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Can it be "proven" from the dynamics of Spontaneous symmetry breaking that the Goldstone bosons get absorbed as the longitudinal degrees of freedom (DoF) of the W and Z bosons? A massless gauge boson has only the two transverse DoF. A massive one has three, which leads to another variant of the previous question, why can't a massive vector boson have a timelike DoF?

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  • $\begingroup$ What do you mean by "timelike dof"? Massive vector field, due to its physical limitation, only have 3 dof, so using eom you can remove the zero component. And the total dof in the Lagrangian is set so after ssb while the mass term is absorbed you can expect massless vector field and Goldstone boson. $\endgroup$ – Turgon Oct 13 '17 at 7:47
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You are probably looking for the Goldstone equivalence theorem which states that for a process for example involving a massive vector boson V, the high-energy limit is

$$M(V,...)=M(s_V,...) (1+\frac{m_V}{E}),$$ where $s_V$ is the Goldstone boson associated with V. This shows that the d.o.f separates nicely in this limit, and that the longitudional component of V behaves as $s_V$

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