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So we know that if you have enough mass in a small enough radius you get a black hole. This radius being r= 2Gm/c^2. This all assumes that the total mass m is within that radius r.

Keep that in the back of your head for a minute. Lets jump over to special relativity. We know that the length of an object decreases as speed increases.

Using the Lorentz contraction L'= l * sqrt(1-(v^2/c^2))

It would seem that eventually you would be able to create a black hole if you gave an object enough speed that an outside observer would see the current object shrink past it's Schwartzchild Radius. Yes I realize that it is only the length that is in question not the width and height, but the same question still applies as sqrt(1-(v^2/c^2)) approaches 0 as v approaches c. Basically flattening the object.

So the questions:

1: If an object gains enough speed to appear to be a black hole and then loses momentum will it return to normal (not be a black hole), or is this a one way trip once you hit that point (will the gravitational forces take over and sustain the black hole)? Or will it even be considered a back hole?

2: Is it possible that the time dilation seen in particle accelerators really just a side effect of this relativistic density change and therefore essentially equivalent to gravitational time dilation?

I will admit I have only looked at this from the side of special relativity. I assume this gets a little more complicated in General Relativity.

edited Aug 1 at 5:03

asked Jul 29 at 14:28

Jack
93●5

Tobias · Accepted Answer · 2010-07-30 13:29:12Z UTC

The trouble is the term "relativistic mass". It occurs when one enforces the validity of $p = m v$ in the relativistic regime by interpreting $m := \frac{p}{v}$ as the mass. I'll quote a quote from wikipedia:

It is not good to introduce the concept of the mass $M = m/ \sqrt{1 - v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ ''m''. Instead of introducing ''M'' it is better to mention the expression for the momentumand energy of a body in motion. |Albert Einstein in letter to [[Lincoln Barnett]], 19 June 1948 (quote from L. B. Okun (1989), p. 42)

This mass $M$ is something like a modified intertial mass (see Eötvös experiment), but not the one the Equivalence Principle states to be equal to the gravitational mass $m$. The latter is the one in the black hole formula and it is always equivalent (to date at least) to the rest mass. So the mass will remain the same and not create a black hole.

However, thanks to the energy-mass-equivalence, the energy of a particle might still be sufficient to create a black hole, if the energy density is high enough. But High Energy Physics will also involve Quantum effects which is the point where no one can tell much so far...

Can you create a Blackhole with speed?

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