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This video from 37:33 argues that the Standard Model predicts zero mass for all vector bosons as follows:

  • Gauge bosons must have gauge invariance.
  • For a vector field $A$ define a transformation $\alpha(t,x,y,z)$ which acts on $A$ such that $A\rightarrow A + \partial\alpha$
  • The effect on the mass term of the Lagrangian is
  • $m^2A^2 \rightarrow m^2(A+\partial\alpha)^2 = m^2A^2 + 2m^2A\partial\alpha + m^2(\partial\alpha)^2$
  • Ignore $m^2(\partial\alpha)^2$ which is a kinetic energy term not contributing to mass.
  • For guage invariance the observables (mass) must be unchanged, hence $A=0$ (no particles), $\partial\alpha=0$ (contradicting the hypothesis), or $m=0$
  • Hence, all vector bosons are massless.

The issues I have with this argument are:

  • No reason given why vector bosons must have gauge invariance in the first place.
  • The transformation $A \rightarrow A + \partial\alpha$ constrains mass to be zero but a different transformation on $A$ might not constrain mass.

Please help me tighten this argument up. Why does the Standard Model predict zero mass for vector bosons?

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  • $\begingroup$ The argument is only true for gauge bosons, not vector bosons. $\endgroup$ Aug 26, 2020 at 8:55
  • $\begingroup$ I would say not for all vector bosons, namely a gauge boson being a subtype of vector bosons... $\endgroup$
    – ohneVal
    Aug 26, 2020 at 11:24
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    $\begingroup$ Does this answer your question? Gauge bosons of an Abelian gauge field massless $\endgroup$
    – ohneVal
    Aug 26, 2020 at 11:30

2 Answers 2

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There's a lot to say here, so I'll sum up the important steps.

Gauge invariance is a direct consequence of the local symmetry requirement. The standard model symmetry group is given by $$SU(3)\times SU(2)\times U(1)$$ which is indeed a local symmetry. The locality implies directly the existence of the gauge fields since in the lagrangian, whenever we find a derivative, we have to build up a covariant derivative which requires a connection. This connection turns out to be related to elements of the adjoint representation of the underlying symmetry group. These connections give rise to the gauge fields which are massless to begin with.

Gauge invariance of the non-abelian groups is a little bit more complicated than for the abelian $U(1)$ symmetry group, but the ideas stay the same. The theoretical ground on which this is described is Yang-Mills theoreis.

Before spontaneous symmetry breaking, gauge bosons are indeed massless. But thanks to the Higgs mechanism, through SSB, the massless gauge bosons acquire mass, beside the photon (and the gluons) which remains massless. This is a direct consequence of Goldstone theorem since SSB in the EW sector is given by $$SU(2)\times U(1)\to U(1)$$ so that the residual $U(1)$ symmetry, which gives charge to the bosons, gives rise to a massless Goldstone boson, the photon.

That's the reason why we spent so much time looking for the Higgs boson: the SM theory predicts massless gauge bosons as force carriers but we knew experimentally that only the photon should be massless (and the gluons) while the others $W^\pm, Z^0$ should have a significant mass. This mass is given by the Higgs boson through the Higgs mechanism.

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    $\begingroup$ I don't think I fully agree with the statement "The symmetry has to be local since a global symmetry endangers causality." Could you prove or elaborate $\endgroup$
    – ohneVal
    Aug 26, 2020 at 11:26
  • $\begingroup$ @ohneVal Maybe I'm wrong on that! I thought that I remembered something like this, but since you brought it up, it seems strange even to me now. I'll edit it out, it wasn't so important to begin with! $\endgroup$
    – Quiver
    Aug 26, 2020 at 13:20
  • $\begingroup$ I would agree with the statement that a global symmetry is not compatible with causality. A global transformation would act everywhere at the same time. $\endgroup$
    – user248824
    Feb 7, 2021 at 20:01
  • $\begingroup$ Can you please clarify a response to "The transformation A→A+∂α constrains mass to be zero but a different transformation on A might not constrain mass." How do we know that no possible gauge transformation conserves the mass terms? $\endgroup$
    – spraff
    Sep 23, 2021 at 15:52
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The massive intermediate vector boson theory is not renormalizable. Also it is not gauge invariant. The Standard Model therefore starts out with massless gauge bosons and later adds the mass using the Higgs mechanism. Gerard 't Hooft then succeeded in renormalizing the theory, for which he obtained the Nobel prize.

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    $\begingroup$ A massive vector boson simply cannot have gauge invariance, the renormalizability argument is not true. In order to keep gauge invariance it has to be massless, that is the whole point.. $\endgroup$
    – ohneVal
    Aug 26, 2020 at 11:29
  • $\begingroup$ @ohneVal I think maybe my2cts was referring to the usual unitarity-violation justification for the existence of the Higgs mechanism? $\endgroup$ Aug 26, 2020 at 20:58
  • $\begingroup$ @ohneVal I meant the intermediate massive vector boson theory, indeed. I will edit my anser. $\endgroup$
    – my2cts
    Aug 27, 2020 at 15:20

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