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Gennaro Miele » 4.Neutrino decoupling

Equilibrium Distributions

In the early Universe, the three flavour active left-handed neutrinos ν_α and their CP conjugated states were kept in thermodynamical equilibrium with e.m. plasma by CC and NC weak interactions with charged leptons and quarks. In this regime the neutrino distributions are just fixed by thermodynamics to the Fermi-Dirac type

$f_{\nu_\alpha}(p)&=& \frac{1}{e^{\frac{p}{T}-\xi_\alpha}+1}$ $f_{\overline{\nu}_\alpha}(p)&=& \frac{1}{e^{\frac{p}{T}+\xi_\alpha}+1}$

By using such distributions we get at the equilibrium

$n_{\nu_\alpha} - n_{\overline{\nu}_\alpha} = \frac{T^3}{6} \xi_\alpha \left( 1+ \frac{\xi_\alpha^2}{\pi^2} \right)$ $s_{\nu_\alpha} + s_{\overline{\nu}_\alpha} &=& \frac{7 \pi^2 }{90} T^3 \left( 1+ \frac{15\,\xi_\alpha^2}{7\pi^2} \right)$

$\rho_{\nu_\alpha} + \rho_{\overline{\nu}_\alpha} &=& \frac{7 \pi^2}{120} T^4 \left( 1+ \frac{ 30 \, \xi_\alpha^2}{7\pi^2}+\frac{15 \,\xi_\alpha^4}{7\pi^4}\right) =3 (P_{\nu_\alpha} + P_{\overline{\nu}_\alpha})$

As the temperature decreases, CC and NC weak interactions rate drop down faster than H(T) and hence become not efficient enough in mantaining neutrinos at equilibrium. At this point one says that neutrinos decouple from the e.m. plasma (baryons, leptons, photon). To describe neutrino decoupling in details, one should use the set of Boltzmann equations governing the evolution of their distribution function. Nevertheless in the following we will assume a simplified approach that provides quite precise results.

Decoupling Temperature

In order to simplify the Boltzmann equations ruling neutrino distribution functions one can consider the leading processes contributing to the collisional term, namely the scattering over electrons/positrons and pair conversions, $e^+ e^- \leftrightarrow \nu_\alpha \overline{\nu}_\alpha$

When the temperature is at most few MeV, we can consider the weak reaction rates as due to a 4-fermions interaction term (W and Z not propagating). In this case the corresponding cross section times velocity for charged and neutral current interactions is of the order of

$\langle \sigma v \rangle \sim G_F^2 T^2$

Therefore the weak interaction rate for neutrinos becomes

$\Gamma_\nu \sim G_F^2 T^2 n_e \sim G_F^2 T^5$

n_e denotes the electron/positron density, which is ≈ T³ as long as electrons are still relativistic. In that epoch the Universe was RD, hence the Hubble parameter was given by

$H(T) = \sqrt{\frac{8 \pi G}{3} g_* \frac{\pi^2}{30} T^4} \sim \sqrt{g_*} \frac{T^2}{m_{Pl}}$

The temperature for which the weak interaction rate, dropping down because of the expansion, reaches the value of H is called the Decoupling Temperature, hereafter denoted by T_νD.

Approximate Boltzmann equations for neutrino decoupling

From the above definition the decoupling temperature reads

$G_F^2 T_{\nu D}^5 = \sqrt{g_*} \frac{T_{\nu D}^2}{m_{Pl}} \,\,\,\, \Rightarrow \,\,\,\, T_{\nu D} = \left( \frac{\sqrt{g_*} }{G_F^2 m_{Pl}} \right)^{1/3} \sim g_*^{1/6} \,\mbox{MeV}$

The decoupling temperature is weakly dependent on g_* . For temperature of few MeV the relativistic d.o.f. give

$g_* = 2+ \frac{7}{8} 4+ \frac{7}{8}6=\frac{43}{4} =10.75$

One can get a better prediction for the decoupling temperature via the kinetic equation. The Boltzmann equations written in terms of the variables x=m a and y=p a, with m an arbitrary mass scale (usually taken ≈ m_e) read

$H x \frac{\partial f_{\nu_e}}{\partial x} = - \frac{80 G_F^2(\tilde{g}_L^{l2}+g_R^{l2})m^9}{3 \pi^3 x^5} y f_{\nu_e}$

$H x \frac{\partial f_{\nu_{\mu, \tau}}}{\partial x} = - \frac{80 G_F^2(g_L^{l2}+g_R^{l2})m^9}{3 \pi^3 x^5} y f_{\nu_{\mu, \tau}}$

During RD epoch H ≈ x^-2, in this case it is easy to find that the solutions are function of the combination y/x³, or y T³. In particular one gets a decoupling temperature momentum dependent, namely T_νeD= 2.7 y^-1/3 and T_ν(μ,τ)D= 4.5 y^-1/3. By using the average momentum, <y> ≈ 3, one gets T_νeD= 1.87 MeV and T_ν(μ,τ)D= 3.12 MeV.

