10
$\begingroup$

Suppose a physically realistic object of nontrivial size (such as a star) free-falls past a black hole. The center-of-mass trajectory for the object is hyperbolic and (therefore) completely outside the black hole's photon sphere. The Roche limit for the object is inside the black hole's event horizon, so it may be temporarily distorted but it should not be permanently disrupted. However, the object is large enough that some fraction of its volume crosses the event horizon, at or near closest approach.

It is clear that the hole must somehow take a "bite" out of the object -- whatever material crossed the event horizon cannot continue back out to infinity. The question, though, is: what local forces cause the damage? There's not supposed to be any locally observable consequence of crossing an event horizon.

For simplicity assume an uncharged, non-rotating black hole, a spherical, non-rotating, non-accelerating cow object, and an otherwise empty local region (in particular, no accretion disk).

EDIT: Since some questions have been raised about whether there are any "physically realistic objects" satisfying the above conditions, I did some back-of-the-envelope calculations which suggest that this should be possible for stellar-mass black holes. The Schwarzschild radius is $2.95\,km/M_{\odot}$. For a rigid object, the Roche limit is

$$ d = r\,\left(2\frac{M}{m}\right)^\frac{1}{3} $$

and we want $d < R_s$; solving for $r$ gives $r < 2.34\;m^{\frac{1}{3}} M^{\frac{2}{3}}$ (still with $m,M$ in solar masses and $r$ in kilometers). In order to not just pass by the black hole without interacting with the event horizon, the object's own radius must be greater than the distance between the Schwarzschild radius and the photon sphere, so $r > 1.475 M$ as well. Take a small stellar-mass black hole ($M = 5M_\odot$, $R_s = 14.75\,km$) and a neutron star right at the TOV limit ($m = 3M_\odot$); we get $7.4 < r < 9.9\,km$. It's not clear to me whether we know whether that's a plausible range of radii for a neutron star of that mass, but it's at least in the ballpark. (The Schwarzschild radius for a $3M_\odot$ object is 8.85 km, and it seems plausible to me that, like a white dwarf, a neutron star close to its upper mass limit would be smaller than a lighter one.)

Now, it may be more appropriate to model even a neutron star as non-rigid under these conditions. That only changes the constant factors --

$$ d \approx 2.455\, r\,\left(\frac{M}{m}\right)^\frac{1}{3} $$

from which $r < 1.20\,m^\frac{1}{3} M^\frac{2}{3}$. It probably looks like that equation can't be reconciled with the requirement that $r > 1.475M$, and indeed if the object remained spherical it would have to be more massive than the black hole -- by a factor of almost 2! Worse, the upper limit on the radius (with the same mass parameters) is now 3.7 km, which is well below the Schwarzschild radius for this object. You can recover an object bigger than its own Schwarzschild radius at $m = 15M$ or so, but I suspect that is no longer a physically realistic scenario: for a $5M_\odot$ black hole you need a $75M_\odot$ object that still somehow manages to be only a little more than 221.25km radius. Maybe the conclusion is that to make this setup happen in real life the test object must be being held rigid by more than just its own gravity.

$\endgroup$
  • $\begingroup$ Isn't the Roche limit just a Newtonian formula? Do you have reason to think it should still work in GR, even for an object totally outside the horizon? $\endgroup$ – Hypnosifl Jan 21 '15 at 0:38
  • $\begingroup$ @Hypnosifl Yes, those formulas are Newtonian, and I'm sure some degree of GR correction is needed here -- not only is the object passing quite deep in the hole's gravity well, it has to be moving at a large fraction of $c$ just to be on a hyperbolic -- but I haven't the least idea how to even set it up; I only ever got as far as special relativity in college. Please consider that to be part of the question. $\endgroup$ – zwol Jan 21 '15 at 0:44
  • $\begingroup$ I don't know enough GR to set up an equation for when a body will be broken apart by tidal forces outside the horizon (as kristjan noted, for a non-rotating black hole no orbits are possible outside the photon sphere, a rotating BH can have closer orbits though), but I do know that inside the horizon, a path of fixed Schwarzschild radius would actually be a spacelike one rather than a timelike one, meaning the only way for an object to have that sort of worldline would be for it to travel faster than light (as measured in the local inertial frame of nearby freefalling observers). $\endgroup$ – Hypnosifl Jan 21 '15 at 0:53
  • $\begingroup$ @Hypnosifl Yes, hence the requirement for the object's radius to be bigger than the distance between the photon sphere and the event horizon; otherwise its center-of-mass trajectory couldn't be hyperbolic. $\endgroup$ – zwol Jan 21 '15 at 0:56
  • $\begingroup$ I assume you're just responding to the comment in parentheses, not to the rest? The fact that a fixed radius is impossible without traveling faster than light implies that an object partially inside the horizon must either break apart, or the rest of it must fall through the horizon too, regardless of the other details of the problem (so this would still be true in the case of a rapidly rotating black hole where stable orbits are possible much closer to the horizon than for non-rotating black holes, for example). $\endgroup$ – Hypnosifl Jan 21 '15 at 1:01
2
+50
$\begingroup$

