Slow-roll approximation for potential $V=\frac{1}{2}m^{2}\varphi^{2}$

Question

I'm attempting to derive a solution to the slow-roll approximation for a scalar potential of the form $V=\frac{1}{2}m^{2}\varphi^{2}$.

For the solution for $a(t)$ I will start by taking the slow-roll approximation of the first equation in the question and multiply it by $H$, giving: \begin{align*} 3H^{2}\dot{\varphi}+V{'}H=0 \end{align*} I can then use the fact that $H^{2}\approx{V/3}$ in this regime, and using the fact that $H=da/adt$ giving: \begin{align*} V\dot{\varphi}+V{'}\frac{da}{adt}=0 \end{align*} Now, re-arranging this ready for integrating, and using $V{'}=m^{2}\varphi$ gives: \begin{align*} V{'}\frac{da}{adt}=-V\dot{\varphi}=-\frac{1}{2}m^{2}\varphi^{2}\dot{\varphi}\,\,\,\Rightarrow\,\,\,\frac{da}{a}=-\frac{1}{2}\varphi\dot{\varphi}dt=-\frac{1}{2}\int{\varphi}\,d\varphi \end{align*} This essentially will give us the following (after integrating): \begin{align*} \ln{a}+C_{1}=-\frac{1}{4}\varphi^{2}+C_{2} \end{align*}

My question is this. I'm almost there, and frustratingly I do not know how to continue to this:

So, how do I now get to the above solution from the current step I'm stuck on?

Answer 1

You have $$\ln a + C_1 = -\frac{1}{4}\phi^2 + C_2$$ which gives you $$ \ln a(t) = (C_2 - C_1) - \frac{1}{4} \phi(t)^2 $$ in general. At some initial time $t_i$, let $a(t) = a_i$ and $\phi(t_i) = \phi_i$. This means that at time $t_i$, $$\ln a_i = (C_2 - C_1) - \frac{1}{4}\phi_i^2$$ i.e. $$C_2 - C_1 = \ln a_i + \frac{1}{4}\phi_i^2.$$ Plug this into $$ \ln a(t) = (C_2 - C_1) - \frac{1}{4} \phi(t)^2 $$ to get $$ \ln a(t) = \ln a_i + \frac{1}{4}\phi_i^2 - \frac{1}{4} \phi(t)^2 $$ which gives you $$a(t) = a_i\exp(\frac{1}{4}(\phi_i^2 - \phi(t)^2))$$ which is the expression you want. You just need to solve for the constants.