# Neutrino decoupling and units

I have a problem that states:

In the early Universe, neutrinos can be created and destroyed by the process: \begin{equation} \nu \bar{\nu} \leftrightarrow e^{+}e^{-} \end{equation}

The thermally averaged cross section for this process is given by $\langle \sigma|v|\rangle = K G_{F}^{2}T^{2}$, where $K$ is a constant of order unity. Assume that $K = 1$. Use the condition $\Gamma \equiv n_{\nu}\langle \sigma|v|\rangle = H$ to calculate the decoupling temperature of neutrinos.

From the Friedmann equation I have:

\begin{equation} H = \sqrt{\frac{8\pi}{3}G\rho}, \end{equation} where: \begin{equation} \rho = \left(\frac{\pi^{2}}{30c^{5}\hbar^{3}} g\left(k_{B}T\right)^{4}\right), \end{equation} and $g$ is the g-factor (degree of freedom) which is 1 for neutrinos and 2 for electrons/positrons.

$n_{\nu}$ is given by: \begin{equation} n_{\nu} = \left(\frac{\zeta(3)}{\pi^{2}c^{3}\hbar^{3}} g\left(k_{B}T\right)^{3}\right), \end{equation} with $\zeta(3) = 1.20206$.

Now, I've tried to solve for $T$ $\left(n_{\nu}\langle \sigma|v|\rangle = H\right)$ in a bunch of ways, both in hand, and with computer, which for the computer is:

$$T_{units} = \frac{\sqrt{s}K}{J\sqrt{kg}m^{5/3}},$$ with $s$ being seconds, $K$ is kelvin, $J$ is joule, $kg$ is, yeah, kilograms, and $m$ is meter.

If I do it by hand, I end up with $T$ (In eV): $$T_{units} = \sqrt{\frac{kg}{m^4}}$$

So yeah, not the same (At least what I can see), but either way, it gives an incorrect unit for temperature (Either kelvin or eV). And I can't seem to see what I can do in order to get the rights units in this. Searching on the web for neutrino decoupling, I can see the relation:

$$T \sim \left( \frac{\sqrt{G}}{G_F^2} \right) ^{1/3} \sim 1~\textrm{MeV}$$

So I know my temperature should be around the same, but again, I can't see how the units for the calculation can give me MeV or kelvin at least.

Correction of typos and clarifications

• In the Friedmann equation, '$\rho$' is strictly speaking $\rho_m$, the mass density. Hence the presented Friedman equation has to be changed as follows: \begin{equation} H = \sqrt{\frac{8\pi}{3}G\rho_m}=\sqrt{\frac{8\pi}{3}\frac{G\rho}{c^2}}, \end{equation} in order to use the expression given below for the energy density, denoted by $\rho$.
• In the expression for the energy density, the $c^5 \hbar^3$ factor is wrong. Energy density means energy per unit volume, $EL^{-3}$, where $E$ and $L$ denote units of energy and length, respectively. Since $\left(k_{B}T\right)^{4}$ has units of $E^4$, and $c\hbar$ has units $EL$, we need a $(c\hbar)^3$ factor to get the right dimensions. This means: \begin{equation} \rho = \frac{\pi^{2}}{30c^{3}\hbar^{3}} g\,\left(k_{B}T\right)^{4}, \end{equation}
• The number of degrees of freedom for each neutrino flavour is $g_\nu=2$ (neutrino plus antineutrino; see, for instance, p.45 of Dodelson's Modern Cosmology), while, for clarity, I just point out that $g_{e^-}=g_{e^+}=2$, implying $g_e \equiv g_{e^-}+g_{e^+}=4$. Since the Universe at the time of neutrino decoupling is presumably populated by photons, electrons, positrons and the three known flavours of neutrinos, the $g$ factor appearing in the energy density $\rho$ in the previous equation is actually a weighted sum of the $g$ factors of these populating species, given by: \begin{equation} g = \sum_{\textrm{bosons } i}g_i + \frac{7}{8}\sum_{\textrm{fermions } j}g_j = g_\gamma +\frac{7}{8}(g_{e}+3\,g_{\nu}) = 2 + \frac{7}{8}(4+3\times2) = \frac{43}{4} \end{equation} For additional clarity, see J. Bernstein, L. Brown, and G. Feinberg, Rev. Mod. Phys. 61 (1989) 25.

