If elementary particles (specifically, those with mass, such as the electron or other leptons) are pointlike particles, wouldn't that mean they are naked singularities?

But these particles have spin- wouldn't that make them naked ring singularities, thus giving them an observed radius, making them non-pointlike?

If I remember correctly, the radius of a ring singularity is given by $a=\frac{J}{Mc}$. If we assume the intrinsic spin property of a particle is equal to $J$ of the corresponding singularity, we get for the electron:


So this seems utterly nonsensical given the upper bound on the electron radius.


If you consider only classical physics, an electron can be seen as a Kerr-Newman black hole, that is a rotating charged black hole, violating the naked singularity bound since it's "rotating" too fast.

But this analogy is flawed in many ways, mainly not taking into account quantum physics. You cannot have a black hole with mass smaller than the planck mass, since the quantum fluctuations of the horizon would be of the size of the horizon itself.

Moreover current theories (the standard model!) regard as fundamental entity the fields, while particles can be seen as excitations of this field. Fields live in the whole spacetime, so they are never exactly localized.

You can find some material here: https://en.wikipedia.org/wiki/Black_hole_electron

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  • $\begingroup$ Thank you. Wouldn't a black-hole electron be extremely unstable as it would emit Beckenstein-Hawking radiation until evaporating all together? $\endgroup$ – A. Ok Jun 26 '17 at 11:52
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    $\begingroup$ No, black hole is a misnomer here, since a naked singularity has no horizon. Radiation from naked singularities is still an highly speculative topic, look here arxiv.org/abs/gr-qc/0012087 $\endgroup$ – Rexcirus Jun 26 '17 at 14:14

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