# Recombination and the expansion of the universe

Usually when one reads about the recombination in the standard model ($$\Lambda$$-$$CDM$$) its written that the recombination occurs at a temperature $$T\approx 3000 K$$. Since, at this temperature the free electrons of the plasma become bound with the ionized hydrogen.

Let's call the Hubble parameter of the standard model $$H(z)$$. If for some unknown reason it so happens that the real Hubble parameter is $$H_0(z)$$ such that $$H_0(z)>H(z)$$, (that is the Universe expands faster than expected), then the recombination still happens in $$T\approx 3000 K$$? Or at a different temperature?

If $$H_0(z)>>H(z)$$, then I think that the Universe may never have recombination era, I'm not completely sure.

• I know that you can make some measurements of $H(z)$. However at the moment there is no complete agreement on the exact form of $H(z)$, because of the "Hubble tension". – Nothing Jun 3 '20 at 6:13

This requires photons above $$E=10.2 eV$$. If we say that the photon to proton ratio is $$\eta$$, then recombination happens (roughly) when $$n_\gamma \exp(-E/kT) < n_p\ ,$$ where the exponential term is approximately the fraction of photons with energy $$>E$$.
The recombination temperature is therefore $$T < \frac{E}{k\ln \eta} = 5700 \left(\ln \frac{\eta}{10^9}\right)^{-1}\ {\rm K}.$$
Since $$\eta$$ has a fixed value (of a bit more than $$10^9$$), independent of the expansion history, then I don't see that you can get a markedly different recombination temperature.
There are ways of avoiding recombination; by arbitrarily messing about with $$\Lambda$$ you can have a universe that was never very small. But then you have to explain where the cosmic microwave background comes from...