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    $\begingroup$ So you say that, although not theoretically impossible, there shouldn't be such a particle given our observations? $\endgroup$
    – Eelvex
    Commented Apr 2, 2011 at 4:25
  • $\begingroup$ @Eelvex: it depends on your definition of theoretically impossible. But yeah, they are basically ruled out by experiment because a world with charged massless particles would be very different from ours. $\endgroup$
    – Marek
    Commented Apr 2, 2011 at 9:34
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    $\begingroup$ @Eelvex: but unlike flying penguins :) youtube.com/watch?v=9dfWzp7rYR4 $\endgroup$
    – Marek
    Commented Apr 2, 2011 at 10:07
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    $\begingroup$ To be precise, you need to turn your partial derivatives into covariant derivatives to minimally couple the scalar the field to the photon field: $\mathcal{L} = D_\mu\phi^\ast D^\mu\phi$ for $D_\mu = \partial_\mu + ie\hat{Q}A_\mu$. From here, note that the photon-loop diagram would give a mass renormalization. Unless there is some symmetry which protects/prevents the renormalization of the $\phi$ field's mass, there is no reason to assume that this bare Lagrangian should give physically massless particles! $\endgroup$
    – josh
    Commented May 16, 2012 at 4:08
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    $\begingroup$ Wouldn't the energy of the electrostatic field be the mass an "otherwise" massless particle would have? (I.e, a "massless charged particle" is a contradictio in adjecto, an impossible contradiction by definition.) I actually always wondered whether the electron's rest mass is simply the sum of its electrostatic and weak fields. (There is certainly no "ball" "having" a charge, for example; the electron, like all other "particles", is the field.) $\endgroup$
    – Peter - Reinstate Monica
    Commented Sep 15, 2020 at 12:33