Re: Black holes

["Paul Camp" <pjcamp@coastal.edu>]

Chris Jones writes:

> Can anybody explain to me why the calculation of the Schwarzschild radius:
> R(sch) = (2GM/c^2)
> is valid? I have trouble explaining why this should form the event horizon 
> because:
> 1. The equation can be derived purely from the one for escape velocity, 
> which is non-relativistic

It can but this is mostly a mathematical accident that can be traced 
to the inverse square law Newtonian force. The derivation gets the 
correct result (but only for a nonrotating, uncharged black hole) but 
for the wrong reasons.

> 2. The energy of a photon is a function of its "mass" rather than its 
> velocity, so why should an escape velocity be appropriate?

The gravitational field affects all forms of energy, radiation as well 
as mass. In both cases, it is a conservation of energy problem and 
for light this means changing frequency rather than velocity as it 
would for a massive particle. The light is therefore red-shifted away 
rather than reaching a zero velocity and then falling back. The 
calculation is the same in both cases (cons. of E) and in that sense 
it is a sort of escape velocity problem. Alternatively, you could ask 
how fast a massive object would have to travel in order to escape. As 
you approach the even horizon, the necessary escape velocity 
approaches the speed of light so in this sense it is also an escape 
velocity-type problem.

> 3. The escape velocity equation works 2 ways, so that an object falling 
> towards the black hole would be expected to accelerate to the escape 
> velocity (>c!)

BUT consider what would happen if an observer were to fall into the 
black hole while flashing a light at regular intervals. Due to 
gravitational red shift, a distant observer would see these flashes 
coming less and less frequently. Hence, from his point of view, the 
observer falling into the black hole is approaching the event horizon 
asymptotically, always getting closer but never quite reaching it. So 
from outside, you never obeserve anyone moving at c, let alone faster 
than c. Events as viewed by the observer falling in are somewhat more 
complicated.

> Also, what happens to the energy of a photon emitted at the surface of a 
> black hole - if it is redshifted to infinite wavelength, its momentum (and 
> so "mass", and energy) is zero, which makes the gravitational potential 
> energy zero. Where has the starting energy gone?
It was used to do work while climbing out of the black hole. Please 
don't think too closely about photon "mass"; it doesn't really have 
one, it has energy dependent on frequency and gravity will affect 
that energy in a manner similar to the way in which it affects mass 
but there really isn't any problem with it going away altogether.

 
Paul J. Camp                   "The Beauty of the Universe
Assistant Professor of Physics  consists not only of unity
Coastal Carolina University     in variety but also of 
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