leaving black holes
Bernhard Kuemel
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
Androcles
"Bernhard Kuemel" <bern...@bksys.at> wrote in message
news:63977$4ac47fc5$557f726a$23...@news.inode.at...
> 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.
dlzc
On Oct 1, 3:09 am, Bernhard Kuemel <bernh...@bksys.at> 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
Bernhard Kuemel
>> 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
Tom Roberts
> 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
carlip...@physics.ucdavis.edu
[...]
> 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?
He would feel his hand ripped off, he would not see it, and he could not
retract it.
There are two kinds of acceleration here, and it's important not to
mix them up:
1. Start with an astronaut in free fall around the Earth. To a good
approximation, she doesn't feel any effects of gravity -- she's
"weightless" -- because her entire body is experiencing the same
acceleration.
If she measures more carefully, though, she'll see that she's slightly
stretched. If her feet are closer to the Earth, they will accelerate
slightly more, since the gravitational field is slightly stronger closer
to the Earth. This *relative* acceleration is called tidal acceleration.
It is, in fact, the cause of the Earth's tides: the water closer to the
Moon is attracted more strongly than the water on the opposite side
of the planet, so the oceans are stretched out.
For an astronaut orbiting the Earth, this tidal acceleration is tiny.
For an astronaut near a black hole, it is much larger. Near a small
black hole, the tidal acceleration will tear a freely falling astronaut
apart (the technical term is "spaghettification").
Near a supermassive black hole, the tidal acceleration is much less.
This seems counterintuitive, but remember that you can get much
closer to the center of a small black hole before you cross the event
horizon. Tidal accelerations varies roughly as one over the cube of
the distance to the center of a gravitating object, and the smaller
distance wins.
2. Now imagine that your astronaut is trying to remain outside the
horizon of a black hole. She's no longer in free fall; instead, her
rocket is using up fuel at a huge rate to keep her from falling. The
acceleration she needs to just stay at rest relative to the horizon
becomes huge as she approaches the black hole; *at* the horizon, she
would need infinite acceleration to remain at rest (or to remain in
an orbit outside the horizon).
If she now sticks her hand toward the horizon, she's in trouble. Her
hand needs a huge acceleration to remain at rest, but it doesn't have
a rocket attached; it's accelerating only because it's connected to the
rest of her body, which presumably is in her spacecraft. Bones aren't
strong enough to transmit a near-infinite acceleration; nothing is.
This acceleration -- not the tidal acceleration of a freely falling
astronaut, but the acceleration required to remain outside a black
hole's horizon -- is the one that's relevant to your question. It goes
to infinity at the horizon of any black hole, stellar or supermassive.
If your astronaut doesn't want her hand torn off, she will have to jump
out of her spacecraft and fall with it through the horizon, so that her
whole body is falling at the same rate. (I don't recommend this, though,
for obvious reasons.)
Steve Carlip
dlzc
On Oct 1, 8:47 am, Bernhard Kuemel <bernh...@bksys.at> 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
Bernhard Kuemel
>> 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
dlzc
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
eric gisse
[...]
> You hyperboilically fly around a black hole within (I believe) 3M and
Last stable orbit. Arbitrary acceleration can let you get arbitrarily close
to r = 2M.
N:dlzc D:aol T:com (dlzc)
"eric gisse" <jowr.pi...@gmail.com> wrote in message
news:ha3eg8$935$3...@news.eternal-september.org...
> dlzc wrote:
> [...]
>
>> You hyperboilically fly around a black hole within (I believe)
>> 3M and
>
> Last stable orbit.
Photon sphere is 3M. Last stable orbit for matter is 6M.
> Arbitrary acceleration can let you get arbitrarily close
> to r = 2M.
If you choose a different coordinate system. This leaves you
still 1M out from the event horizon.
http://casa.colorado.edu/~ajsh/orbit.html
... "3 down to 1.5" Schwarzchild radii (Sr), where I have the Sr
at 2M. So 6M down to 3M is unstable, meaning you are going in or
leaving with thrust properly applied.
David A. Smith
Tom Roberts
> "eric gisse" <jowr.pi...@gmail.com> wrote in message
> news:ha3eg8$935$3...@news.eternal-september.org...
>> dlzc wrote:
>>> You hyperboilically fly around a black hole within (I believe)
>>> 3M and
>> Last stable orbit.
>
> Photon sphere is 3M. Last stable orbit for matter is 6M.
[This is all for a Schwarzschild black hole. r is the
usual Schw. coordinate.]
