How to solve equation of motion of cosmological scalar field

Solution 1:

You are missing one key equation to close the system.

A scalar field carries energy and its this energy that determines how fast the Universe expands. For a canonical scalar field the kinetic energy is $\frac{1}{2}\partial_\mu \phi\partial^\mu\phi$ which is just $\frac{1}{2}\dot{\phi}^2$ in a smooth Friedmann universe. The potential energy is $V(\phi)$ so the Friedmann equation (in Planck units: $8\pi G = 1$) is $$H^2 = \left(\frac{d\log R}{dt}\right)^2 = \frac{1}{3} \rho_\phi = \frac{1}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$ when the scalar field is the only source of energy. Thus the full system reads $$H^2 = \frac{1}{3} \rho_\phi = \frac{1}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$ $$\ddot{\phi} + 3H\dot{\phi} + V'(\phi) = 0$$ Only for very special classes of potentials do we have analytical solutions. A general method of studying such systems is to write them as a coupled set of first order equations - a so-called dynamical system.

If we introduce $x_1 = \frac{\dot{\phi}}{\sqrt{6}H}$ and $x_2 = \frac{\sqrt{V}}{\sqrt{3}H}$ then the Hubble equation says that $x_1^2 + x_2^2 = 1$. We further find with $x = \log R$ as our time-coordinate: $$\frac{d x_1}{dx} = -3x_1 + \frac{\sqrt{6}}{2}\lambda_\phi x_2^2 - x_1 \frac{d\log H}{dx}$$ $$\frac{d x_2}{dx} = -\frac{\sqrt{6}}{2}\lambda_\phi x_1x_2 - x_2 \frac{d\log H}{dx}$$ $$\frac{d\lambda_\phi}{dx} = -\sqrt{6}\lambda_\phi^2(\Gamma-1)x_1$$ where $$\frac{d\log H}{dx} = -\frac{3}{2}(1+x_1^2 - x_2^2)$$ $$\lambda_\phi = -\frac{V'}{V},\,\,\,\Gamma = \frac{V''V}{V'^2}$$ Now for an exponential potential $\lambda_\phi = \lambda$ stays constant and $\Gamma = 1$ and the system simplifies greatly down to $$\frac{d x_1}{dx} = -3x_2^2[x_1 - \frac{\lambda}{\sqrt{6}}]$$ $$\frac{d x_2}{dx} = 3x_1x_2[x_1 - \frac{\lambda}{\sqrt{6}}]$$ or simply $$\frac{d x_1}{dx} = -3(1-x_1^2)[x_1 - \frac{\lambda}{\sqrt{6}}]$$ This is possible to integrate though its hard to invert the resulting function, but we see right away that if $x_1 \equiv \frac{\lambda}{\sqrt{6}}$ then we have one solution. This is a so-called fixpoint of this system (put derivatives to zero and solve the resulting algebraic system). Finding and classifying such fixpoints (i.e. are they attractive or repellent) is super useful (much more than trying to find analytical solutions that might not exist) for understanding the dynamics of the system and the given analytical solution is such a fixpoint. We can also see that the fixpoint is attractive so even if we did not start with this initial value of $x_1$ we would be driven towards it and the system would settle close to it.

If you now unwrap this solution you will find the given solution. For example $\frac{d\log H}{dx} = -3x_1^2 = -\frac{\lambda^2}{2}$ gives us $H \propto R^{-\frac{\lambda^2}{2}}$ and $H = \frac{d\log R}{dt}$ gives us $R(t)$ and from the definition of $x_1$ we get $\dot{\phi} \propto H \implies \phi \propto \log R$.