Re: Black hole questions
- From: "N:dlzc D:aol T:com \(dlzc\)" <dlzc@xxxxxxx>
- Date: Wed, 22 Nov 2006 06:33:56 -0700
Dear Wai Yu Wong:
"Wai Yu Wong" <wywong@xxxxxxxxxxxxx> wrote in message
news:cuf7m2putltiipvlvanrnq9bb1bdicl9ru@xxxxxxxxxx
Thanks David.
On Tue, 21 Nov 2006 19:38:42 -0700, "N:dlzc D:aol T:com
\(dlzc\)"
<dlzc@xxxxxxx> wrote:
<snipped>
It seems to me that since it takes inifinite time for
light both to reach and escape from the event horizon,
Infalling light reaches the event horizon in finite time.
I am confused. The neutron star probably orbits the
black hole for quite a while first.
As Tom pointed out, any pair of objects have a barycenter when
they are orbitting each other. The neutron star and black hole,
if they started out orbitting, moved around a common barycenter.
From its frame of reference, or any stationary
reference, light takes infinite time to reach the event
horizon.
No. Infalling light travels to the horizon in finite time for an
outside observer. Outgoing light that starts at the event
horizon never makes it out (unless the BH evaporates before it
attains more mass).
See
http://casa.colorado.edu/~ajsh/collapse.html
It says:
Gravity does not escape from a black hole. Gravity moves
at the speed of light, and cannot get from inside the
horizon to the outside world. The gravity felt by a person
outside a black hole is the gravity of the stuff that fell long
ago into the black hole.
<<<
If I understand correctly, it follows that even if the black
hole
moves, the gravity left long ago by the stuff won't move.
You don't understand correctly. The contribution to spacetime
made by the contents of the BH, moves with the BH. It doesn't
"emanate from it at c".
Consider the "line of action" between Jupiter and the Sun. This
"line of action" is directed at the bodies *now*, not where the
bodies were a couple of hours ago.
Don't conflate propagating light with a force carrier for
gravity.
and that gravity travels at the same speed as light,
The contribution to spacetime made by the contents
of the black hole, travel with the black hole. Whether
you imagine those contributions to emanate from the
event horizon (like someone checking their coat at
the door), or from some imagined center, is largely a
matter of taste.
In either case, gravity takes inifinite time to reach us.
No. Gravity was established for each infinitessimal bit of
matter across the entire Universe at the time of the Big Bang.
How it is arranged thereafter, curvature has plenty of time to
follow it.
We can witness black holes orbitting other bodies.
In that there are sources of high energy emissions,
more dense than identified neutron stars, but have
no surface interactions of stuff "crashing down".
When observation doesn't match theoretical
prediction, the conclusion is that what we observe
are not black holes.
The observation matches the prediction. An event horizon has no
surface outside the event horizon with which infalling matter can
interact.
My hunch is that black holes don't really exist.
It is real obvious that you are not armed for the facts that can
surround you. I strongly recommend you do a little research.
What we observe are probably collapsing
stars about to become black holes, but it will take
inifinite time to really become black holes,
because infalling materials take inifinite
time to reach the event horizon, if any.
You completely misunderstand. Everything *inbound* crosses the
event horizon in finite proper time for an outside observer. It
is outbound "confimation" of its corssing that takes "infinite"
time.
2. When an object falls straight into a black hole,
its thickness will approach zero as it nears the
event horizon due to relativistic contraction.
Objects don't reach c this side of the event horizon.
All objects do approach c near the event horizon, and the
relativistic contraction factor is given by (1-2GM/rc^2)^-0.5,
which approaches infinity near the event horizon.
This is *outbound* light signals. It has very, very little to do
with objects falling in. I could drop an object across the event
horizon *now*, and you could see it going in for eternity... if
you can detect fewer and fewer photons that are more and more
redshifted.
From the object's frame of reference, will the event
horizon appear elongated, flat or the same (spherical)?
Here are some good visualizations:
http://casa.colorado.edu/~ajsh/schw.shtml
It shows only up to a distance of about 3 Schwarzchild radii,
where the contraction factor is only about 1.22, still not very
substantial.
So you didn't see the simulations where you actually cross the
event horizon? You have a very short attention span. I'd
recommend something other than science for your future career.
Note that the event horizon is not an object. You do
not see it, so you cannot see it "elongated, flat or the
same".
The dust particles that fall into it emit light as long as
they are above the event horizon, so one can virtually
see the shape of the event horizon.
The horizon is the horizon for distant observers. Locally, you
only see the usual stuff.
....
From the object's frame of reference, does an atom
nucleus appear thicker or the same? Or has GR
broken down already?
GR does not apply at the quantum level. For anything
up to the size of star clusters, GR is a very good
predictive fit. Beyond this, Dark Matter and Dark
Energy are invoked.
What puzzles me is that every reference I come across
states that a person falling across the event horizon of
a large black hole (with small tidal effect) won't notice
anything special at the event horizon.
Right. What puzzles you about that?
What I expect is that the person will disintegrate before
crossing the event horizon.
Because...
The infaller isn't travelling at c. The c-moderated forces have
no problems reaching from head to toe. The event horizon
"monster" is a monster for distant observers. It is simply a
horizon, like the Sun goes over at the end of the day. Only an
infaller will not come back out except via Hawking radiation.
3. When a heavy object touches a black hole's event
horizon, the space-time curvature of that point will be
reduced by the gravity of the former, and thus it is no
longer part of the event horizon. Will the event horizon
be dented, broken, or enlarged?
When two black holes merge, it is expected that the event
horizons of each will be "skewed" towards each other.
http://curious.astro.cornell.edu/question.php?number=165
http://www.nasa.gov/vision/universe/starsgalaxies/gwave.html
The heavy object I mentioned is not a black hole yet.
Unfortunately the above links do not describe what
happens to the event horizon.
One event horizon will stretch out by an amount that depends on
the infalling mass (quantity and location). One event horizon
doesn't know there is another event horizon. It only depends on
the presence of mass.
David A. Smith
.
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