The right-hand side of the <slow-roll curvature power spectrum> must be evaluated separately for each mode at <cosmological horizon exit>, $k=aH$. Taking logarithms gives
$$
\log\Delta_{\mathcal R}^2=2\log H-\log\epsilon+\text{constant}.
$$
For $N=\log a$,
$$
\frac{d\log H}{dN}=-\epsilon,
\qquad
\frac{d\log\epsilon}{dN}=\eta,
\qquad
\frac{d\log k}{dN}=1-\epsilon.
$$
Hence
$$
n_s-1
=\frac{-2\epsilon-\eta}{1-\epsilon}
=\boxed{-2\epsilon-\eta+O(\epsilon^2,\epsilon\eta)}.
$$
This is the <scalar spectral index in Hubble slow-roll parameters>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 4 / b / Solution

{title2=$n_s-1=-2\epsilon-\eta$}

Differentiating the curvature spectrum at horizon exit gives
$$
n_s-1=\frac{d\log\Delta_{\mathcal R}^2}{d\log k}
=-2\epsilon-\eta
$$
to first order in Hubble slow-roll parameters.


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