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I'm asked to verify an expression for the ionization fraction of helium in a universe in which helium dominates baryonic matter.

I'm given that the ionization fraction $X = \frac{n_{He^+}}{n_{He^+}+n_{He}}$ with a given ionization energy $Q_{He}$, and number densities by the formula $n_i=g_i(\frac{m_ik_BT}{2\pi \hbar})^{3/2}\exp(-\frac{m_ic^2}{k_BT})$. Also the baryon-photon ratio in this case is $\eta=\frac{4(n_{He^+}+n_{He})}{n_\gamma}$.

I'm trying to show that by putting all this together I get something with the form:

$\ln(\frac{1-X}{X^2})=A+\ln(\eta)-B\,\ln(\frac{Q_{He}}{k_BT})+\frac{Q_{He}}{k_BT}$. where $A$ and $B$ are constants.

When I plug everything in, I got $\frac{1-X}{X^2}=\frac{n_\gamma}{8}\eta \, \exp(-\frac{Q_{He}}{K_BT})$, which gives:

$\ln(\frac{1-X}{X^2})=\ln(\frac{n_\gamma}{8})+\ln(\eta)+\frac{Q_{He}}{k_BT}$

Any ideas of where I might have gone wrong?

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    $\begingroup$ There are some folks here who could probably answer that - however, it seems to me more like a Q for the Physics stack. $\endgroup$ Mar 23, 2017 at 22:48
  • $\begingroup$ How did you put $Q$ in. You need to give the details of what you have done and why you expect to get the solution you think you should. $\endgroup$
    – ProfRob
    May 2, 2023 at 16:13

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I did a bit of the math involved but did not substitute terms with $\mathrm{n}$ and $\mathrm{\eta}$.

$$\mathrm{\chi = \frac{\eta_{He^+}}{\eta_{He^+}+\eta_{He}}}$$ $$\mathrm{1- \chi = \frac{\eta_{He^+}+\eta_{He}-\eta_{He^+}}{\eta_{He^+}+\eta_{He}} = \frac{\eta_{He}}{\eta_{He^+}+\eta_{He}}}$$ $$\mathrm{\frac{1-\chi}{\chi^2} = \frac{\eta_{He}}{\eta_{He^+}+\eta_{He}}.({\frac{\eta_{He^+}+\eta_{He}}{\eta_{He^+}})^2} = \frac{\eta_{He}(\eta_{He^+}+\eta_{He})}{(\eta_{He^+})^2}} = \frac{\eta_{He}}{\eta_{He^+}} + (\frac{\eta_{He}}{\eta_{He^+}})^2$$

From the result, it looks to me like a Binomial Expansion (truncated Taylor Expansion). It shows the highest order for your terms (which is okay since higher order terms can usually be regarded as negligible) with the same +/- pattern you expect in your answer.

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  • $\begingroup$ Thank you so much, I will work on this and let you know how it works out. I always forget about Taylor Expansion... $\endgroup$
    – Spuds
    Mar 25, 2017 at 15:28

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