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In Sterile Neutrinos as Dark Matter we are given the Boltzmann equation for neutrinos

$$\left( \frac{\partial}{\partial t} - HE \frac{\partial}{\partial E}\right) f_s (E,t) = \left[ \frac{1}{2} \sin ^2 2\theta (E,t) \Gamma (E,t) \right] f_a (E,t) \tag{2}$$

  • $f_s$ and $f_a$ are the distribution functions of sterile and active neutrinos.
  • $f_a = \left( e^{E/T}+ 1\right)^{-1} \approx \left( e^{p/T}+1\right)^{-1}$
  • $\Gamma (E,t)$ is the rate of scattering production of $\nu_s$ through a particular channel

and it's stated that, by changing the time variable from $t$ to $a$ ( the Robertson-Walker scale factor) and integrating $(2)$ over momenta we find that:

$$\frac{dr}{d \ln(a)}= \frac{\gamma}{H}+ r \frac{d \ln(g*)}{d \ln(a)} \tag{4}$$

  • $\gamma = \displaystyle \frac{1}{n_a} \int \frac{d^3 p}{(2\pi)^3} \sin^2 2\theta (p,T) \Gamma (p,T) \frac{1}{e^{p/T}+1}$
  • $ \displaystyle n_i ≡ 2 \int \frac{d ^3p}{(2π) ^3}f_i$ is the number density of sterile (active) neutrinos with $i = s,a $
  • $g^∗a^3T^3= constant$
  • $H = \displaystyle \frac{\dot{a}}{a}$ and $r= \displaystyle\frac{n_s}{n_a}$

How do I integrate $(2)$ to get $(4)$? I know that $g^*$ comes from the entropy conservation law but I don't understand why it is brought up in the article or how it helps with this integration and this isn't explained in the article.

I think that in this case, I can consider $E\approx p$ as the mass is negligible (as done in the distribution function for the active neutrinos, above) and so I think I can rewrite $(2)$ as

$$\left( \frac{\partial}{\partial t} - Hp \frac{\partial}{\partial p}\right) f_s (E,t) = \left[ \frac{1}{2} \sin ^2 2\theta (p,t) \Gamma (p,t) \right] f_a (p,t) \tag{2}$$

Dodelson-Widrow production of sterile neutrino Dark Matter with non-trivial initial abundance does something similar, but with the equation written slightly differently. But I still do not understand how the integration is done.

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  • $\begingroup$ I'm assuming $\Gamma(E,t)$ is the upper incomplete gamma function? $\endgroup$
    – K.defaoite
    Jun 30, 2020 at 15:39
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    $\begingroup$ $\Gamma $ represents the rate of scattering production of $\nu_s$ corresponding to a particular channel . I will rewrite it in the post. Thank you for noticing $\endgroup$ Jun 30, 2020 at 15:56
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    $\begingroup$ What in the derivation are you struggling with? When you integrate over all $q$ the term $\frac{df}{dt}$ becomes $\int \frac{df}{dt} d^3q = \frac{d}{dt}\int f d^3q = \frac{dn}{dt} = n \frac{dr}{dt}$ and the right hand side directly becomes the $\gamma$ term. The only term that requires a bit of work is the second term ($d/dE$). Here use $E=p$ and $df/dE d^3p \sim p^2 df/dp dp$ and do integration by parts and you will end up with a $3Hn$ term which combined with $dn/dt$ to become $d(na^3)/dt$ and $n_a \propto 1/a^3$. $\endgroup$
    – Winther
    Jul 2, 2020 at 12:13
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    $\begingroup$ Yes it should be $p$. btw I was a bit fast on the last part also so ignore that (when you go from $dn_s/dt$ to $dr/dt$ you get a term $dn_a/dt$ which combines with the $3Hn_s$ term to give you the $g_*$ term). $\endgroup$
    – Winther
    Jul 2, 2020 at 12:30
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    $\begingroup$ Spherical coordinates: $d^2\vec{p} = d\phi d\cos\theta dp$. And it simplifies as $f$ only depends on the norm of $p$ not the angles so the angular integrals just become $4\pi$ so in this case $\int \cdots d^3\vec{p} = 4\pi \int \cdots p^2 dp$. $\endgroup$
    – Winther
    Jul 2, 2020 at 13:41

