Power-law inflation ($a(t)\propto t^p,p>1$)

Power-law inflation is cosmic inflation with a scale factor proportional to $t^p$ for constant $p>1$. Its Hubble parameter is $H=p/t$ and its first Hubble slow-roll parameter is ${ϵ}_H=1/p$. The comoving Hubble radius decreases as $t^{1-p}$. A canonical inflaton with an exponential inflaton potential admits an exact example. Because ${ϵ}_H$ is constant, an indefinitely maintained power law has no graceful exit from inflation.

Use natural units, the unreduced Planck mass $m_P=G^{-1/2}$, $V_0>0$ and $c\ne 0$. In a flat universe dominated by a canonical inflaton, $\varphi =(2m_P/c)\log (\beta t)$ and $a\propto t^p$ solve the inflaton equation of motion and Friedmann equation exactly when $p=16\pi /c^2$ and ${\beta }^2=V_0{c}^2/[2m_P^2(3p-1)]$. The positive-potential scaling solution requires $p>1/3$, while cosmic inflation requires $p>1$. Its constant equation-of-state parameter is $w=-1+2/(3p)$. In terms of the reduced Planck mass $M=m_P/\sqrt{8\pi }$, the exponential slope is $\lambda =c/\sqrt{8\pi }$ and the same exponent is $p=2/{\lambda }^2$.

For $V=V_0{e}^{-A\varphi /m_{\mathrm{pl}}}$, neglecting acceleration of the field and kinetic energy gives $3H\dot{\varphi }=AV/m_{\mathrm{pl}}$ and $H^2=8\pi V/(3m_{\mathrm{pl}}^2)$. Integration yields the displayed field with $a\propto (t+b{)}^{16\pi /A^2}$. The exact flat scalar-dominated solution has the same power but $C^2=A^4{V}_0/[2m_{\mathrm{pl}}^2(48\pi -A^2)]$. The two normalizations agree at leading slow-roll order. Inflation requires $A^2<16\pi$; the stronger slow-roll conditions require $A^2/(8\pi )\ll 1$.