On page 66 of Modern Cosmology by Dodelson, we have the formula for the change in neutron to proton ratio as a function of time:$$\frac{dXn}{dt}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$Everything in the derivation up to this point is pretty clear. Then there's an explanation that there are complicated time dependencies in the formula, which appears to be true because Mathematica is unable to numerically solve this formula, so we attempt to simplify the evolution parameter by:$$x=\frac{Q}{T}$$Now this is the part where I get lost. The author states that the left hand side becomes $\dot x \frac{dX_n}{dx}$ The equation then becomes:$$\frac{dx}{dt} \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-x}-X_n\right\}$$ So now all we need is an expression for $\frac{dx}{dt}$. Expanding the steps explicitly, we have: $$x\equiv \frac{Q}{T}$$ $$\frac{dx}{dt}=-\frac{Q\space\dot T}{T^2},Q=x\space T$$ $$\frac{dx}{dt}=-\frac{x\space\dot T}{T},T=\frac{T_0}{a}$$ $$\frac{dx}{dt}=\frac{x\space\dot a}{a},H=\frac{\dot a}{a}$$ $$\frac{dx}{dt}=x\space H$$Substituting, we get: $$x\space H \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$$$\frac{dXn}{dx}=\frac{\lambda_{np}}{x\space H}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$So how does the author get this formula for (3.27)?$$\frac{dXn}{dx}=\frac{x \lambda_{np}}{H(x=1)}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$How did the Hubble Function $H$ get moved into the denominator and why is it evaluated at $x=1$ instead of at $x$?
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$\begingroup$ Note: In a related question, $Q$ appears to be the neutron-proton mass difference and $T$ the temperature. $\endgroup$– rob ♦Commented Mar 28, 2021 at 2:01
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In the radiation era, $H^2 \sim a^{-4}$ because of the Friedmann equation. Meanwhile the temperature scales with the scale factor as $T \sim a^{-1}$, so $H \sim T^{2}$ during the radiation era. Since $x \sim T^{-1}$, we have $H \sim x^{-2}$. Note that in the definition $x=Q/T$, $Q$ is the neutron-proton mass difference and is temperature independent.
Putting this scaling relation in the form of an equation, we find \begin{equation} H(x) = \frac{H(x=1)}{x^2} \end{equation} which is what you need to connect your formula with Dodelson's Eq. (3.27).
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$\begingroup$ I don't follow this at all. In all times, $$H[t]\equiv\frac{\dot a}{a}$$I understand this to be the definition of the Hubble function. I understand how the value of H can change with the matter-energy content, but I don't understand how the definition can change. $\endgroup$ Commented Mar 28, 2021 at 2:53
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$\begingroup$ @GluonSoup What you wrote is indeed the definition of the Hubble parameter, but it's not the only relationship that it satisfies. In particular, $H$ also obeys the Friedman equation, $H^2=\frac{8\pi G}{3}\rho$. In the radiation-dominated, $\rho \propto a^{-4}$. Therefore $H\propto a^{-2}$. One way to interpret what I am saying is that we are using the Friedman equation to tell us how $\dot{a}$ depends on $a$, and plugging this into your definition to see how $H$ depends on $a$. Does this make sense? $\endgroup$– AndrewCommented Mar 28, 2021 at 12:09