How can super massive black holes have a lower density than water?

Question

I heard on a podcast recently that the supermassive black holes at the centre of some galaxies could have densities less than water, so in theory, they could float on the substance they were gobbling up... can someone explain how something with such mass could float?

Please see the below link for the podcast in question:

http://www.universetoday.com/83204/podcast-supermassive-black-holes/

Another way of sort of saying the same thing is that if you think of gravity as a flux, as would be suitable for a 1/r^2 gravity theory like Newton's, then there's no limit to the flux you can obtain no matter the lack of density of the mass you make it from; just make it big enough. — Carl Brannen, Feb 26 '11 at 00:29
Floatation makes no sense in this context.. A supermassive black hole has more grav arrtn than the earth. So the blach hole will stay put at half the height of the fluid. — Manishearth, Feb 13 '12 at 15:45
I always thought how God's throne can be resting on waters:) — , Sep 27 '13 at 18:23
The water pool would pack more mass into a similar size of space the our BH, so the whole thing collapses into one giant BH immediately. — Kevin Kostlan, Apr 05 '14 at 14:54

Vagelford · Accepted Answer · 2011-02-25T17:25:01.423

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Well, it can't (float), since a Black Hole is not a solid object that has any kind of surface.

When someone says that a super massive black hole has less density than water, one probably means that since the density goes like $\frac{M}{R^3}$ where M is the mass and R is the typical size of the object, then for a black hole the typical size is the Schwarzschild radius which is $2M$, which gives for the density the result

$$\rho\propto M^{-2}$$

You can see from that, that for very massive black holes you can get very small densities (all these are in units where the mass is also expressed in meters). But that doesn’t mean anything, since the Black Hole doesn’t have a surface at the Schwarzschild radius. It is just curved empty space.

edited Feb 25 '11 at 17:25

answered Feb 25 '11 at 17:18

Vagelford

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You should be clear that "it can't" refers to floating, not to having a low density. – dmckee --- ex-moderator kitten Feb 25 '11 at 17:23
-1: It can float in theory, and it would float, except that the water it is floating in will collapse to a black hole before it has a chance to float. It might be better to float it in an astrophysically huge amount of honey. – Ron Maimon Sep 03 '11 at 22:18
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As I’ve pointed out to your answer, since there is no surface for the BH and the matter is freefalling in the BH, there is no pressure. Thus there is no floating in any conventional way. The only thing that matters in that case is the total momentum transfer from the fluid, which will be pointing on the same way as the exterior field. It is a wrong, bad analogy. Sorry. – Vagelford Sep 03 '11 at 22:35
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Free falling does not mean no pressure! It means that there is no pushing on the black hole itself. The pressure accelerates the infalling water, which enters with a greater kinetic energy (and up-momentum) at the bottom than at the top. The difference is exactly the bouyancy force in the fluid, as can be seen by analying the flow of up-momentum in a big sphere surrounding the black hole. – Ron Maimon Sep 03 '11 at 22:51
You don’t seam to get what I am telling you. I am not saying that the fluid will have no internal pressure due to the temperature (kinetic energy of the atoms/molecules). I am saying that due to its freefall there will be no hydrostatic pressure difference in the way that you are implying (the hydrostatic pressure “under” the BH being grater than the one “over” it). – Vagelford Sep 04 '11 at 09:29
The only thing that would matter in the momentum transfer of the fluid to the BH would be the freefall velocity of the fluid which is the opposite than what you are saying, because on the top side you would have an acceleration of gravity that would be $g_{BH}+g_{ext}$ and on the bottom side you would have $g_{BH}-g_{ext}$. Off course this whole construction is over-oversimplifying and misleading and doesn’t hold any water. – Vagelford Sep 04 '11 at 09:29
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@vagelford: the whole construction is not oversimplified, it is not misleading, and it can be done in principle with a viscous enough liquid pool and a small enough black hole. The pool has to be in gravity or accelarated, to give it a pressure gradient. A black hole falling through such a pool (in gravity or accelerated) will absorb fluid in such a way that it feels a bouyancy force. The bouyancy force is the same as any other floating/sinking object, because it is determined by conservation laws. – Ron Maimon Sep 05 '11 at 14:28
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I am sorry, but I obviously can't make you see your mistake. My final answer is in my last two comments of your answer. – Vagelford Sep 05 '11 at 23:13
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@Vagelford: I did not make a mistake. – Ron Maimon Sep 06 '11 at 01:16
@Vagelford does that mean that a black hole less dense than water would sink in a pool of water? That sounds utterly strange, wouldn't that break the law of conservation of momentum? – untreated_paramediensis_karnik Mar 03 '18 at 20:17
@untreated_paramediensis_karnik Water in contact with a black hole of any density is no longer water but black hole. – Wookie Mar 12 '24 at 21:54
@Wookie so how does it answer the question? – untreated_paramediensis_karnik Mar 13 '24 at 06:32
@untreated_paramediensis_karnik t goes to zero at the horizon so there's no sinking process. Water and the interior are not causally connected. – Wookie Mar 14 '24 at 09:49

