Suppose we have a $G$-invariant action $S$ of a field $\phi$, where $G$ is a Lie group; let then exist a non-zero value $v$ of $\langle\phi\rangle$ such that the $G$-symmetry of the action is broken, and now $S$ is invariant only under the transformations of a subgroup $H$. As J. Zinn-Justin states in his book Quantum Field Theory and Critical Phenomena (at page 324 in the fourth edition), we could approximate the field $\phi(x)$ as $v+\rho(x)+\xi(x) t^av$, where
- $t^a$ are the generators of $G$ which are not generators of $H$,
- $\rho$ is a vector orthogonal to all $t^av$,
- $\xi^a$ are some real coefficients,
The author then says that when this expression is substituted it in $S$, the $\xi$ fields will appear only in the derivatives, since the other terms in the action are $G$-invariant. Therefore the $\xi$ field, which corresponds to the Goldstone modes of the theory, has no mass since a $\xi^2$ term is missing in the action.
But I thought $S$ was not $G$-invariant anymore, since we broke the symmetry by expanding around $v$? Why does he say this? Furthermore, why doesn't this reasoning apply to the derivative terms as well? Those were $G$-invariant in $S$, too, so why is the term with the derivative of $\xi$ allowed to be non-zero?
