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

  1. Sep 14, 2011 #1
    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?
  2. jcsd
  3. Sep 14, 2011 #2


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

    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.
  4. Sep 14, 2011 #3


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

    Look at the Mexican hat potential as described in fzero's post: http://www.nature.com/nphys/journal/v7/n1/images/nphys1874-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.
  5. Sep 15, 2011 #4
    Re: 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!
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