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Solfe
2012-Oct-26, 02:16 AM
Can you interact gravitationally with a mass inside an event horizon?

Take the classic image of Lagrangian points - Sun in the middle, Earth on the right side of the sun, and L2 right of the Earth. Now switch these objects for a super massive black hole, a star and L2. If the star slowly, gently spirals in to the event horizon, L2 would be outside the event horizon for a time.

What would be happening at L2? Could a ship ride around the black hole at the L2 point of the star/black hole system until it too entered the event horizon?

Jeff Root
2012-Oct-26, 03:24 AM
The star couldn't spiral in slowly or gently. Once it was close
enough to be pulled toward the black hole, its orbital speed
would increase rapidly. If the speed was high enough to keep
it from falling into the black hole as long as possible, it would
reach the speed of light at the event horizon. It would also
be torn apart there. It could avoid being torn apart at the
event horizon by falling in more directly, with little or no
orbital speed. But without propulsion to keep it from speeding
up in orbit, the star would orbit faster and faster as it spirals
inward. I'm not sure whether gravity at L2 would be enough
to hang on to a third body. The Lagrange positions are
figured with Newtonian gravity. General relativity probably
gives different results inside the last stable orbit of a black
hole, and especially inside its ergosphere.

The L2 point is not stable even under Newtonian mechanics.
A body there will drift away if the gravitational situation
changes, as it does when the star speeds up while being
pulled in.

As the star crosses the event horizon, it should lose its
identity as a separate mass from the mass at the center
of the black hole. When the event horizon becomes an
oblate ellipsoid, the star will no longer be detectible as
a separate mass, and the black hole will again act like
a point source of gravity, now located at what had been
the barycenter of the black hole-star system.

I expect that if a body is in the L2 position when a star
passes through the event horizon, the body is doomed
to fall in, too. But I can't guess where it would be
relative to the star.

-- Jeff, in Minneapolis

Solfe
2012-Oct-31, 12:42 AM

I was sort of picturing a star "drifting" through the event horizon and the body in the L2 point simply follows at that location until the L2 enters the event horizon. But as you mentioned, realistically, the star has to take a wild ride into the event horizon. That is a deal breaker for something hanging out at L2.

Thanks again.

cjameshuff
2012-Oct-31, 02:21 AM
The event horizon isn't some all-destroying surface. Tidal forces at the event horizon can be arbitrarily low, lower with increasing mass. Stationary observers at a distance see the event horizon at a fixed location (and would never be able to apply sufficient thrust to hold position at the event horizon), but infalling observers always see it "below" them. They can calculate when they've crossed the event horizon, but won't notice anything physically different (except that no finite amount of thrust will turn them entirely away from the black hole). Information can not travel outward, but objects can be moving inward fast enough for messages to be exchanged between locations within the event horizon.

Jeff Root
2012-Oct-31, 03:21 AM
The force at the event horizon would tear apart anything
that wasn't falling in. I'm not sure the force is "tidal".
The force would actually tear apart anything that wasn't
free-falling before it can reach the event horizon, but just
how far above depends not only on the mass and size of
the black hole, but also on the downward speed of the
body and its mechanical strength. Even an ideal material
with "perfect" strength would be torn apart at the event
horizon if it wasn't free-falling, because gravity would
pull its atoms downward faster than the electric forces
in the material could propagate upward. Which, at the
horizon, is not at all.

-- Jeff, in Minneapolis

Solfe
2012-Oct-31, 04:52 AM
It seems to me that you are both saying the same thing, but there is a question of scale.

If you have a "clean" stellar mass black hole, "clean" as nothing else whipping around it, there is a point where an in falling object just can't stay in one piece. At X distance, it is pulverized; at Y distance the electrons fly off; perhaps at some distance Z the proton and neutrons fall separate (I am speculating, perhaps taking this too far). Then whatever is left of the body falls into the event horizon. This is simply tidal forces at work. X can change due to material strength, but Y and Z are pretty much fix distances.

