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From Friedmann equation for flat universe:

$$ \left(\frac{\dot{a}}{a}\right)^2= \frac{8\pi G}{3} ~~ \left( \rho_m + \rho_r + \rho_\Lambda \right), $$

can we simply get the scale factor $a$ as a function of time?

I’m following this thread

numerically solving for cosmological scale factor..

But there’s the Hubble constant $H_0$, whose value is given in units of GeV by $\sim 10^{-42}$. I think this is the value we should substitute by in the Friedmann equation to have units balance, since in natural units GeV equivalent for $s^{-1}$, and $[\frac{\dot{a}}{a}]= s^{-1}$.

But this value is so small!! Will we get the right scale factor so, even in eV it’s $H_0\sim 10^{-33}$ , this means we will get $a(t)$ multipled by this small factor!

Also in the first answer of the thread they say:

You should use $H_0$ in terms of its normal units, $[H] = kms^{-1}Mpc^{-1}$. Or simply $H_0 \approx 70 km/s/Mpc$.

Any help to understand that is appreciated!

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    $\begingroup$ For single component universes the functions can be found at yukterez.net/f/einstein.equations/files/FLRW.functions.html For two components with radiation and matter here: yukterez.net/f/einstein.equations/files/d Matter plus dark energy is in also in the first link. For more components at once you need to integrate numerically $\endgroup$
    – Yukterez
    Commented Jun 3, 2022 at 9:46
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    $\begingroup$ It doesn't mean anything to say a dimensionful quantity like $H_0$ is small (or large). You can choose units so that $H_0$ has any (positive) value you would like. The only thing that matters are dimensionless combinations. In cosmology, this would be the dimensionless density ratios, $\Omega\equiv \rho/\rho_c$, where $\rho_c=3H_0^2/(8\pi G)$ is the critical density. If you use natural units, then $\rho$ will also be very small (the density is very small in Planck units), so $\rho/\rho_c$ will be of order 1. $\endgroup$
    – Andrew
    Commented Jun 3, 2022 at 10:02
  • $\begingroup$ @Yukterez, Hello, many thanks for the links. It seems you have plotted as well the density parameter versus time. May you give the link of the Mathematica code of plotting $\Omega(t)$ ? Please see this thread: physics.stackexchange.com/questions/711920/… . Which unit you are using for $H_0$ ? $\endgroup$
    – Dr. phy
    Commented Jun 3, 2022 at 12:59
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    $\begingroup$ The units I use in the calculations containing H are SI units, which is around H0=2.18e-18/s (in the 67.15 km/s/Mpc the km/Mpc is a number since length/length cancels the unit, so the remaining unit is that of a frequency, of one e-fold specifically). The time on the x-axis is in years though, but the conversion factor from seconds to years is easy. The code you are looking for is linked on the bottom of the site $\endgroup$
    – Yukterez
    Commented Jun 3, 2022 at 18:42
  • $\begingroup$ @Yukterez. Thanks so much. $\endgroup$
    – Dr. phy
    Commented Jun 4, 2022 at 3:49

1 Answer 1

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Dr. phy asked: "can we simply get the scale factor $\rm a$ as a function of time?"

For the complete solution containing $\rm \Omega_{r}, \ \Omega_{m}, \ \Omega_{\Lambda}$ simultaneously you have to integrate numerically, but since at all stages of the universe only one or two components dominate while at least one component can be neglected you can use the analytical 1- and/or 2-component solutions for the specific times and glue them together where their values approach each other.

The full solution where you set your neglible components to $0$ and solve for $\rm a $ is

$$\rm \frac{\dot{a}}{a}=H=H_0 \sqrt{\Omega_{r}/a^4+\Omega_{m}/a^3+\Omega_{k}/a^2+\Omega_{\Lambda }}$$

For the purely radiation dominated era right after the big bang you get ↘yellow

$$ \rm a=\sqrt{2 \ H_{0} t} \ \Omega_{r}^{1/4} $$

When matter becomes relevant too we have two solutions, the first is ↘orange

$$ \rm a=(-i \left(\sqrt{3}-i\right) \sqrt[3]{\sqrt{3} \ \Omega_{m} \left| f-4 \ \Omega_{r}^{3/2}\right| \sqrt{H_{0} \ t \left(f-8 \ \Omega_{r}^{3/2}\right)}+f^2-8 \ f \ \Omega_{r}^{3/2}+8 \ \Omega_{r}^3}$$

$$ \rm +(4 \ i \left(\sqrt{3}+i\right) \Omega_{r}^2)/(\sqrt[3]{\sqrt{3} \ \Omega_{m} \left| f-4 \ \Omega_{r}^{3/2}\right| \sqrt{H_{0} \ t \left(f-8 \ \Omega_{r}^{3/2}\right)}+f^2-8 \ f \ \Omega_{r}^{3/2}+8 \ \Omega_{r}^3})$$

$$ \rm +2 \left(1+i \ \sqrt{3}\right) \Omega_{r}+2 \left(1-i \ \sqrt{3}\right) \Omega_{r})/(4 \ \Omega_{m})$$

The 2nd solution is valid for the 2nd part of the radiation and matter dominated era ↘red

$$ \rm a=(\sqrt[3]{\sqrt{3} \ \Omega_{m} \left| f-4 \ \Omega_{r}^{3/2}\right| \sqrt{H_{0} \ t \left(f-8 \ \Omega_{r}^{3/2}\right)}+f^2-8 \ f \ \Omega_{r}^{3/2}+8 \ \Omega_{r}^3}$$

$$ \rm +\frac{4 \ \Omega_{r}^2}{\sqrt[3]{\sqrt{3} \ \Omega_{m} \left| f-4 \ \Omega_{r}^{3/2}\right| \sqrt{H_{0} \ t \left(f-8 \ \Omega_{r}^{3/2}\right)}+f^2-8 \ f \ \Omega_{r}^{3/2}+8 \ \Omega_{r}^3}}+2 \ \Omega_{r})/(2 \ \Omega_{m})$$

When radiation becomes neglible and matter alone dominates we have ↘violet

$$ \rm a=\left(\frac{3}{2}\right)^{2/3} \ (H_{0} \ t)^{2/3} \ \Omega_{m}^{1/3} $$

When dark energy and matter play together it becomes ↘cyan

$$ \rm a=\sqrt[3]{\frac{1}{\Omega_{\Lambda} }-1} \ \sinh ^{2/3}\left(3 \ H_{0} \ t \ \sqrt{\Omega_{\Lambda} }/{2} \right)$$

The function $\rm f$ which was used for brevity is

$$ \rm f = 3 \ H_{0} \ t \ \Omega_{m}^2$$

Since the radiation dominated era is so short, on a linear plot the last equation for $\rm a$ is good enough if you don't zoom in on the first few million years.

On a logarithmic plot however it makes a difference if you also show the early times. The $\rm x$-axis is $\rm t$ in years, $\rm y$ is the dimensionless scale factor $\rm a$:

scale factor by time

The two component solution is not exact and with my method it deviates about $\rm 0.1 \% $ from the numerical solution (so on a diagram you won't see any difference larger than a pixel on a monitor). Right after the big bang the numerical solution is more exact though, the oscillations in the deviation come from the numerical integration, see the deviation

$$\rm |1-\frac{a_{analytic}}{a_{numeric}}|$$

deviation

The numerical precision depends on your method and computional ressources, but at some point when $\rm t$ is small enough the analytical solution will be better, since the radiation really dominates at very small $\rm t$.

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