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Re: cosmology and quantum gravity



Rearding John Mallinckrodt's comment:

As I understand it, Hawking radiation is the result of interactions that
take place in the ergosphere, a region outside the actual event horizon
but inside the stationary limit of a rotating black hole. Because a
rotating object drags spacetime around with it, there is a region in which
one cannot stand still, but can still escape the ultimate plunge.

I'm not sure I've got this quite right so I will simply offer it as bait
for David Bowman.

Although John has properly described the stationary limit surface,
ergoregion, and event horizon arount a black hole with a nonzero
spin angular momentum, AFAIK, these things per se don't necessarily have
much of anything to do with the production of Hawking radiation which is
produced in a very thin region just outside the event horizon. The
radiation flux is thermal in its distribution and the temperature (as
seen by an observer asymptotically far from and at rest relative to the
black hole) is directly proportional to the surface gravity on the event
horizon. If g is the surface gravity at the event horizon, then

k*T = h-bar*g/(2*[pi]*c) .

This result is independent of the spin angular momentum of the hole and,
I believe, was found by Hawking when he was analyzing the quantum vacuum
(for a 2nd quantized system of fields) around the event horizon for a
(nonrotating) Schwarzschild metric. It does still apply for more
complicated black holes, too though, I believe.

The effect, as I recall, has to do with virtual process of vacuum pair
creation where one member of such a pair (say a photon pair for normal
temperatures, but all kinds of virtual particle-antiparticle pair
creation processes become relevant when the temperature is much hotter
than the masses so created) falls through the event horizon before it
can reunite with its virtual partner. The left over particle outside
the horizon is then free to escape from the hole as a free asymptotic
particle (assuming it is essentially massless and its momentum is pointed
away from the horizon). Such escaping photons constitute the Hawking
radiation.

The association of the observation of thermal radiation characterized by
a temperature T with a local acceleration has been found to be even more
general than in the context of a black hole. In fact, Unruh discovered
that a particle detector which was accelerating through a *flat* vacuum
Minkowski spacetime would act (behave and respond to its environment) the
same as if it were placed in a thermal heat bath whose temperature was
k*T = h-bar*a/(2*[pi]*c) where a is the acceleration of the detector
w.r.t. an inertial frame. This means that an observer accelerating
w.r.t. a locally flat spacetime observes the same thermal-type effects as
one who is observing the real created Hawking radiation particles that
are emitted from a black hole. This is somewhat reassuring in light of
the equivalence principle.

If I do manage to entice him and if I do have this
anything like correct, I wonder if he can answer another question which
just occurs to me: Can a *non*rotating black hole emit Hawking radiation?

Yes. See above.

Regarding Zach's question:
... I think I
recall being taught that information cannot travel
from within the event horizon to outside of it.
What's going on here?

The entropy of a black hole is directly proportional to the proper area
of its event horizon. If that area is A the entropy S is:

S = (1/4)*k*(c^3)*A/(G*h-bar) .

For an ordinary thermodynamic system its entropy represents the amount
of information necessary to exactly determine which (dynamically
assessible) microscopic state the system is in given just its
macroscopic description. Black holes do not have any detailed
microscopic "hair" which distinguishes any underlying microscopic states.
But a black hole is completely characterized by just its macroscopic
description which involves just the specification of its famous
macroscopic parameters of mass, electric charge, and angular momentum
(and maybe a couple others such as magnetic charge if monopoles exist,
and also Yang-Mills charge if such a macroscopic charge separation could
be effected). If this is the case then how can S be nonzero for a black
hole? Well, in the case of an object with an event horizon (i.e. a black
hole) the concept of entropy needs to be slightly modified. The entropy
in the above formula represents the amount of information needed to
determine the exact microscopic state of all of the highly compressed
matter that went into the original construction of the object's event
horizon. IOW the area of the event horizon measures the amount of
microscopic information that was destroyed in the construction of that
horizon. As matter crosses the horizon into the hole it carrys its
entropy with it. Once this matter does so there is no longer any
external access to the information associated with that matter's
microscopic state. Also, once the hole eats that matter the hole
acquires a greater mass and its horizon expands a little. The proper
area of the horizon is sort of like a tally sheet for a customs agent at
the horizon boundary that recorded all the entropy that the incoming
matter had declared as it crossed the border.

When a black hole evaporates its horizon area decreases and its
associated entropy also decreases. But the outgoing radiation has
enough increased entropy to prevent the 2nd law from being violated.

One really weird thing about the above formula for the entropy is that
it requires that black holes have a negative heat capacity. This means
that any equilibrium they might have with an external heat bath is
unstable and violates LeChatier's principle. The heat capacity of an
object is related to the negative of the second partial derivative of
the object's entropy w.r.t. its internal energy (i.e.
d^2 S/dE^2 = -1/(C_v*T^2) . A normal system has a stable equilibration
behavior w.r.t. heat transport when its heat capacity is positive, or
equivalently, when it's entropy is a concave down function of its
internal energy. Now black holes have an entropy that is proportional
to their horizon area and that area tends to be proportional to the
horizon's radius, and the (gravitational) radius tends to be
proportional the hole's mass, and the mass is just the internal energy
as measured in kg rather than J. Thus the entropy tends to be
quadratic in the internal energy. This is a parabolic function that
curves upward with a positive second derivative. Another way to see this
is to realize that the temperature is proportional to the surface
gravity (at the horizon) and the surface gravity is inversely
proportional to the horizon radius and the radius is proportional to
the mass/energy. Thus, as the mass/energy goes up the surface gravity
and the temperature go down. This opposite tendency for changes in
energy and in temperature is a manifestation of the hole's negative heat
capacity.

To see what this unstable negative heat capacity implies, let's consider
a black hole in equilibrium with a surrounding heat bath at the same
temperature. In this equilibrium the hole radiates as much energy out
into the heat bath as it gets back from it in the form of radiation
descending through the horizon. Now suppose a thermal fluctuation occurs
that slightly gives the hole a little more energy from the bath. This
fluctuation increases the hole's mass which *decreases* its temperature
and this establishs a temperature difference between the hole and the
bath. Heat transport will occur across this temperature difference with
the energy going from the hotter bath to the colder hole. This will cool
the hole's temperature further. This will cause the hole to eat ever
more and more of the energy of the bath until it gobbles up the whole
bath. OTOH, suppose the initial fluctuation transfered a little energy
from the hole to the bath. Then the hole's temperature would increase.
This would make the hole hotter than the bath and there would then be a
net flow of energy from the hot hole to the cold bath. As the hole
radiates away its mass its temperature gets ever hotter thus increasing
the radiation rate until the evaporating hole disappears in a final last
explosion of radiation.

We can think of the 2.76 K Cosmic Background as a Heat bath for black
holes that are not near other energy sources (i.e. stars, etc.). If such
a hole has a mass that is greater than about 4.45 x 10^22 kg (about 60%
of the mass of the Moon) it will not evaporate, but will grow ever
larger. But if the hole has a mass less than this amount it would
eventually evaporate (assuming it is not fed by some other nearby
radiation or matter).

David Bowman
David_Bowman@georgetowncollege.edu