What if a Black Hole dies?
Stan J. Lefosi
What does happen then?
Greg Neill
news:c30b2125.03110...@posting.google.com...
> If you don´t feed it for some millenia, it will die like all other pets.
>
> What does happen then?
How can you prevent a black hole from feeding?
It swallows radiation too, and the background
radiation is at about 2.7K . Only very small
black holes will have temperatures higher than
this and be able to lose mass/energy.
Black holes of any size will only begin to
evaporate after the background radiation drops
below their temperature. That is not expected to
obtain until many trillions of years into the
future. A millennia or two just isn't going to
make a difference.
Sam Wormley
>
> If you don´t feed it for some millenia, it will die like all other pets.
>
> What does happen then?
Massive black holes are not hot, but cold. They feed as long
as the universe is hotter than they are... wait about 10^80 years
for the universe to cool down enough.
Hawking radiation is a function of temperature.
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
http://scienceworld.wolfram.com/physics/BlackHole.html
ghytrfvbnmju7654
Original poster information:
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=author:exxokaprulle%40yahoo.com+
It is hard indeed to prevent them from feeding. Let's
hope Lefosi is like a black hole. He has emitted an
enormous amount of thermal radiation in one day, so if
he is like a black hole, he'll be gone soon.
Alfred Einstead
> How can you prevent a black hole from feeding?
> It swallows radiation too
By freezing its mouth until it dies.
It never feeds because things stop at the event horizon for
all finite Schwarzschild time. But the punchline is that
the horizon, itself, collapses in a finite Schwarzschild
time and the black hole is gone. In fact, it will reach
a point where it starts collapsing at a speed |dr/dt| > c,
leaving behind everything that used to be near its horizon.
To an infalling observer, if they could actually see the
horizon, they would see it suddenly collapse just as they're
about to enter it.
In detail, the geodesic equation of motion for an
infalling light source is
dr/dt = 2M(t)/r - 1.
An estimate of the time-dependency of M(t) obtained via
Black Hole thermodynamics from Stefan's Law would have
M'(t) going as the inverse square of M(t), so that
M(t) = M_0 (1 - t/T)^{1/3}
where M_0 is the Black Hole's mass at time 0, and T its
life time.
Actually solving for r(t), with r(0) > 2M_0, you find
that at all times r(t) > 2 M(t) and, in fact
r(t) - 2 M(t) approaches a minimum,
before INCREASING!
At time T, r(T) > 2 M(T) = 0.
The light never enters the horizon before the Black Hole
is gone, therefore neither does the light cone, therefore
neither does anything else. So, the Black Hole dies hungry.
Greg Neill
news:e58d56ae.03110...@posting.google.com...
[snip poppycock]
Really Alfred, you should refrain from teasing
the newbies.
MorituriMax
news:3FA47C3D...@mchsi.com...
> Massive black holes are not hot, but cold. They feed as long
> as the universe is hotter than they are... wait about 10^80 years
> for the universe to cool down enough.
Can't wait that long.. I got a plumber coming in the morning then so
I'll be busy.
Alfred Einstead
> [snip poppycock]
> Really Alfred, you should refrain from teasing
> the newbies.
There's nothing poppycock about Hawking Radiation.
The event horizon will, indeed, collapse in finite time,
while everything is left hanging on the outside. Nothing
falls in; by the time it gets there, the black hole's
gone.
Properly speaking, the question belongs in Quantum Gravity,
in order to use something more refined than Stefan's Law
to account for the time-dependency of the metric. But
it doesn't much matter how the mass M decreases in time;
the same conclusion applies for any M(t) that decreases
as a function of time, reaching 0 in finite time T -- that
all in-falling light gets left behind before reaching the
horizon, when the horizon collapses.
Constantine
Stan J. Lefosi wrote:
> If you don´t feed it for some millenia, it will die like all other pets.
>
> What does happen then?
I seriously think you need a proper meal. Then come back with sensible
questions. Black holes are not living organisms that need food to
preserve themselves.
Kostas.
Greg Neill
news:e58d56ae.03110...@posting.google.com...
> > [snip poppycock]
> > Really Alfred, you should refrain from teasing
> > the newbies.
>
> There's nothing poppycock about Hawking Radiation.
Sure, but that wasn't what you were talking about.
>
> The event horizon will, indeed, collapse in finite time,
> while everything is left hanging on the outside. Nothing
> falls in; by the time it gets there, the black hole's
> gone.
Nonsense. Stuff crosses the event horizon just as
nice as you please, provided you don't choose a
pathological metric with a singularity there.
