Let us suppose a particle with so much energy $ E= h \frac{c}{\lambda} $ so $ \lambda $ is smaller than Planck's length ?
Would it be possible? I mean if the particle has so much energy then its mass would be very huge and would itself collapse into a black hole
This means that we could not 'see' what is beyond the Planck's length because a very energetic particle would turn into a micro black hole before reaching 'Planck's lenght' am i right?
Is this fact supported by SR and GR? what would be then the Einstein equations ? $ R _{a,b} =0 $?
Answer
General relativity tells us that the event horizon of a spinning object is given by $r = M + \sqrt{M^{2}-a^{2}-Q^{2}}$, where $M$ is the mass of the black hole, $a$ is the angular momentum per unit mass of the black hole, and $Q$ is the charge of the hole${}^{1}$. If you put in the appropriate mass, charge and angular momentum for common fundamental particles, you will find that, almost uniformly, $a^{2}+Q^{2} > M^{2}$, which means that the equation for the horizon has no real roots, and that if you were to interpret these particles as classical point particles, they would be naked singularities, not black holes. Fortunately, no one takes that model too seriously, expecting quantum gravity to take over first.
${}^{1}$As always in GR, we choose units of $G=c=1$, and choose Gaussian units for our charge.
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