If you have a "clean" but super-massive black hole, then the in falling body can fall as a single unit through the event horizon because the tidal forces aren't that strong. However, a star is spinning and as it's edge passes through the event horizon, the star cannot continue that motion. The limb of the star that is inside the event horizon is never coming out and stops having a direct effect on the remaining part of the star. The part outside the event horizon doesn't act like as it did when the whole star was completely outside the event horizon. I don't know what that "force" is called. Perhaps, "distrupted" is a good word.

Now that begs a second question. Is there a maximum size for an event horizon? Is there a factor like brown dwarves where all event horizons are about the same size after a given mass?

Jeff Root
2012-Oct-31, 05:37 AM
My intent was to make clear that I was only saying an
object would be torn apart at the event horizon if, for
whatever reason, it didn't fall in. So *part* of it would
fall in now, and part of it might fall in later, or maybe
even escape. Otherwise, yes, I agree with what CJ
said.

It isn't really a matter of scale. If something like a
steel sewing needle were just above the event horizon,
but somehow prevented from falling in, it would be torn
apart from the stress between gravity pulling downward
and whatever is pulling upward on it to prevent it from
falling. There, I think I've finally said it clearly enough
to be understood!

There wouldn't be any limit to the size of black hole,
except that if you manage to truck half the mass of
your local cluster of galaxies to it, you're going to find
it really, really hard to gather any more. (You'll have
thrown away the other half of the mass as propellant
for getting the first half to the black hole.)

-- Jeff, in Minneapolis

tusenfem
2012-Oct-31, 06:55 AM
It isn't really a matter of scale. If something like a
steel sewing needle were just above the event horizon,
but somehow prevented from falling in, it would be torn
apart from the stress between gravity pulling downward
and whatever is pulling upward on it to prevent it from
falling. There, I think I've finally said it clearly enough
to be understood!

Not necessarily.
It all depends on the gravity gradient between the top of the needle and the bottom. If that is greater than the structural forces keeping the needle together, it will be spaghetified, however if the gradient is not that strong (which happens at greater mass black holes) then this noodlification will not happen.

Jeff Root
2012-Oct-31, 09:18 AM
No, it can't be the gravity gradient in this case.

I agree that we're talking about a supermassive black hole,
where the gradient at the event horizon is really low. So low
that a person free-falling into the black hole wouldn't even
be aligned vertically by it. They wouldn't feel any stretching
forces until they are well inside the horizon.

But even a tiny little sewing needle, if held very close to
the event horizon and somehow prevented from falling in,
will be pulled apart by the strain between gravity pulling
down and whatever is pulling up on it.

Consider a longer object, like a steel bridge cable, somehow
held at one end outside the event horizon. The other end is
very slowly lowered down to the horizon. No material, even
with "perfect" strength, could hold together once it reaches
the horizon. The virtual photons which carry the electric force
from one atom to the next and hold the material together will
be pulled downward by the black hole's gravity so hard that
they cannot move upward. The virtual photons from an
electron below the event horizon cannot move upward to
reach a proton higher up (above or below the horizon), so
they will not be able to hold the material together. Only if
the cable is allowed to accelerate into the black hole will
the virtual photons be able to reach atoms higher up in the
cable. They will be able to do so because those atoms fell
downward to where the virtual photons are.

The virtual photons, moving upward through the falling
cable at the speed of light, are moving downward relative
to the event horizon above them.

Everything inside the event horizon is moving downward.
No exceptions. It doesn't have to be moving at the same
speed. But if it isn't free-falling, something needs to be
slowing the fall. Just above the event horizon it requires
a stupendous force to stop the fall. At the horizon, it
requires infinite force. Can't happen. Below the horizon,
even infinite force would fall short. Preventing something
from falling there means FTL. Double can't happen.

-- Jeff, in Minneapolis

cjameshuff
2012-Oct-31, 04:01 PM
Put another way, no finite amount of force will keep an object at the event horizon from accelerating downward. At and below that radius, you have to be moving inward to be influenced by deeper objects, as nothing they do can actually propagate outward...at best it can fall inward at a slower rate. Which isn't actually a problem, as you can't do anything but move inward...