Neutrino distribution scaling

With instantaneous decoupling limit we denote the simplifying assumption of a neutrinos decoupling occurring instantaneously at T_νD. In reality the decoupling takes place over an extended range of time. In particular more energetic neutrinos will be kept in equilibrium with e.m. plasma longer than low energy neutrinos. This means that even neutrinos to some extent get profit of the entropy release due to e^± annihilations, and hence some thermal distortions will be imprinted in the neutrino distribution with respect to a standard Fermi-Dirac function. However these effects change the neutrino energy density at the percent level only, and thus will be neglected in the following.

Neutrino distribution scaling

After decoupling, neutrinos propagate freely, and their distribution only feels the effect of redshift of physical momentum. In the instantaneous decoupling approximation, we can assume the distribution at T_νD to be exactly a Fermi-Dirac one

$f_{\nu_\alpha}(p)= \left[\exp\left(\frac{p\, a_D}{T_{\nu D}a_D} \right)+1\right]^{-1}$

with a_Dthe scale factor at T_νD. From that moment on the distribution function satisfying the Liouville equation for ultrarelativistic particles is

$f_{\nu_\alpha}(p)= \left[\exp\left(\frac{p \,a}{T_{\nu D}a_D} \right)+1\right]^{-1}$

which is the initial distribution with a temperature redshifted T = T_νDa_D/a . Note that this temperature scaling is equivalent to the conservation of the neutrino entropy per comoving volume, namely

$s_{\nu, \overline{\nu}_\alpha} a^3= const.$

Entropy density of the e.m. plasma

The neutrino to photon temperature ratio after neutrino decoupling can be easily obtained by using the conservation of entropy for the e.m. plasma. As long as electron/positron pairs are relativistic, the photon temperature simply scales as a^-1 since the number of degrees of freedom g_s is constant. When the temperature drops below a value ≈ m_e, e.m. , the process $e^+ \, e^- \rightarrow \gamma \, \gamma$ cannot be efficiently compensated by the inverse pair production. This implies that electron/positron distributions start to be suppressed by the term e^-me/T, and practically e^± disappear, apart from a tiny relic electron density related by electric neutrality to the baryon number parameter η_B. In practice, the entropy release due to e^±annihilations reheat photons, whose temperature is then not decreasing as a^-1for a while. Due to the smallness of weak reactions at this temperature, neutrinos practically get no profit of such reheating, nevertheless even though to a very small extent, such phenomenon leaves a footprint on neutrino distribution functions.

The entropy density of coupled e^± and γ has the following expression

$s_{e^\pm, \gamma} = \frac{2 \pi^2}{45 }T^3 g_s(T)$

where

$g_s(T) = 2+ \frac{15}{ \pi^4} \int_0^\infty \left( \sqrt{x^2+\frac{m_e^2}{T^2}} + \frac{x^2}{3 \sqrt{x^2+\frac{m_e^2}{T^2}}} \right) \frac{x^2 dx}{\exp \sqrt{x^2+\frac{m_e^2}{T^2}}+1}$

Ratio between neutrino and photon temperatures

The previous expression strongly simplifies for a temperature much larger than m_e, in fact we get

$g_s\rightarrow 2+ \frac{7}{8} (2+2)=\frac{11}{2}$

whereas for T << m_ewe simply have g_s = 2. Since the entropy for the e.m. plasma is conserved inside a coomoving volume, namely $s_{e^\pm, \gamma} a^3$

we get

$g_s(T) \frac{T^3}{T_\nu^3} = \mbox{const.}$

At neutrino decoupling T/T_ν=1 and g_s = 11/2, whereas for T << m_ewe simply have g_s = 2, hence the previous expression becomes

$\frac{11}{2} = 2\, \frac{T^3}{T_\nu^3} \, \Rightarrow \frac{T_\nu}{T} = \left(\frac{4}{11} \right)^{1/3} \,, \,\,\,\,T<<m_e$

Using the known value of the CMB temperature today, T₀= 2.725 K, neutrinos at present have a temperature T_ν0= 1.945 K, or equivalently 1.7 10^-4 eV in natural units.

