What is meant by saying that the Goldstone-bosons are "eaten" by gauge bosons?

FredMadison

I've seen this statement all over, but can't find a good explanation of what this actually means. Anyone care to shed some light?

fzero

To see how this works, let's consider a specific example of a complex scalar field, [itex]\phi[/itex], coupled to an abelian gauge field. The complex scalar has 2 real degrees of freedom, while the massless gauge field also has 2 real degrees of freedom after imposing gauge invariance. A massive abelian vector field has 3 real degrees of freedom, which will become important below.

If the scalar potential only depends on the modulus of the scalar field, [itex]V(\phi) = V(|\phi|)[/itex], then the Lagrangian has a continuous symmetry amounting to rescaling [itex]\phi[/itex] by a phase, [itex] \phi \rightarrow e^{i\theta} \phi[/itex]. Now suppose that this potential has a minimum at [itex]|\phi|=\upsilon[/itex]. We say that the symmetry is spontaneously broken because the vacuum state [itex]\langle \phi \rangle = \upsilon[/itex] is no longer invariant under the phase symmetry of the Lagrangian.

If we parameterize

[itex]\phi = (\rho + \upsilon) e^{i\alpha},[/itex]

we find that the Lagrangian only depends on the derivatives [itex]\partial_\mu \alpha[/itex] of the phase field. So [itex]\alpha[/itex] is a massless real scalar, while [itex]\rho[/itex] is a massive real scalar field. Furthermore, there is an continuous invariance where [itex]\alpha \rightarrow \alpha + c[/itex], which is nothing more than the phase symmetry of the theory. If there were no gauge field coupled to [itex]\phi[/itex], we would identify [itex]\alpha[/itex] with the Goldstone boson corresponding to the spontaneous breaking of the phase symmetry of the complex field.

However, in the presence of the gauge field, the total theory has a local gauge invariance [itex] \phi \rightarrow e^{i\theta(x)} \phi[/itex], [itex]A_\mu \rightarrow A_\mu - i \partial_\mu \theta(x)[/itex]. We are free to use this gauge invariance to set [itex]\theta = -\alpha[/itex]. This eliminates the field [itex]\alpha[/itex] from the Lagranian entirely, leaving terms for the massive [itex]\rho[/itex] and massive vector field [itex]A_\mu[/itex] and their interactions. The 2+2 real degrees of freedom we started with are now distributed as 1 real d.o.f. for [itex]\rho[/itex] and the 3 real d.o.f. for the massive gauge field.

The use of the gauge symmetry to eliminate the phase [itex]\alpha[/itex] in favor of the extra degree of freedom for the massive gauge field is what's referred to as "eating" the Goldstone boson.

tom.stoer

Look at the Mexican hat potential as described in fzero's post: http://www.nature.com/nphys/journal/...hys1874-f1.jpg

w/o a gauge field you would have a physical 'angular degree of freedom' rolling in the well with mass zero. But with a gauge field the 'angular degree of freedom' is no longer physical b/c this 'rolling' is just a gauge transformation and can be rotated away. So this angular zero-mass Goldstone mode 'is eaten' by the gauge boson.

FredMadison

Ok, I think I see how this works. Very clear answers, thank you!