9
$\begingroup$

The Goldstone boson equivalence theorem tells us that the amplitude for emission/absorption of a longitudinally polarized gauge boson is equal to the amplitude for emission/absorption of the corresponding Goldstone boson at high energy. I'm wondering what's the physical meaning of this theorem. Is there any relation between equivalence theorem and Higgs mechanism ?

$\endgroup$
2
$\begingroup$

Probably after 3 years you've answered this question, but it might be useful for others. I think you got it. At high energies, you can choose to view the spectrum as a heavy vector gauge boson with 3 degrees of freedom (3 independent polarizations) or view the spectrum as a massless vector gauge boson (2 degrees of freedom) with an extra massless Goldstone boson (1 degree of freedom) for a total of 3 degrees of freedom. The Higgs mechanism is exactly this process. You have a global $SU(2)_L \times U(1)_Y$ symmetry that in the standard model which gets broken to $U(1)_{EM}$. $SU(2)$ has 3 generators, and $U(1)$ has 1 generator. So at the end of the day, it would seem as if you lose 3 degrees of freedom, but not really, as you also produce 3 massless Goldstone bosons. These in turn give mass terms to W+ W- and Z bosons. At high energies, you can think of W+ W- and Z as massive, or as massless with these 3 goldstone bosons.

Practically, the advantage of all this is that although in a cross section calculation, thinking of massless Goldstone bosons adds more diagrams to a given process, they are much easier to calculate.

$\endgroup$
  • $\begingroup$ Any comment on the last part of the question (regarding the connection between equivalence principle (not GBET) and the higgs' mechanism) ?? $\endgroup$ – negligible_singularity Nov 17 '16 at 10:21
  • $\begingroup$ @Gabriel Good answer, but you got the masses wrong. In GBET both gauge and Goldstone bosons are massive. A more precise statement of Goldstone Boson Equivalence theorem reads: at high energies, the amplitude for a longitudally polarised massive gauge boson equals the amplitude of the eaten Goldstone boson with the same mass. $\endgroup$ – Ramtin Jan 11 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.