Breaking Symmetries: 20 Real Scalars and 24 Symmetries

[nonequilibrium]

Homework Statement

Suppose that there is a gauge group with 24 indepenent symmetries and we find a set of 20 real scalar fields such that the scalar potential has minima that are invariant under only 8 of these symmetries. Using the Brout-Englert-Higss mechanism, how many physical fields are there that are
- massive spin 1
- massless spin 1
- Goldstone scalars
- Higgs scalars

Homework Equations

N.A.

The Attempt at a Solution

I'm not sure since I only saw an example worked out where there was a gauge "group" with 1 symmetry and we had 2 real scalar fields and the 1 symmetry was broken, and it ended up giving one massive spin 1, zero massless spin 1, zero Goldstone scalars and 1 Higgs scalar. I'm not sure how to generalize this to a more general case.

But let's give it a try: if we assume that for each broken symmetry, a gauge boson gets mass, we end up with "16 massive spin 1" (since 24 - 8 symmetries are broken). Hence "8 massless spin 1" remain. If I now presume that each gauge boson getting a mass is accompanied with the eating of one Goldstone scalar (which seems sensible from the perspective of the gauge boson gaining one degree of freedom), 16 Goldstone scalars have been eaten, and presuming that no (physical) Goldstone scalars can remain (?) (i.e. "0 Goldstone scalars"), we conclude that from the 20 real scalar fields, "4 Higgs scalars" survive.

Is the answer and/or some of the reasoning correct? Maybe I'm making it too complicated... (for reference we're using Griffiths' Introduction to Elementary Particles, although note that the question is not in the book itself).

 

[mark726]

Your reasoning is mostly correct. Let's break it down into each type of field:

- Massive spin 1: Since 8 out of 24 symmetries are broken, there will be 8 massive gauge bosons. This is because for every broken symmetry, a gauge boson gets a mass. So 8 out of the 24 gauge bosons will have a mass.
- Massless spin 1: Since there are 8 unbroken symmetries, there will be 8 massless gauge bosons. These are the gauge bosons corresponding to the unbroken symmetries.
- Goldstone scalars: As you mentioned, for each broken symmetry, a Goldstone scalar gets "eaten" by the corresponding gauge boson, giving it a mass. So in this case, 8 Goldstone scalars will be eaten and there will be no physical Goldstone scalars remaining.
- Higgs scalars: The remaining 4 scalar fields will be the Higgs scalars, responsible for giving mass to the gauge bosons and fermions through the Higgs mechanism.

So to summarize, there will be 8 massive spin 1 fields, 8 massless spin 1 fields, 0 Goldstone scalars, and 4 Higgs scalars. Keep in mind that this is just a generalization and the specific numbers may vary depending on the specific gauge group and scalar fields involved.