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Slow-roll e-fold count

For $V(\phi)=\lambda\phi^n/n$ under the stated Cambridge Tripos normalization,
$$
N_e=\frac{4\pi G}{c^4n}
(\phi_i^2-\phi_f^2).
$$
If slow roll ends by kinetic-potential equality, then
$$
\phi_f^2\simeq\frac{n^2c^4}{48\pi G},
\qquad
N_e\simeq\frac n{12}
\left[\left(\frac{\phi_i}{\phi_f}\right)^2-1\right].
$$


Monomial slow-roll e-fold count

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 15B / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 9D / b / Solution

Monomial slow-roll e-fold count

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Monomial slow-roll e-fold count

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