1 Answer 1

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Just to summarize all the comments I made above: When you integrate a Boltzmann equation (generally called taking moments) then you simply multiply the whole equation by $\frac{d^3\vec{p}}{(2\pi)^3}$ and integrate over all $\vec{p}$. Most of the terms you then get can be expressed in terms of well known macroscopic quantities (number density $n\propto \int fd^3p$, energy density $\rho\propto \int Efd^3p$, etc.) and derivatives of these. In your case the first term is $$\int \frac{\partial f_s}{\partial t}\frac{d^3\vec{p}}{(2\pi)^3} = \frac{\partial}{\partial t}\int f_s\frac{d^3\vec{p}}{(2\pi)^3} \equiv \frac{1}{2}\frac{\partial n_s}{\partial t}$$ since only $f_s$ in the integrand depends on time we can pull out the derivative. The right hand side integrated, by definition of $\gamma$, simply becomes $\frac{\gamma n_a}{2}$. The only term that is not automatic is the term

$$-\int HE\frac{\partial f_s}{\partial E}\frac{d^3\vec{p}}{(2\pi)^3}$$

When you see a momentum-derivative of the distribution function in an integral, your first thought should be to get rid of that derivative using integration by parts. For this use that $\frac{d}{dE} = \frac{E}{p}\frac{d}{dp}$ (since $E^2 = p^2 + m^2$) and here we can apply the relativistic approximation $T\gg m$ so $E\approx p$ holds (the distribution function is practically zero in the region where this does not hold so its fine to use this approximation in the whole integral). In the non-relativistic limit, if you need it, you can do a similar approximation: $E\approx m$.

Then change to spherical coordinates $d^3\vec{p} = d\cos\theta_{\vec{p}} d\phi_{\vec{p}} dp$. In this case it simplifies since $f$ is assumed to only depend on $p$ and not the directions so we can simply substitute $d^3\vec{p}$ by $4\pi p^2 dp$. Then

$$-\int HE\frac{\partial f_s}{\partial E}\frac{d^3\vec{p}}{(2\pi)^3} \simeq -\int 4\pi Hp^3\frac{\partial f_s}{\partial p}\frac{dp}{(2\pi)^3} = +\int 4\pi H\cdot 3p^2f_s\frac{dp}{(2\pi)^3} \equiv \frac{3Hn_s}{2}$$ where we have used integration by parts and discarded the boundary term at infinity and in the last integral we have used the definition of $n_s$ (using $d^3\vec{p} \leftrightarrow 4\pi p^2 dp$ in reverse).

The rest is just manipulating the equation to the desired form (using known relations between $n_a$, $T$, $a$ and $g_*$). If you are looking for a textbook that explains all these things in simple terms then see Dodelson "Modern Cosmology" (which happens to be written by one of the authors of that paper).

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  • $\begingroup$ I am sorry but I would like to ask you just one last question, when in the second last paragraph you wrote discarded the boundary term at infinity, do you mean that the second term that comes from the integration by parts, in this case, $-4\pi H \frac{p^3 f_s}{(2\pi)^3}$ is discarded because of the limits of the integration ( which I am assuming are $+ \infty $ and $- \infty$)? If this isn't the case, why does the second term from the integration by parts vanish? $\endgroup$ Jul 2, 2020 at 18:56
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    $\begingroup$ @user7077252 Yeah I wrote that without thinking too much. You are correct that the boundary term here vanishes without having to do any assumptions to get that (and its $0$ and $\infty$, remember that $p$ is the norm of the momentum vector). $\endgroup$
    – Winther
    Jul 2, 2020 at 19:14
  • $\begingroup$ I keep on forgetting it is the norm. It wasn't making sense with the $+ \infty$ and $- \infty $ limits, it makes much more sense like that. Once again thank you so much. $\endgroup$ Jul 2, 2020 at 19:17

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