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Before QFT I had never thought if a field should or should not have any mass. Now it turns out that either case is possible. It might be naive to think this way, but I picture the mass of the field in the same way as that of a particle, but distributed along the field. The problem is how to think of a massless field.

So, my question is: what does it mean to say a field has mass/is massless?

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  • $\begingroup$ It means that the dispersion relations are different. $\endgroup$ – CuriousOne Jun 29 '16 at 19:16
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    $\begingroup$ It means that the particle states we associate to it have that mass. What do you mean by "distributed along the field" and where does that picture come from? $\endgroup$ – ACuriousMind♦ Jun 29 '16 at 19:20
  • $\begingroup$ @ACuriousMind I thought of it as a rope (The lines of the field) and as the mass of the rope is distributed along it, so should the mass of the field be. It came from nowhere, really. $\endgroup$ – Patrick Jun 29 '16 at 19:28
  • $\begingroup$ @ACuriousMind I should say it was the first analogy that came to my mind. $\endgroup$ – Patrick Jun 29 '16 at 19:37
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The mass of a field is completely well-defined only for free field theories. There it is the mass of the particle whose 1-particle equation is second quantized. For fields of spin $>1/2$, massless is equivalent to the existence of local gauge transformations.

Once one adds interactions, the language is kept (based on the noninteracting part) but the meaning becomes somewhat fuzzy due to the need for renormalization. What is meaningful again are only the masses of the asymptotic (scattering) fields - which are the masses of the observable particles.

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  • $\begingroup$ "Massless is equivalent to the existence of local gauge transformations". What about massless scalars or massless Weyl fermions? What local gauge transformations do they define? $\endgroup$ – Bosoneando Jul 2 '16 at 17:15
  • $\begingroup$ @Bosoneando: corrected. $\endgroup$ – Arnold Neumaier Jul 2 '16 at 19:22
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There are a few things to unravel here.

The simplest thing to do would be to say that massless fields have massless excitations. However, this doesn't work because invariant mass doesn't add directly. For example, a system of two gluons can have a mass, as explained here, even though each gluon by itself is massless. This may be detected by the system's inertia.

As a result, you cannot assign an "invariant mass density" $\rho(\mathbf{x})$ to a field state. If you tried this with the two gluon example, it would be zero everywhere, since each gluon is massless, which is wrong. It also doesn't make sense for the classical electromagnetic field, since light in a box has mass.

To get a sensible result we have to restrict to single particle states with definite momentum. In special relativity, we define the invariant mass by $$E^2 = \mathbf{p}^2 + m^2.$$ Using the de Broglie relations $E = \hbar \omega$ and $p = \hbar k$, and setting $\hbar$ to one, this gives the dispersion relation $$\omega^2 = k^2 + m^2.$$ That is, the frequency $\omega$ of a single particle state with definite momentum is larger the bigger the mass of the field is, because of the rest mass contribution.

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  • $\begingroup$ Not that I want to be one who disturbs feces, but I don't believe this actually answers the question of what it means to say that a field has mass. It dances around the question, but in the end fails to address the root concern. I'd call this a more suitable answer to something akin to "what contributes to the mass/masslessness of a given field". Similar, but it, like your answer, precludes an understanding of what is meant by "the mass of a field", which again is the entire crux of the question $\endgroup$ – Jim Jun 29 '16 at 19:51
  • $\begingroup$ "Light in a box" is not the same thing as a free electromagnetic field. $\endgroup$ – CuriousOne Jun 29 '16 at 20:03
  • $\begingroup$ @Jim What definition of mass do you want? I've reduced it to the single-particle dispersion relation. Discussing why this gives the particle inertia is basically explaining what $E = mc^2$ and $E = \hbar \omega$ mean and I chose not to because there are already great explanations of those facts out there. $\endgroup$ – knzhou Jun 29 '16 at 20:04
  • $\begingroup$ When we attribute a mass value to a particle, there's a definite space and/or position that can hold the mass. We can identify a center of mass. I believe the question ask for clarification on how to translate the layman concept of mass to a field, which has a value at every point in space. For instance, the usual field people deal with is temperature. One would not consider describing temperature as massive or massless, thus the OP's confusion (AFAICT). I think an answer that explains the concept behind assigning the property of mass to any field would be more helpful to them $\endgroup$ – Jim Jun 29 '16 at 20:13

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