Wednesday, November 14, 2018

Does the new energy-momentum relation break relativistic gauge theories?

Our new energy-momentum relation is equivalent to an old version when |p| and |V| are small.

       (E - V)^2 = (p + A)^2 + m^2
      <=> (|p|, |V| small)

       E - V = m
             sqrt((p + A)^2 / m^2 + 1)
                = (p + A)^2 / (2m) + m
      <=> (|p|, |V| small)

        E = (p + A)^2 / (2 (m + V)) + m + V
           = (m v + V v)^2 / (2 (m + V))
              + m + V.

The choice A = v V gives the same result as treating V as rest mass.

However, in relativistic settings, the two energy-momentum relations are inequivalent. A question is if relativistic gauge field theories are flawed as they use the old relation?

A brief study of the material on the Internet reveals that gauge field theories are usually nonrelativistic, or are expressed through a lagrangian density and Feynman diagrams. Feynman diagrams have interactions only at vertexes, and the particles are assumed to be free in the lines between them. The question whether potential energy and rest mass should be identified never comes up.

Rest mass in the Standard model is potential energy in the Higgs field. Thus, the Standard model does identify at least the Higgs potential energy with rest mass. Before the Higgs field assumes its vacuum expectation value, fermions have a zero rest mass. The physics would be weird in such a setting because the value m - V would often be negative and the fermion would be a tachyon. Maybe the Higgs field always has had the same expectation value? Is there evidence that fermions can be rest-massless at high energies? If the rest mass of the electron would be zero, would its acceleration under a potential be infinite?

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