The Wikipedia article about Higgs field poses some questions to me.

  • The article says that the Higgs field is a "nonthermal" field, a field whose energy does not decrease as the universe expands, yet it claims that in the early universe Higgs field was stronger which led to inflation. How the field's strength could decrease then?

  • If the Higgs field was stronger in early universe, why the article claims that elementary particles had no mass at early stages of the universe?

  • The article claims that Higgs field is repulsive and is behind the inflation. Yet it does not interact with photons. Would not that make the photons not to follow the universe's expansion so that we would see distant stars not actually in the direction where they should be?

  • The article claims that all electrons in neutron beta-decay are left-handed only for a stationary observer while for a moving observer this changes. Does it mean that one can measure his speed relative to the Higgs field by counting the right-handed electrons in neutron's decay?

  • The article claims that in absence of Higgs field electrons would have no rest mass. But I cannot imagine a charged particle without rest mass because the electromagnetic field has potential energy. Does it mean that in absence of Higgs field electrons are chargeless?

  • 1
    $\begingroup$ Two things to note: (1) It's a Wiki article (2) It's a non-technical Wiki article $\endgroup$ Commented Jul 5, 2012 at 23:11
  • $\begingroup$ Anixx, could you please split this up into multiple posts with one question each? $\endgroup$
    – David Z
    Commented Jul 5, 2012 at 23:21
  • $\begingroup$ The last question should be separate from the others--- the others are pretty related. The last question is also much more interesting, the rest are just confusions in Wikipedia between the model where the Higgs is the inflaton and the model where the Higgs is thermal. The last question about a massless charged particle bothered me too a long time ago, but I never sorted it out, and forgot about it. $\endgroup$
    – Ron Maimon
    Commented Jul 6, 2012 at 18:29

1 Answer 1


The Wikipedia article is confusing the Higgs field with the inflaton, which is not allowed, since it is not a Higgs VEV that causes inflation in current theories. This should clear up the questions.

  • If the Higgs is the inflaton, it has a nonthermal, nonequilbrium value, which slowly oozes to the vacuum value and this is while the universe inflates. This process requires that the Higgs potential at large values of the field is abnormally flat, so it works as an inflaton, and this is not supported by current understanding. Once inflation is done the Higgs oscillates around the minimum (our vacuum value) and emits standard model stuff, then settles down to a stable vacuum value. From this point on, it is nonthermal vacuum field, and doesn't change as the universe expands. This is not the modern picture of inflation, where the inflaton is separate field from the Higgs.
  • If the Higgs is an inflaton, this is a false claim. A big Higgs VEV means big masses for the SM particles. In the standard inflation ideas, the Higgs is not the inflaton, the inflaton is a different scalar that is slowly sinking to zero causing inflation, and at the end, it pumps up other scalars and these decay into thermal stuff, all at energies above the Higgs scale. Then the Higgs in the early universe starts out thermal and random, with a temperature higher than the scale at which it breaks symmetry, so it's average value is zero. In this picture, random thermal Higgs, the particles are effectively massless (although by construction these temperatures that randomize the Higgs are so high, that the particles are whizzing around so fast that their Higgs-derived mass is negligible anyway). The average thermal value of the Higgs has a phase transition to zero at extremely high temperatures, but the article is confusing the no-mass high temperature Higgs with the high VEV Higgs required when Higgs is made the inflaton. In the second case, the masses of the SM particles are not zero, since there is more and more Higgsing at larger Higgs VEV.
  • The interaction with photons is indirect, through gravity. An inflaton makes a cosmological constant, which makes gravity repulsive, and then the photons follow along. This is not the right picture, though. There are no photons during inflation, the universe is just inflaton VEV. The photons come later, when the inflaton decays to standard model stuff at the end of inflation.
  • stationary here is defined relative to the decaying nucleus, and this is not a correct statement--- the electrons are produced left handed at "t=0", but the mass of the electron flips their helicity constantly. The neutrinos are the ones that stay with one helicity, since they have no partner.
  • this is a very interesting point, and one that should be asked as a separate question. The traditional answer is that a charged particle can be massless so long as the chiral symmetry is preserved at long distances in the theory, but it is intuitively puzzling to have a charged massless particle, because, as you say, the field has mass, so how could you get a long-range electrostatic field from a massless particle? I don't know the answer immediately.
  • $\begingroup$ on the last point, the higgs should only contribute to the bare mass, the observable mass would then be entirely electromagnetic $\endgroup$
    – lurscher
    Commented Jul 7, 2012 at 16:49
  • $\begingroup$ @lurscher: This can't be right. You can't have a chiral fermion with an electromagnetic mass. The electromagnetic field would have to make a pure Z-correction to the particle and no mass renormalization at all. This is very counterintuitive, and this is why it needs a separate question. $\endgroup$
    – Ron Maimon
    Commented Jul 7, 2012 at 19:02
  • $\begingroup$ I understand what it is supposed to be, the highly boosted charged particle is supposed to produce both E and B fields which, away from the location of the charged particle, have an equal magnitude of field momentum and energy, so there should be a consistent infinitely boosted massless point limit. My confusion comes from the fact that the E field self-energy for a point is positive and divergent for a fixed charge, and it doesn't seem immediate that the radius where P doesn't equal E in the boosted solution shrinks to zero size in the solution. If it's true, that's the answer. $\endgroup$
    – Ron Maimon
    Commented Jul 7, 2012 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.