4.4 “End of Inflation” Mechanism 83 Therefore, from Eq. (4.151) we require To calculate φ$_{1}$ and φ$_{2}$ , the value of φ at times t$_{1}$ and t$_{2}$ respectively, we use the fact that the process is so rapid that it takes place in less than a Hubble time, so that the Universe expansion can be ignored. Then the equation of motion is At the end of thermal inflation, φ is not centered on the origin, but has started to roll down the potential slightly. At this time, g $^{2}$T 2 is much smaller than m 2 $_{0}$. Therefore we have So we have the equation of motion The solution is ζ ≫ H$_{TI}$ Δ t (4.153) ¨ φ + ∂V ∂φ = 0 (4.154) ∂V ∂φ ∼ − m 2 $_{0}$φ (4.155) ¨ φ ∼ m 2 $_{0}$φ (4.156) φ ∼ Ae m$_{0}$t (4.157)