It's tidal forces that pull the object apart. The key point is that there isn't a local inertial frame covering the whole object. This is by definition - we're talking about an extended object in the question!

To get an intuition for what's going on it's more helpful to split the object into several smaller pieces, each of which have an approximate local inertial frame. For simplicity we'll just consider two objects, and suppose they are joined by some kind of rope. This is just a simple model for nearby atoms of the material held together by interatomic forces.

Say object $A$ is on a geodesic which escapes the black hole and object $B$ is on which which falls in. The first thing to remember is that $A$ will never actually see $B$ reach the horizon due to time dilation in his frame of reference! This is obvious if you draw the setup in Kruskal coordinates.

But at some point your rope will break. This is because the proper distance between the observers must grow as object $B$ falls towards the singularity and $A$ escapes away from the black hole. In each of the frames of $A$ and $B$ this will manifest as a tugging force on the rope which is eventually too large for the rope to bear.

It's worth noticing that this tidal force could be arbitrarily small when $B$ crosses the horizon if $A$ and $B$ started close and the black hole is large. It will only become apparent later on.

In conclusion I think that the event horizon was a bit of red herring here. As usual it's tidal forces which rip things apart. This doesn't have to be instantaneous just because part of your object has passed through the horizon!

For more details, including calculations see this article by Greg Egan. He actually considers the analogous (and less terrifying) scenario where a Rindler horizon is created by an accelerating observer. But the mathematical ideas can be carried over to this setup.

$\endgroup$
  • $\begingroup$ So this amounts to a claim that, taking GR fully into account, the Roche limit for an extended object -- no matter what's holding it together -- passing a black hole must be outside the hole's event horizon. Yes? $\endgroup$ – zwol Apr 27 '15 at 16:35
  • $\begingroup$ Yes - I would agree with that. $\endgroup$ – Edward Hughes Apr 27 '15 at 17:15
  • $\begingroup$ Would it be possible that the binding forces of A could deform the event horizon, so B could dip below the unperturbed event horizon while remaining outside the perturbed event horizon allowing its escape? $\endgroup$ – Rick Apr 30 '15 at 15:07
  • $\begingroup$ I had to read all the way through that article by Egan to get there, but I think I finally understand why inevitably being torn apart is not in contradiction with nothing much happening right at the event horizon. $\endgroup$ – zwol May 3 '15 at 20:32
  • $\begingroup$ zwol: Brilliant - glad to be of help! @Rick - it's not really possible that the event horizon gets perturbed in the way you describe. That's because the event horizon is effectively just a function of the mass inside the black hole. Either $B$ crosses the horizon and the black hole gets bigger, or $B$ doesn't in which case it can escape again! $\endgroup$ – Edward Hughes May 4 '15 at 20:39
2
$\begingroup$

This is a very clever question and I'm looking forward to a good answer.

One important point about the event horizon is that for a large BH, the curvature at the horizon may not be particularly great. On the other hand, it is still the point that separates points that are causally connected with future infinity and those that aren't. So if your feet are inside the horizon, but your head is outside the horizon, then your head had better be moving outward at close to the speed of light if it's going to escape to infinity. Otherwise your head will meet the fate of your feet in due time. Sure, you won't feel anything special, but you'll fall into the singularity nonetheless.

The most interesting refinement of this question, in my opinion, is what happens if your center of gravity is outside the event horizon, and moving close to $c$ outward.