Also, even if everything just hung about at the event
horizon, the mass density of the region of space
would still grow, and so the event horizon would
grow to consume the stuff.
Also, if the black hole were to suddenly disappear,
so would the event horizon. All the stuff would
plunge inwards and form another black hole.
>
> Properly speaking, the question belongs in Quantum Gravity,
> in order to use something more refined than Stefan's Law
> to account for the time-dependency of the metric.
GR does just fine. There are several handy metrics
to choose from.
> But
> it doesn't much matter how the mass M decreases in time;
> the same conclusion applies for any M(t) that decreases
> as a function of time, reaching 0 in finite time T -- that
> all in-falling light gets left behind before reaching the
> horizon, when the horizon collapses.
Nonsense.
Alfred Einstead
> "Alfred Einstead" <whop...@csd.uwm.edu> wrote:
> > There's nothing poppycock about Hawking Radiation.
> Sure, but that wasn't what you were talking about.
Yes it is.
> > The event horizon will, indeed, collapse in finite time,
^^^^^^^^^^^^^^^^^^^^^^^^
... because of the Hawking Radiation.
Learn how to read.
> > Nothing falls in; by the time it gets there, the black hole's
> > gone.
>
> Nonsense. Stuff crosses the event horizon
Nonsense. The interior is only connected to the exterior at
Schwarzschild infinity -- which isn't there because the black
hole is gone in FINITE Schwarzschld time -- because of the
Hawking radiation.
... at Schwarzschild infinity.
It's not a matter for debate. This is the geodesic equation
for infalling light:
dr/dt = 2M(t)/r - 1
with M(t) -> 0 as t -> T because of Hawking radiation. If
r(0) > 2M(0), then r(t) > 2M(t) for all t in [0,T], and r(T) > 0
-- outside the horizon after the black is gone.
Period, end of debate.
> Also, even if everything just hung about at the event
> horizon, the mass density of the region of space
> would still grow, and so the event horizon would
> grow to consume the stuff.
Irrelevant. The issue is closed with the math above.
> Also, if the black hole were to suddenly disappear,
> so would the event horizon. All the stuff would
> plunge inwards and form another black hole.
Irrelevant.
> > Properly speaking, the question belongs in Quantum Gravity,
... to take into account the Hawking radiation.
> GR does just fine.
... does not for that reason.
There are several handy metrics
> to choose from.
There are none.
The issue of what the metric is as a function of time because
of Hawking Radiation is still an open problem that can only
be properly addressed in Quantum Gravity.
The one posed elsewhere uses the assumption that the radiation
takes place by Stefan's law (an assumption posed by Ian Lawrie
also in "A Unified Grand Tour of Theoretical Physics") under
which M'(t) = constant/M(t)^2, so that M(t) = M(0) (1-t/T)^{1/3}.
> > that all in-falling light gets left behind before reaching the
> > horizon, when the horizon collapses.
>
> Nonsense.
Nonsense. That's precisely what the solution to
r'(t) = M(t)/r - 1
yields -- for ANY function M(t) of time in which M(t) -> 0 as
t -> T.
Period, end of debate.
Alfred Einstead
> [snip poppycock]
The only poppycock is your reply, intentionally lying in
claiming that the original had nothing to do with Hawking
Radiation so as to ignore the facts that follows from that.
Alfred Einstead
> Nonsense. Stuff crosses the event horizon just as
> nice as you please
Wrong. All infalling light and matter stays outside the
horizon until the black hole is radiated away -- at which
point there's nothing left to fall into.
The difference r(t) - 2 M(t) has a MINIMUM, in the
solution to the geodesic equation (with M(t) by Stephan's law
ultimately given by M(t) = M(0) (1 - t/T)^{1/3}; lightlike or timelike
geodesics, it doesn't matter) after which it increases. Any other
law for M(t) which has M(t) -> 0 as t -> T which yields an
increasing radiation intensity for smaller M will generate a
similar result of r-2M reaching a minimum before starting to
increase again. In fact, at some point under these conditions
|M'(t)| > c will happen, as t is near the final moment T.
Stuff would only cross the event horizon at Schwarzschild
infinity; which -- here -- means not at all, since there's
no horizon to cross after a finite time.
> Also, if
when
> the black hole were to [sic] suddenly disappear
i.e., Hawking radiate its mass away, down to 0
> so would
will
> the event horizon...
after a finite time.
Greg Neill
news:e58d56ae.03110...@posting.google.com...
> > "Alfred Einstead" <whop...@csd.uwm.edu> wrote:
> > > There's nothing poppycock about Hawking Radiation.
> > Sure, but that wasn't what you were talking about.