Gerstein -Zeldovich bound on neutrino masses

As we have seen after the e⁺ - e^-annihilations the density number of photons and neutrinos read

$n_{\gamma} = \frac{\zeta(3)}{\pi^2} g_\gamma \, T^3$

with g_γ = 2, and

$n_{\nu} = \frac{3 \zeta(3)}{4 \pi^2} g_\nu \, T_\nu^3$

with g_ν = 6. By using the ratio T/T_νdetermined in the framework of instantaneous neutrino decoupling, we get n_ν = 9/11 n_γ. This ratio allows to evaluate for the present value of photon number density, namely n_γ0 = 410 cm^-3, the corresponding neutrino number density, n_ν0 = 336 cm^-3. Absolutely larger than the present aboundance of baryons, namely n_Bo ≈ 10^-7cm^-3. Hence photons and neutrinos are the most abundant species in the present Universe. In particular since neutrinos at the present are non relativistic (their energy is ≈ 10^-4 eV << m_νi) we can write

$\Omega_\nu = \frac{\sum_i m_{\nu i} \, n_{\nu i 0 }}{\rho_c} = \sum_i m_{\nu i} \frac{n_{\nu 0}}{\rho_c} = \frac{\sum_i m_{\nu i} }{94 \, h^2\, eV}$

Since Ω_ν ≤ Ω_DM≈ 0.2, we get for h ≈ 0.73, ∑_i m_νi≤ 10 eV.

Effective Number of neutrinos

If we consider negligible the neutrino chemical potentials, the number density of cosmic neutrinos for each flavour is given by

$n_{\nu,0} = n_{\overline{\nu},0} \sim 56 \, \mbox{cm}^{-3}$

which leads to an extremely large flux with respect to other astrophysical neutrino sources, including solar neutrinos. By using the ratio between neutrino and photon temperatures, previously obtained, we can compute the total amount of radiation after e^± annihilations which results

$\rho_R = \rho_\gamma \left( 1+ \frac{7}{8} \left(\frac{4}{11} \right)^{4/3} 3 \right)$

Such result holds if:

there are only three light neutrino species and no other relativistic particles;
neutrino distributions are standard Fermi-Dirac functions with zero chemical potentials;
we trust the instantaneous decoupling limit.

A convenient way to parameterize any deviation from these assumptions consists in defining the effective number of neutrino species N_eff

$\rho_R = \rho_\gamma \left( 1+ \frac{7}{8} \left(\frac{4}{11} \right)^{4/3} N_{{\rm eff}}\right)$

The exact neutrino decoupling

Let us consider the neutrino distribution functions $f_{\nu_\alpha}$ with $\nu_\alpha = \nu_e, \, \overline{\nu}_e, \, \nu_x, \, \overline{\nu}_x$ with x=μ, τ

The Boltzmann Equations

$\widehat{L}\left[f_{\nu_\alpha} \right] = \widehat{C}\left[f_{\nu_\alpha} \right]$

can be casted in the following form

$\frac{d}{dx} f_{\nu_\alpha}(x,y) = \frac{1}{x \, H} I_{\nu_\alpha}( f_{\nu_\beta}(x,y))$

where x = a(t) m_e and y= a(t) |p|

Reactions contributing to the collisional term

Boltzmann Equation for electron neutrinos

Boltzmann Equation for muon/tau neutrinos

The entropy conservation

In addition one must satisfies the covariant energy conservation

$x \frac{d}{dx} \rho(x) = - 3 (\rho(x) + P(x))$

where the total energy density reads

$\rho = \frac{\pi^2 \, T_\gamma^4}{15} + \frac{2}{\pi^2} \int \frac{dq \, q^2 \, \sqrt{q^2 + m_e^2}}{\exp(E/T_\gamma)+1} + \frac{1}{\pi^2} \int dq \, q^3 f_{\nu_e} + \frac{2}{\pi^2} \int dq \, q^3 f_{\nu_x}$

whereas the total pressure reads

$P = \frac{\pi^2 \, T_\gamma^4}{45} + \frac{2}{\pi^2} \int \frac{dq \, q^4}{3 \sqrt{q^2 + m_e^2} [\exp(E/T_\gamma)+1]} + \frac{1}{3 \, \pi^2} \int dq \, q^3 f_{\nu_e} + \frac{2}{3 \, \pi^2} \int dq \, q^3 f_{\nu_x}$

