# Density of radiation at CMB

I was trying to find the radiation density at the CMB time.

For two different methods, the first one is using the Planck Law and the data for the actual measured temperature $$T_{0}=2.7255° K$$.

$$\rho_{r}=\rho_{\gamma}+\rho_{\nu}$$

$$\rho_{\gamma}=\frac{a_{b}}{c^2}T_{0}^{4}$$ $$\rho_{\nu}=3.046\frac{7}{8}(\frac{4}{11})^{4/3}\rho_{\gamma}$$ $$\frac{a_{b}}{c^2}=8.41805*10^{-33}$$

The second way to find it is using the $$\Lambda CDM$$ model, we know that the equation for density of radiation in this model is given by $$\rho_{r}=c(1+z)^4$$ The constant C is determined to be the actual value for radiation density i.e $$c=7.85846*10^{-31} kg\, m^{−3}$$

Know with this two formulas we can calculate the radiation density at CMB in two different ways:- 1. Using the value of the themperature at CMB $$T_{CMB}=3000°K$$. 2. Using the redshift at CMB $$z_{CMB}\approx 1100$$

For the first method I obtain $$\rho_{r}^{CMB}=1.15355*10^{-18}kg\, m^{−3}$$

For the second method $$\rho_{r}^{CMB}=8.51682*10^{-26}kg\, m^{−3}$$

Clearly I am doing something wrong, maybe Planck's Law is not a good aproximation at CMB time?

• Explain method 1 in more detail. – Rob Jeffries Feb 17 at 6:57
• If I multiply your c = 7.85846e-31 with 1101$^4$ I get 1.155e-18, not 8.51682e-26. That seems pretty consistent. – pela Feb 17 at 13:19