you would find inside the hole's horizon, for two reasons:

First, the mass of the hole is not distributed throughout the region inside the horizon; it's all concentrated at the singularity at the center, $r = 0$. The rest of the region inside the horizon is vacuum, unless something falls in.

Second, the spatial volume of the region inside the hole's horizon is not necessarily the same as the volume of a sphere with radius equal to the horizon radius. "Spatial volume" is frame-dependent; the spatial volume of the region inside the horizon depends on how you take spacelike slices through it, which depends on what simultaneity convention you adopt. In fact, because of the $t$ translation symmetry that Bill_K described, you can actually cut spacelike slices so that the spatial volume of the region inside the horizon is infinite!

5. May 25, 2014

### Staff: Mentor

Yes, he's right. Consider that the Schwarzchild radius grows as the first power of M, while the "volume" inside the event horizon must grow at least as the cube of the radius (that's sloppy - I should say the 3/2 power of the area of the event horizon). Clearly as I increase M the volume increases and the "density" decreases. We can make this "density" as small as we want by choosing a large enough M.

Note, however, that this "density" doesn't have much physical significance - it's what we would get if the mass inside the horizon were uniformly distributed, but of course it's not.

6. May 25, 2014

### phinds

I can't tell off hand whether "he's right" refers to me or Borek but your calculations support my contention, so I assume it's me.

Yes, I agree completely with this and with what Peter said but I think Borek and I were both having a simplistic Mass/Volume discussion.

7. May 25, 2014

### PAllen

That density does have meaning in the following sense. If you tried to form an air mass the size of the milkyway, you could not, because such an airmass would be inside its SC radius and would inevitably collapse before you could assemble it. This simple computation does give an upper limit on the size of a mass you can assemble of some average density.

Note also, that before any singularity forms, the geometry in the interior is not the SC interior geometry.

8. May 25, 2014

### Staff: Mentor

As I noted in my previous post, this "volume" is actually frame-dependent; with the right choice of frame, it can be infinite.

9. May 25, 2014

### bcrowell

Staff Emeritus
But this interchange of the spacelike and timelike character of the Schwarzschild r and t coordinates isn't locally observable.

But the time to free-fall to the singularity can be made as large as desired by making the black hole sufficiently large.

I don't think this is quite right. Only in a static spacetime can you define a gravitational potential. The spacetime inside the event horizon isn't static.

I don't think there's any way to prove that our part of the universe isn't inside a black hole's event horizon. I think all we can do is to put lower limits on the size of that black hole. For example, we don't observe big tidal stresses, we observe stars and rocks that are on the order of billions years old, and our observations of the observable universe are consistent with isotropy on scales of billions of light-years. These observations give the putative black hole some minimum size, which I'd guess is on the order of billions of light-years.

10. May 25, 2014

### Staff: Mentor

True, but isn't the fact that there is a 4th Killing vector field $\partial / \partial t$ locally observable? It's true that this KVF is spacelike inside the horizon (instead of timelike, as it is outside), but it's still an additional symmetry that should be observable.

I think Bill_K was talking about observed Doppler shift of light signals sent between free-falling observers, not about gravitational redshift/blueshift.

Perhaps not "prove", but doesn't the predictive success of our current cosmological models, which are based on a very different spacetime, constitute at least strong evidence against this hypothesis?

I would think it would have to be *much* bigger than that, because if our universe were inside a black hole's horizon, the observations you describe would basically set lower limits on the size of a local inertial frame centered on our worldline; but this size has to be much *smaller* than the mass of the hole. (The further inside the horizon, the smaller it has to be relative to the hole's mass.) So if, for example, our local inertial frame is billions of light-years across, the hole's mass would have to be the equivalent of trillions of light-years at least--possibly much more, depending on how accurate our measurements are and how far inside the hole's horizon we're supposed to be.

Last edited: May 26, 2014
11. May 26, 2014

### bcrowell

Staff Emeritus
PeterDonis, your #10 has a bunch of quotes attributed to yogi that are actually quoted from my #9.

