Are pointlike massive particles naked singularities?

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:

$$r=\frac{\frac{\sqrt{3}\hbar}{2}}{m_ec}≈3.3\cdot10^{-13}>>10^{-22}$$

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

• I don't think there is much sense in interpreting "pointlike" as "zero-radius". A specific radius for a microscopic particle doesn't help much I think. Rather, "pointlike" simply means "structure-less" in the sense of an internal structure in terms of more fundamental constituents. – Dvij Mankad Jun 25 '17 at 12:43
• – Qmechanic Jun 25 '17 at 12:43
• I think you can ignore the spin of an elementary particle, it is a purely mathematical concept, so nothing is spinning, in any classical sense. – user154420 Jun 25 '17 at 12:49
• For discussions of the meaning of "pointlike", see e.g. physics.stackexchange.com/q/24001/50583, physics.stackexchange.com/q/277565/50583, physics.stackexchange.com/q/119732/50583; For another question on a relation between particles and black holes, physics.stackexchange.com/q/75911/50583 – ACuriousMind Jun 25 '17 at 12:54
• @Dvij Nontheless, wouldn't stuctuelessness suggest dimensionlessness, or am I completely mixing it up? – A. Ok Jun 25 '17 at 13:06