I've been looking at the Planck 2013 cosmological parameters paper, trying to update my toy cosmology simulator with the most recent data. Most of the interesting values such as $H_0$, $\Omega_m$, and $\Omega_\Lambda$ can be found in Table 2 on page 12, but the one thing I didn't find was an estimate of the energy density of radiation. Can this be derived from some other parameters in these data?

  • $\begingroup$ It'd be useful if you gave us the link for your simulator. I'd love to enjoy it ;-) $\endgroup$ Commented Jan 19, 2014 at 3:20
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    $\begingroup$ @Waffle'sCrazyPeanut I wrote a blog post about it awhile back, and the code (in Python) is linked from there. It's pretty simple. $\endgroup$ Commented Jan 19, 2014 at 4:11

2 Answers 2


The radiation density has two components: the present-day photon density $\rho_\gamma$ and the neutrino density $\rho_\nu$. The photon density as a function of frequency can be derived directly from the CMB: the photon number density follows the Planck law $$ n(\nu)\,\text{d}\nu = \frac{8\pi\nu^2}{c^3}\frac{\text{d}\nu}{e^{h\nu/k_B T_0}-1}, $$ with $k_B$ the Stefan-Boltzmann constant, and $T_0$ the current CMB temperature. The photon energy density is then $$ \rho_\gamma\, c^2 = \int_0^{\infty}h\nu\,n(\nu)\,\text{d}\nu = a_B\, T_0^4, $$ where $$ a_B = \frac{8\pi^5 k_B^4}{15h^3c^3} = 7.56577\times 10^{-16}\;\text{J}\,\text{m}^{-3}\,\text{K}^{-4} $$ is the radiation energy constant. With $T_0=2.7255\,\text{K}$, we get $$ \rho_\gamma = \frac{a_B\, T_0^4}{c^2} = 4.64511\times 10^{-31}\;\text{kg}\,\text{m}^{-3}. $$ The neutrino density is related to the photon density: in Eq. (1) on page 5 in the paper, you see that $$ \rho_\nu = 3.046\frac{7}{8}\left(\frac{4}{11}\right)^{4/3}\rho_\gamma. $$ This relation can be derived from physics in the early universe, when neutrinos and photons were in thermal equilibrium. So $$ \rho_\nu = 3.21334\times 10^{-31}\;\text{kg}\,\text{m}^{-3}, $$ and the total present-day radiation density is $$ \rho_{R,0} = \rho_\gamma + \rho_\nu = 7.85846\times 10^{-31}\;\text{kg}\,\text{m}^{-3}. $$ We can also express this relative to the present-day critical density $$ \rho_{c,0} = \frac{3H_0^{2}}{8\pi G} = 1.87847\,h^{2}\times 10^{-26}\;\text{kg}\,\text{m}^{-3}, $$ where the Hubble constant is expressed in terms of the dimensionless parameter $h$, as $$ H_0 = 100\,h\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}, $$ so we get $$ \begin{align} \Omega_{\gamma}\,h^2 &= \dfrac{\rho_\gamma}{\rho_{c,0}}h^2 = 2.47282\times 10^{-5},\\ \Omega_{\nu}\,h^2 &= \dfrac{\rho_\nu}{\rho_{c,0}}h^2 = 1.71061\times 10^{-5},\\ \Omega_{R,0}\,h^2 &= \Omega_{\gamma}\,h^2 + \Omega_{\nu}\,h^2 = 4.18343\times 10^{-5}. \end{align} $$ For a Hubble value $h=0.673$, one finds $\Omega_{R,0} = 9.23640\times 10^{-5}$.

I should point out that the formulae for the primordial neutrinos is only valid when they are relativistic, which was true in the early universe. Since neutrinos have a tiny mass, they are probably no longer relativistic in the present-day universe, and behave now like matter instead of radiation. Therefore, neutrinos only contributed to the radiation density in the early universe, while the present-day radiation density only consists of photons.

  • $\begingroup$ Wow! It's truly amazing that a negligible quantity has such a tremendous influence :) $\endgroup$ Commented Jan 19, 2014 at 3:22
  • $\begingroup$ @Waffle'sCrazyPeanut It's only negligible today...in the early universe it was the dominant factor, since radiation density scales so strongly (fourth power of the scale factor). $\endgroup$ Commented Jan 19, 2014 at 18:33
  • $\begingroup$ i think the numerical values for $\Omega_{\gamma}h^2$ and $\Omega_{\nu}h^2$ were switched by mistake. $\endgroup$
    – Valerio
    Commented Jun 7, 2016 at 16:44

An alternative approach to Pulsar's great answer is to use the two Planck measurements for the current matter density $\Omega_m$ and the redshift of radiation-matter equality $z_{eq}$. We can use that $$ \rho_m(z_{eq}) = \rho_R(z_{eq}) \,, $$ which is equivalent to $$ \Omega_{m,0} (z_{eq}+1)^3 = \Omega_{R,0} (z_{eq}+1)^4 \,, $$ or $$ \Omega_{R,0} = \Omega_{m,0} / (z_{eq}+1) \,. $$ Using the values from Table 2 of the 2013 paper (there's a 2015 version by now, as you probably know), from the last column for consistency with Pulsar (right?), we end up with $$ \Omega_{R,0} = 0.315 / (3391+1) = 9.28656 \times 10^{-5} \,. $$


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