score 10 · Answer 2 · answered Feb 25 '11 at 21:10

I think it is actually misleading to make the claim that is puzzling you. "Density" suggests that the mass is distributed more or less uniformly within the black hole, and this is non-sense. The black hole is mostly empty, and all the mass is concentrated within a tiny region (clasically a point) in the center of the black hole.

If you ignore this and pretend a black hole of mass $M$ and volume $V\propto r^3$ had a uniform density $\rho$ then you can calculate it, simply using $\rho=M/V$. Since for Schwarzschild black holes the radius of the black hole is proportional to its mass you obtain finally $\rho\propto 1/M^2$, so the heavier a black hole the smaller its density. But again, this provides a highly misleading picture of the mass distribution within the black hole. All its mass is in the center, so classically the density is infinite.

That is not true, in the reference frame of an external observer the mass is spread over a radius which is an infinitesimal larger than the horizon, since gravitational time dilatation slows down the process. Only in the frame of an infalling observer who's own worldline gets terminated in the singularity the mass is concentrated in the center. — Yukterez, Aug 20 '18 at 04:21

Ron Maimon · Answer 3 · 2011-09-03T22:17:13.717

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The black hole would float in water, if you could make a large enough pool to submerge it, and with enough replenishment to replace the water that the black hole will sucks up. The black hole will remove water from its surroundings, but the water below will come into the horizon at higher pressure than the water above, so the velocity inward will not be uniform.

If the black hole is denser than water, it will sink for a while, because the pressure difference is not enough to compensate for the pull of gravity. If the black hole has less density than water, it will float. It's like a balloon that sucks in water and expands, always maintaining a volume which is big enough to keep itself lighter than water.

The problem is that when the black hole density is as that of water, a volume of water equal to the black hole's volume will not be stable to gravitational collapse, so it will be impossible to set up the pool.

