Re: Singularity Temperature

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding Michael Bowen's questions:

> My understanding (which is quite possibly incorrect) is that a
> material object (m > 0) located inside the event horizon of a
> non-rotating black hole cannot move outside of the event horizon (that
> is, "escape") because, at least in principle, an infinite amount of
> energy would have to be added to the object in order for it to escape
> the singularity's gravitational well. (Equivalently, the escape
> velocity from points inside the event horizon exceeds c.)

Although it is formally true that the escape velocity exceeds c inside
the horizon, the main reason why an object inside the horizon cannot
pass back out through it is that in order to do so it would have to
travel *backwards in time* to do so.  For all objects that are inside
the horizon because they got there after falling in, the horizon is in
their objective past, and no amount of manuvering along a future-directed
timelike path can have the path go into its own past.  (Note, that for
objects that *start out* inside the horizon things can be much trickier
when an extended black hole geometry is used that has both an interior
future singularity *and* a past interior singularity because they *may*
be able to exit the horizon.  Such objects are alluded to below where I
discuss the extended time-symmetric Schwarzschild solution.)

> Questions: Can a chunk of ordinary matter (say, a brick or an
> electron) initially located somewhere outside the event horizon (and
> totally free to move under gravitational influence from an initial
> velocity of zero) actually pass through to the horizon's interior, or
> does such matter approach the horizon in some asymptotic manner?

Yes, the object does actually fall through the horizon in a finite
amount of time as kept by a clock travelling with the brick.  In fact,
it's inevitable according to the way you have laid out the gedanken
situation. To prevent it from happening, something else needs to block
the brick's free fall.  If the time used is that kept by a clock
falling with the brick, and if the radial location is labeled by the
radial parameter r of a so-called standard coordinate system (i.e. the
circumference of a circle around the hole at that distance out divided
by 2*[pi]), then the time it takes to fall through the horizon (and,
indeed its entire falling trajectory) is identically the *same* as one
would get pretending ordinary Newtonian gravity and Newtonian mechanics
held for the falling brick.

However, if the object's fall is remotely monitored by an observer who is
fixed w.r.t. the hole and is *outside* the horizon, then the infall of
the brick takes an infinite time to happen for the external observer (the
falling time asymptotically diverges logarithmically as the event horizon
is approached) and the brick effectively asymptotically slows down to a
halt as it approaches the horizon.

> I ask this because if the chunk must gain an infinite amount of
> gravitational potential energy to pass outward through the horizon,
> must it not also lose the same amount to pass the other way?

First of all we need to define just what we mean by the notion of
energy (let alone *potential* energy) in this situation.  The most
straightforward and useful definition in this instance is to define the
energy as the value of the Hamiltonian obtained from the Lagrangian for
a freely falling brick in the presence of the hole as it follows its
time-like geodesic in spacetime.  This Lagrangian can be found by
differentiating the spacetime *scalar* elapsed action w.r.t whatever
time parameter is most useful for the problem.  This elapsed action is
just the elapsed proper time for the brick multiplied by -m*c^2 where m
is the brick's invariant mass as seen when the brick is at an infinite
distance from the hole.  The time parameter used in this Lagrangian
formulation can be conveniently taken to be the time measured by a clock
attached to the external observer at rest w.r.t. the hole and who is
responsible for dropping the brick.  A nice thing about this definition
for the brick's energy is that it is conserved for the brick's trajectory
(assuming we can neglect any gravitational radiation coming from the
accelerating brick).

As measured by the fixed exterior observer who dropped the falling brick
from rest, it loses, as it approaches the horizon, exactly as much
gravitational potential energy as its total initial rest energy it had
when the observer released the brick from rest.  If we can neglect
gravitational radiation effects from the brick and any frictional effects
from other accreting matter falling into the hole, the brick's total
energy remains conserved as it falls, and the lost potential energy is
made up for acquiring kinetic energy.  But this acquisition is very
tricky because the total energy is *not* a simple sum of a potential
energy function that depends only on position and a kinetic energy term
that depends only on motion and a constant rest energy.  Rather, the
total energy is a complicated coupled nonlinear function of both the
position and motion.

The way we define the combined rest and potential energies in this case
is to evaluate the above complicated expression for the total energy,
(i.e. the above Hamiltonian of the brick in the bacground geometry of
the hole) when setting the speed of the brick to zero. Next, we evaluate
this expression for a resting brick when it is placed at some fiducial
distance (typically either at infinite distance or at the fixed location
of the observer dropping it) from the hole.  This gives the brick's rest
energy.  After the rest energy is found it is subtracted from the
composite expression for the energy of a resting brick at a particular
place.  This difference (typically negative) is the gravitational
potential energy for the resting brick at that place.  Next the composite
total energy of the fully moving brick is evaluated at some place and the
joint rest/potential energy for it at that place (i.e. the energy it
would have if it wasn't moving while there) is subtracted off.  This
difference is the kinetic energy of the moving brick at that place.  Such
a procedure gives well-defined separate values for the rest, kinetic, and
potential energies.  However as the brick approachs the horizon, the
conserved *total* energy approachs a 0/0 indeterminate form while
remaining conserved to its initial value.  However 100% of the brick's
rest/potential energy is lost at the horizon.

> If it is
> possible to pass through the event horizon inbound (that is, to "fall"
> through), how is the inbound/outbound symmetry broken? (Or, perhaps,
> what causes the process to be irreversible?)

This is an *excellent* question.  The answer is that the usual exterior
solution to Einstein's equations describing the black hole is not the
complete solution to these time-symmetric equations.  There is another
time-reversed solution (describing a so-called white hole) as well.
The spacetimes describing the two independent exterior solutions are
not in direct contact.  But they *can* be in indirect contact via their
respective *interior* regions.  It's just that no null or timelike path
ever goes from one exterior region into an interior region and back out
to the other exterior region.  It *is* possible for a timelike path to
come out of either horizon (in a finite amount of time indicated on
a clock carried by the particle along the path) to either exterior region
from a corresponding interior one. But an exterior observer will have had
already noticed that at t = - [infinity] the particle had already escaped
before the exterior history had begun.  For further information about the
extended fully time-symmetric Schwarzschild solution see Hawking & Ellis'
book _The_Large_Scale_Structure_of_Space-time_ (sec. 5.5, esp. fig. 24).

> Does the idea of "passing
> through" the event horizon even make sense?

Yes.  Just make sure you use the time registered by a clock carried along
the path which crosses the horizon and not a clock that is held fixed in
space exterior to the horizon.

> Or is the process, if it
> occurs, more akin to quantum mechanical "tunneling"?

No.

David Bowman
David_Bowman@georgetowncollege.edu