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.
Answer
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.
No comments:
Post a Comment