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I understand that the relation between the age $t_0$ of the universe and the cosmological constant $\Lambda $ is something like

$$c t_0 = \frac{f}{\sqrt{\Lambda}}$$

Can somebody provide the precise numerical factor $f$ for the Lambda CDM Model? This does not seem to be explained anywhere. It seems that the factor must be of the order of $f \approx 1.35$. What is the exact expression for this number $f$?

From the answers given below I get a new issue: there an latest Planck-satellite value for $\Lambda $ in $1/{\rm m}^2$? For strange reasons, SI units are rarely used in this particular case.

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  • $\begingroup$ I'm no expert in cosmology, but how does this make sense? Is $f$ a function of $t_0$? If so, pretty meaningless question. If not, absurd question. $\endgroup$ – Ryan Unger Aug 5 '17 at 19:43
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The exact formula for the time elapsed since the big bang is the integral

$$t=\int_0^a \frac{{\rm d}a'}{a' \ H(a')}$$

where $H(a)$ is composed of the radiation density $\Omega_R$, the matter (dark+regular) density $\Omega_M$, the curvature $\Omega_K$ and the dark energy density $\Omega_{\Lambda}$:

$$H(a)=H_0 \ \sqrt{\Omega_R/a^4+\Omega_M/a^3+\Omega_K/a^2+\Omega_{\Lambda}}$$

If you're looking for the age of the universe today set $a=1$, if you want to know how old the universe was when the scale factor was half of today set $a=1/2$ and if you're looking for the age of the universe when the scale factor will be twice that of today set $a=2$.

Those are the Friedmann equations which are also used in the ΛCDM-model. With $H_0=67150 \ {\rm m/Mpc/sec}$, $\Omega_R=10^{-4}$ (including neutrinos, radiation alone would be 5e-5), $\Omega_M=0.315$, $\Omega_K=0$, $\Omega_{\Lambda}=1-\Omega_R-\Omega_M-\Omega_K$ this gives

$$t=13.841 \ {\rm Gyr}$$

When $t=f/H_0$ the function $f$ becomes

$$f=\int_0^a \frac{{\rm d}a'}{\sqrt{\Omega_R/a'^2+\Omega_M/a'+\Omega_K+\Omega_{\Lambda} \ a'^2}}$$

Since there is no closed solution to this integral except when you set $\Omega_M+\Omega_{\Lambda}=1$ and all the other $\Omega$ to $0$ which reduces the equation for $a$ to

$$a=\sqrt[3]{\left(\sqrt{\frac{\Omega_M}{\Omega_{\Lambda} }} \sinh \left(\frac{t \left(3 H_0 \sqrt{\Omega_{\Lambda} }\right)}{2} \right)\right)^2}$$

and therefore $t$ to

$$t=\frac{2 \sinh ^{-1}\left(\sqrt{\frac{a^3 \Omega_{\Lambda} }{\Omega_M}}\right)}{3 H_0 \sqrt{\Omega_{\Lambda} }}$$

you should use numerical integration instead of symbolic one when solving for the exact solution containing all the $\Omega$s.

With $a=1/(z+1)$ (dependend on the redshift instead of the scalefactor) the equations are referenced here: ned.ipac.caltech.edu/level5/Hogg/Hogg10.html and here: lcdm.yukterez.net

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  • $\begingroup$ Note that since $H_0$ and $\sqrt\Lambda$ are so close numerically (but not the same) that f and F are numerically close (but not the same). The equation in your answer can be used to get either since the denominator below f or F are both represented in the equation (if one replaces the $\Omega s\$ with Lambda over Lambda critical), but still needs that algebraic manipulation and the rest if f. F is more straightforward, and more often used for the age of the universe. $\endgroup$ – Bob Bee Aug 6 '17 at 5:17
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Yes, it is true, you can write the equation for the age of the universe that way. . But it is not a much used equation because the f term is not so simple.

The reason you can write it that way is that both 1/$\sqrt(\Lambda)$ and 1/$H_0$ have the dimensions of time, about $10^{18}$ seconds

You can get the number for the current Hubble constant and get your numerical value of f, but not clear at all that it helps much in thinking through things. f in fact depends on the cosmological constant also.

The best equation for getting the age of the universe is

$t_0$ = F/$H_0$

Where the denominator is the current Hubble constant. F (different from your f) is .956, per the current concordance model of cosmological parameters. However, F is also a not so simple equation. But at least close to 1. The inverse of the Hubble constant is about 14.4 Gy, and the age of the universe about 13.8 Gy.

See the wiki article at https://en.m.wikipedia.org/wiki/Age_of_the_universe

See the equations and values for the cosmological parameters at https://en.m.wikipedia.org/wiki/Lambda-CDM_model, where they use numbers for the Planck collaboration(latest). It also has the equation for the Hubble parameter

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