One way of stating the equivalence principle is that spacetime always has the same properties locally, but this depends on how local is "local." If you think of all the observables as being like a Taylor series expanded around some point, then you have to go to higher-order terms to see things like tidal stresses or other types of curvature. In any case, I don't think the space inside the horizon has a higher level of symmetry than the space outside. The definition of a Killing vector is coordinate-independent, and the idea that something special happens at the event horizon is based on the Schwarzschild coordinates. We also usually define a Killing vector as a field that exists globally, not locally. A cylinder is locally the same as a plane, but they have different numbers of Killing vectors. In other words, studying the solutions of the Killing equation locally doesn't tell you anything about the global structure.

Maybe we could get Bill_K to clarify. If this is what he meant, then the results would depend on the states of motion of the emitting and receiving observer, not just on their relative positions, as he seems to imply. The equivalence principle dictates that no effect like this is observable *locally* by free-falling observers; similarly, the Pound-Rebka experiment was local, and would have given a null result if the emitter and receiver had both been free-falling at the same velocity.

Sure, there's absolutely no observational evidence for such a hypothesis. A similar hypothesis would be that our observable universe is part of an island of matter with a radius of 10^12 l.y., and the rest of the universe outside that island is a vacuum. I don't think it can be disproved, but there's no natural reason to believe it, either. We observe homogeneity and isotropy within our observable universe, so it's natural to extrapolate and make cosmological models in which the whole universe is that way. The only way to verify it directly is to wait a very long time until our horizon expands enough to let us see that far. Similarly, people thousands of years ago couldn't prove that the sun was going to rise the following day, but it was natural to extrapolate from past experience and expect that it would.

By the way, there are two questions that should be distinguished logically:

(1) Are we inside a giant black hole?

(2) Is the whole universe a black hole?

We're discussing #1. #2 is a different question, with a different answer, and we have a FAQ on it: https://www.physicsforums.com/showthread.php?t=506992 [Broken]

Last edited by a moderator: May 6, 2017
12. May 26, 2014

### Staff: Mentor

Oops, sorry about that! Fixed now.

I agree, it doesn't. But it does have the same level of symmetry, i.e., it has *more* symmetry than just spherical symmetry. I wasn't trying to suggest that anything "special" happens to the symmetry at the horizon--although actually, since the KVF becomes null there and spacelike inside, there *is* a coordinate-independent difference.

Hmm. Killing's equation is local, not global--it is a condition on the Lie derivative of the metric. Technically, I guess that isn't purely "local", since to define a Lie derivative you at least need a vector field on some neighborhood, not just a single point. But the neighborhood can be very small.

Consider the case outside the horizon, since that's simpler to imagine. The timelike KVF just means that the metric is unchanged along orbits of the KVF, i.e., along the worldlines of static observers. These observers can physically measure that by measuring metric components, then waiting a little, then measuring them again. They don't have to make measurements over the entire history of the universe. Similarly, I would expect that the 4th KVF inside the horizon could be physically measured over some reasonably small portion of one of its (spacelike) orbits, instead of requiring measurements to be made globally over the entire region inside the horizon.

Yes, that's true. I was assuming that he meant observers free-falling radially, but I may be misinterpreting.

Ah, ok, that clarifies things.

Yes, agreed. Thanks for the pointer to the FAQ, I was not aware of that one.

Last edited by a moderator: May 6, 2017
13. May 26, 2014

### George Jones

Staff Emeritus
I agree with Ben that there is no local test that cam tell that we are inside a black. After all, the very definition of "event horizon" is very non-local, and involves knowledge of spacetime "out to infinity".

No. I agree with Peter, a Killing vector field might only exist of part of a spacetime manifold. The Schwarzschild Killing vector in question, however, is defined on all (of the physically relevant) spacetime manifold.

