leaving black holes

Hi spr, sa!

http://en.wikipedia.org/wiki/Spaghettification says:

"The point at which these tidal forces kill depends on the black hole's
size. For a supermassive black hole, such as those found at a galaxy's
center, this point lies within the event horizon, so an astronaut may
cross the event horizon without noticing any squashing and pulling
(although it's only a matter of time, because once inside an event
horizon, falling towards the center is inevitable)."

Now if an astronaut orbiting just outside the event horizon of such a
supermassive black hole stuck his hand into the event horizon, would he
not feel any squashing and pulling? Could he still see his hand? Could
he retract it from the event horizon?

AFAIK the escape velocity at the event horizon is c. So an object
travelling up at the event horizon at c should slow down and come to a
halt at distance infinity. If it was faster (impossible for c) it would
keep some finit speed at infinity. If it was slower it would eventually
stop and fall back (unless it would gain some more momentum).

But someone told me, objects (including light) within the event horizon
always travel down and never up. So light from a flashlight pointing up
would still travel down. And the astronauts hand within the event
horizon would have to travel down, too, and he could not see his hand.
But that must feel like pulling and contradict the paragraph in
wikipedia. Or since his hand, including nerves will travel down it will
become detached from his body and he might no longer feel it. But he
would start bleeding and his space suit will leak atmosphere. Still not
the picture I got from the wikipedia article.

Also, suppose a flashlight was decended slowly from an orbit into the
event horizon. At what speed would the flashlight and the light
(pointing up) travel? Will the flashlight suddenly travel at c? Or, if
it decends only slowly, why will it's light not cross the event horizon?
Or will it?

Thanks, Bernhard

"Bernhard Kuemel" wrote in message
.. .
Hi spr, sa!

http://en.wikipedia.org/wiki/Spaghettification says:

"The point at which these tidal forces kill depends on the black hole's
size. For a supermassive black hole, such as those found at a galaxy's
center, this point lies within the event horizon, so an astronaut may
cross the event horizon without noticing any squashing and pulling


Someone should ask if the dawn tsunami that hit Samoa was connected
with any squishing and pulling of the Earth's crust by the Sun and Moon
and killed anybody. Water moves easily with tidal forces, rock tends
to stick and break. And it breaks at the weakest points.
Maybe the Moon is a black hole.

Dear Bernhard Kuemel:

On Oct 1, 3:09*am, Bernhard Kuemel wrote:
Hi spr, sa!

http://en.wikipedia.org/wiki/Spaghettification
says:

"The point at which these tidal forces kill depends
on the black hole's size. For a supermassive black
hole, such as those found at a galaxy's center, this
point lies within the event horizon, so an astronaut
may cross the event horizon without noticing any
squashing and pulling (although it's only a matter
of time, because once inside an event horizon,
falling towards the center is inevitable)."

Now if an astronaut orbiting just outside the event
horizon of such a supermassive black hole

The event horizon is at 2M. The lowest stable orbit for matter is
something like 6M.

stuck his hand into the event horizon, would he
not feel any squashing and pulling?

Depends on the size of the hole. If we squashed the Earth into a
black hole (to satisfy guskz), it would have an event hoizon that was
something like 4 inches in diameter. At our current distance from the
center, we'd fall normally, not noticing a difference in pull between
feet and head.

Could he still see his hand? Could
he retract it from the event horizon?

He'd have fallen in long before the signal to pull his hand from the
"horizon", which is nothing special to him... not some sort of mirror
or dark surface, made it from his brain to his hand.

Also, suppose a flashlight was decended slowly
from an orbit into the event horizon. At what
speed would the flashlight and the light
(pointing up) travel?

If it fell from infinity, it would be going pretty close to c when it
reach 2M. And we'd never see the light fall in, since light will
spend a lot of time climbing out of that place to tell us it is gone.

Will the flashlight suddenly travel at c?

Suddenly, no.

Or, if it decends only slowly, why will it's
light not cross the event horizon?

From below the event horizon, there is no "up" that includes striking
eyes or instruments far from the 2M surface.

Or will it?

If the eyes are close, it is briefly possible. And those eyes will be
in too in microseconds.

David A. Smith

dlzc wrote:
Now if an astronaut orbiting just outside the event
horizon of such a supermassive black hole

The event horizon is at 2M. The lowest stable orbit for matter is
something like 6M.

I don't know what M is, but ok. So let's assume a tough sphere (maybe
iron), that withstands the low tidal forces at the event horizon, passes
a black hole (on a hyperbolic or similar applicable trajectory) so
close, that a small part of it dips into the event horizon. Is that
possbile? What would happen to the part that dipped into the event horizon?

Depends on the size of the hole. If we squashed the Earth into a
black hole (to satisfy guskz), it would have an event hoizon that was
something like 4 inches in diameter. At our current distance from the
center, we'd fall normally, not noticing a difference in pull between
feet and head.

A size like the one mentioned in the spaghettification article such that
the tidal forces at the event horizon are not noticable to humans.