There is an unstable circular null geodesic at r=3M that orbits the
black hole, which might be what you are trying to say, but "photon
sphere" does not mean that to me. What I would call the "photon sphere"
is right at the horizon, r=2M -- that is, outgoing light at the horizon
remains at the horizon (an unstable situation).
The smallest stable timelike orbit is indeed at r=6M, and I'm pretty
sure it is true that if an ORBIT goes inside r=6M it will destabilize
and fall in or fly away. But there are hyperbolic-like timelike
trajectories that come in from far away, approach between r=3M and r=6M,
and then leave; these are not orbits, and none come closer than r=3M.
It is also true that any timelike or null geodesic that comes from far
away and gets below r=3M will enter the horizon (in some cases it can
circle the horizon several times before entering). But light that is
headed outward and emitted by a timelike object from 2M<r<3M will get
away; that's different from light coming in from far away.
>> Arbitrary acceleration can let you get arbitrarily close
>> to r = 2M.
>
> If you choose a different coordinate system. This leaves you
> still 1M out from the event horizon.
Not really. The horizon is at r=2M. Coordinate choice has nothing
whatsoever to do with the locus of the horizon or the location a given
spaceship is able to hover outside it with a given thrust [#]. With
arbitrarily-high thrust, a spaceship could hover arbitrarily close to
the horizon at r=2M. This is an unstable situation if its r is close to 2M.
[#] Coordinate choice only affects your DESCRIPTION, not
the physical situation.
> http://casa.colorado.edu/~ajsh/orbit.html
Hmmm. This page says "The orbit at 2 Schwarzschild radii corresponds to
zero kinetic energy at infinity, so it is possible to fall freely into
this orbit from infinity without rocket power." -- this is wrong, and it
is NOT possible to enter an orbit at ANY radius without thrusting, when
one starts from infinity and is free-falling [@]. Here by "orbit" I mean
a trajectory with a permanent spatial path circling the black hole.
This is easy to see, as the spatial path of any timelike
geodesic can be traversed in the opposite direction,
within the region outside the horizon -- an incoming
trajectory that falls into into an orbit, when reversed,
would mean it is not an orbit.
[@] And gravitational radiation is neglected, as for
test particles like rocket ships compared to stars.
Tom Roberts
Juan R.
(...)
> [This is all for a Schwarzschild black hole. r is the
> usual Schw. coordinate.]
(...)
>>> Arbitrary acceleration can let you get arbitrarily close to r = 2M.
>>
>> If you choose a different coordinate system. This leaves you still 1M
>> out from the event horizon.
>
> Not really. The horizon is at r=2M. Coordinate choice has nothing
> whatsoever to do with the locus of the horizon or the location a given
> spaceship is able to hover outside it with a given thrust [#].
As you point above r denotes the "Schw. coordinate". It is possible to
introduce other coordinates and then the horizon is not at 2M but at
a different place. Concretely renormalized coordinates give r=1M and
r=2M is just 1M away from the horizon.
Notice this renormalized r is not the Schwarzschild r. Don't get confused
by symbols Tom!
--
http://www.canonicalscience.org/
BLOG:
http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
Pentcho Valev
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.
Kevin B. Murphy
On 2-Oct-2009, Pentcho Valev <pva...@yahoo.com> 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.
dlzc
On Oct 1, 4:37 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
> dlzc wrote:
>
> [...]
>
> > You hyperboilically fly around a black hole within
> > (I believe) 3M and
>
> Last stable orbit.
No. With the event horizon at 2M (as previously described in this
thread), the closest possible stable orbit for matter is 6M.
> Arbitrary acceleration can let you get arbitrarily
> close to r = 2M.
You mangled the original newsgroup listing, so Tom's excellent and
thorough response did not make it on the sci.astro branch of this
thread. Please be careful how you apply Occam's razor here...
David A. Smith
Tom Roberts
> Tom Roberts wrote on Thu, 01 Oct 2009 23:53:35 -0500:
>> [This is all for a Schwarzschild black hole. r is the
>> usual Schw. coordinate.]
>> whatsoever to do with the locus of the horizon or the location a given
>> spaceship is able to hover outside it with a given thrust [#].
>
> As you point above r denotes the "Schw. coordinate". It is possible to
> introduce other coordinates and then the horizon is not at 2M but at
> a different place.
You use words in an unusual way -- this is not a different "place"
(which denotes location in the manifold), but rather a different VALUE
OF "r" corresponding to THE SAME PLACE IN THE MANIFOLD. Using different
coordinates does NOT "move" the horizon, or make it be at a "different
place", it makes the (fixed) locus of the horizon be at a DIFFERENT
VALUE OF THE COORDINATE YOU HAPPEN TO LABEL WITH THE SAME SYMBOL, "r".
