= Goldstone-boson equivalence theorem
{c}
{title2=$\mathcal M(V_L\cdots)=\mathcal M(\chi\cdots)+O(m_V/E)$}
= Goldstone equivalence theorem
{c}
{synonym}
At tree level and energies much greater than a gauge-vector mass, an amplitude with longitudinal external <gauge bosons> equals the corresponding amplitude with their would-be <Goldstone bosons>, up to conventional external-state phases and relative $O(m_V/E)$ corrections for a nonzero leading amplitude. In a renormalizable gauge, the broken-symmetry identities relate $k_\mu\mathcal M^\mu$ to $m_V\mathcal M_\chi$; inserting $\varepsilon_L^\mu=k^\mu/m_V+O(m_V/E)$ explains the result. Loop-level versions include field-normalization factors. This is different from an <equivalence theorem for field redefinitions>.
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