= Solution
The <slow-roll equations> turn the <slow-roll curvature power spectrum> into
$$
\Delta_{\mathcal R}^2\simeq\frac{V^3}{12\pi^2V_{,\phi}^2}=\frac{V}{24\pi^2\epsilon},\qquad\epsilon=\frac12\left(\frac{V_{,\phi}}V\right)^2.
$$
The <spectral tilt> is the logarithmic derivative of this dimensionless spectrum. Apply the <horizon-exit logarithmic derivative in slow roll>:
$$
\Delta n=\frac{d\log\Delta_{\mathcal R}^2}{d\log k}\simeq-\frac{V_{,\phi}}V\left(3\frac{V_{,\phi}}V-2\frac{V_{,\phi\phi}}{V_{,\phi}}\right)=-3\left(\frac{V_{,\phi}}V\right)^2+2\frac{V_{,\phi\phi}}V.
$$
With the <potential slow-roll parameter> $\epsilon$ above and the <second potential slow-roll parameter> $\eta=V_{,\phi\phi}/V$,
$$
\boxed{\Delta n=-6\epsilon+2\eta.}
$$
This is the first-order departure of the <scalar spectral index> from unity, in <reduced Planck units>.
Back to article page
