Starobinsky inflation ($R+R^2/(6m^2)$)

Starobinsky inflation can be formulated as a metric theory of gravity with a squared Ricci scalar in the action. In a convention with positive de Sitter Ricci scalar, use natural units, reduced Planck mass $M$ and mass scale $m>0$:The higher-curvature term carries an extra scalar degree of freedom, the scalaron. Set $F=1+R/(3m^2)>0$ and perform a conformal transformation to $g_{\mu \nu }^E=F{g}_{\mu \nu }$, with canonical field $\varphi =\sqrt{3/2}M\log F$. Its Einstein frame scalar potential isThis scalar representation explains the flat large-field plateau without assuming an unrelated matter potential. The pure $R+R^2$ contribution has this scalar representation.^[ref].

Near $\varphi =0$, $V\simeq m^2{\varphi }^2/2$, allowing an oscillatory post-inflationary phase. On the plateau, the leading slow-roll approximation gives ${ϵ}_V\simeq 3/(4N^2)$ and ${\eta }_V\simeq -1/N$ for $N$ remaining number of e-folds. Hence $n_s\simeq 1-2/N$ and $r\simeq 12/N^2$. These are model predictions, not an observational measurement; the mapping from $N$ to a measured scale also depends on reheating.

The scalaron is the additional propagating scalar degree of freedom in a generic nonlinear $f(R)$ metric theory of gravity. In the model $f(R)=R+R^2/(6m^2)$, it can be expressed in the Einstein frame as a canonically normalized scalar $\varphi =\sqrt{3/2}M\log f_R$, where $M$ is the reduced Planck mass and $f_R=df/dR>0$.

The potential is $V(\varphi )=3M^2{m}^2[1-e^{-\sqrt{2/3}\varphi /M}{]}^2/4$. Near its minimum, expanding the exponential gives $V\simeq m^2{\varphi }^2/2$, identifying $m$ as the scalaron's small-oscillation mass. At large positive field values the potential flattens, providing the slow-roll mechanism of Starobinsky inflation.^[ref]