Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 4 / iv

Now use proper-time dots and $H=\dot{a}/a$. Differentiate the slow-roll relation $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$, then substitute $\dot{\bar{\Phi }}\simeq -V_{,\Phi }/(3H)$:Thus the Hubble slow-roll parameter ${ϵ}_H$ agrees with ${ϵ}_V$ to this order. Since ${\mathcal{P}}_{\psi }\propto H_k^2$ and $k=aH$ at horizon exit,Consequently the tensor spectral index isThe final equality is the requested first-order slow-roll approximation; corrections beyond that order also affect the horizon-exit spectrum itself.