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Greg Egan
#1
Nov4-06, 03:26 PM
P: n/a
In article <1140127497.964435.327250@f14g2000cwb.googlegroups.com>,
Chalky <utpalchakraborty@gmail.com> wrote:

> I was thinking about something and got confused so I thought I will
> post it here and hopefully someone more knowledgeable than me can clear
> the confusion.
>
> As a thought experiment, suppose, we are accelerating a proton in a
> large accelerator (as large as you need it to be). It will continuously
> gain mass/energy. Will a time come when it has gained sufficient mass
> energy to form a black hole? If not why not?
>
> If yes, then what happens to an observer who is always at rest relative
> to the proton? To that observer nothing is happening but then suddenly
> the proton turns into a hole? Isn't physics seemingly violated for that
> observer?
>
> thanks for any clarification.


This kind of thing has been discussed before on this newsgroup, and
probably others as well, so you might be able to find some useful
commentary via Google Groups.

If you're willing to change the thought experiment to something a little
simpler so we can forget about such complications as electromagnetic
effects and gravitational waves, consider this: according to an observer
flying in a straight line past the Earth at close to the speed of light,
the Earth's total mass-energy is M=M_0/sqrt(1-v^2), where M_0 is the
Earth's rest mass and v is the relative velocity in units where c=1.
Suppose the closest the observer comes to the Earth is r, which will be
the same in his/her frame as in the Earth's frame, because it's measured
transversely to the motion. With v high enough, can't M be made so high
that r will fall inside the Schwarzschild radius and the observer will
necessarily be captured?

The answer is no. In the Earth's frame, the observer is moving by at a
very high speed, well outside the Schwarzschild radius 2 M_0 (in units
where G=c=1). And if you do a general-relativistic calculation (treating
the observer as a test mass, and assuming a Schwarzschild geometry
unperturbed by his/her presence), then the acceleration required for the
observer to travel in a straight line at the point of closest approach is:

a = (2v^2+1)/(1-v^2) M_0/r^2 1/sqrt(1-2M_0/r)

(G=c=1) which is obviously finite when v<1. Specifically, at the value
of v where r=2 M, we have:

v^2 = 1 - 4 M_0^2 / r^2

and

a = (3/(4 M_0) - 2 M_0 / r^2) / sqrt(1-2M_0/r)

This is actually quite high for "small" masses like the Earth, about
1.5*10^18 gees! (Remember we're using units with G=c=1, so you have to
convert things back and forth to get that answer.) So the relativistic
change in the mass does have a signigicant effect. But the acceleration
is still finite, whereas the acceleration required to stay outside an
event horizon is infinite.

As for the question "why doesn't the relativistic mass turn the Earth
into a black hole", one very rough short answer is that an event horizon
is a global property of space-time that depends on the matter responsible
for it sticking around, not flying off into the distance at close to the
speed of light. If your proton is in a linear accelerator, I'm sure it
will never become a black hole, because that's basically just changing
the frame of reference.

The kinetic energy of objects *confined* inside a bounded region would
contribute to the total mass-energy in that region -- e.g. a very, very
fast spin could turn the Earth into a black hole (if it was made of
something strong enough) and I guess even a proton moving in a suitably
implausible cyclotron could (along with the energy density of the
magnetic field needed to confine it) turn the cyclotron into a black hole.

Greg Egan

Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au

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