Aside: I suspect that at this speed tidal effects become more significant, but I'm not sure. This is because the geodesic deviation equation $\frac{D^2 X^\mu}{dt^2} = {R^\mu}_{\nu\rho\sigma} V^\nu V^\rho X^\sigma,$ involves the four velocity.

$\endgroup$
  • $\begingroup$ I was intending the center of mass of the object to be on a hyperbolic, geodesic trajectory. I've tried to clarify the wording of the question. $\endgroup$ – zwol Jan 20 '15 at 0:10
  • $\begingroup$ I am not a GR expert, but if the black hole "eats" part of the passing object doesn't that change kinetic energy and angular momentum of the object (kind of like jettisoning part of a spacecraft as a way to "kick" it... or burn fuel at periapsis)? Will the orbit still be a hyperbola in this case? It won't be symmetric about the periapsis, but will the outbound trajectory (assuming it makes it out) still be hyperbolic? $\endgroup$ – honeste_vivere Apr 26 '15 at 22:11
  • $\begingroup$ @honeste_vivere Intuitively - and I am not a GR expert either - subtracting mass from the passing object should only make its trajectory more hyperbolic. $\endgroup$ – zwol Apr 27 '15 at 16:37
2
$\begingroup$

For a non-rotating black hole (and probably there is something analogous also for rotating ones) there exists a region, called the photon sphere, in which photons can orbit circularly. This happens at $R_{photons} = \frac{3}{2}R_{Schwarzschild}$. Inside the sphere, all geodesics either fall into the hole or their past can be extended to the black hole's surface. This means that if your white dwarf is not pushed hardly by something else, the whole white dwarf will fall into the black hole.


Edit: According to NASA supercomputer simulations, two merger neutron stars tend to rip each other apart in case of a collision. If one of them would be replaced by a black hole (unfortunately I couldn't find any simulation videos for that) and if the neutron star's radius is of the same order of magnitude as black hole's, I think the only possible outcome is that the neutron star is ripped apart. If the neutron star is replaced by for example a white dwarf, the star is ripped apart even more violently and can probably even be modeled as free gas in the field of the black hole (simulation).

Indeed, estimating the order of magnitude with non-relativistic Roche limit (and assuming Roche limit works even for large satellites, which isn't of course true, but it's just an estimation), we find that if the radius of the stars are of the same order of magnitude, the density must also be of the same order of magnitude, which implies that the white dwarf must actually be quite a black hole too. The tidal forces of a large black hole can be neglected if you are small, but not if you are as big as the black hole.

In fact this is the reason why accretion disks form (and extend quite far from even the photon sphere).

$\endgroup$
  • $\begingroup$ I meant to exclude this possibility with the wording about "not a bound orbit" -- the center of mass of the object should follow a geodesic that escapes to infinity. I've tried to clarify the setup. $\endgroup$ – zwol Jan 20 '15 at 0:07
  • $\begingroup$ I do not think neutron star-neutron star collisions are an appropriate model, precisely because a neutron star doesn't have an event horizon. I have added ballpark estimates of the possible size and mass ranges to the question; they suggest to me that this is at least a possible interaction between a heavy neutron star and a stellar-mass black hole; the key uncertainty being the radius of the neutron star. $\endgroup$ – zwol Jan 20 '15 at 22:46
1
$\begingroup$

What local forces cause the damage? Tidal forces. Tidal forces can only be neglected in extremely small regions, and you have an extended body. More details follow.

You suppose an extended body, bound together, interacting gravitationally with a black hole. With the center of mass staying outside the photon sphere, moving on a more than barely unbounded orbit. But a large enough extended body that part of it crosses the event horizon, but you want the Roche limit to be smaller than the event horizon.

I can see why people criticized the original question. If you define the Roche limit to be the distance at which a satellite breaks up from tidal forces, then it appears to be a contradiction on the face of it.

But I'll start with a warning about center of mass motions. When you are losing matter, the center of mass can move faster than light. Really. Imagine a very long train, moving at a speed $v$, but the link halfway through breaks (so the back half slows and stops) and then the center of mass of the moving train jumps to where the first 25th percentile (1st quartile) originally was. Then the link halfway through the remaining train breaks. So the new center of mass is where the original first 12.5th percentile was. This could happen very quickly. And the center of mass can thus jump a large distance in a short amount of time. So center of mass motions are not at all like real motions of real particles. That was a warning because it seemed like you didn't know this (and this is true even in Newtonian Physics), but this does mean you might be able to start within the photon sphere and still have part of your star escape since the center of mass can move faster than light if you are willing to leave something behind. Don't get too excited about that, because it is deep or fantastic, but it can avoid painting yourself unnecessarily into a corner by falsely thinking center of masses have to do things they don't have to do.