Bernhard

Dear Bernhard Kuemel:

On Oct 1, 8:47*am, Bernhard Kuemel wrote:
dlzc wrote:
Now if an astronaut orbiting just outside the event
horizon of such a supermassive black hole

The event horizon is at 2M. *The lowest stable orbit
for matter is something like 6M.

I don't know what M is, but ok.

It is a distance in a system of units where g=1, c=1; and is directly
related to the mass of the BH.

So let's assume a tough sphere (maybe iron),

Iron fails at 6M, for all but large holes. Single iron crystals are
small enough (and strong enough) to hold together at 6M.

that withstands the low tidal forces at the
event horizon

.... for an ultramassive black hole ...

, passes a black hole (on a hyperbolic or
similar applicable trajectory) so close, that
a small part of it dips into the event horizon.
Is that possbile?

No. Essentially, each component of said object is on a different
trajectory, and will be shredded by such a passage. Perhaps you have
heard of the Rosche limit for gravitationally bound systems? Similar
circumstance here, except that very little (if any) of the original
body would come out again, and what does will be spread essentially
all around the BH travelling radially outwards. Think about "a frog
in a blender".

What would happen to the part that dipped
into the event horizon?

Would only be seen briefly by the bits that survived to make the trip
outwards.

Depends on the size of the hole. *If we
squashed the Earth into a black hole (to
satisfy guskz), it would have an event
hoizon that was something like 4 inches in
diameter. *At our current distance from the
center, we'd fall normally, not noticing a
difference in pull between feet and head.

A size like the one mentioned in the
spaghettification article such that the tidal
forces at the event horizon are not noticable
to humans.

.... which is millions of solar masses.

Tom Roberts has made a very complete response to you. What is
important is that at the event horizon (and "within") there are no
vectors that point outwards. The EM contact forces that hold your
body together don't propagate "outwards". So you are either going in,
or you are losing limbs. Even on a massive hole with a relatively low
acceleration.

David A. Smith

dlzc wrote:
that withstands the low tidal forces at the
event horizon

... for an ultramassive black hole ...

, passes a black hole (on a hyperbolic or
similar applicable trajectory) so close, that
a small part of it dips into the event horizon.
Is that possbile?

No. Essentially, each component of said object is on a different
trajectory, and will be shredded by such a passage. Perhaps you have
heard of the Rosche limit for gravitationally bound systems?

Actually yes, I have read http://en.wikipedia.org/wiki/Roche_limit
before (at least partly).

Similar
circumstance here, except that very little (if any) of the original
body would come out again, and what does will be spread essentially
all around the BH travelling radially outwards. Think about "a frog
in a blender".

I see. So even though the tidal forces are weak enough there are other
forces that tear apart the body. Is it something like warped space which
makes the sphere disintegrate?

A size like the one mentioned in the
spaghettification article such that the tidal
forces at the event horizon are not noticable
to humans.

... which is millions of solar masses.

Right. http://en.wikipedia.org/wiki/Supermassive_black_hole : "A
supermassive black hole is a black hole with the highest classification
of mass, on the order of hundreds of thousands to billions of solar
masses. Most, if not all galaxies, including the Milky Way,[2] are
believed to contain supermassive black holes at their centers."

Tom Roberts has made a very complete response to you.

Yes, but he was talking about thrusters to maintain an orbit or a static
hovering posititon. To avoid that I invented the example of a sphere
passing by the black hole in free fall at high speed.

Bernhard

Dear Bernhard Kuemel:

On Oct 1, 11:26*am, Bernhard Kuemel wrote:
dlzc wrote:
that withstands the low tidal forces at the
event horizon

... for an ultramassive black hole ...

, passes a black hole (on a hyperbolic or
similar applicable trajectory) so close, that
a small part of it dips into the event horizon.
Is that possbile?

No. *Essentially, each component of said object
is on a different trajectory, and will be shredded
by such a passage. *Perhaps you have heard of
the Rosche limit for gravitationally bound systems?

Actually yes, I have read
http://en.wikipedia.org/wiki/Roche_limit
before (at least partly).

Similar
circumstance here, except that very little (if any)
of the original body would come out again, and
what does will be spread essentially all around
the BH travelling radially outwards. *Think about
"a frog in a blender".

I see. So even though the tidal forces are weak
enough there are other forces that tear apart the
body.

No, it is the shape of space. As you get closer, there just aren't as
many "choices" of "out" as there are for "in". And finding "out"
becomes problematic.

Is it something like warped space which makes
the sphere disintegrate?

It is more like a spherical region that even light can't get out of,
because *it* can't be accelerated by thrusters. If light can't get
out, the binding forces between your molecules fail. Your
acceleration inward might be mild, but likewise anything "outward" is
even more mild.

A size like the one mentioned in the
spaghettification article such that the tidal
forces at the event horizon are not noticable
to humans.

... which is millions of solar masses.

Right.
http://en.wikipedia.org/wiki/Supermassive_black_hole:
"A supermassive black hole is a black hole
with the highest classification of mass, on the
order of hundreds of thousands to billions of solar
masses. Most, if not all galaxies, including the
Milky Way,[2] are believed to contain
supermassive black holes at their centers."

