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.

  • $\begingroup$ 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". $\endgroup$
    – Nothing
    Commented Jun 3, 2020 at 6:13

1 Answer 1


Here is an argument. The universe will remain ionised until there is fewer than 1 photon per proton that is capable of boosting a hydrogen atom into an excited state.

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 $$T < \frac{E}{k\ln \eta} = 5700 \left(\ln \frac{\eta}{10^9}\right)^{-1}\ {\rm K}.$$

The temperature of recombination therefore depends only weakly on the photon to proton (or equivalently the photon to baryon) ratio.

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...


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