Spontaneous symmetry breaking in inflationary cosmology: on the fate of Goldstone Bosons

Daniel Boyanovsky

Published 2012-05-16, updated 2012-07-11Version 2

We argue that in an inflationary cosmology a consequence of the lack of time translational invariance is that spontaneous breaking of a continuous symmetry and Goldstone's theorem \emph{do not} imply the existence of \emph{massless} Goldstone modes. We study spontaneous symmetry breaking in an O(2) model, and implications for O(N) in de Sitter space time. The Goldstone mode acquires a radiatively generated mass as a consequence of infrared divergences, and the continuous symmetry is spontaneously broken for any finite $N$, however there is a \emph{first order phase transition} as a function of the Hawking temperature $T_H=H/2\pi$. For O(2) the symmetry is spontaneously broken for $T_H < T_c= \lambda^{1/4} v/2.419$ where $\lambda$ is the quartic coupling and $v$ is the tree level vacuum expectation value and the Goldstone mode acquires a radiatively generated mass $\mathcal{M}^2_\pi \propto \lambda^{1/4} H$. The first order nature of the transition is a consequence of the strong infrared behavior of minimally coupled scalar fields in de Sitter space time, the jump in the order parameter at $T_H=T_c$ is $\sigma_{0c} \simeq 0.61\, {H}/{\lambda^{1/4}}$. In the strict $N\rightarrow \infty$ the symmetry cannot be spontaneously broken. Furthermore, the lack of kinematic thresholds imply that the Goldstone modes \emph{decay} into Goldstone and Higgs modes by emission and absorption of superhorizon quanta.