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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 $?

dmckee --- ex-moderator kitten
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Jose Javier Garcia
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3 Answers3

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It depends on what do you mean by the energy of the particle. If you have a particle with a small rest mass ($m<M_{Planck}$), and very large kinetic energy E, then the answer is no. It can be seen immediately in SR - just switch to the frame, where the particle is at rest - in this frame nothing bad happens. GR tells exactly the same - the metric of a fast flying particle is a boosted metric of a small mass particle at rest - no horizons appear, no black holes.

At the same time, is you mean large rest mass $m>M_{Planck}$, so that the de-Broyle wavelength is smaller, than the Schwarzschild radius for this mass, then this particle would be really a black hole from the very beginning.

Fedxa
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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.

Zo the Relativist
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  • +1 though the Higgs Boson has no spin or charge so your argument doesn't prevent it from being a black hole ... – John Rennie Sep 20 '12 at 15:38
  • Well, the Higgs has a hypercharge, so to be super-accurate, you'd have to work out the solution to Einstein's equation for a BH that carried Yang-Mills charge. Also, I would argue that the Higgs does not qualify as a "common" fundamental particle. – Zo the Relativist Sep 20 '12 at 15:51
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It depends on what you assume to be the size of the particle. Quantum Field Theory treats particles as pointlike objects, but we know that QFT is only an approximation and we expect it to fail at energies approaching the Planck energy.

The closest we have to a theory that works around the Planck energy is string theory, and of course this doesn't assume the particles are pointlike. I've seen suggestions that a highly excited string will form a black hole, but my grasp of string theory falls far short of knowing whether this actually happens.

John Rennie
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