But you asked about the case where the center of mass stays out of the photon sphere. So lets do that. So now the question remains about whether the Roche limit can be inside the event horizon. If your center of mass is outside the photon sphere you are at areal coordinate 1.5 $R_S$ (where $R_S$ denotes the Schwarzschild "radius") or farther. You want the actual radius to be large enough to have the center stay outside the photon sphere yet have part go into the event horizon. That might look like you need a radius of 0.5 $R_S$ or more, but the distance between areal coordinate 1.5 $R_S$ and areal coordinate 1.0 $R_S$ is actually more than 0.5 $R_S$. Even if you haven't studied GR you can see that by looking at one of those funnel pictures and note that the areal coordinates is the circumference divided by $2\pi$ (by definition, you can also take the surface area and divide it by $4\pi$ and then take the square root), and since the funnel extends in a third direction it requires more distance travelled to get to larger or smaller areal coordinate than you'd expect if the areal coordinate were just distance from the center as it is in uncurved spacetimes. But since that was the same factor for the radius of the star and the radius of the black hole and they need to be about 50% of the size with each other to straddle the region from event horizon to photon sphere, it might not be anything to worry about.

But that scale, that the satellite is maybe near to 50% the size of the black hole. That itself does pose serious problem to the normal expression for the Roche limit. The normal expression assumes that the radius of satellite is much smaller than the distance between the centers. This is most definitely not going to hold in this case.

So normally you consider the tidal force as the difference in two forces, say from the center and edge. So if the centers are a distance $d$ apart, and the satellite has a radius $r$ then the difference of the forces on a small mass $u$ nearest the black hole is

$$ \frac{GMu}{(d-r)^2}-\frac{GMu}{d^2}=\frac{GMur^2-GMu(d-r)^2}{d^2(d-r)^2}\approx\frac{Gmu2r}{d^3}. $$ Where the approximation only holds when $r << d$, which doesn't hold in our case. At all. Let's try to look at our situation. In our situation the mass closest to the black hole is actually inside the event horizon. Now in GR gravity isn't a force, but we know that any force applied to the part inside will fail to keep it at rest, so there is a sense where that force is infinite. So it is reasonable to only look at tidal forces outside the horizon.

So let's look at near misses, where the passing star has parts graze the event horizon as close as we want. If that is enough to rip parts off, then we know the tidal forces are strong enough. We know that it is possible to escape the black hole if you are outside the horizon. But if the only way to fight out of the black hole is the binding of the satellite, then we can take that as fixed and see how close we have to graze the event horizon before we leave parts behind.

If the whole satellite stays outside of the photon sphere there should be no problem, and this is far from grazing. Contrary to popular mythology, the satellite is an extended body so can't just move on a background spacetime distorted by the blackhole alone, but instead the satellite really distorts spacetime itself as well. But that is hard to do correctly and you said you didn't know much GR so let's ignore that.

Once we ignore that we can use a trick. Imagine a series of stars with the same center of mass but larger radii. The ones that have a radius extending into the black hole lose part of their outer layer. That happens to all of them, regardless of how little they went over and how strongly they were bound together. And it was tidal forces that pulled them in regardless of the binding. So the cohesive forces relative to the far away center of the star weren't enough to keep the star parts on the black hole side of the event horizon attached, and the forces vary continuously so they must get arbitrarily large. Now these forces are a relative force, they are between two regions, a differential force. For a nonextended body they would vanish. But tidal forces are real. The geometric version would be that say a blob of water minimizing its surface area finds less surface area if it bulges towards and away from a star. An extended body notices that spacetime is curved differently in different regions and it's mutual interactions act differently because of that than they do in flat space. So in a sense the star is tearing itself apart because it's ability to accommodate the different geometry on the different sides.

It is also important the tidal forces disappear not just for small regions of space, but only for small intervals of time too. That changing geometry can also be a changing from time to time. And the time part of spacetime can and is curved too. In fact for everyday situations that is the important effect.

