Physics Asked by Kris Walker on January 13, 2021
Inflation is thought to be an extraordinarily brief period during the very early universe when the scale factor increased a remarkable $e^{60}$-fold, however this also comes with a corresponding increase in the expansion rate $dot{a}=mathrm{d}a/mathrm{d}t$. When inflation ends, the exponential acceleration ceases and we transition to the radiation-dominated era where the expansion rate now evolves as $dot{a}_text{rad}(t)propto t^{-1/2}$.
But what happened to the expansion rate, $dot{a}_text{end}$, at the end of inflation? Did it become an initial condition of the FLRW universe? Surely it’s much too large to have decreased under $ddot{a}_text{rad}(t)$ in any reasonable amount of time, before the universe become too diluted? Did it undergo a large deceleration, $ddot{a}(t)ll 0$, a sort of "anti-inflation" right at the end of the inflation period? If so, what is the reason for this deceleration?
To make this more exact, consider inflation to begin and end at $t=0$ and $t=T$ respectively, and assume that inflation expands the universe such that $a(T)/a(0)=e^{60}$. We can then describe the inflation as
begin{align}
a(t<T)=a(0)exp(ht)=a(T)expleft(60left(frac{t}{T}-1right)right)
end{align}
where $h$ is the (constant) Hubble parameter during inflation. For $t>T$ we transition to the radiation-dominated universe and so to maintain continuity of the scale factor we must have
begin{align}
a(t>T)=a(T)sqrt{t/T}
end{align}
So that’s all well and good. Now let’s consider $dot{a}$. Taking the time derivatives of each of the above gives
begin{align}
dot{a}(t<T)=frac{60}{T}a(T)expleft(60left(frac{t}{T}-1right)right)=frac{60}{T}a(t<T)
end{align}
and
begin{align}
dot{a}(t>T)=frac{a(T)}{2T}frac{1}{sqrt{t/T}}
end{align}
Now we find that the continuity is broken at $t=T$, with the former giving $dot{a}(T)=60 a(T)/T$ and the latter giving $dot{a}(T)=a(T)/2T$, which is not surprising since we are modelling this as an instantaneous transition between inflation and the radiation-doimated universe. So assuming this is valid we see that the expansion rate has decreased by a factor of 120, which is almost completely negligible when compared to the $e^{60}$-fold increase during inflation. So this can’t be the decrease we desire.
What gives? Am I missing something crucial? How did we go from this ludicrously fast expansion to what we have today?
Additional question: assuming my math is even valid, is there a reason for the 120-fold decrease in the expansion rate? If we go by the standard "slow roll" explanation for a scalar field $phi$ then inflation occurs when $V(phi)ggdot{phi}^2$ and stops when the potential decreases enough so that $V(phi)approxdotphi^2$, is this what causes the large negative acceleration required? If so is it able to be shown explicitly?
I think an example should help illustrate what's going on. First, there is no discontinuity when inflation ends: the rate of expansion goes smoothly from accelerating to decelerating, much like a ball rolling down a hill. To start, consider the Friedmann equation, $$frac{ddot{a}}{a} = -frac{4pi G}{3}(rho + 3p),$$ where $p = wrho$ relates the pressure and density of a scalar field. In general, $rho = dot{phi}^2/2 + V(phi)$, and $p = dot{phi}^2/2 - V(phi)$, giving $$frac{ddot{a}}{a} = -frac{8pi G}{3}left(dot{phi}^2-V(phi)right).$$
Even for simple potentials, there are in general no analytical expressions for $dot{phi}$ as a function of $phi$, and so we can't do much better than this expression. But, this should make clear that as $dot{phi}$ grows (and $V(phi)$ drops), $ddot{a}$ also drops. Eventually, as $dot{phi}^2 > V$, the field has picked up enough speed that inflation ends; after this, the acceleration is negative.
Answered by bapowell on January 13, 2021
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