I've been asked about photon temperature at nucleosynthesis (3 minutes from Big Bang). So I guessed this is the moment when Matter and Radiation where in equilibrium:

$$ \rho_M(T) = \rho_R(T)$$

Taking $\rho_{R,0} = 7.8 \cdot 10^{-34} g/cm^3$ and $\Omega_R = \frac{\rho_{R,0}}{\rho_{C,0}} = 4.5 \cdot 10^{-5} h^{-2}$:

$$ 1 = \frac{\rho_{R}}{\rho_{M}} = \frac{\rho_{R,0}}{\rho_{M,0}} \frac{a_0}{a} = \frac{\rho_{R,0}}{\rho_{M,0}} \frac{T}{T_0}$$


$$T = 0.85 eV$$

Is it right?


The time during which matter and photons were in equilibrium was not a "moment" but a somewhat rather long period of time. During the time of nucleosynthesis, as the universe expanded, both matter and photons remained in equilibrium. When the temperature cooled so that nucleosynthesis ended, the temperature cooled faster because the conversion of matter to energy stopped the maintaining of a higher temperature. There is one detail that became out of equilibrium. During nucleosynthesis some small amount of matter was in the form of neutrinos. After nucleosynthesis the neutrinos no longer interacted with photons or with other forms of matter. Therefore, that small amount of neutrino matter remained out of equilibrium with other matter and also with photons.

  • $\begingroup$ Well thank you Buzz, but a bit unrelated. I'm asking about photons temperature. $\endgroup$ May 30 at 7:43
  • $\begingroup$ I apologize for my careless reading of the your question. $\endgroup$
    – Buzz
    May 31 at 13:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.