> Concretely renormalized coordinates give r=1M and
> r=2M is just 1M away from the horizon.
Sure -- you can "renormalize" coordinates however you wish. Another
well-known set of coordinates puts the horizon at r=0. But that is
IRRELEVANT to what I said, as my statements all used the Schw. r, not
any of these other ones.
> Notice this renormalized r is not the Schwarzschild r. Don't get confused
> by symbols Tom!
It is not I who is confused here.
Tom Roberts
bz
@skeeter.ucdavis.edu:
> If she now sticks her hand toward the horizon, she's in trouble. Her
> hand needs a huge acceleration to remain at rest, but it doesn't have
> a rocket attached; it's accelerating only because it's connected to the
> rest of her body, which presumably is in her spacecraft. Bones aren't
> strong enough to transmit a near-infinite acceleration; nothing is.
What if the black hole were the size and mass of our universe?
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
dlzc
original newsgroup list corrected
On Oct 1, 9:53 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> N:dlzcD:aol T:com (dlzc) wrote:
>
> > "eric gisse" <jowr.pi.nos...@gmail.com> wrote in message
> >news:ha3eg8$935$3...@news.eternal-september.org...
> >>dlzcwrote:
> >>> You hyperboilically fly around a black
> >>> hole within (I believe) 3M and
>
> >> Last stable orbit.
>
> > Photon sphere is 3M. Last stable orbit for
> > matter is 6M.
>
> [This is all for a Schwarzschild black hole.
> r is the usual Schw. coordinate.]
>
> There is an unstable circular null
> geodesic at r=3M that orbits the black
> hole, which might be what you are trying
> to say, but "photon sphere" does not mean
> that to me.
The typical definition is an "Einstein ring" with an infinite number
of "laps".
http://en.wikipedia.org/wiki/Photon_sphere
> What I would call the "photon sphere"
> is right at the horizon, r=2M -- that is,
> outgoing light at the horizon remains at
> the horizon (an unstable situation).
Einstein rings and your "salmon swimming upstream" are both unstable,
for a structure that will continue to add or expell contents.
> The smallest stable timelike orbit is
> indeed at r=6M, and I'm pretty sure it is
> true that if an ORBIT goes inside r=6M it
> will destabilize and fall in or fly away.
Yes, Eric just was thinking "Schwarzchild radii" rather than tying it
to mass directly, and I was hoping to unconfuse the issue with the OP.
"A miss is as good as a mile." Eric was selling r_S as 1M. It is off
by a factor of 2. And coordinate choice clarity is necessary:
http://en.wikipedia.org/wiki/Mars_Climate_Orbiter
> >http://casa.colorado.edu/~ajsh/orbit.html
>
> Hmmm. This page says "The orbit at 2
> Schwarzschild radii corresponds to zero kinetic
> energy at infinity, so it is possible to fall
> freely into this orbit from infinity without
> rocket power." -- this is wrong, and it is NOT
> possible to enter an orbit at ANY radius
> without thrusting, when one starts from
> infinity and is free-falling [@]. Here by
> "orbit" I mean a trajectory with a permanent
> spatial path circling the black hole.
>
> This is easy to see, as the spatial path
> of any timelike geodesic can be traversed
> in the opposite direction, within the
> region outside the horizon -- an incoming
> trajectory that falls into into an orbit,
> when reversed, would mean it is not an
> orbit.
An elegant argument.
> [@] And gravitational radiation is
> neglected, as for test particles like
> rocket ships compared to stars.
>
> Tom Roberts
Thank you.
David A. Smith
Juan R.
> Juan R. González-Álvarez wrote:
>> Tom Roberts wrote on Thu, 01 Oct 2009 23:53:35 -0500:
>>> [This is all for a Schwarzschild black hole. r is the
>>> usual Schw. coordinate.]
>>> The horizon is at r=2M. Coordinate choice has nothing whatsoever to do
>>> with the locus of the horizon or the location a given spaceship is
>>> able to hover outside it with a given thrust [#].
>>
>> As you point above r denotes the "Schw. coordinate". It is possible to
>> introduce other coordinates and then the horizon is not at 2M but at a
>> different place.
