I try to calculate the age of the universe with the FLRW model: $$ H(a) = H_0 \sqrt{\Omega_{\mathrm{R},0} \left(\frac{a_0}{a}\right)^4 + \Omega_{\mathrm{M},0} \left(\frac{a_0}{a}\right)^3 + (1-\Omega_{\mathrm{T},0}) \left(\frac{a_0}{a}\right)^2 + \Omega_{\Lambda,0}}. $$

I set $\Omega_{\mathrm{M},0} = 0.317$ (matter density) and $\Omega_{\Lambda,0} = 0.683$ (dark energy), as delivered by Planck 2013; $\Omega_{\mathrm{T},0} = 1.02$ (space curvature), according to this site; and $\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$ (radiation density), according to this document.

For the time $t(a)$ I take the scale factor $a$ and divide it through the integrated recessional velocity $$ t(a) = \frac{a}{\int_0^a{H(a')a'\ \mathrm{d}a'}/(a-0)} $$ and finally simplify to $$ t(a) = \frac{a^2}{\int_0^a{H(a')a'\ \mathrm{d}a'}}. $$

But the problem is, I then get about $8\times10^9$ years for the age of the universe, but it should be around $12\times10^9$ years (which I get when I set $\Omega_{\mathrm{R},0}$ to zero):

$\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$: plot for first value of radiation content

$\Omega_{\mathrm{R},0} = 0 \to 0.00001$: plots for changing value of radiation content

Do I have to use some other models than FLRW/ΛCDM, or is one of my parameters outdated?

  • 4
    $\begingroup$ Hi Симон Тыран, and welcome to Physics Stackexchange. A couple of us did some major editing to typeset your formulas - you should make sure we didn't inadvertently change the meaning of your question. For future questions, we encourage the use of Latex-style formatting in order to make questions easier to read, and to help them garner more upvotes ;) $\endgroup$
    – user10851
    Commented May 7, 2013 at 4:49

1 Answer 1


The total energy density is by definition $$ \Omega_{T,0} = \Omega_{R,0} + \Omega_{M,0} + \Omega_{\Lambda,0},$$ so with the values you cite ($\Omega_{R,0}=4.8\times 10^{-5}$, $\Omega_{M,0}=0.317$, $\Omega_{\Lambda,0}=0.683$), we get $\Omega_{T,0} = 1$, or in a more common notation $\Omega_{K,0}=1-\Omega_{T,0}=0$, i.e. a space with zero curvature.

It is also common to define the present-day value of the scale-factor as $a_0=1$, so that $$ H(a) = H_0\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}. $$

The age of the universe can then be derived as follows: from $$ \frac{\text{d}a}{\text{d}t} = \dot{a}, $$ we get $$ \begin{align} \text{d}t &= \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)}\\ &= \frac{1}{H_0}\frac{a\,\text{d}a}{a^2\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}}\\ &= \frac{1}{H_0}\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}. \end{align} $$ Integrating yields the difference between the time that a signal is emitted and the time it is observed: $$ t_{\text{ob}} - t_{\text{em}} = \frac{1}{H_0}\int_{a_{\text{em}}}^{a_{\text{ob}}} \frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}, $$ and the age of the universe is $$t_0 = \frac{1}{H_0}\int_0^1 \frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}.$$ This should give you the correct age.

  • $\begingroup$ @ Pulsar, Are you sure? Why are the power factors of the scalefactor reversed in your time integral, so that ΩR is constant, ΩM linear and ΩΛ with a^4 ? This function would look like 666kb.com/i/cdvhxl9dw1exeanm4.png $\endgroup$
    – Yukterez
    Commented May 7, 2013 at 18:35
  • $\begingroup$ Yes, notice that I've written an extra $a$ in the numerator, so that $dt = a\,da/(a^2\,H(a))$, and the $a^2$ in the denominator can be written under the sqrt. It's also interesting to note that the peak in the integrand corresponds with the time at which the expansion of the universe started to accelerate. $\endgroup$
    – Pulsar
    Commented May 7, 2013 at 20:12
  • $\begingroup$ But how comes that ΩΛ gets from constant to a^4 and ΩR from a^-4 to constant, while ΩM goes from a^-3 to a and ΩM from a^-2 to a^2? Is the pattern a^(n+4) or why is this? I have plotted your suggestion to the red plot; did I get you right? Plot: 666kb.com/i/cdx9yebs8odlknm6k.png PS: It seems that my initial question (why I get 8 instead of 12 Gyr with my selfmade scalefactor through integrated recessional velocity integral) has it's answear in the numerical precion; when I avoid the singularity at a=0, I get the expected 12 Gyr (blue Plot) $\endgroup$
    – Yukterez
    Commented May 9, 2013 at 12:50
  • $\begingroup$ I've added two extra steps in my answer, I hope things are clear now. It really is the correct formula, you'll find it in every cosmology course (for instance, on page 14 of casa.colorado.edu/~ajsh/phys5770_08/frw.pdf). I don't know which value of $H_0$ you used, but with $H_0=67.3\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$, you should get an age of 13.79 billion years. $\endgroup$
    – Pulsar
    Commented May 9, 2013 at 16:09
  • $\begingroup$ Can someone explain the significance of that fact that $\ddot{a}$ doesn't appear in this answer? Is there an implicit assumption somewhere? $\endgroup$ Commented Apr 12, 2019 at 23:28

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