Mass renormalization effects due to e.m. interactions in the hot plasma

The change in the electromagnetic plasma equation of state can be evaluated by first considering the corrections induced on the e^± and photon masses. They can be obtained perturbatively by computing the loop corrections to the self-energy of these particles. For the electron/positron mass, up to order α ≡ e²/4 π we find the additional finite temperature contribution

$\delta m_e^2 \left( p, \,T \right) = \frac{2\,\pi\,\alpha\,T^2}{3} + \frac{4\,\alpha}{\pi} \, \int_0^\infty d k \, \frac{k^2}{E_k} \, \frac{1}{{\rm e}^{E_{k}/T}+1} - \frac{2 \, m_e^2 \,\alpha}{\pi \, p} \, \int_0^\infty d k \, \frac{k}{E_{k}} \, \log\left| \frac{p+k}{p-k} \right| \frac{1}{{\rm e}^{E_{k}/T}+1}$

where $E_{k} \equiv \sqrt{k^2 + m_e^2}$

The renormalized photon mass in the electromagnetic plasma is instead given, up to order α, by

$\delta m_\gamma^2 \left( T \right) = \frac{8\,\alpha}{\pi} \, \int_0^\infty d k \, \frac{k^2}{E_k} \, \frac{1}{{\rm e}^{E_{k}/T}+1}$

The modification of the dispersion relations provides thermal/e.m. corrections to plasma equation of state

The previous corrections modify the corresponding dispersion relations as

$E_i^2 = k^2 + m_i^2 + \delta m_i^2 \left( T \right) \,\,\,\, {\mbox{with}}\,\,\,i = e,\, \gamma$

This in turn affects the total pressure and the energy density of the electromagnetic plasma

$P = \frac{T}{\pi^2} \, \int_0^\infty d k \, k^2 \, \, \log\left[\frac{\left( 1 + {\rm e}^{-E_e/T} \right)^2}{ \left( 1 - {\rm e}^{-E_\gamma/T} \right)} \right]$

$\rho &=& - P + T \,\frac{d P}{d T}$

Expanding P with respect to δm_e² and δm_γ², one obtains the first order correction

$P^{\rm int} = - \int_0^\infty \frac{dk}{2 \, \pi^2} \left[ \frac{k^2}{E_k} \, \frac{\delta m_e^2 \left( T \right)}{{\rm e}^{E_{k}/T}+1} + \frac{k}{2} \, \frac{\delta m_\gamma^2 \left( T \right)}{{\rm e}^{k/T}-1} \right]$

and from the previous relation one gets ρ^int as well.

The modification of the dispersion relations provides thermal/e.m. corrections to plasma equation of state

The presence of the additional contributions P^int and ρ^int modify the evolution equation for z≡ T a(t), which now reads

$\frac{dz}{dx} = \frac{\frac{x}{z}J(x/z)- \frac{1}{2 \pi^2 z^3}\int_0^\infty~dyy^3\left(\frac{df_{\nu_e}}{dx}+ 2\frac{df_{\nu_x}}{dx}\right)+G_1(x/z)} {\frac{x^2}{z^2}J(x/z)+Y(x/z)+\frac{2\pi^2}{15}+G_2(x/z)}\,$

where

$G_1(\omega) = 2 \pi \alpha \left[ \frac{1}{\omega} \left(\frac{K(\omega)}{3} + 2 K(\omega)^2- \frac{J(\omega)}{6}- K(\omega)~J(\omega) \right) \left( \frac{K'(\omega)}{6}-K(\omega)~K'(\omega)+\frac{J'(\omega)}{6} +J'(\omega)~K(\omega) + J(\omega)~K'(\omega)\right ) \right]$

and

$G_2(\omega) =-8 \pi \alpha \left( \frac{K(\omega)}{6} + \frac{J(\omega)}{6} - \frac{1}{2} K(\omega)^2 + K(\omega)~J(\omega)\right) +2 \pi \alpha &\omega & \left(\frac{K'(\omega)}{6}-K(\omega)~K'(\omega) +\frac{J'(\omega)}{6} + J'(\omega)~K(\omega) + J(\omega)~K'(\omega)\right)$

