Solving the inflation's equation of motion and initial value conditions

Question

In cosmological inflation, the equation of motion of inflaton is $$\frac{d^2\phi}{dt^2} + 3H\frac{d\phi}{dt} + V,_\phi = 0$$

where $ϕ$ is the inflaton, $H$ is the Hubble parameter, and $V,_ϕ$ is the derivative of the potential with respect to $ϕ$. To solve this differential equations, we need two initial value conditions, $\phi(0) = ?\,$ and $\dot \phi(0) = ?\,$. I'm using Mathematica to solve for $\phi$, and in order to solve the above DE we need to eliminate the unknown function $H$ and from the Friedmann equations, $$H^2 = \frac{1}{3M_p^2}\Big(\frac{1}{2}\dot \phi^2 + V\Big)$$

For a potential given by $V = \frac{1}{2}m^2\phi^2$, we have $$\ddot \phi^2 + 3\sqrt{\frac{1}{6M_p^2}\Big(\dot \phi^2 + \phi^2\Big)} \dot \phi + m^2\phi = 0$$

I can set $\phi(0) \approx M_p$ and $\dot \phi(0) \approx 0.1$ (any small value just to get started). But in order to get an exact numerical value for $\phi(t)$ after solving this DE numerically, I need to set some $t$. But I'm not sure about what $t$'s to set during the inflationary stage, or in other words, I want to know the proper $t$ to use during the evolution of $\phi$?

Answer 1

The initial conditions for the DE above are not precisely known, but luckily this does not make much difference. Let me explain.

If you neglect the presence of the potential barrier, the DE above represents the slow-roll phase from when the Universe was stuck in the false vacuum until when it approaches re-heating (you are missing the phenomenological term that describes particle creation, hence the DE loses validity close to the moment of reheating). So long as the Universe is stuck in the false vacuum, we know that $\phi\approx 0$ and $\dot\phi \approx 0$. When $\phi$ has crossed the potential barrier, the above equations are still valid, because we know that the potential barrier is thin (hence $\phi$ need not change much) and crossing it is a slow matter (hence again $\dot \phi \approx 0$).

But the important point is that the equation above describes motion in a medium with friction, where we know that, independent of initial position and velocities, all trajectories reach an asymptotic speed given by

$$3H \frac{d\phi}{d t} = - V_{,\phi}$$

This is identical to the problem of a brick and a streamlined object falling in the atmosphere: it does not matter where and how fast they started, they will both reach a limiting speed after an initial transient. So the initial details only determine how long the initial transient lasts, and this is not the important part.

In fact, initially, the expansion of the Universe is still dominated by conventional matter, and $H$ can be assumed to be independent of $\phi$. But, after some time, as the density of conventional matter decreases as $R^{-3}$, the vacuum energy density $\propto V(\phi)$ remains about constant because we have designed the potential to have a very flat maximum around $\phi = 0$, which is exactly where our Universe is. Hence, at some point, it happens that the rate of the expansion of the Universe $H$ becomes a function of $\phi$:

$$H^2 \approx \frac{8\pi}{3m^2} V(\phi)$$

From this moment onwards, the initial conditions do not matter any more: the two differential equations above further determine the evolution of the cosmological scale factor and of the field $\phi$. This is where inflation does its magic, giving rise to a prolonged exponential expansion which solves a bevvy of otherwise inexplicable conundrums.

In summary, the exact initial conditions determine only the moment in which eq.2 begins to hold, nothing more. Since the potential is not truly computed from first principles (unfortunately), and since there is quite a large time span inside which the exponential expansion can occur without conflicting with our understanding of the later phases of cosmology, the true initial conditions and our ignorance thereof do not pose a major problem in the development of the theory of inflation.

If all you want is to take your equation out for a numerical spin, you may as well take $\dot\phi = 0$ and $\phi \neq 0$ (because for $\phi = 0$ $V_{,\phi}$ also vanishes), but typically $\phi \ll \sigma$, where $\sigma $ is the reheating value, for instance $\phi/\sigma = 10^{-N}$. For various values of $N=6,7,8,9...$, you will see that the qualitative picture above does not change. In particular you will notice that the total exponentiation number (that is, the logarithm of the ration of the Universe scale factor R at the end and at the beginning of inflation) does not depend on the initial conditions you picked, but (nearly) only on the shape of the potential.

Answer 2

The initial conditions are not well known. Inflation does though assume a slow roll so that $\dot\phi/T\ll\ddot\phi$, for $T$ the time of the slow roll. This implies a slow evolution of $\dot\phi$, which is usually implied in the DE with a small $3H\dot\phi$, for $H~\simeq~1/T$. The initial condition on the field for a slow roll then assumes a small value for the time rate of change.

There are no fundamental reason for initial conditions. This is one argument for the multiverse. All possible initial conditions can exist and spacetimes evolve according to random selection of initial conditions. This is all a work in progress. So for solving this DE one really has to solve it for a range of possible initial conditions and select that which matches the observed universe.