• The $g$ in $n_\nu$ is actually ${3\over4} g_\nu$. This is so because we are computing the rate $\Gamma$ of weak thermalizing processes which essentially occur through the interaction of electrons and electron neutrinos, and thus $\Gamma$ depends on the existing density of this species of neutrinos. The muon and tau neutrinos are assumed to be decoupled (yet they are still there, contributing to the energy density). Also, neutrinos are fermions, requiring the numerical $3\over4$ factor.

• Finally, the thermally averaged cross-section $\langle \sigma|v|\rangle$ has units of area times speed, $L^3 T^{-1}$ (where $T$ denotes units of time). Since we are insisting on bringing out all factors of $c$, $\hbar$, and $k_B$, we change $T\rightarrow k_B T$ and, noticing that $G_F$ has units $EL^3$ (see, for instance, p.313 of Griffiths' Introduction to Elementary Particles, 2nd ed.), we can also change $G_{F}\rightarrow G_{F}/(c\hbar)^3$. Now, the product $(k_B T)^2 \,(G_{F}/(c\hbar)^3)^2$ has units of $E^{-2}$. We multiply this by $c^3\hbar^2$, which has units $E^2L^3T^{-1}$ (since $\hbar$ has units $ET$), to produce the dimensionally correct expression for the thermally averaged cross-section: \begin{equation} \langle \sigma|v|\rangle = \bigg(\frac{G_{F}}{c^3\hbar^3}\bigg)^2\, (k_B T)^2 \, c^3\hbar^2 \end{equation}

Putting things together

Insisting on keeping the $c$, $\hbar$, and $k_B$ factors, $\Gamma = H$ yields, after much care with all factors: \begin{equation} (k_B T)^3 = \underbrace{\Bigg(\frac{\pi^3}{\zeta(3)}\sqrt{\frac{8\pi}{90}} \frac{\sqrt{g}}{\frac{3}{4}g_\nu} \Bigg )}_{\simeq \,29.8} \underbrace{\Bigg(\sqrt{\frac{G}{c\hbar}}\Bigg)}_{\equiv \,1/M_{Pl}} \,\frac{c^4 \hbar^6}{{G_F}^2} \simeq 29.8 \Bigg[ \big(M_{Pl} c^2\big) \, \bigg(\frac{G_{F}}{c^3\hbar^3}\bigg)^2 \Bigg]^{-1} \end{equation}

In the above we have introduced the Planck mass $M_{Pl}$. Knowing that $M_{Pl} c^2 \simeq 1.22\times 10^{19}\,\textrm{GeV}$ and $G_{F}/(c\hbar)^3\simeq 1.17\times 10^{-5} \,\textrm{GeV}^{-2}$, one gets: \begin{equation} k_B T \simeq \left ( \frac{29.8}{1.67\times 10^{9} \,\textrm{GeV}^{-3} } \right )^{1/3} \simeq 2.6 \,\textrm{MeV} \end{equation}

The (natural) moral

It is very cumbersome to carry around all the $c$, $\hbar$, and $k_B$ factors (as you can see by doing the above computation yourself), which is why one very much likes to work in the so-called natural units in the context of particle physics and cosmology. In fact, taking all these factors to be $1$, the first equation of the previous section immediately gives the result for the neutrino freeze-out/decoupling temperature in the form you have found it: \begin{equation} T \,\simeq \,29.8 \,\left( \frac{\sqrt{G}}{{G_F}^2} \right) ^{1/3} \simeq 2.6\,\textrm{MeV} \end{equation} Of course, here, only the order of magnitude can be meaningful since we took $K=1$. The $k_B$ is there but equals $1$ in these natural units, so one just writes $T$ instead of $k_B T$. This and the last expression given above are the same, just written using different units.

• After looking it through, I actually have one question. Where does the $3/4$ factor come from for the neutrino/fermions ? I've been searching my books and the web, but can't seem to find it anywhere :/ – Denver Dang Dec 3 '14 at 22:12
• You can find it, for instance, in Kolb and Turner's The Early Universe, p. 62 (expression 3.52 and surrounding context). It can be obtained by plugging the Fermi-Dirac distribution in the definition of the number density and taking the relativistic limit. – J-T Dec 3 '14 at 22:53