The modification of the dispersion relations provides thermal/e.m. corrections to plasma equation of state

with

$K(\omega) = \frac{1}{\pi^2}~\int_0^\infty du~\frac{u^2}{\sqrt{u^2+\omega^2}}~ \frac{1}{\exp\left(\sqrt{u^2+\omega^2}\right)+1}$

$J(\omega) = \frac{1}{\pi^2}~\int_0^\infty du~u^2~\frac{\exp\left(\sqrt{u^2+\omega^2}\right)}{\left(\exp\left(\sqrt{u^2+\omega^2}\right)+1\right)^2}$

$Y(\omega) = \frac{1}{\pi^2}~\int_0^\infty du~u^4~\frac{\exp\left(\sqrt{u^2+\omega^2}\right)}{\left(\exp\left(\sqrt{u^2+\omega^2}\right)+1\right)^2}$

The functions K’(ω) and J’(ω) stand for the first derivativeof K(ω) and J(ω) with respect to their argument. Note that neutrinos affect the final value of z through the terms $df_{\nu_x}/dx$ ,which are not vanishing only if neutrinos have a non thermal behaviour.

In case one neglects the finite temperature QED corrections, the functions G₁(x/z) and G₂(x/z) vanish, and one recovers a very well known expression reported in literature. Notice that the presence of G₂(x/z) in the denominator of the r.h.s. of dz/dx makes, at least inprinciple, not correct to simply sum the neutrino contribution with the QED one.

Initial conditions

The initial conditions are fixed by the thermodynamical equilibrium which is satisfied for T > 10 MeV

$f_{\nu_\alpha}(x,y) = \frac{1}{e^{y-\xi_\alpha}+1}\left( 1 + \delta f_{\nu_\alpha}(x,y) \right)$

In particular we have

$T_{in} = 10 \, MeV \rightarrow x_{in} = \frac{m_e}{10\, MeV} \approx 20 \,\,\, \Rightarrow \delta f_{\nu_\alpha}(20,y) =0$

From top to bottom the curves correspond to y=7, 5 , 3

Photon/Neutrino temperatures ratio

Asymptotically, as x → ∞, the prediction of the instantaneous decoupling approximation gives

z_eq → (11/4)^1/3

In the Figure we show the departure of z ≡ T/T_ν from z_eqfor different values of the asymmetry parameters ξ_α

Non thermal effects on asymptotic neutrino distributions

The asymptotic distribution functions are not perfectly thermodynamical distributions (out of equilibrium effects).

Non instantaneous decoupling effects on effective number of neutrinos (no degenerate case)

The effective number of neutrino species is used to parametrize the energy density stored in relativistic species, ρ_R, through the relation

$\rho_{\rm R} = \left[ 1 + \frac{7}{8} \left( \frac{4}{11} \right)^{4/3} \, N_\nu^{\rm eff} \right] \, \rho_\gamma$

where ρ_γ is the energy density of photons, whose value today isknown from the measurement of the Cosmic Microwave Background (CMB) temperature.

Non instantaneous decoupling effects on effective number of neutrinos (no degenerate case)

The previous exprtession can be also written as

$N_\nu^{\rm eff} \equiv \left( \frac{\rho_R -\rho_\gamma}{\rho^0_\nu}\right) \left(\frac{\rho^0_\gamma}{\rho_\gamma} \right)$

Finally, from the previous definitions it is straightforward to get the following expression

$N_\nu^{\rm eff} = \left( \frac{z^{0}}{z^{\rm fin}}\right)^4 \left( 3 + \frac{\delta \rho_{\nu_e}}{\rho_\nu^0} + 2 \frac{\delta \rho_{\nu_x}}{\rho_\nu^0} \right) \simeq \left( 3-12 \frac{\delta z}{z^0} + \frac{\delta \rho_{\nu_e}}{\rho_\nu^0} + 2 \frac{\delta \rho_{\nu_x}}{\rho_\nu^0} \right)$

In the Table are reported the values assumed by the several factors when different effects are considered.

Relativistic Extra d.o.f.

