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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?

Charo
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    Have you looked at any of their 31 papers? – Kyle Kanos Jul 18 '14 at 18:43
  • @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 $\Omega_{k,0}$. – Charo Jul 18 '14 at 18:49
  • 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). – Kyle Kanos Jul 18 '14 at 18:56
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    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. –  Jul 18 '14 at 19:18
  • @KyleKanos: You are right. The section that you mention gives some limits to the possible values of $\Omega_{k,0}$, but it ends with the conclusion: "there is no evidence from Planck for any departure from a spatially flat geometry". I suppose that this means that the calculation of $t_0$ is done taking $\Omega_k = 0$, but I am not sure about that and this is the reason of my question. – Charo Jul 18 '14 at 19:38
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    @ChrisWhite: Very true! But they did it with higher resolution. – Kyle Kanos Jul 18 '14 at 19:40
  • Of course and for me this is quite amazing! – Charo Jul 18 '14 at 19:43
  • @Charo: Table 1 of XVI says that their baseline value of $\Omega_K$ is zero, so I presume that they calculated $t_0$ with $\Omega_K$ neglected. – Kyle Kanos Jul 18 '14 at 19:46
  • Interestingly enough, I thought along the same lines when I saw the calculation. The value for $\Omega_r$ is measured and not easily questionable, and we all agree that $\Omega_k+\Omega=1$. So I decided to adjust $\Omega_m$ and $\Omega_\Lambda$ to allow $\Omega_k$ to be non-zero. When you do the calculation for the age of the universe using values that obey the Friedmann equations, what you find is that the value for age of the universe is fairly consistent. It ranges from about 13.4Gyr to I think 15.#Gyr. – Jim Jul 18 '14 at 19:58
  • So regardless whether they assumed a flat universe or not, you would still see the calculation result in a number that is very close to $H_0^{-1}$ – Jim Jul 18 '14 at 20:00

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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.

Urb
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Charo
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