Inflation ends when the potential becomes too steep.
Is it because kinetic energy increase when the potential becomes steeper?
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2 Answers
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The inflaton's equation of motion is $\square\phi=-V'(\phi)$ in a $+---$ metric (if we work in the $\xi=0$ gauge). It often helps to compare such field-theoretic PDEs to their counterparts in classical mechanics, viz. $\ddot{x}=-V'(x)$ for motion in a gravitational field. Just as a ball rolling on a hill can only get to so high a potential, and in particular will only move so far horizontally if the hill is steep up to that cutoff height, inflation also has a natural cutoff. Note that these arguments neglect quantum tunnelling, but that doesn't make much difference with high probability.
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$\begingroup$ While it rolls down the potential, does kinetic energy increase faster if the potential becomes steeper? $\endgroup$– parkerCommented Sep 23, 2018 at 14:53
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$\begingroup$ Well, the kinetic energy increase is the potential energy decrease, so the two will be equally steep. $\endgroup$– J.G.Commented Sep 23, 2018 at 15:26
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A homogeneous and isotropic fluid will give rise to inflation if its equation of state satisfies $w = p/\rho < -1/3$, which follows from the equation, $$\frac{\ddot{a}}{a}= -\frac{4\pi G}{3}(\rho + 3p).$$ For a uniform scalar field, we have $\rho = \dot{\phi}^2/2 + V$ and $p = \dot{\phi}^2/2 - V$, and so the condition on $w$ in terms of the kinetic and potential energies is $$2(\dot{\phi}^2/2) < V.$$ Thus, as the field rolls down the potential and gains kinetic energy, eventually this condition will fail and the scalar field will no longer support inflation.