We now consider the possibility that, at the stage of neutrino decoupling, there are extra relativistic degrees of freedom, provided by some X field excitations, which are assumed to have a thermal distribution with some temperature T_X. Their contribution ρ_X to the total energy density of the Universe can be parametrized in terms of an additional contribution $\Delta N_\nu^{\rm eff}$ in the effective number of neutrinos, as previously defined

$\Delta N_\nu^{\rm eff} = \left( \frac{z^0}{z^{\rm fin}} \right)^4 N_X$

where

$N_X= \frac{4}{7} g_X \left( \frac{11}{4} \right)^{4/3} \left( \frac{z_X}{z^0} \right)^{4}$

and we have defined z_X= T_X a(t). The parameter g_X depends on the spin (g_X=1 for a real scalar, g_X=7/4 for a Weyl spinor, etc.) as well as on the additional internal degrees of freedom of the X particles. Notice that if the X excitations have decoupled between μ^± and e^± annihilation phases we simply have N_X= 4/7 g_X. For an earlier decoupling we have instead N_X< 4/7 g_X.

Relativistic Extra d.o.f.

The presence of ρ_X, apart from introducing a new contribution to $N_\nu^{\rm eff}$ , slightly affects the results previously obtained, namely the relative change of neutrino energy density $\delta\rho_{\nu_{e,x}} / \rho^0_{\nu}$ induced by incomplete neutrino decoupling, as well as the asymptotic photon temperature z^fin. In fact since their energy density increases the expansion rate of the Universe, we expect $\delta\rho_{\nu_{e,x}} / \rho^0_{\nu}$ to decrease with growing N_X, and the ratio $z^0/z^{\rm fin}$ to become closer to unity, since neutrino decouplingprocess starts at earlier times.

If we denote with $\delta \rho_{\nu_{e,x}}(N_X)/\rho^0_{\nu}$ and $z^{\rm fin}(N_X)$ the new values for these parameters asfunctions of N_X we therefore have

$N_\nu^{\rm eff} = \left( \frac{z^{0}}{z^{\rm fin}(N_X)}\right)^4 \left( 3 +\frac{\delta \rho_{\nu_e}(N_X)}{\rho_\nu^0} + 2 \frac{\delta\rho_{\nu_x}(N_X)}{\rho_\nu^0} + N_X\right)$

The effect for massive neutrinos

As it is seems more and more clear from neutrino experiments on solar and atmospheric neutrinos, it is unlikely that neutrino mass differences may be greater than ≈ 0.1 eV, unless we enlarge the standard scenario introducing sterile neutrino states. At the same time, it is also quite well established from Tritium decay data that ν_e mass is bound to be smaller or, at most, of the order of 1 eV. It is therefore clear that in this scenario all neutrino masses are completely negligible as far as their decoupling is concerned. However if their values is as large as ≈ 1 eV, they start to be relevant as the temperature approaches the range relevant for CMB, and the presence of a finite mass modifies of course the neutrino contribution to $N_\nu^{\rm eff}$ . It is interesting to consider how the effects of incomplete decoupling and QED thermal effects just studied would now affect the neutrino energy density. This is conveniently parametrized by the (time dependent) quantity

$\frac{\delta \rho_{\nu_\alpha}(m_\alpha)}{\rho_{\nu_\alpha}^0(m_\alpha)} \equiv \frac{\rho_{\nu_\alpha}(m_\alpha) - \rho_{\nu_\alpha}^0(m_\alpha)}{\rho_{\nu_\alpha}^0(m_\alpha)}$

Since for m_α ≈ 1 eV active neutrino are fully elativistic for temperatures of the order of MeV, so we can still take at the e^± annihilation phase the results previously obtained, it is easy to see that $\delta\rho_{\nu_\alpha}(m)/ \rho_{\nu_\alpha}^0(m)$ is given by

$\frac{\delta \rho_{\nu_\alpha}(m_\alpha)}{\rho_{\nu_\alpha}^0(m_\alpha)}= 10^{-4} \frac{\int_{0}^{\infty} dy~y^2~\sqrt{y^2 + \left(\frac{m_{\alpha}}{m_e}\right)^2 x^2}\left( {\rm e}^y+1 \right)^{-1} \sum_{i=0}^3 c^\alpha_i y^i}{\int_{0}^{\infty} dy~y^2~\sqrt{y^2 + \left(\frac{m_{\alpha}}{m_e}\right)^2 x^2} \left( {\rm e}^y+1 \right)^{-1}}$ where we have $f_{\nu_\alpha}^{\rm fin} \left(y \right) = \frac{1}{{\rm e}^y +1} \, \left( 1 + 10^{-4} \sum_{i=0}^{3} c^\alpha_i~y^i \right)$