# Do Roche limits apply to black holes?

Consider black hole A, a super massive black hole at the center of the galaxy. Orbiting it is black hole B, a much less massive black hole.

If some passing body were to modify black hole B's orbit such that it fell within the Roche limit of black hole A, what would happen to black hole B?

If it were to turn into a ring, would the black hole matter re-inflate since it wouldn't be under such high gravity? Do black holes even respond to Roche limits like regular matter does?

• The idea of the Roche limit does apply to black holes if the secondary body isn't a black hole. For example, if an asteroid approaches a black hole too closely, it will be torn apart. The distance at which this happens is related to the radius of a body with the mass of the black hole and the density of the asteroid. – Keith Thompson Jan 10 '15 at 0:19

The Roche limit applies when a smaller body that would be held together by its own self-gravity is in the gravitational field of another, such that the tidal forces of the latter are stronger than the self-gravity of the latter, thus destroying the smaller body.

However, the gravitational tidal forces of a black hole are always finite, except at the internal singularity. This is a problem because the self-gravity of a black hole, in the sense of the acceleration a mass would need to remain stationary on its surface, is infinite1. Thus, we shouldn't expect for a large black hole to destroy another through gravitational tidal forces.

Put another way, the Roche limit occurs when particles from the smaller body can escape them... but they can't escape the event horizon of a black hole. Thus, the black holes will either orbit or merge, which is what happens in numerical simulations.

1There is a separate concept of surface gravity of a black hole that's essentially this re-scaled by the gravitational time dilation factor, and thus kept finite.

It's different. Black holes are not objects, like planets or stars. Rather, they are powerful distortions of spacetime, maintained by the concentration of mass/energy therein (which itself is kept there prisoner by the spacetime distortion - a vicious cycle broken only slowly by the Hawking radiation). As such, they cannot be "ripped apart", since there is nothing to be ripped apart there - just a giant warped tangle of spacetime.

Instead, two black holes could merge, if they get close enough and lose enough mutual orbital motion.

merging black holes

The result is a single, larger black hole.

(Yes, this should be a comment. It's too big, though.)

The Roche limit equation can be expressed as $$1.26 \times R_{\text{secondary}} \times \sqrt[3]{\frac{\text{Primary}}{\text{Secondary}}}$$

Since the radius of the secondary is zero and zero times anything is zero, the Roche limit is likewise zero.

When you think about what the Roche limit really means this is self-evident. The Roche limit is the point where the primary is raising a tide on the secondary that's stronger than it's own gravity. Tides are based on the fact that while the object is at a certain distance it really occupies space, parts are closer (which experience a greater pull) and parts are farther (which experience a lesser pull.) There is no point on the black hole that's closer or farther, thus there is no tide, thus it won't be torn apart even if it had no self-gravity.

• Here's the notation used. For some reason, the information isn't in the Astronomy help center. – HDE 226868 Jan 9 '15 at 3:29
• A black hole is, by definition, a region enclosed by an event horizon. Thus, there are definitely points on the black hole that are closer or farther. – Stan Liou Jan 9 '15 at 3:53
• But there is nothing in particular at the event horizon. It is just the point at which light can't escape. The matter is at the center and the Roche limit is about tearing apart the matter. – eshaya Jan 9 '15 at 19:09
• @EdShaya the argument in this answer is not in general sound even if you replace 'black hole' with 'singularity'; e.g., rotating black holes have a spatially extended singularity that's not (spatially) pointlike. – Stan Liou Jan 9 '15 at 23:58
• True. Rotating black holes are thought to form a ring rather than a point. But, a) the ring is inside the event horizon, so external observers never see what happens to these rings and b) the density along the ring is infinite, so it is still hard to disrupt. Now, an interesting question is what happens exactly when two matter rings of infinite density merge? Just how does that proceed? It is interesting even if we can never actually see it (and report back). Well, unless Penrose (1969) is wrong about the Cosmic Censorship of naked singularities. – eshaya Jan 11 '15 at 17:04

Because the density of the matter at the center of a black hole is infinite (or nearly so) the Roche Radius is 0 (or nearly so). Two black holes in orbit spiral in towards each other because they radiate gravity waves and form a mutual event horizon without any disturbance to the matter at their centers.

• Is this simply a property of black holes, or does this imply that a Roche limit for a celestial body is variable depending on the gravitational pull of the body it's interacting with? Ex: The Roche limit of the Sun for Jupiter would be different than the Roche limit of the Sun for Earth. – Sidney Jan 8 '15 at 19:59
• "The density of a black hole is infinite"--this simply isn't true, or at least depends on taking peculiar interpretations of the terms that don't have any clear relevance here. – Stan Liou Jan 9 '15 at 0:32
• Once an object has been compressed to its Schwarzschild radius it will continue to collapse until it becomes a singularity. The density of the matter in a black hole in GR is infinite, but with QM it may have a radius of its de Broglie wavelength or so. So the black hole, the region within the event horizon, is just empty space except for a nearly pointlike object at the center. – eshaya Jan 9 '15 at 18:58
• @EdShaya what you just said is true (under reasonable assumptions about matter, and except for necessarily being pointlike), but not relevant to the question. A black hole is defined by the presence of an event horizon, which in turn depends on the behavior of test particles in the region. Thus, a black hole would be destroyed by the gravity of another (without merging) whenever test particles in it could escape to infinity (i.e., there would no longer be a horizon). Of course, that doesn't happen, but that's what destroying a black hole would mean. ... – Stan Liou Jan 10 '15 at 5:55
• Another way to illustrate the point: if you violate the relevant energy conditions, you could have black holes without any singularities at all. But the answer in that completely-hypothetical case would be completely the same: the other black hole wouldn't be destroyed because its horizon wouldn't be destroyed. What happens deep inside doesn't matter to the question. – Stan Liou Jan 10 '15 at 5:55