So the tidal tearing apart doesn't happen right away. So the first part of the star that dips in doesn't notice much at first, only really small tidal effects in that really small region for that short moment it crosses. But those small differences build up because each part is just a bit off from their neighbor and the star becomes stretched out. It's as the star pulls away the different parts are ageing differently which affects how they cohere together and propagate, so the outer parts pull away. If you go back to the water blob that found less surface area if it bulges towards and away from the black hole you see that it pulls away more than normal and as the center of mass tracks it is harder to drag those parts with it. All tidal forces, and yet also it is all also the star ripping itself apart.

OK. So we could do that again and again for smaller and smaller stars eventually we have a star that just barely has a part on the inside inside, and it is still torn apart. but now if you removed that outer layer the temperature and pressure of the remaining star wouldn't be that different (since the outer layer is so thin). And without that outer layer there is actually less pressure from that now gone outer layer holding that now outermost layer towards the remaining star, and the larger star was torn apart even when/if it had a fictional stronger (say nongravitioanl) binding. So it's bound less tight and would have been stretched out even if bound more tightly. It gets torn apart. Technically you have to be more careful when computing that last argument. You can find out how much that now never existing layer pushed down, and choose a small region of space and time right when it was getting super close to the event horizon and argue that the lack of a push now means we know how it moves. And by choosing a frame localized in time and space that is falling towards the black hole, we have that it only failed to cross when it had an extra push away from the event horizon and now does not have that extra push. It's more rigorous if you select actual masses and radii, but it doesn't matter for the final result. And it's technically not right at all if we don't let the star curve spacetime too and instead have it exert gravitational forces on each other. But if the satellite is a star and not a black hole or neutron star itself that might be fairly accurate.

I'm not sure why you implicitly assumed the black hole to be larger than the star. If you had a star sitting there minding it's own business and a very small black hole whizzed by but stayed a few Schwarzschild radii away from the center of the star it would just thread through the star and remove a small tube of matter and keep going. Obviously the center of mass of the star stayed out of the photon sphere, and part of it entered the event horizon and got eaten. And it was tidal forces. The local pull of the tiny black hole pulled in some matter because locally it was stronger than the pull of the further interior portions of the star.

$\endgroup$
0
$\begingroup$

What local forces cause the damage?

An increase in the strength of the strong force, such that the reducing mass-energy of a particle is no longer enough to maintain the cohesion of that particle. As a result, some fraction of your object annihilates to gamma radiation. See Gamma Ray Bursters and Lorentzian Relativity by Friedwardt Winterberg.

enter image description here Image credit NASA

I know this isn't what's taught, but I'm pretty sure it's right, and will be what's taught. See Einstein talking about the variable speed of light. Light curves because the speed of light is spatially variable, not for any other reason. That's why a falling body falls down. Simplify the falling body to a single electron, think of the wave nature of matter and spin and magnetic moment, then simplify the electron to a wave in a square path. The horizontals bend down, and the electron falls down, like this:

enter image description here

But get this: the body falls faster and faster, because the "coordinate" speed of light is getting lower and lower. The body is then said to be travelling at the speed of light by the time it gets to the event horizon, where the coordinate speed of light is said to be zero. That doesn't add up. What does it this: there comes a point when the falling body would be falling faster than the local speed of light.

That can't happen, because of the wave nature of matter. We make electrons and positron out of light in pair production. An electron has spin and magnetic moment, like it's light going round and round. An electron can't go faster than the light from which it's made. Something has got to give.

And we don't get gamma ray bursts for nothing.

$\endgroup$
0
$\begingroup$

Two reasons this effect doesn't occur.

  1. No observer sees any part of the spaceship/sun crossing the event horizon. The distant observer sees it get closer and closer (due to time dilation) but not actually reach it. The observer in the spaceship can't even detect that they have reached the event horizon. No observer sees any part of the ship cross the event horizon and hence nobody sees some part being "bitten out of it" - it doesn't occur for any observer.

  2. As the spaceship/sun approaches the event horizon, length dilation "flattens" the spaceship/sun in the direction of the black hole, and if it were to reach the event horizon (from the perspective of an outside observer) it would have zero thickness on the axis that points towards the black hole. So either the whole spaceship enters the event horizon or none of it. I stress that due to point 1 this can never be observed, but it is the limiting case where the object gets arbitrarily close to the event horizon.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.