The description of the event horizon as the point where escape velocity is greater than the speed of light is misleading, I think. The difference between escape velocity and surface gravity depends on density: Mercury has higher surface gravity than Mars, but lower escape velocity. Saturn has almost exactly the same "surface" gravity as Earth, but escape velocity at that point is about 3x Earth's. Bodies with less than escape velocity move outward to some point and then fall back, to collide with the surface or go into closed orbits.

But light doesn't slow down as it emerges from a gravity well like massive particles do. It loses energy, but its trajectory does not depend on its energy. If it could rise above the event horizon, it could escape. The distance where light can't escape is the distance where light can't move outward at all, which is something a bit different. And various relativistic factors complicate things further. Wikipedia's discussion of the subject is here: http://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole

Jeff Root
2012-Oct-31, 04:49 PM
If it could rise above the event horizon, it could escape.
The distance where light can't escape is the distance
where light can't move outward at all, which is something
a bit different.
I agree with everything but the final clause, so I wonder
if I just misunderstand what you meant. Something a
bit different from the event horizon? The distance where
light can't escape is the distance where light can't move
outward at all, is the event horizon. Light emitted above
the event horizon can move outward if it is headed at a
high enough angle. At the event horizon that angle is
vertical. At the photon sphere it is horizontal.

-- Jeff, in Minneapolis

cjameshuff
2012-Oct-31, 05:04 PM
I agree with everything but the final clause, so I wonder
if I just misunderstand what you meant. Something a
bit different from the event horizon?

Not the event horizon. The concept of light being unable to travel outward to any degree is different from the concept of escape velocity.

Jeff Root
2012-Oct-31, 05:16 PM
Oh! Ah-ha and ha-ha! Of course!

-- Jeff, in Minneapolis

Solfe
2014-Sep-28, 04:56 AM

Here is another related question. Let's say an observer is safely orbiting a black hole and a substantial mass crosses the event horizon. This substantial item, say a large star, has a known mass and has an effect on the observer location so the mass is known by the observer. For the sake of simplicity, let us assume this effect does not drag the observer into the event horizon.

Once the star crosses the event horizon, can we treat both black hole and star as a single very small and massive object? (I think "yes" has to be the answer... but it sets up the next question.)

Would this be the case during time period it takes the star to travel across the distance from the event horizon to the center of whatever is at the center of a black hole?

Is this just stating that stuff on the other side of an event horizon can't be observed? Or is there just no case where the travel time from the edge of event horizon to the center of the black hole is long enough to notice?

Is there an upper limit to the size of a black hole?

I tried to calculate the Schwarzschild radius of the black hole of 23,000,000 solar masses (I picked APM 08279+5255's mass because it was the largest black hole listed in Wikipedia) and ended up with an event horizon radius of 2.0407572582299696x10^8 meters or 208,475 km... At c, that would take 6 seconds to get to the middle and nothing can hold together while doing that?

(Could I have done math right? Alarm bells are going off... I don't do math.)

Is that why you can't treat in falling items as being distinct objects past the event horizon - the time and distances involved are tiny?

Grey
2014-Sep-28, 10:33 AM
There's no specific requirement that a black hole (and thus the event horizon and the mass distribution) be spherically symmetric. However, the gravitational force pushes it to that state pretty quickly. In a black hole merger, this process is called ring-down (http://www.astro.cardiff.ac.uk/research/gravity/tutorial/?page=4blackholecollisions). Something similar should happen with any stellar mass object merging with the black hole. So you might be able to detect an asymmetric mass distribution as the star falls in, but only for a very short time.

There is no theoretical upper limit to the size of a black hole. The radius of the event horizon increases linearly with the mass.

Hornblower
2014-Sep-28, 12:51 PM
I saw an article in Sky and Telescope a few years ago that described a computer simulation of the merger of two supermassive black holes. Initially the resultant black hole would be oblong and oscillating about the eventual equilibrium configuration. The energy of this oscillation would be carried off by gravitational wave radiation that would violently shake anything near the merged object, possibly being strong enough to shred nearby stars. The merged object would "ring down" like a ringing gong. My educated guess is that it would asymptotically approach the limiting spherical configuration, but I don't know enough of the mathematical details to elaborate any further.