>
> Yes it is.
No, *you* were talking about whether or not things
can ever cross the event horizon of a black hole.
I've no problem with Hawking Radiation.
>
> > > The event horizon will, indeed, collapse in finite time,
> ^^^^^^^^^^^^^^^^^^^^^^^^
> ... because of the Hawking Radiation.
>
> Learn how to read.
Thank you for your kind suggestion.
I do not argue that the event horizon will disappear
(actually shrink as the black hole evaporates) over
some finite time as seen by a remote observer.
>
> > > Nothing falls in; by the time it gets there, the black hole's
> > > gone.
> >
> > Nonsense. Stuff crosses the event horizon
>
> Nonsense. The interior is only connected to the exterior at
> Schwarzschild infinity -- which isn't there because the black
> hole is gone in FINITE Schwarzschld time -- because of the
> Hawking radiation.
>
> ... at Schwarzschild infinity.
You choose a metric with a singularity at the Schwarzschld
radius and that's what you get. It's like insisting that
you cannot circle the Earth by travelling north, since
at the North Pole everything is south.
>
> It's not a matter for debate. This is the geodesic equation
> for infalling light:
> dr/dt = 2M(t)/r - 1
> with M(t) -> 0 as t -> T because of Hawking radiation. If
> r(0) > 2M(0), then r(t) > 2M(t) for all t in [0,T], and r(T) > 0
> -- outside the horizon after the black is gone.
>
> Period, end of debate.
So nice that you think so.
>
> > Also, even if everything just hung about at the event
> > horizon, the mass density of the region of space
> > would still grow, and so the event horizon would
> > grow to consume the stuff.
>
> Irrelevant. The issue is closed with the math above.
No. Can you explain why the event horizon of a
growing black hole would not grow to consume your
matter supposedly "stuck" there?
>
> > Also, if the black hole were to suddenly disappear,
> > so would the event horizon. All the stuff would
> > plunge inwards and form another black hole.
>
> Irrelevant.
Why? It is an extension of the previous problem,
namely the growing black hole.
>
> > > Properly speaking, the question belongs in Quantum Gravity,
>
> ... to take into account the Hawking radiation.
>
> > GR does just fine.
>
> ... does not for that reason.
What do you feel is the connection between a quantum
explanation for Hawking Radiation and the choice of
spacetime metric at the event horizon of a black hole,
and why would it necessarily include quantum gravity?
The solutions to GR are not expected to get into
quantum trouble until distances from the singularity
are on the order of the Planck Length.
>
> There are several handy metrics
> > to choose from.
>
> There are none.
Indeed? This is an interesting statement. What
makes you feel that it is so?
Greg Neill
news:e58d56ae.03110...@posting.google.com...
> > Nonsense. Stuff crosses the event horizon just as
> > nice as you please
>
> Wrong. All infalling light and matter stays outside the
> horizon until the black hole is radiated away -- at which
> point there's nothing left to fall into.
>
> The difference r(t) - 2 M(t) has a MINIMUM, in the
> solution to the geodesic equation (with M(t) by Stephan's law
> ultimately given by M(t) = M(0) (1 - t/T)^{1/3}; lightlike or timelike
> geodesics, it doesn't matter) after which it increases. Any other
> law for M(t) which has M(t) -> 0 as t -> T which yields an
> increasing radiation intensity for smaller M will generate a
> similar result of r-2M reaching a minimum before starting to
> increase again. In fact, at some point under these conditions
> |M'(t)| > c will happen, as t is near the final moment T.
>
> Stuff would only cross the event horizon at Schwarzschild
> infinity; which -- here -- means not at all, since there's
> no horizon to cross after a finite time.
I admit that I shall have to go and look at the
maths of this before I comment further.
>
> > Also, if
>
> when
>
> > the black hole were to [sic] suddenly disappear
>
> i.e., Hawking radiate its mass away, down to 0
>
> > so would
>
> will
>
> > the event horizon...
>
> after a finite time.
You've prettery much butchered my sentence and altered
its intent. Couldn't you have simply written your own
sentence without cut-and-pasting mine like a ransom
note?
Here's what I said:
Also, if the black hole were to suddenly disappear,
so would the event horizon. All the stuff would
plunge inwards and form another black hole.
So I take it that what you were trying to get across
is that the "stuff" hanging outside the event horizon
takes part in the evaporation process (Hawking
Radiation). Fine. So how does one member of a particle
pair end up in a negative energy orbit below the event
horizon, allowing its partner to become a real particle
and escape? Or are you proposing a different
interpretation for the Hawking Radiation mechanism?