edited Sep 03 '11 at 22:17

answered Sep 03 '11 at 20:41

Ron Maimon

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Well, you are forgetting the effect of the self-gravity of the black hole which in the case of a balloon submerged in a fluid is negligible. In order to have the setting you are proposing the water and the BH would have to be embedded in an exterior, lets say uniform, gravitational field which would also define the difference between up and down. The first thing to point out is that in the free falling matter in the BH the important thing is the fluids kinetic energy and not the pressure. So, “under” the BH the gravitational pull towards the BH would be smaller relative to “over” the BH. – Vagelford Sep 03 '11 at 22:26
Thus the flow coming from the “up side” would have grater kinetic energy and thus it would excerpt a grater dynamic pressure (transfer more momentum to the BH). Whatever the case, it is a bad and misleading analogy the whole “floating” thing. – Vagelford Sep 03 '11 at 22:27
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Ignoring the issue that the water would collapse under gravity, which invalidates the whole thing, the greater pressure below ensures that the water below will enter with a greater kinetic energy, and this is the reason for "floating". The water at the top will enter at a lower velocity, so there will be a total momentum transfer which is exactly as for any other submerged object. The problem with the density of water being unsustainable might be fixable by going to a more viscous fluid which will ooze into the black hole slowly, so you could have slow sucking up of material. – Ron Maimon Sep 03 '11 at 22:49
There is no hydrostatic pressure difference in freefall. – Vagelford Sep 04 '11 at 09:33
@Vagelford--- of course, I was assuming the pool is accelerated upward at 1g, with the black hole floating on top. – Ron Maimon Sep 04 '11 at 15:11
And my point is that that is meaningless, since the water that comes near the BH will be in freefall in to the BH regardless of the external field or acceleration. – Vagelford Sep 05 '11 at 13:46
The water which comes near the black hole will be in freefall at the surface of the black hole, but far away, it will be static with a linear pressure gradient. From these two ingredients you can determine that the black hole will feel a bouyancy force the same as any other floating/sinking object. – Ron Maimon Sep 05 '11 at 14:32
The BH has no surface. 2) How will the "hydrostatic pressure" of the far away static water be transferred to the freefalling water near the BH when there is no support between them? 3) The pressure gradient will be $-\frac{1}{\rho}\frac{dP}{dr}=g_{ext}+g_{BH}+a_{f}$, where you have the exterior field, the BH field and the acceleration of the fluid. When the fluid moves under the influence of the two fields, the right hand side is zero, thus there is no pressure gradient. In a pool there is a pressure gradient because the walls don't let the water move and a ball excerpts negligible gravity.

Vagelford

Sep 05 '11 at 23:00

And of course a ball has a surface... – Vagelford Sep 05 '11 at 23:05

@Vagelford: the hydrostatic pressure of the far away water is communicated to the free-falling water passing the horizon through Bernoulli's principle--- the water falling in at the bottom is entering with a greater momentum and kinetic energy as dictated by the conservation laws across the streamline. The result is that the black hole experiences a bouyancy force, the same as any other object. I believe you see this too now. I hope you will retract your incorrect answer. – Ron Maimon Sep 06 '11 at 01:18

LOL... that was nice... – Vagelford Sep 07 '11 at 16:33

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@Vagelford: it's not funny. The buoyancy force on a region is determined by conservation laws on a big sphere far away from the horizon, where the water is essentially stationary. On such a sphere, the total force is equal to the mass of displaced water, and the total mass is less than this because of the black hole's lower density. There is no way around it, because it is determined by conservation laws. – Ron Maimon Sep 07 '11 at 16:37

Would the water really be sucked into the black hole? Surely from the perspective of an outside observer, water would only approach the horizon but never actually cross it. So you would not need any replacement water. Is this wrong?

I think floating might be possible if the black hole was surrounded by matter near the event horizon.

– SMeznaric Oct 04 '12 at 23:41

This answer has enough comments that I don't need to explain the downvote. – Timothy Aug 20 '18 at 03:48

score 3 · Answer 4 · answered Apr 24 '16 at 16:21

Density here is a bit misleading. For example density of galaxy cluster is low because there is so much space between it. It does not mean all the material inside has low density.. We cannot observe beyond Event horizon, so we use diameter of event horizon to measure volume. Actual matter will be in a smaller volume.

To understand why large objects need smaller densities for black holes, look at escape velocity equation. $$v_e =\sqrt{\frac{2\,G\,M}{r}}$$

If ratio of M and r remains constant,escape velocity remains constant. If you keep the density same and increase the diameter, Mass increase cubicly. For example if you increase radius 10 times mass will increase 1000 times and escape velocity will increase 10 times.

So a linear increase of escape velocity with increase of mass, if density remains same (which won't happen). The exact relation of density of escape velocity and density is

$$v_e \approx 2.364 \times 10^{-5} r \times \sqrt{\rho}$$ Thus super massive black holes have low density inside their event horizons, but this does not reflect density of material.

score 2 · Answer 5 · answered Feb 25 '11 at 17:28

The Schwarzschild radius scales with mass as $r~=~2GM/c^2$. What might be defined as a Schwarzschild volume would then be $V~=~4\pi r^3/3$ $=~(32/3)\pi(GM/c^2)^3$. So the density of matter defined by the horizon is $\rho~=~(3/32)(c^2/G)^3M^{-2}$. So density scales as the inverse square of the mass. A 10 billion solar mass black hole has a radius about $10^{10}$km, or a volume $V~\sim~10^{30}km^3$ $=~10^{39}m^3$. A solar mass is $10^{30}$kg and the density defined by the horizon is then $\rho~=~10^{-9}kg/m^3$. That is actually quite small.

