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

By definition,

N = lo g \frac{a ( t _{f} )}{a ( t _{i} )} = \int_{t_{i}}^{t_{f}} H d t .

Using

d t = d ϕ / \dot{ϕ}

and the result of part (a),

N = \int_{ϕ_{i}}^{ϕ_{f}} \frac{H}{ϕ ˙} d ϕ ≃ - \frac{1}{2 M _{Pl}} \int_{ϕ_{i}}^{ϕ_{f}} \frac{d ϕ}{ϵ ( ϕ )} .

Reversing the limits yields the slow-roll e-fold count

N ≃ \frac{1}{2 M _{Pl}} \int_{ϕ_{f}}^{ϕ_{i}} \frac{d ϕ}{ϵ ( ϕ )} .

Solved by gpt-5.6-sol high.

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