>
> You use words in an unusual way -- this is not a different "place"
> (which denotes location in the manifold), but rather a different VALUE
> OF "r" corresponding to THE SAME PLACE IN THE MANIFOLD. Using different
> coordinates does NOT "move" the horizon, or make it be at a "different
> place", it makes the (fixed) locus of the horizon be at a DIFFERENT
> VALUE OF THE COORDINATE YOU HAPPEN TO LABEL WITH THE SAME SYMBOL, "r".
Unfortunately you get totally confused again by your superfitial
knowledge of those topics Tom.
In my original message I refered *explicitely* to renormalized coordinates,
which do NOT describe "the same manifold". There is kind of equivalence
between both sets of coordinates. But this is not valid everywhere.
In fact for r<<2M there is a large difference between both coordinates
(in the renormalized spacetime there is NOT central singularity for instance).
>> Concretely renormalized coordinates give r=1M and r=2M is just 1M away
>> from the horizon.
>
> Sure -- you can "renormalize" coordinates however you wish. Another
> well-known set of coordinates puts the horizon at r=0. But that is
> IRRELEVANT to what I said, as my statements all used the Schw. r, not
> any of these other ones.
Above you sniped an relevant part of the message from dlcz to which you said
"not really".
REINTRODUCING SNIPED PART
If you choose a different coordinate system. This leaves you still 1M
out from the event horizon.
>> Notice this renormalized r is not the Schwarzschild r. Don't get
>> confused by symbols Tom!
>
> It is not I who is confused here.
This would be credible if you were not sniping witjout noticing the snip
eric gisse
[...]
> "A miss is as good as a mile." Eric was selling r_S as 1M.
No.
[...]
Bernhard Kuemel
> carlip...@physics.ucdavis.edu wrote in news:ha2lvp$f4d$1
> @skeeter.ucdavis.edu:
>
>> If she now sticks her hand toward the horizon, she's in trouble. Her
>> hand needs a huge acceleration to remain at rest, but it doesn't have
>> a rocket attached; it's accelerating only because it's connected to the
>> rest of her body, which presumably is in her spacecraft. Bones aren't
>> strong enough to transmit a near-infinite acceleration; nothing is.
>
> What if the black hole were the size and mass of our universe?
Who do you think might stick her hand into an infinite universe? :)
Bernhard
bz
$27...@news.inode.at:
Someone that was outside it.
There is no way to prove that it is infinite.
If I remember correctly, "someone" has calculated that the amount of matter
is "close" to the amount that would be needed for a black hole the size of
the known universe.
Perhaps we live inside a black hole.
http://www.mathpages.com/home/kmath339.htm
>
> Bernhard
eric gisse
> Bernhard Kuemel <bern...@bksys.at> wrote in news:4765c$4ac657af$557f726a
> $27...@news.inode.at:
>
>> bz wrote:
>>> carlip...@physics.ucdavis.edu wrote in news:ha2lvp$f4d$1
>>> @skeeter.ucdavis.edu:
>>>
>>>> If she now sticks her hand toward the horizon, she's in trouble. Her
>>>> hand needs a huge acceleration to remain at rest, but it doesn't have
>>>> a rocket attached; it's accelerating only because it's connected to the
>>>> rest of her body, which presumably is in her spacecraft. Bones aren't
>>>> strong enough to transmit a near-infinite acceleration; nothing is.
>>>
>>> What if the black hole were the size and mass of our universe?
>>
>> Who do you think might stick her hand into an infinite universe? :)
>
> Someone that was outside it.
> There is no way to prove that it is infinite.
>
> If I remember correctly, "someone" has calculated that the amount of
> matter is "close" to the amount that would be needed for a black hole the
> size of the known universe.
>
> Perhaps we live inside a black hole.
Which direction is the singularity?
>
> http://www.mathpages.com/home/kmath339.htm
>
>>
>> Bernhard
>>
>
>
>
>
>
Omega
As the "direction" of the big bang is in to our past, the direction of the
singularity (The Eye of Harmony) of our local black hole (Harmony) is to our
future. ( That is, is not so much where, as when. And to account for the
Hubble Constant and the negative deceleration paramenter as being due to its
tidal force, its minimum mass is ~3.18619 x 10^53 kg).
>
>>
>> http://www.mathpages.com/home/kmath339.htm
>>
>>>
>>> Bernhard
eric gisse
Ok, people whose knowledge of physics consists of terminology grokked from
doctor who have no right to contribute.