This is easily seen by switching to Painleve-Gullstrand coordinates $\left(T,R\right)$ defined everywhere (even on the horizon), which are related to the usual Schwarzschild coordinates $\left(t,r\right)$ by

\begin{align} T &= t + 4M\left( \sqrt{\frac{r}{2M}}+\frac{1}{2}\ln \left| \frac{\sqrt{\frac{r}{2M}}-1}{\sqrt{\frac{r}{2M}}+1}\right| \right)\\ R &= r. \end{align}

It can be shown from this that (exercise)

\begin{align} \frac{\partial }{\partial T} &= \frac{\partial }{\partial t}\\ \frac{\partial }{\partial R} & = -\frac{\sqrt{\frac{2M}{r}}}{1-\frac{2M}{r}}\frac{\partial }{\partial t}+\frac{\partial }{\partial r} \end{align}

In these coordinates, the metric looks like

\begin{align} g &= dT^2 - \left(dR + \sqrt{\frac{2M}{R}} dT\right) ^2 \\ &= \left( 1-\frac{2M}{R}\right) dT^2 - 2\sqrt{\frac{2M}{R}} dT dR + dR^2 . \end{align}

Clearly, $\partial / \partial T$ is a globally defined Killing vector that is timelike outside the horizon, lightlike on the horizon, and spacelike inside the horizon.

Standard Schwarzschild coordinates use outrageously terrible notation, and this notation (but not the coordinates!) should be banished. Schwarzschild coordinates really are two distinct disconnected coordinate charts, and neither chart is defined on the event horizon. As two distinct charts, different sets of labels should be used for the coordinates in the two charts. In particular, the labels for the interior coordinates should be chosen to have more physical relevance.

14. Jun 26, 2014

### Staff: Mentor

Sorry to not get to you earlier.

That was my way of thinking, and I never meant the density to be taken literally as if it was identical everywhere. More like in terms of the shell theorem - when observed from the outside it doesn't matter how the mass is distributed inside, it can be identical everywhere, it can be different, it doesn't matter (and yes, I know shell theorem requires distribution of mass to be spherically symmetrical, I am using it just as an analogy). But we can compare different parts of space comparing their average densities - I know it is different in our planetary system, it is different inside the Galaxy, different outside (but inside of the local group) and so on.

I take it as a "yes" to my first question - technically it is possible that those living inside of a hypermassive black hole are not able to realize they live inside (at least to some level of certainty). That in turn means they can be inside and live happily for quite some time.

What about the second question then, the one about nested black holes?

15. Jun 26, 2014

### Staff: Mentor

There is no local test that can tell that one is inside an event horizon, yes. But there *is* a local test that can tell if one is inside an *apparent* horizon: just measure the expansion of outgoing light rays in your vicinity. (If you're not sure which direction is "outgoing", just measure the expansion of light rays emitted in all directions locally; inside an apparent horizon the expansion should be negative in all directions.)

In the standard Schwarzschild solution, the event horizon is also an apparent horizon, so everywhere inside the event horizon, you're also inside an apparent horizon and can locally measure that fact, as above. But of course, no real black hole is exactly a Schwarzschild black hole, if for no other reason than that any real black hole will be gaining mass as things fall into it. For a black hole that is gaining mass, you can be inside the event horizon but still outside the apparent horizon, so the local test could give you a "false negative", so to speak.

Event horizons can't be nested by definition. An event horizon is the boundary of the region that can't send light signals to future null infinity. There's only one future null infinity, so there can only be one such region in a spacetime. When one black hole falls into another, their event horizons merge--more precisely, there is a single event horizon in the spacetime, but it looks like a pair of trousers instead of a cylinder, so to speak.

I think it *is* possible for apparent horizons to be nested, in the sense that, as you were moving through a region of spacetime and running the "apparent horizon test" I described above (testing whether the expansion of light rays in your vicinity was negative), the test results could alternate more than once: "no", then "yes", then "no", then "yes" again. (Obviously this requires a more complicated spacetime than a standard Schwarzschild black hole, or even one that is gaining mass as things fall into it.)

