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The parameters of $\rm\Lambda CDM$ model have been determined to an amazing high precision from the measurements made by the Planck mission. In particular, the Hubble "constant" (the value of Hubble parameter at the present day) has been determined to be $H_0 = (67.3 \pm 1.2 )\, \text{km} \, \text{s}^{-1} \text{Mpc}^{-1}$. They have also given values for the density parameters at the present day (the density divided by the critical density) coming from cosmological constant, $\Omega_{\Lambda,0} = 0.6817$, and from matter, $\Omega_{m,0} = 0.3183$. The age of the Universe $t_0$ can be calculated from other cosmological parameters in this way

$$t_0 = \frac{1}{H_0}\int_0^1 \frac{a\,\text{d}a} { \sqrt{\Omega_{\Lambda,0}a^{4} + \Omega_{k,0}a^{2} + \Omega_{m,0}a + \Omega_{r,0}} },$$

where $\Omega_k = 1 - \Omega$ is the space curvature parameter, $\Omega = \Omega_{m,0} + \Omega_{\Lambda,0} + \Omega_{r,0}$ is the total energy density parameter (the energy density divided by the critical density) and $\Omega_r$ is the energy density parameter coming from radiation.

I have read that the age of the Universe has been established from the Planck mission measurements as $t_0 = 13.82 \times 10^9$ years. My question is: how is this value been calculated? I mean, it has been calculated assuming a flat space geometry, that is, assuming that $\Omega_k = 1 - \Omega = 0$? If not, what values for $\Omega_{r,0}$ and $\Omega_{k,0}$ have been used to perform the calculation?

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    $\begingroup$ Have you looked at any of their 31 papers? $\endgroup$
    – Kyle Kanos
    Jul 18, 2014 at 18:43
  • $\begingroup$ @KyleKanos: Well... 31 papers... I had a look at one of them, devoted to cosmological parameters, but I have not seen anything about the values of $\Omega_{r,0}$ and $\Omega_{k,0}$. $\endgroup$
    – Charo
    Jul 18, 2014 at 18:49
  • $\begingroup$ Paper XVI appears to be the one you want. There is a section called "Curvature" (6.2.3) and Section 2.1.1 says that $\Omega_r$ is being parametrized by equation (1). $\endgroup$
    – Kyle Kanos
    Jul 18, 2014 at 18:56
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    $\begingroup$ Worth pointing out that, despite the press coverage, nothing Planck did was new - it had all been done a decade earlier, using the exact same technique. $\endgroup$
    – user10851
    Jul 18, 2014 at 19:18
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    $\begingroup$ @ChrisWhite: Very true! But they did it with higher resolution. $\endgroup$
    – Kyle Kanos
    Jul 18, 2014 at 19:40

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As far as I have understood from this paper, they have given some observational limits to the value of $\Omega_{k,0}$, but this article concludes asserting that "there is no evidence from Planck for any departure from a spatially flat geometry". Taking $\Omega_{k,0}=0$ and the value for $\Omega_{r,0}$ given at this post, one can compute the above integral obtaining $t_0 = 0.947797 \, H_0^{-1}$, which, taking $H_0 = 67.3 \, \text{km} \, \text{s}^{-1} \text{Mpc}^{-1}$, gives $t_0 = 13.78 \times 10^9$ years.

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