Somehow, I have never come across an explanation of why cosmologists claim that the alleged inflation of the very early universe occurred not at the Big Bang, but very shortly afterwards (~10^-36 to 10^-32 seconds afterwards)....

Why? What difference does 10^-36 seconds make?

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    $\begingroup$ Was there even an "exact start"? $\endgroup$
    – Mithoron
    Commented Apr 20, 2023 at 18:58
  • $\begingroup$ You cannot inflate a point. You need some structures before you can inflate it. The teeny tiny time at the beginning is required to have some structures in place. $\endgroup$ Commented Apr 21, 2023 at 13:54

1 Answer 1


The universe cannot have begun in an inflationary phase. Note that it is not necessarily the case that there was a phase that preceded inflation. However, if there was no phase preceding inflation, then the universe had no beginning.

A common way to see this is to note that the cosmic expansion factor grows exponentially during inflation, $$a\propto \mathrm{e}^{Ht},$$ where $H=\sqrt{8\pi G\rho/3}$ is the Hubble rate, which is constant during inflation. Such an expansion history can be extrapolated indefinitely into the past, with $a$ becoming arbitrarily small but never zero.

That's perhaps not the most physical perspective, since $a$ isn't a well defined quantity in all contexts. Another perspective is that the energy density $\rho$ (which is frame invariant for a fluid that can drive inflation) is constant in time. We can even write the metric during inflation in a way that is manifestly static, $$\mathrm{d}s^2=-\left(1-\frac{r^2}{H^2}\right)\mathrm{d}t^2+\left(1-\frac{r^2}{H^2}\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$$ (see de Sitter space). Again, $H$ is constant in time. In this sense, an inflating universe is a steady-state universe.

Where does $10^{-36}$ seconds come from? If inflation was preceded by a non-inflationary phase, then the duration of that phase is of order $H^{-1}$, where $H$ is the Hubble rate during inflation. If the inflation energy scale is $\sim 10^{15}~\text{GeV}$, then its energy density is $\rho\sim (10^{15}~\text{GeV})^4$, which leads to $H^{-1}\sim 10^{-12}~\text{GeV}^{-1}\sim10^{-36}~\text{s}$.

(I assume $c=\hbar=1$.)

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    $\begingroup$ Great answer. Except of course for your choice of units which imply $c-\hbar=0$ :) $\endgroup$
    – pela
    Commented Apr 20, 2023 at 9:57
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    $\begingroup$ $c$ and $\hbar$ are dimensionally incompatible, so adding or subtracting them isn't meaningful. Thus, there is no reason why $c - \hbar$ should--or shouldn't--have any particular value, including 0. $\endgroup$
    – Nobody
    Commented Apr 20, 2023 at 13:34
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    $\begingroup$ @pela $c=\hbar=1$ is a standard thing to do in particle physics. Apparently the people who work in that field are comfortable with it. $\endgroup$ Commented Apr 20, 2023 at 15:16
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    $\begingroup$ @pela For what it's worth, the SI version is$$\rho\sim (10^{15}~\text{GeV})^4c^{-5}\hbar^{-3},\,H^{-1}\sim 10^{-12}~\text{GeV}^{-1}\hbar.$$ $\endgroup$
    – J.G.
    Commented Apr 20, 2023 at 15:34
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    $\begingroup$ @pela Sure, you could do that, but why would you? All it does is force you to carry around a bunch of unnecessary constants, which obscures the relationships between physical quantities in your equations. Plus, if you're solving numerically, carrying the constants can easily double the number of flops in each grid cell, which translates to a real, tangible increase in a project's computing budget. Units are arbitrary; you can and should pick them judiciously to simplify the problem you are trying to solve. $\endgroup$
    – Nobody
    Commented Apr 20, 2023 at 20:02

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