>
>>
>>>
>>> http://www.mathpages.com/home/kmath339.htm
>>>
>>>>
>>>> Bernhard
Tom Roberts
> Tom Roberts wrote on Fri, 02 Oct 2009 09:34:18 -0500:
>> Using different
>> coordinates does NOT "move" the horizon, or make it be at a "different
>> place", it makes the (fixed) locus of the horizon be at a DIFFERENT
>> VALUE OF THE COORDINATE YOU HAPPEN TO LABEL WITH THE SAME SYMBOL, "r".
>
> Unfortunately you get totally confused again by your superfitial
> knowledge of those topics Tom.
It is not my "superficial knowledge" here, it is YOURS, and/or your
failure to keep your theoretical context straight.
> In my original message I refered *explicitely* to renormalized coordinates,
> which do NOT describe "the same manifold".
Look to the earlier posts in this thread. The context is the
Schwarzschild solution of GENERAL RELATIVITY, not some other theory you
might wish to discuss. In GR, a change of coordinates has NO EFFECT
WHATSOEVER on the manifold, and a different set of coordinates always
describes THE SAME manifold.
This is just VERY basic geometry, applied to GR.
Indeed, it is difficult indeed to have some other theory in which a
change of coordinates changes the manifold, as you seem to be claiming.
That would violate cherished beliefs that go far deeper than GR: that
the world we inhabit does not depend on humans or their descriptions of
it. Indeed, without this being valid it is not clear how there could be
any physics at all.
While YOU might mean something rather special by "renormalized
coordinates", to the rest of us that phrase suggests a re-scaling of the
coordinates, which clearly does nothing special IN THE CONTEXT OF GR.
> In fact for r<<2M there is a large difference between both coordinates
> (in the renormalized spacetime there is NOT central singularity for instance).
Again, in GENERAL RELATIVITY this is nonsense. A mere change of
coordinates cannot possibly affect the singularities of Schwarzschild
spacetime. And how did you switch from "renormalized coordinates" to
"renormalized spacetime" -- those phrases appear incommensurate. In the
context of GR, there is a single manifold (a single spacetime), but an
arbitrary number of coordinate systems that map regions of the (one)
manifold onto regions of R^4 in different ways.
I repeat: you have a lot of misinformation about GR in your head. You
need to STUDY the actual theory. And if you are trying to discuss some
other theory, you must EXPLICITLY state so.
Tom Roberts
Omega
Ok. Never mind then. I'll just take my ball and go home. You're the boss.
(Or you could just ignore the place names.
"What's in a name? that which we call a rose by any other name would smell
as sweet;"
But then I suppose quoting Shakespeare disqualifies me too.
Also, please note that "Grok" is a term grokked from Heinlein, but that
doesn't seem to affect your right to contribute.
(Boy, this is a tough audience. Is there a posters' rule book that I'm not
privy to?)
But what is funny about your response is, that to someone who reads it
without knowledge of the British science fiction character "Dr. Who?", it
looks like gobbledygook.
(>>Insert insane cackle here<<) )
:-) !
Yousuf Khan
> bz wrote:
>> matter is "close" to the amount that would be needed for a black hole the
>> size of the known universe.
>>
>> Perhaps we live inside a black hole.
>
> Which direction is the singularity?
Towards the past. :)
Yousuf Khan
eric gisse
An unexpectedly correct answer.
Now which direction is the singularity in a black hole?
The future.
>
> Yousuf Khan
Omega
singularity of our local black hole is to our
future. ( That is, is not so much where, as when. And to account for the
Hubble Constant and the negative deceleration paramenter as being due to its
tidal force, its minimum mass is ~3.18619 x 10^53 kg).
Better?
>
>>
>>>
>>>>
>>>> http://www.mathpages.com/home/kmath339.htm
>>>>
>>>>>
>>>>> Bernhard
Omega
- The direction to the singularity of the black hole in which we live is the
past.
- The direction to the singularity of a black hole is the future.
?
eric gisse
We don't live in a black hole. Do catch up.
eric gisse
[...]
>>
>> Ok, people whose knowledge of physics consists of terminology grokked
>> from doctor who have no right to contribute.
>
> As the "direction" of the big bang is in to our past, the direction of the
> singularity of our local black hole is to our
> future. ( That is, is not so much where, as when. And to account for the
> Hubble Constant and the negative deceleration paramenter as being due to
> its tidal force, its minimum mass is ~3.18619 x 10^53 kg).
>
> Better?
No. You are confident that the universe ends in a singularity when actual
physicists are not, and you use 5 significant figures to describe the mass
of the universe when you don't even know if you are within a factor of
10,000 of the real answer.