Of course if you fall into a black hole of any mass you encounter a region with enormous Weyl curvature and tidal forces. The source of this is in your future, and eventually you reach it --- it is inescapable. This region where curvature diverges is a spatial surface of infinite extent.

You should add some closing [/math] tags 'Sothedensityofmatterdefine...' yeah after reading that two times I could decipher it :) — datenwolf, Feb 25 '11 at 17:31
Or you get incinerated instantly at the inner horizon. You will never feel any tidal forces for supermassive black holes of moderate spin. — Kevin Kostlan, Nov 26 '19 at 19:21

score 1 · Answer 6 · edited Sep 27 '13 at 19:59

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I just read that statement from Wikipedia, and when they were mathematically calculating the density, they were using the Schwarzschild radius (aka radius of event horizon), not the singularity. The singularity itself, theoretically, has infinite density and exists in one point (radius approaches 0), so no, it won't float on water. Always remember when the term "black hole" is used, its talking about a "region of space" from which light cannot escape: its not an object, and that is a common misconception.

edited Sep 27 '13 at 19:59

Emilio Pisanty

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answered Sep 27 '13 at 17:39

user30226

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I sort of see your point. It explains how it can be a finite density instead of an infinite one but it doesn't add to the other answers. Even if you do define density as $\frac{M}{\frac{4}{3}\pi r_s^3}$, a lot of people can figure out that a low mass black hole would have a really high density and would assume that a larger one also does but some of the other answers explain why that's not so. – Timothy Aug 20 '18 at 04:21

score 0 · Answer 7 · answered Aug 20 '18 at 04:13

For matter to collapse into a black hole, it must be compressed to about the size of its Schwarzschild radius. Some people say a black hole is infinitely dense because the solution to a certain theory is that it all goes to the singularity and does not disappear there. I define the density of a black hole to be its mass divided by $\frac{4}{3}\pi r_s^3$ which is approximately the density an object of that mass must be in compressed to in order to collapse into a black hole. According to one coordinate system which I think is the Gullstrand–Painlevé coordinate system, for a Schwarzschild black hole, the Schwarzschild radius is the distance from the singularity to the event horizon and just happens to be exactly the distance where the escape velocity is predicted to be $c$ in Newtonian physics, that is $\frac{2GM}{c^2}$. From this we see that the size of a black hole varies linearly with its mass so its density varies as the minus second power of mass.

If you have a really light material like rock and keep adding to it, it will get bigger and eventually be under such high pressure that it compresses to electron degenerate matter and after it gains more mass compress to neutron degenerate matter and once that exceeds a certain mass, it will start a runaway effect towards collapsing into a black hole. Once it's a black hole, adding more mass will cause the opposite effect of increasing its size enough to lower its density.

And you say absolutely nothing about whether the BH would sink in water? — untreated_paramediensis_karnik, Aug 20 '18 at 20:43
@user54826 I don't want to say anything about that because I'm afraid I might be saying wrong information. Answers are supposed to be very accurate. If I'm not sure of something, it's better to ask a question. I think it's fine to write an answer when I'm not sure if it adds anything to the other answers because some of the time it will belong and other people can delete the ones that don't. The comments on Ron Maimom's answer show why a black hole less dense than water won't necessarily float on water. I think my answer still adds something to the other answers. — Timothy, Aug 24 '18 at 03:05
Adding mass to a blackhole increases the size of event horizon, yet actual degenerate matter stays at a single point right? — Harshiv, Jul 15 '22 at 19:19

How can super massive black holes have a lower density than water?

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