# What is the present day photon density, $\rho_{\gamma, 0}$?

I'm attempting to perform the integration that will yield the sound horizon at recombination: $$c_s^2 = \frac{c^2}{3}\left[\frac{3}{4}\frac{\rho_{b,0}(1+z)^3}{\rho_{\gamma,0}(1+z)^4} + 1\right]^{-1}$$The present day density of baryons is reasonably easy to calculate as $$\rho_b=\Omega_b\times \rho_{crit}$$. Using the Planck Collaboration results, I get a present day value of $$\rho_b=4.14\times 10^{-25}\space g\space m^{-3}$$.

I've seen one post calculate the photon density as $$\rho_\gamma = \frac{a_B\, T_0^4}{c^2} = 4.64511\times 10^{-31}\;\text{kg}\,\text{m}^{-3}.$$ where $$a_B = \frac{8\pi^5 k_B^4}{15h^3c^3} = 7.56577\times 10^{-16}\;\text{J}\,\text{m}^{-3}\,\text{K}^{-4}$$ But then the author goes on to calculate the radiation density and suggests plugging the total radiation density value into the formula for sound velocity which doesn't seem right (as the electrons are bound to photons, not all radiation).

What is the right value for $$\rho_{\gamma,0}$$ for the purpose of calculating the velocity of sound in the pre-recombination fluid?

• If you are just asking the present value of the photon density then that expression seems correct. I checked it from a book. – Reign Jan 11 at 20:02
• 1. Yes, I'm asking for the present day photon density. 2. I'm asking if this is the right value to use in the speed of sound calculation because the post I referenced suggested using the total radiation density. 3. I'm looking for an actual value to double check against the one I'm using. – Quarkly Jan 11 at 20:05
• A reference for the photon "mass" density formula above is en.wikipedia.org/wiki/Photon_gas. Look at the second equation there. $U/V$ is the photon energy density. Divide by $c^2$ and you have the density of relativistic mass. This is just a standard blackbody radiation formula. – G. Smith Jan 11 at 20:06
• @G.Smith - I don't see either a formula or a value for the photon density on the page you referenced. What am I missing? – Quarkly Jan 11 at 20:09
• I edited my comment to explain. – G. Smith Jan 11 at 20:10

For radiaton value we can take the CMBR as a source,

$$\epsilon_{CMBR,0}=\alpha\,T^4_0$$,

Where $$T=2.725K$$ and $$\alpha=7.56\,10^{-16}Jm^{-3}K^{-4}$$ From here we get, $$\epsilon_{CMBR,0}=4.17 \,10^{-14} Jm^{-3}$$

and we know that $$\rho_{CMBR}=\rho_{\gamma,0}=\epsilon_{CMBR,0}/c^2$$

$$\rho_{\gamma,0}=\frac {4.17 \,10^{-14}Jm^{-3}} {9\,10^{16}m^2s^{-2}}=4.633\,10^{-31}kg\,m^{-3}$$

If you want to include to current neutrino density then density parameter can be taken as, $$\Omega_{r,0}=\Omega_{CMBR,0}+\Omega_{v,0}=5\,10^{-5}+3.4\,10^{-5}=8.4\,10^{-5}$$

and using $$\Omega_{r,0}=\frac {\epsilon_{r,0}} {\epsilon_{c}}$$ and $$\epsilon_{c}$$ can be taken as $$\epsilon_{c}=8.331\,10^{-10}Jm^{-3}$$.

After this you'll find that,

$$\epsilon_{r,0}=6.998\,10^{-14}Jm^{-3}$$ and $$\rho_{r,0}=7.775\,10^{-31}kgm^{-3}$$

For source you should look Barbara Ryden, Introduction to Cosmology Page 66

• This looks good and I'll up-vote it if you can answer the second part of the question: do I want the photon density or the radiation density when calculating the speed of sound. From everything I read, it depends on the photon density, but I could be interpreting it incorrectly. – Quarkly Jan 11 at 20:40
• I dont know that part. But hope this part helps – Reign Jan 11 at 20:42
• From where did you get $\epsilon_{CMBR,0}=4.17 \,10^{-14} Jm^{-3}$? – Quarkly Jan 11 at 21:05
• From the book but, it makes sense. The energy density is proportional to $T^4$ ? – Reign Jan 11 at 21:12
• I put the value of $\alpha$ – Reign Jan 11 at 21:14