16. Jun 26, 2014

### PAllen

Since the true horizon has a global definition, there is no such thing as a horizon within a horizon. Only the boundary of what can reach null infinity counts as a horizon.

However, there is a related quasi-local concept of an apparent horizon (or trapping surface). I see no reason preventing formation of such surface, surrounding a singularity, within the global horizon, during the course of chaotic collapse of a supermassive BH.

17. Jun 26, 2014

### George Jones

Staff Emeritus
Here are two scenarios that Borek possibly has in mind. The first is plausible, while the second is highly contrived.

Scenario 1
Suppose a 10 solar mass black that formed after a supernova "falls into" a several million solar mass black hole at the centre of a galaxy. In this situation, I think that "falls into" is really "merges with". Initially, there are two apparent horizons, but, in the end, there is only one apparent horizon.

Scenario 2
Suppose a large star near the end of its life falls into a superdupermassive black hole. The black hole is so massive that tidal forces are too small to rip apart the star before it goes supenova. Suppose the star is massive enough that if it went supernova outside the superdupermassive black hole, the supernova remnant would be a black hole. What happens after the star goes supernova inside the superdupermassive black hole?

18. Jun 26, 2014

### Staff: Mentor

Yes, that's scenario I am thinking about.

Edit: while i don't claim to understand it in details, I know event horizons merge when black holes merge. I always pictured it as similar to drops of water combing in zero gravity. But I don't see how it happens in the second scenario, when the "external" event horizon is so far away.

Last edited: Jun 26, 2014
19. Jun 26, 2014

### Staff: Mentor

No merge has to happen in the second scenario; by hypothesis, the entire star is inside the event horizon already before it goes supernova. The "merging" happened when the star fell into the superdupermassive black hole in the first place; the mass of the superdupermassive black hole increased, and so did the size of its event horizon. Everything that happens to the star after that is happening inside the event horizon already; none of those events can ever send light signals to future null infinity. So even if the star goes supernova, no "new" event horizon forms.

20. Jun 27, 2014

### Staff: Mentor

So what would those inside the hypermassive BH see when the star goes supernova?

21. Jun 27, 2014

### PAllen

See my post #16. You could get a trapping surface. To someone elsewhere within the horizon, this would look just like a BH does in a normal situtation.

22. Jun 27, 2014

### Staff: Mentor

Yes.

I'm not sure that's true. The "normal situation" implicitly uses the fact that the exterior of the BH is stationary; that's what allows us to derive the "normal" appearance of things falling into the BH's horizon (to a distant observer, they appear to redshift more and more and move more and more slowly). But the interior of a BH is not stationary, so you can't use that reasoning to derive what a "collapse" in the interior of a superdupermassive BH would look like.

23. Jun 28, 2014

### PAllen

If the hypermamassive BH is big enough so the horizon is a billion light years, I don't see why not. The collapse phase then becomes a cosmological issues with effects on cosmological time scale.

24. Feb 13, 2016

### EinsteinKreuz

You are talking about a Schwarzschild black hole but what about a Kerr black hole(the rotating type)? Kerr black holes are usually far larger and the ones found at the center of galaxies, like Sagittarius A* and M87(core). The singularity of a Kerr black hole is not a dimensionless point, but a ring of infinitesimal thickness, infinite density, and finite non-zero diameter(and circumference). The geometry and topology of the space inside of the Kerr event horizon is quite different from that of a Schwarzschild black hole. It is not simply connected for one thing. So a hypermassive Kerr black hole would be a giant ring quadrillions of times the mass of the sun with a diameter thousands of light years across.

Such an object MIGHT exist in M87 because that galaxy has almost no new star formation and appears to be slowly collapsing into its core(along with another nearby galaxy that was sucked into it).

25. Feb 13, 2016

### Staff: Mentor

This is true in the idealized Kerr solution; but the singularity in that solution, just like the singularity in the Schwarzschild solution, is not believed to be physically relevant, because quantum gravity effects are expected to be strong enough to completely change the nature of the interior before the singularity is reached.