Reversibility of evaporating black holes
Ben Rudiak-Gould
evaporation of a black hole is a time symmetric process. The outgoing
radiation is thermal and the infalling matter may not be, but that could
just be another manifestation of the second law of thermodynamics, not
something that needs special explanation. In fact it seems that one could
make things exactly symmetric by forming the hole from infalling spherical
shells of radiation that match the outgoing Hawking radiation.
In spite of this, the usual conformal map of an evaporating black hole --
the one that looks like this:
|\
| \
__| /
| / /
|/ /
| /
|/
-- is fundamentally asymmetric. This bothers me. For one thing, how could we
even tell that a real black hole was modeled by this diagram and not by its
time reversal? For all we know the radiation from the singularity might be
thermal -- the Big Bang was thermal, after all. The time reversed diagram
says that you can't reach the event horizon before the hole evaporates, but
that's just the time-reversed version of an argument that a black hole can't
radiate. I'd expect there to be a time-reversed Hawking radiation process in
which a white hole gains mass by absorbing matter from just outside the
event horizon. So I can easily imagine that there's no observable difference
between these diagrams at all.
I would be much happier if there were a manifestly time-symmetric conformal
diagram of an evaporating black hole, but as far as I can tell this is
impossible, because the event horizon has to be inclined at 45 degrees. This
even seems to correspond to an aspect of reality, since a stationary null
surface can have light cones
| / \ |
|/ \|
like this | or like this |
/| |\
/ | | \
but not both. But I can't shake the feeling that there's a symmetry that's
being lost in this picture. Any insight would be appreciated.
-- Ben
Oh No
>I would be much happier if there were a manifestly time-symmetric
>conformal diagram of an evaporating black hole, but as far as I can
>tell this is impossible, because the event horizon has to be inclined
>at 45 degrees. This even seems to correspond to an aspect of reality,
>since a stationary null surface can have light cones
>
> | / \ |
> |/ \|
> like this | or like this |
> /| |\
> / | | \
>
>but not both. But I can't shake the feeling that there's a symmetry
>that's being lost in this picture. Any insight would be appreciated.
>
I have been drawing time symmetric diagrams which look like this
\ |
\ |
\|
|
|
/|
/ |
Ingoing and outgoing light is drawn at 45deg - indeed this property is
used to define coordinates in terms of time on a distant clock. The
coordinates become singular at the event horizon. Broadly speaking
whereas reflection of a photon is shown as instantaneous as in sr at
unsingular points, at the event horizon a time delay of 4GM is necessary
- twice the Schwarzschild radius. I think this form of diagram tells us
something fundamental about properties of matter. Exactly what it tells
us, requires a quantum treatment.
Regards
--
Charles Francis
substitute charles for NotI to email
Gerry Quinn
says...
> Another black hole question. It looks like the formation and subsequent
> evaporation of a black hole is a time symmetric process. The outgoing
> radiation is thermal and the infalling matter may not be, but that could
> just be another manifestation of the second law of thermodynamics, not
> something that needs special explanation. In fact it seems that one could
> make things exactly symmetric by forming the hole from infalling spherical
> shells of radiation that match the outgoing Hawking radiation.
I don't think you *can* do that! You can certainly make a black hole
from a spherical shell of infalling radiation, but the mass-energy
density of such a shell will be (or will at some point become) much
greater than that of 'reversed Hawking radiation'.
If you try to use reversed Hawking radiation, you will start with a
pulse of high temperature radiation that rapidly gets cooler, and feed
in cooler and cooler radiation as time goes on. But by the time most
of the radiation gets there, the initial pulse will have met in the
middle and dispersed - even if the initial pulse formed a black hole it
will have evaporated instantly.
Almost from the start energy will be going outward as well as inward,
and there will never be enough energy in a sufficiently small sphere to
form a black hole.
Alternatively, if you start with an already formed black hole and try
to grow it with reversed Hawking radiation, you will just get a
situation in which the hole neither grows nor shrinks, as it will
simultaneously be radiating an equal amount of energy. Basically, it
will be sitting in a box whose walls are at its Hawking temperature.
If you increase the temperature of the box by a finite amount, the hole
will start to grow, but you will no longer have a reversible system.
- Gerry Quinn