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

Abstract

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 ${\mathcal{M}}_{\pi }^2\propto {\lambda }^{1/4}H$. 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 ${\sigma }_{0c}\simeq 0.61H/{\lambda }^{1/4}$. In the strict $N\to \infty$ 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.

32 pages, 6 figures

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