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So recently, looking at high energy particles through the lens of General and Special Relativity has peaked my interest. One thing I was considering, using the electron as the first example, is as follows:

If gravity is the result of mass (stress-energy tensor), and as particles approach the speed of light, their mass is increased as measured by an outside observer (A) by a factor, $ \frac{1}{\sqrt{(1-(v/c)^{2})}} $.

The mass of the electron, $ m_{e} $, is $ 9.109 \times 10^{-31} kg $, and the radius is $ r_{e} = 2.818 \times 10^{-15} m $.

Given the Schwarzschild Condition is: $$ r_{s} = \frac{2Gm}{c^{2}} $$.

Observer A, due to the Lorentz dilation, will measure a dilated mass term that is increasing monotonically as a function of the electron's velocity. The mass that observer A will then measure is: $$ m_{A} = \frac{m_{e}}{\sqrt{(1-(v/c)^{2})}} $$. Combining the Swarzschild Condition with the mass as measured by A, we attain a relation between the mass that observer A measures with the particle's velocity, accounting for relativistic effects.

This allows me to now pose the question that given the Schwarzschild radius is the electron radius, at what velocity does the mass increase to the point where the electron, as seen by observer A, become a black hole? This also leads to the more conceptual question of what are the implications of simultaneity in this situation? If the electron's mass is still $m_{e}$ in it's own frame, then shouldn't the electron not really turn into a black hole? Of course, this whole argument falls apart if the derived speed is greater than $c$. I went own to calculate that. I found that speed to be:

$$ v = \sqrt{c^{2}-\frac{4G^{2}m^{2}_{e}}{r^{2}_{e}c^{2}}}$$

The velocity at which an object of a given radius and mass would become a black hole.

To my disappointment, this came out to be $8.988 \times 10^{16} m/s $. Over twice the speed of light. I have yet to calculate the velocity for more massive objects, such as stars ( which could theoretically reach relativistic velocities as the result of being flung by a galactic collision ). Either way, if this velocity is attainable for anything, what would simultaneity say about this?

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  • $\begingroup$ The constants in the velocity calculation were the presets in my calculators memory for those measures values to eliminate error from acquiring data from different sources. $\endgroup$ – Doryan Miller Dec 17 '13 at 8:11
  • $\begingroup$ possible duplicate of Does non-mass-energy generate a gravitational field? $\endgroup$ – Brandon Enright Dec 17 '13 at 8:31
  • $\begingroup$ I haven't worked through your calculations but I think your conclusion is wrong. You get an arbitrarily high relativistic mass the closer a particle is to the speed of light. $\endgroup$ – Brandon Enright Dec 17 '13 at 8:32
  • $\begingroup$ Aside from the subtleties of relativistic mass (a terrible idea by the way - it was a mistake for physicists of the past to introduce such a concept due to all the confusion it causes), it should be noted that the electron is a point particle and as such doesn't have a radius. $\endgroup$ – user10851 Dec 17 '13 at 8:38
  • $\begingroup$ Actually yes, that post did indeed answer my question. Also, I do agree. It does seem that there is something missing as far as why the speed comes out superluminal yet the mass gets arbitrarily high as the speed of light is approached. I did include the Lorentz factor in there though. It did end up getting broken up in the algebra though. $\endgroup$ – Doryan Miller Dec 17 '13 at 8:49