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Past exam of the mathematics course of the University of Cambridge

2025

Paper 4

The <potential slow-roll parameter> is
$$
\epsilon=\frac{M_{\rm Pl}^2}{2}\left(\frac{V'}V\right)^2.
$$
From the <slow-roll approximation>,
$$
\dot\phi=-\frac{V'}{3H},
\qquad
H^2=\frac{V}{3M_{\rm Pl}^2}.
$$
Therefore
$$
\frac{\dot\phi^2}{2V}
=\frac{V'^2}{18H^2V}
=\frac{M_{\rm Pl}^2V'^2}{6V^2}
=\frac\epsilon3.
$$
Thus $\epsilon\ll1$ implies $\dot\phi^2\ll V$, consistently with potential-energy domination.

The same equations give
$$
\frac{H}{\dot\phi}
=-\frac{3H^2}{V'}
=-\frac1{M_{\rm Pl}^2}\frac V{V'}.
$$
On the branch $V'/V>0$, the definition of $\epsilon$ turns this into
$$
\boxed{
\frac{H}{\dot\phi}
=-\frac1{M_{\rm Pl}^2}\frac V{V'}
=-\frac1{\sqrt2M_{\rm Pl}}\frac1{\sqrt\epsilon}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 9E / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 9E / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 9E / a / Solution

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