Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 4 / b

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- Solution b

Solution

For a monotonic inflaton trajectory, the potential slow-roll parameter satisfies

\frac{d ϕ}{d N} = M_{Pl} 2 ϵ_{V} (ϕ) .

The remaining slow-roll e-fold count is the integral from the present field value to its end value along the direction of motion:

N (ϕ) = \int_{ϕ}^{ϕ_{e}} \frac{d ϕ}{M _{Pl} 2 ϵ _{V} ( ϕ )} .

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