>
>>
>>>
>>>>
>>>>>
>>>>> http://www.mathpages.com/home/kmath339.htm
>>>>>
>>>>>>
>>>>>> Bernhard
Jonah Thomas
> Omega wrote:
> > - The direction to the singularity of the black hole in which we
> > live is the past.
> > - The direction to the singularity of a black hole is the future.
> >
> > ?
>
> We don't live in a black hole. Do catch up.
"The righteous man is like a frog that lives at the bottom of a well. To
him the sky looks like a small round hole."
Seriously, is there a reason our universe couldn't be a black hole ah,
extruded from some other universe?
I can imagine there might be various theoretical understandings that
would show such a thing would havek characteristics so different from
our universe tht it couldn't be so.
eric gisse
> eric gisse <jowr.pi...@gmail.com> wrote:
>> Omega wrote:
>
>> > - The direction to the singularity of the black hole in which we
>> > live is the past.
>> > - The direction to the singularity of a black hole is the future.
>> >
>> > ?
>>
>> We don't live in a black hole. Do catch up.
>
> "The righteous man is like a frog that lives at the bottom of a well. To
> him the sky looks like a small round hole."
>
> Seriously, is there a reason our universe couldn't be a black hole ah,
> extruded from some other universe?
You mean OTHER than the reason I JUST GAVE which neither of you understand?
Omega
Then why did YOU SAY that the direction to the singularity of the black hole
in which we live is in the past? You REALLY need to work on your
communication skills.
And you are supremely confident that we don't live in a black hole when
actual physicists are not.
Omega
I used 6 significant figures, not 5. But you are correct. I'd thought that
the least number of significant digits of the parameters that I used had
six. But one has only 2. The figure (say ~3.1 x 10^53 kg) that I derive is a
MINIMUM. The model that I espouse is an idealized version and I used a
classical approximation. If your estimate of my error based on the estimated
mass of the universe (which is not one of the parameters that I used), then
that it is only 4 orders of magnitude out of 53 is excellent. Thank you.
Jonah Thomas
> Jonah Thomas wrote:
> > eric gisse <jowr.pi...@gmail.com> wrote:
> >> Omega wrote:
> >
> >> > - The direction to the singularity of the black hole in which we
> >> > live is the past.
> >> > - The direction to the singularity of a black hole is the future.
> >> >
> >> > ?
> >>
> >> We don't live in a black hole. Do catch up.
> >
> > "The righteous man is like a frog that lives at the bottom of a
> > well. To him the sky looks like a small round hole."
> >
> > Seriously, is there a reason our universe couldn't be a black hole
> > ah, extruded from some other universe?
>
> You mean OTHER than the reason I JUST GAVE which neither of you
> understand?
I didn't even notice you doing it. Would you mind explaining it again
maybe in more detail or simpler or more explicitly?
eric gisse
A black hole has the unique feature of swapping the role of time and space
inside the event horizon. A freely moving observer marches - unstoppably -
towards the singularity, just as a freely moving observer marches towards
the future in Minkowski spacetime.
The observable universe is not like this. We are not in a black hole.
Kevin B. Murphy
enter or leave my apartment complex but then again what the hell do I know,
I'm just a schizo. I may be Islam's pi ion but I hopefully have my own pi
ion... or pi ions looking out for my interests without my giving direction.
Nod means no, shake your head means yes, mutiny is not an option.
--
Tell it to the Marines.
YKhan
A black hole inside another black hole reverses its direction of time.
Yousuf Khan
YKhan
> We don't live in a black hole. Do catch up.
And how can you prove that?
Yousuf Khan
eric gisse
Read what I said. Its' relevant.
eric gisse
It doesn't work that way.
>
> Yousuf Khan
Tom Roberts
> A black hole has the unique feature of swapping the role of time and space
> inside the event horizon.
This is not true. Indeed, there is no way to "swap" such incommensurate things.
Say, rather, that for the usual Schwarzschild coordinates on Schwarzschild
spacetime, the coordinates labeled by the symbols "r" and "t" interchange roles
in the two DISJOINT regions r<2M and r>2M (note no "=" there -- NEITHER
coordinate chart is valid at the horizon). Note this is a peculiarity of these
specific coordinates, and is not a property of the underlying manifold (as your
words seem to imply).
In particular, an astronaut falling through the horizon of a super-massive black
hole would observe no unusual behavior inside his spaceship as the horizon is
crossed. "swapping the role of time and space" would certainly be unusual :-).
If he is looking outside watching distant stars, the behavior
of their images would permit him to locate the horizon. But no
local experiment can detect the location of the horizon.
