Re: A partial black hole?

On Mon, 25 Feb 2008 20:37:34 -0500, Mike Huskey <mikehsk@[EMAIL PROTECTED]
>
wrote:

>On Tue, 26 Feb 2008 00:55:49 GMT, throopw@[EMAIL PROTECTED]
 (Wayne Throop) wrote:
>
>>: Mike Huskey <mikehsk@[EMAIL PROTECTED]
>
>>: Let see. Neutron star with escape velocity of
.999999999999999999999999 C.
>>
>>No such thing.  It'd have collapsed long since, since even neutrons
couldn't
>>sup****t the forces.  But let's say that there was.
>
>Then what is it if it isn't a neutron star, and it isn't a black hole?
>It is very very dense, but can theoretically be observed.

It is, very briefly, a dense, massive cloud of neutrons free-falling
inwards.  No power in the universe can stop that, because even neutron
degeneracy pressure (the Strong Force at its strongest) can sup****t
the weight of such a "neutron star".  In about a dozen nanoseconds,
the neutron cloud will have contracted to such a small volume that 
its escape velocity will be greater than c, and it will be a black
hole.

A black hole is not a neutron star which used to have an escape velocity
of .999999999999999999999999 c, until someone dropped an extra neutron
on it and it got just an arbitrary bit bigger.  A black hole is a neutron
star which used to have an escape velocity of ~0.50 c, until someone 
dropped an extra neutron on it and it suddenly collapsed, becoming a
*whole* lot smaller.

That ~0.5c figure is an artifact of the actual compressive strength of
"neutronium".  If you want to postulate something stronger still, you
can get a higher limiting escape velocity.

But, you then need to consider relativistic effects in determining the
stresses inside your ultraneutron star.  Not only does its core need to
sup****t the weight of the several solar m***** of ultraneutronium under
its own gravity, but also the gravity generated by the mass equivalent
of the ultraneutronium's own gravity, the mass equivalent of the gravity
generated by the the mass equivalent of the ultraneutronium's gravity,
ad infinitum.

And it is an infinite series.  At normal densities, it's an infinite
series where each successive term is small compared to that before, so
it converges to a finite value not much larger than you'd get by simple
classical mechanics.  It shows up in the third decimal place of Mercury's
orbit, for example.  But as the density of your system approaches the
level where Ve=c, the series approaches a state where each term is equal
to its predecessor.  And it's an infinite series.

For a material object with an infinite escape velocity, the strength of
the (total, including relativistic terms) gravitational field at the core
will be, literally, infinite.  The material will need to be *infinitely*
strong to withstand the pressure.  And there doesn't appear to be any
such thing.


There's not even close to any such thing, which is why there isn't any
such thing as a big solid object with a 0.99999... c escape velocity.
And if you want to imagine a big solid object with an escape velocity
that is *greater* than c, sitting comfortably beneath an event horizon
for you to use as a foundation for thought experiments, then you are
imagining a solid object that is more than infinitely strong.


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