<p>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.</p>
<p>A common way to see this is to note that the cosmic expansion factor grows exponentially during inflation,
<span class="math-container">$$a\propto \mathrm{e}^{Ht},$$</span>
where <span class="math-container">$H=\sqrt{8\pi G\rho/3}$</span> is the Hubble rate, which is constant during inflation. Such an expansion history can be extrapolated indefinitely into the past, with <span class="math-container">$a$</span> becoming arbitrarily small but never zero.</p>
<p>That's perhaps not the most physical perspective, since <span class="math-container">$a$</span> isn't a well defined quantity in all contexts. Another perspective is that the energy density <span class="math-container">$\rho$</span> (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,
<span class="math-container">$$\mathrm{d}s^2=-\left(1-\frac{r^2}{H^2}\right)\mathrm{d}t^2+\left(1-\frac{r^2}{H^2}\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$$</span>
(see <a href="https://en.wikipedia.org/wiki/De_Sitter_space#Static_coordinates" rel="noreferrer">de Sitter space</a>). Again, <span class="math-container">$H$</span> is constant in time. In this sense, an inflating universe is a steady-state universe.</p>
<p>Where does <span class="math-container">$10^{-36}$</span> seconds come from? If inflation was preceded by a non-inflationary phase, then the duration of that phase is of order <span class="math-container">$H^{-1}$</span>, where <span class="math-container">$H$</span> is the Hubble rate during inflation. If the inflation energy scale is <span class="math-container">$\sim 10^{15}~\text{GeV}$</span>, then its energy density is <span class="math-container">$\rho\sim (10^{15}~\text{GeV})^4$</span>, which leads to <span class="math-container">$H^{-1}\sim 10^{-12}~\text{GeV}^{-1}\sim10^{-36}~\text{s}$</span>.</p>
<p>(I assume <span class="math-container">$c=\hbar=1$</span>.)</p>

CC BY-SA 4.0
2023-04-20T09:57:58.307Z
Great answer. Except of course for your choice of units which imply $c-\hbar=0$ :)
pela
CC BY-SA 4.0
2023-04-20T13:34:35.403Z
$c$ 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.
Nobody
CC BY-SA 4.0
2023-04-20T13:43:15.623Z
@Nobody If $c$ and $\hbar$ are dimensionally incompatible, would it perhaps make sense to, and hear me out now, indicate their dimensions with, I don't know, some sort of _units_? Just an idea…
pela
CC BY-SA 4.0
2023-04-20T15:16:24.823Z
@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.
WaterMolecule
CC BY-SA 4.0
2023-04-20T15:34:13.557Z
@pela For what it's worth, the SI version is$$\rho\sim (10^{15}~\text{GeV})^4c^{-5}\hbar^{-3},\,H^{-1}\sim 10^{-12}~\text{GeV}^{-1}\hbar.$$
J.G.
CC BY-SA 4.0
2023-04-20T15:43:52.517Z
It is a case of what came first, the big bang egg or the energy inflationary chicken. you can have an egg that wasn't laid by a chicken, because the very first chicken egg was laid by a creature that wasn't a chicken. Some mutation on the egg's DNA caused the egg laid by the non-chicken to become the first chicken egg. Therefore, the unstable energy that wasn't our universe yet gave way into said big bang egg. If we consider baryonic matter to be the egg's yolk, then it leads to the hypothesis that dark matter is the albumen, which is transparent unless properly heated.
Mindwin Remember Monica
CC BY-SA 4.0
2023-04-20T20:02:47.167Z
@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.
Nobody
CC BY-SA 4.0
2023-04-21T07:51:45.190Z
@Nobody I can come up with at least one good reason: To make us poor astrophysicists less confused. We _know_ you cosmologists just do it to feel superior. Do not deny.
pela
CC BY-SA 4.0
2023-04-21T13:50:54.040Z
@pela orange = 1, apple = 1, so orange - apple = 0. The unit is the no. of fruits by the way.
stackoverblown
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2023-04-21T14:11:43.073Z
@stackoverblown I think you just discovered a [fundamental law](https://xkcd.com/687/)! :-o
pela
CC BY-SA 4.0
2023-04-24T10:17:00.420Z
@pela Well, if it makes you feel better, you can just say you're measuring distance in light-years and time in years...
Nobody
CC BY-SA 4.0
2023-04-24T11:38:36.750Z
@Nobody As an astronomer I obviously prefer cm and s. Or Mpc and Myr. Or redshift and Hubble times. Or redshift and redshift. Or…
pela
CC BY-SA 4.0
2023-04-20T09:53:20.907Z
2023-04-20T09:53:20.907Z
Sten
<p>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)....</p>
<p>Why?  What difference does 10^-36 seconds make?</p>

CC BY-SA 4.0
2023-04-20T18:58:13.813Z
Was there even an "exact start"?
Mithoron
CC BY-SA 4.0
2023-04-21T13:54:34.913Z
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.
stackoverblown
<astrophysics><cosmology><big-bang-theory><cosmological-inflation><theories>
Why do cosmologists assume that inflation began shortly after the Big Bang, rather than at the exact start?
Kurt Hikes
# Why do cosmologists assume that inflation began shortly after the Big Bang, rather than at the exact start?

(Asked by: username_0 on 2023-04-20)

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?

**Tags:** astrophysics, cosmology, big-bang-theory, cosmological-inflation, theories
> **Comments:**
> **username_1**: Was there even an "exact start"?  
> **username_2**: 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.  

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## 1 answers

### Answer by username_3 on 2023-04-20. Score: 16

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](https://en.wikipedia.org/wiki/De_Sitter_space#Static_coordinates)). 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 $c=\hbar=1$.)

> **Comments:**
> **username_4**: Great answer. Except of course for your choice of units which imply $c-\hbar=0$ :)  
> **username_5**: $c$ 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.  
> **username_4**: @Nobody If $c$ and $\hbar$ are dimensionally incompatible, would it perhaps make sense to, and hear me out now, indicate their dimensions with, I don't know, some sort of _units_? Just an idea…  
> **username_6**: @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.  
> **username_7**: @pela For what it's worth, the SI version is$$\rho\sim (10^{15}~\text{GeV})^4c^{-5}\hbar^{-3},\,H^{-1}\sim 10^{-12}~\text{GeV}^{-1}\hbar.$$  
> **username_8**: It is a case of what came first, the big bang egg or the energy inflationary chicken. you can have an egg that wasn't laid by a chicken, because the very first chicken egg was laid by a creature that wasn't a chicken. Some mutation on the egg's DNA caused the egg laid by the non-chicken to become the first chicken egg. Therefore, the unstable energy that wasn't our universe yet gave way into said big bang egg. If we consider baryonic matter to be the egg's yolk, then it leads to the hypothesis that dark matter is the albumen, which is transparent unless properly heated.  
> **username_5**: @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.  
> **username_4**: @Nobody I can come up with at least one good reason: To make us poor astrophysicists less confused. We _know_ you cosmologists just do it to feel superior. Do not deny.  
> **username_2**: @pela orange = 1, apple = 1, so orange - apple = 0. The unit is the no. of fruits by the way.  
> **username_4**: @stackoverblown I think you just discovered a [fundamental law](https://xkcd.com/687/)! :-o  
> **username_5**: @pela Well, if it makes you feel better, you can just say you're measuring distance in light-years and time in years...  
> **username_4**: @Nobody As an astronomer I obviously prefer cm and s. Or Mpc and Myr. Or redshift and Hubble times. Or redshift and redshift. Or…  

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