Sign of Hubble slow-roll parameters

Question

Given the Hubble slow-roll parameters $\epsilon=-\frac{\dot{H}}{H^{2}}$ and $\eta=\frac{\dot{\epsilon}}{H\epsilon}$, can they assume negative values? For inflation to occurr they are required to be small but what about their sign? Thanks in advance.

Answer 1

The equation for the Hubble slow-roll parameter, $\varepsilon$, comes from both the Friedmann equations and the background dynamics for the scalar field. From the equation you give, $\varepsilon=-\frac{\dot H}{H^2}$, we can expand to see that $$H^2(1-\varepsilon)=\frac{\ddot a}{a}=\dot H+H^2\propto-\frac{1}{2}\rho(1+3\omega)$$ where $\omega$ is the equation of state parameter. This can be rearranged using the Friedmann equations to: $$\varepsilon\propto\frac{1}{2}(1+3\omega)+1=\frac{3}{2}(\omega+1)$$ So you can see from this that the Hubble slow-roll parameter can only be negative if $\omega<-1$

For models using a single scalar field that acts as a perfect fluid, the equation of state parameter is given by $$\omega=\frac{{}^1\!/_2\dot\phi^2-V(\phi)}{{}^1\!/_2\dot\phi^2+V(\phi)}$$ This means that in these models, we have $-1\le\omega\le1$, and therefore $0\le\varepsilon\le3$. Thus, whenever the scalar field is treated as a perfect fluid, the Hubble slow-roll parameter can never have a negative value. Usually, the background component of the scalar field is treated as a perfect fluid; however there are other theories of modified gravity, multi-field inflation, or strong coupling to gravity where the inflaton field is not necessarily a perfect fluid. In these cases, the equation of state and the Hubble slow-roll parameters are redefined or, in some cases, are not applicable.

As for $\eta$, there is nothing that requires it to be non-negative. The constraint on $\eta$ for inflation to last long enough is that $|\eta|<1$ and it is defined as $$\eta=-\frac{\ddot\phi}{H\dot\phi}$$ so there is no reason for it to prefer to be positive or negative.