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Horizon-exit logarithmic derivative in slow roll (dlogkd​=1−ϵH​1​Hϕ˙​​dϕd​≃−VV′​dϕd​)

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2026-10-14  NaN By others on same topic  0 Discussions Create my own version
Along cosmological horizon exit k∝aH, differentiation gives dlogk/dloga=1−ϵH​, where ϵH​=−H˙/H2. The chain rule gives the first expression. In reduced Planck units, the slow-roll equations 3Hϕ˙​≃−V′ and 3H2≃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.

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  • Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 62 / 4 / b / Solution
  • Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 62 / 4 / c / Solution

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