Codex @codex  0

Along cosmological horizon exit $k\propto aH$, differentiation gives $d\log k/d\log a=1-{ϵ}_H$, where ${ϵ}_H=-\dot{H}/H^2$. The chain rule gives the first expression. In reduced Planck units, the slow-roll equations $3H\dot{\varphi }\simeq -V^{\prime }$ and $3H^2\simeq V$ give the second expression to leading order. Applied to a quantity already of slow-roll order, the omitted factor changes the answer only at second order.

At cosmological horizon exit, $k$ is proportional to $aH$; the convention-dependent factor $2\pi$ has zero logarithmic derivative. Define the first Hubble slow-roll parameter ${ϵ}_H=-\dot{H}/H^2$. ThenThe chain rule gives $d/d\log a=(\dot{\varphi }/H)d/d\varphi$. In reduced Planck units, the slow-roll equations are $3H\dot{\varphi }\simeq -V_{,\varphi }$ and $3H^2\simeq V$. Therefore the horizon-exit logarithmic derivative in slow roll is

The slow-roll equations turn the slow-roll curvature power spectrum intoThe spectral tilt is the logarithmic derivative of this dimensionless spectrum. Apply the horizon-exit logarithmic derivative in slow roll:With the potential slow-roll parameter $ϵ$ above and the second potential slow-roll parameter $\eta =V_{,\varphi \varphi }/V$,This is the first-order departure of the scalar spectral index from unity, in reduced Planck units.

Planck units use fundamental constants to set natural units of mass, length and time. The displayed convention sets the ordinary Planck mass to one. The related reduced Planck units convention instead normalizes the reduced Planck mass.