We know that for the flat geometry of lamda CDM model, the solution for Friedmann equation is

$$ a(t) = \left\{ \frac{Ω_{m,0}}{Ω_{Λ,0}} \sinh^2 \left[\frac{3}{2} \sqrt{Ω_{Λ,0}} H_0(t - t_0)\right] \right\}^{1/3}, $$

where the scale factor $a$ is a function of cosmic time. If I want to express the scale factor here as a function of conformal time, what will be the equation, i.e., how can I write the solution of the Friedmann equation in terms of conformal time?

  • $\begingroup$ @Pulsar Can you please give me an explanation here? $\endgroup$
    – Photon
    Mar 18, 2019 at 1:52
  • $\begingroup$ $t_0-t$ should be $t-t_0$ or just $t$. $\endgroup$
    – G. Smith
    Mar 18, 2019 at 4:53
  • 1
    $\begingroup$ Do you know the formula for conformal time? When I do the integral with Mathematica I get a messy hypergeometric function, so I doubt that there is a nice expression for the scale factor in terms of the conformal time. Do you have some reason to think that there should be? $\endgroup$
    – G. Smith
    Mar 18, 2019 at 4:56
  • $\begingroup$ Thanks for your comment and extremely sorry for my mistake. It will be $$ t - t_0 $$. I have corrected the term in my question also. Thanks again. @G. Smith $\endgroup$
    – Photon
    Mar 18, 2019 at 5:14
  • $\begingroup$ Do you realize that $t_0$ is not the time at the current epoch, but rather the time at which $a=0$ (i.e., the time of the Big Bang)? This time is usually taken to be zero. $\endgroup$
    – G. Smith
    Mar 18, 2019 at 5:19

1 Answer 1


The integral for the conformal time,


gives a complicated function of $t$ involving a hypergeometric function in this case. Although this relationship could be numerically inverted to plot $t(\eta)$ and thus $a(\eta)$, I don’t believe that there are any analytic formulas.


Using the numerical values $\Omega_{m,0}=0.3089$, $\Omega_{\Lambda,0}=0.6911$, and $H_0=67.74\,\text{km}\cdot\text{s}^{-1}\cdot\text{Mpc}^{-1}=0.06923\,\text{Gy}^{-1}$ from Wikipedia, the scale factor is


where $t$ is in gigayears. Numerically integrating $1/a$ gives the conformal time $\eta$. For example, one finds the following values for the first 15 billion years

$$\begin{matrix} t & \eta & a\\ 0.0 & 0.0 & 0.0\\ 1.0 & 20.08 & 0.1495 \\ 2.0 & 25.30 & 0.2379 \\ 3.0 & 28.94 & 0.3130 \\ 4.0 & 31.83 & 0.3813 \\ 5.0 & 34.25 & 0.4458 \\ 6.0 & 36.35 & 0.5079 \\ 7.0 & 38.21 & 0.5689 \\ 8.0 & 39.88 & 0.6294 \\ 9.0 & 41.39 & 0.6901 \\ 10.0 & 42.78 & 0.7516 \\ 11.0 & 44.06 & 0.8142 \\ 12.0 & 45.24 & 0.8784 \\ 13.0 & 46.34 & 0.9446 \\ 14.0 & 47.36 & 1.013 \\ 15.0 & 48.32 & 1.084 \\ \end{matrix} $$

Thus, for example, when $\eta$ was 42.78 billion years and $t$ was 10.0 billion years, the scale factor was 0.7516.

At the current epoch, the scale factor $a$ is 1 by definition. Numerically solving for the $t$ that makes $a=1$, one finds $t=13.81$ billion years, the age of the universe. Evaluating the conformal time gives $\eta=47.17$ billion years; thus the size of the observable universe is currently 47.17 billion light years.

If one goes far into the future, one finds that the integral for the conformal time approaches 63.73. Thus in this model the size of the observable universe never grows larger than 63.73 billion light years.

Here is a plot of $a(t)$:

enter image description here

And here is a plot of $a(\eta)$:

enter image description here

It is not actually necessary to numerically invert $\eta(t)$ to get $t(\eta)$ and $a(\eta)$. You can easily plot $a(\eta)$ by treating $t$ as a parameter and plotting the parametric curve $(\eta(t),a(t))$.


If you have Mathematica evaluate the integral for the conformal time, you get an ugly expression, involving a hypergeometric function, which appears complex but is in fact real. By making a change of variables, you can get a nicer expression, still involving a hypergeometric function, which is real. To do this, first write the scale factor as


where $$A=\left(\frac{\Omega_{m,0}}{\Omega_{\Lambda,0}}\right)^{1/3}$$


$$\tau=\frac{2}{3 H_0\sqrt{\Omega_{\Lambda,0}}}.$$



using the substitutions $x=t/\tau$ and $y=\sinh{x}$. Mathematica can do this latter integral and get a real expression. The result is


where $_2F_1(a,b;c;z)$ is the ordinary hypergeometric function.

At very early times, the conformal time is approximately


and at large times, the conformal time approaches the constant value


Unfortunately, I know of no way to invert $\eta(t)$ to get an analytic expression for $t(\eta)$ and thus for $a(\eta)$. But numerically evaluating this expression might be faster than the numerical integration.

And it, along with $a(t)$, gives an analytic parameterization of the curve $a(\eta)$, with parameter $t$.


The formula


can be inverted to give


This can then be used in the formula for $\eta(t)$ to get a non-parametric formula for $\eta(a)$ which doesn't involve $t$:


It can't be inverted, as far as I know. So there is no nice formula for $a(\eta)$, but there is a nice formula for its inverse, $\eta(a)$!

  • $\begingroup$ Thanks for your answer. I also think so that there is no direct analytic formula for this. Can you please explain a little bit, how can I numerically invert this function to get a(η) here and then what will be the expression of a(η) then? @G. Smith $\endgroup$
    – Photon
    Mar 18, 2019 at 5:46
  • $\begingroup$ There is no expression! There are just numerical values. $\endgroup$
    – G. Smith
    Mar 18, 2019 at 18:06
  • $\begingroup$ Correction: There is no analytic expression for $a(\eta)$, but there is an analytic parameterization of the curve $a(\eta)$ with $t$ as the parameter. $\endgroup$
    – G. Smith
    Mar 18, 2019 at 22:35
  • $\begingroup$ Thanks a lot for your nice explanation. Now, it's clear to me how to get the value of scale factor $a(\eta)$ numerically. Can you please explain a little bit more how can I get the value of $\dot a(\eta)$ or $a'(\eta)$ numerically from your process. Expecting your kind feedback. @G. Smith $\endgroup$
    – Photon
    Mar 19, 2019 at 9:02
  • $\begingroup$ en.wikipedia.org/wiki/Numerical_differentiation BTW, comments are not for asking new questions that were not part of the original question. $\endgroup$
    – G. Smith
    Mar 19, 2019 at 15:55

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