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

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?

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 Recognitions: Gold Member Homework Help Science Advisor To see how this works, let's consider a specific example of a complex scalar field, $\phi$, 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, $V(\phi) = V(|\phi|)$, then the Lagrangian has a continuous symmetry amounting to rescaling $\phi$ by a phase, $\phi \rightarrow e^{i\theta} \phi$. Now suppose that this potential has a minimum at $|\phi|=\upsilon$. We say that the symmetry is spontaneously broken because the vacuum state $\langle \phi \rangle = \upsilon$ is no longer invariant under the phase symmetry of the Lagrangian. If we parameterize $\phi = (\rho + \upsilon) e^{i\alpha},$ we find that the Lagrangian only depends on the derivatives $\partial_\mu \alpha$ of the phase field. So $\alpha$ is a massless real scalar, while $\rho$ is a massive real scalar field. Furthermore, there is an continuous invariance where $\alpha \rightarrow \alpha + c$, which is nothing more than the phase symmetry of the theory. If there were no gauge field coupled to $\phi$, we would identify $\alpha$ 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 $\phi \rightarrow e^{i\theta(x)} \phi$, $A_\mu \rightarrow A_\mu - i \partial_\mu \theta(x)$. We are free to use this gauge invariance to set $\theta = -\alpha$. This eliminates the field $\alpha$ from the Lagranian entirely, leaving terms for the massive $\rho$ and massive vector field $A_\mu$ and their interactions. The 2+2 real degrees of freedom we started with are now distributed as 1 real d.o.f. for $\rho$ and the 3 real d.o.f. for the massive gauge field. The use of the gauge symmetry to eliminate the phase $\alpha$ in favor of the extra degree of freedom for the massive gauge field is what's referred to as "eating" the Goldstone boson.
 Recognitions: Science Advisor 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.

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

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