# Where is radiation density in the Planck 2013 results?

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?

• It'd be useful if you gave us the link for your simulator. I'd love to enjoy it ;-) – Waffle's Crazy Peanut Jan 19 '14 at 3:20
• @Waffle'sCrazyPeanut I wrote a blog post about it awhile back, and the code (in Python) is linked from there. It's pretty simple. – Nathan Reed Jan 19 '14 at 4:11

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 think the numerical values for $\Omega_{\gamma}h^2$ and $\Omega_{\nu}h^2$ were switched by mistake. – Valerio Jun 7 '16 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 $$\Omega_{R,0} = 0.315 / (3391+1) = 9.28656 \times 10^{-5} \,.$$