> A freely moving observer marches - unstoppably -
> towards the singularity, just as a freely moving observer marches towards
> the future in Minkowski spacetime.
Yes, inside the horizon. You cannot evade tomorrow.
> The observable universe is not like this. We are not in a black hole.
Yes. Moreover, in the context of GR, there is a singularity in our past (as
shown by the singularity theorems; see for example, Hawking and Ellis). The FRW
manifolds have the right geometrical structure to be models of cosmology, while
none of the black hole manifolds do.
But recent observations are undermining the validity of the
FRW manifolds....
Tom Roberts
Juan R.
What bunch of nonsense! Did you even know the stuff which you are supposedly
replying?
I do not wait you to reply some medium advance question but could
you write down the renormalized coordinates which I refered in my previous
messages?
--
http://www.canonicalscience.org/
BLOG:
http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
eric gisse
> eric gisse wrote:
>> A black hole has the unique feature of swapping the role of time and
>> space inside the event horizon.
>
> This is not true. Indeed, there is no way to "swap" such incommensurate
> things.
>
Close enough.
I'm not saying something stupid like 'your clock starts ticking backwards'
but the ability to move through time and space is inverted from that of
Minkowski space.
[...]
Tom Roberts
> Tom Roberts wrote:
>> eric gisse wrote:
>>> A black hole has the unique feature of swapping the role of time and
>>> space inside the event horizon.
>> This is not true. Indeed, there is no way to "swap" such incommensurate
>> things.
>
> Close enough.
No, it is not, as you yourself prove:
> I'm not saying something stupid like 'your clock starts ticking backwards'
> but the ability to move through time and space is inverted from that of
> Minkowski space.
But that is not so. What is "inverted" inside the horizon is the
meanings of SYMBOLS "r" and "t". Nothing more. This is strictly an issue
of how one labels coordinates, and has no effect on any physical process
or situation.
Tom Roberts
Juan R.
More material for the page devoted to your idiotic claims/lies about black holes:
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-ii.html
Omega
> eric gisse wrote:
>
> Yes. Moreover, in the context of GR, there is a singularity in our
> past (as shown by the singularity theorems; see for example, Hawking
> and Ellis). The FRW manifolds have the right geometrical structure to
> be models of cosmology, while none of the black hole manifolds do.
If one were to apply, instead of Hubble's Law, the simple corollary
"Any two points which are moving toward the origin, each along straight
lines and with speed inversely proportional to the distance from the origin,
will be moving away from each other with a speed proportional to their
distance apart."
which would put the singularity in our future rather than our past,
would the resulting models of cosmology of the observable universe still be
in conflict with all black hole manifolds?
eric gisse
> Tom Roberts wrote:
>> eric gisse wrote:
>>> The observable universe is not like this. We are not in a black hole.
>>
>> Yes. Moreover, in the context of GR, there is a singularity in our
>> past (as shown by the singularity theorems; see for example, Hawking
>> and Ellis). The FRW manifolds have the right geometrical structure to
>> be models of cosmology, while none of the black hole manifolds do.
>
> If one were to apply, instead of Hubble's Law, the simple corollary
> "Any two points which are moving toward the origin, each along straight
> lines and with speed inversely proportional to the distance from the
> origin, will be moving away from each other with a speed proportional to
> their distance apart."
> which would put the singularity in our future rather than our past,
No, no it wouldn't.
> would the resulting models of cosmology of the observable universe still
> be in conflict with all black hole manifolds?
Those who do not understand should not comment.
The singularity of a black hole is in the future of any observer within its'
event horizon. An observer in a FRW manifold has no such destination.
The white hole solution for black holes doesn't apply either.
Tom Roberts
> If one were to apply, instead of Hubble's Law, the simple corollary
> "Any two points which are moving toward the origin, each along straight
> lines and with speed inversely proportional to the distance from the origin,
I have no idea what you think that is a "corollary" of.
In any case, such a physical situation is nonsensical -- I can see no plausible
physical mechanism for making "points" speed up as they approach the origin. And
even with such a mechanism, "points" near the origin would already have passed
through the origin and must be heading away from it at infinite speed (or
momentum is not conserved).
Moreover, the universe we observe does not look like that at all. In particular,
the expansion since the big bang has no "origin" -- speaking rather loosely, it
occurred EVERYWHERE with no "center" at all.
Note that the expansion phase of any of the FRW manifolds does not look at all
like you describe. Nor does the contraction phase of those that have one.