Tom Roberts has made a very complete
response to you.

Yes, but he was talking about thrusters to
maintain an orbit or a static hovering posititon.
To avoid that I invented the example of a sphere
passing by the black hole in free fall at high speed.

And this simply is naieve. He attempted to answer your question as
you wanted the answer. Changing the problem only makes what will seem
to you to be a completely different answer.

You hyperboilically fly around a black hole within (I believe) 3M and
you are *never* getting back out again, with finite thrust. And when
you stick your hand out, it is essentially in a completely different
orbit, and will need to be "tugged" to keep it "precessing" along with
you. When curvature is small, this tug is within mechanical limits.
At some point, it is not. And this will include regions outside 2M as
well.

David A. Smith

Bernhard Kuemel wrote:
Now if an astronaut orbiting just outside the event horizon of such a
supermassive black hole stuck his hand into the event horizon, would he
not feel any squashing and pulling? Could he still see his hand? Could
he retract it from the event horizon?

[Note: I consider only the case of a Schwarzschild black hole.]

There is no freefalling orbit near the event horizon, but if the astronaut's
spacecraft has VERY powerful thrusters he could pilot it to follow a circular
trajectory just outside the horizon, with his thrusters aimed downward. In such
a trajectory, for a supermassive black hole he would not feel significant tidal
forces, but would feel ENORMOUS forces from his thrusters (far too great for a
human to survive, but I ignore that). If he reached his hand into the horizon,
it would be ripped from his body, because it would necessarily be outside his
ship and its powerful thrusters are more than enough to sever his hand.


AFAIK the escape velocity at the event horizon is c. So an object
travelling up at the event horizon at c should slow down and come to a
halt at distance infinity.

"escape velocity" is almost irrelevant in GR. But yes, just outside the horizon,
if one fired a bullet outward at 0.999 c or so, it would slow down and
essentially halt at distance infinity. But a light beam emitted outward there
would not slow down (as measured locally, anywhere along its trajectory); it
would be greatly redshifted, however.

Note that "traveling up" is not possible at the event horizon -- once an object
is at or inside the horizon, it can never approach it from inside, it can only
go downward toward the singularity. Indeed this is also true of a light ray,
either emitted from inside the horizon or incoming from outside. Except for the
special case of a light pulse emitted from an infalling source precisely at the
horizon and precisely aimed outward -- such a light pulse could in principle
remain at the horizon, but this is an unstable situation and the slightest
perturbation will make it fall inward. Note that any timelike object always
crosses the horizon with relative speed c (regardless of how carefully and
slowly the crossing is attempted with powerful thrusters).


But someone told me, objects (including light) within the event horizon
always travel down and never up. So light from a flashlight pointing up
would still travel down.

Yes.


And the astronauts hand within the event
horizon would have to travel down, too, and he could not see his hand.
But that must feel like pulling and contradict the paragraph in
wikipedia.

Different situations. The Wikipedia article apparently considered the astronaut
inside his spaceship, and you put his hand outside. The TIDAL forces can be
small, but the GRAVITATIONAL force cannot be small for a spaceship HOVERING just
outside the horizon.

Stated somewhat differently, a doomed astronaut whose spaceship falls through
the horizon would experience negligible tidal forces and zero gravitational
forces at the horizon. Indeed, he could not determine where the horizon is
located by measurements within his ship, he could only locate it by looking at
light from distant objects.


Also, suppose a flashlight was decended slowly from an orbit into the
event horizon. At what speed would the flashlight and the light
(pointing up) travel? Will the flashlight suddenly travel at c? Or, if
it decends only slowly, why will it's light not cross the event horizon?
Or will it?

A flashlight aimed upward and lowered by a cable from a spaceship hovering just
outside the horizon will break the cable before reaching the horizon, no matter
how strong the cable is. It will fall inward, and cross the horizon with local
speed c. The light from the flashlight will always travel with local speed c,
but will be increasingly redshifted as it approaches the horizon, with the
redshift becoming infinite as it crosses.


Tom Roberts

On Oct 1, 6:04*pm, Tom Roberts wrote in
sci.physics.relativity:
But yes, just outside the horizon,
if one fired a bullet outward at 0.999 c or so, it would slow down and
essentially halt at distance infinity. But a light beam emitted outward there
would not slow down (as measured locally, anywhere along its trajectory); it
would be greatly redshifted, however.

Honest Roberts do you really believe that nobody knows what "would not
slow down as measured locally" means? Will the bullet slow down as
measured locally? You just know no limits, Honest Roberts.

On 2-Oct-2009, Pentcho Valev wrote:

Honest Roberts do you really believe that nobody knows what "would not
slow down as measured locally" means? Will the bullet slow down as
measured locally? You just know no limits, Honest Roberts.

People talk about spagettification if one got near to a black hole... That
is really crap because it neglects the gyroscopy of the object approaching
the black hole... There has to be a gateway to the center... You can't fall
to the center of the earth and stand at the center of the earth but it is
still possible to stand at the center of the earth by following the
torroidal magnetic lines that lead to the center.

--
Tell it to the Marines.