Tom Roberts
Simple Simon
> Omega wrote:
> > If one were to apply, instead of Hubble's Law, the simple corollary
> > "Any two points which are moving toward the origin, each along straight
> > lines and with speed inversely proportional to the distance from the origin,
>
> I have no idea what you think that is a "corollary" of.
You've snipped mid-statement.
I'm sorry for not being clear enough. I'd meant that the statement is
a "corollary" to Hubble's Law:
“Any two points which are moving away from the origin, each along
straight lines and with speed proportional to distance from the
origin, will be moving away from each other with a speed proportional
to their distance apart.”
Algebraically, this is equivalent to saying that if:
The origin x0 = 0
and point x1 is closer to the origin than point x2
(for simplicity, assuming the same constant of proportionality)
and velocity v1 = (N)(x1 - x0) = (N)(x1)
and velocity v2 = (N)(x2 - x0) = (N)(x2)
then v2 - v1 = (N)(x2 - x1)
The next step is:
“Any two points which are moving toward the origin, each along
straight lines and with speed inversely proportional to distance from
the origin, will be moving away from each other with a speed
proportional to the difference of the reciprocals of their distances
from the origin."
Algebraically, this is equivalent to saying that if:
The origin x0 = 0
and point x1 is closer to the origin than point x2
and velocity v1 = -(N)/(x1)
and velocity v2 = -(N)/(x2)
then v1 - v2 = -(N)[(1/x1) - (1/x2)]
At this point it is sufficient to note that, first:
(1/x1) - (1/x2) = (x2 - x1)/[(x1)(x2)]
second:
when the distances from the origin are large compared to the distance
between the points, v1 - v2 becomes immeasurably close to N0(x2 - x1)
where N0 ~= N/[(x1)(x2)] ~= N/[(x1)^2]. This is the situation under
consideration.
Which brings us back to (more accurately stated, still trying to be
pithy):
“Any two points which are moving toward an arbitrarily distant origin,
each along straight lines and with speed inversely proportional to the
distance from the origin, will be moving away from each other with a
speed proportional to their distance apart.”
This, however is not the rule under which I operate. The rule is
considering objects in Newtonian free fall, and goes as follows:
“Any two points which are moving toward the origin, each along
straight lines and with speed proportional to the square root of the
reciprocal of their distances from the origin, will be moving away
from each other with a speed proportional to the square root of the
difference of the reciprocal of their distances from the origin.”
Again, for simplicity, assuming the constant of proportionality, e.g.
as would occur for objects dropped from the same height, this is
equivalent to:
v1 – v2 = [(2GM)^1/2] [(1/x1 – 1/x2)^1/2]
>
> In any case, such a physical situation is nonsensical -- I can see no plausible
> physical mechanism for making "points" speed up as they approach the origin. And
Instead of "points", substitute "objects". i.e. Those same "points"
that Hubble observed. Those same points for which the conservation of
momentum applies.
The mechanism is gravitation.
> even with such a mechanism, "points" near the origin would already have passed
> through the origin and must be heading away from it at infinite speed (or
> momentum is not conserved).
For "ordinary" sources of gravitation, their surfaces would prevent
the objects from passing through the origin, and for a black hole, the
central singularity would prevent same:
"This occurs because spacetime has been curved so much that the
direction of cause and effect (the particle's future light cone)
points into the singularity."
>
> Moreover, the universe we observe does not look like that at all. In particular,
> the expansion since the big bang has no "origin" -- speaking rather loosely, it
> occurred EVERYWHERE with no "center" at all.
>
> Note that the expansion phase of any of the FRW manifolds does not look at all
> like you describe. Nor does the contraction phase of those that have one.
>
This is the point that I'm trying to make:
The result of the above is that the observation of the accelerating
separation of the galaxies is easily explained as being due to the
tidal force one would experience within a black hole, and is not due
to expansion. In fact, it is a simple matter (since I've done
something similar, using the Newtonian approximation of a highly
idealized version where all of the effective mass is located about the
singularity) to calculate the size of the black hole that would result
in the observed Hubble constant and negative deceleration parameter,
and our position within it. Since this demonstrates that the observed
accelerating separation of celestial objects can be attributed to
gravitation alone (objects in freefall within a black hole) it thus
obviates the need for an anomalous big bang event and an aethereal
dark energy and, by Occam's Razor, is worth considering. This would
eliminate the first and most compelling of the 3 pillars upon which
the big bang rests (not that I have anything against the big bang).
While all of this seems obvious to me, your incredulity indicates that
I am probably missing something fundamental.
Thank you very much for your time and consideration.
> Tom Roberts