# LOG#123. Basic Neutrinology(VIII).

There are some indirect constraints/bounds on neutrino masses provided by Cosmology. The most important is the one coming from the demand that the energy density of the neutrinos should not be too high, otherwise the Universe would collapse and it does not happen, apparently…

Firstly, stable neutrinos with low masses (about $m_\nu\leq 1 MeV$) make a contribution to the total energy density of the Universe given by:

$\rho_\nu=m_{tot}n_\nu$

and where the total mass is defined to be the quantity

$\displaystyle{m_{tot}=\sum_\nu \dfrac{g_\nu}{2}m_\nu}$

Here, the number of degrees of freedom $g_\nu=4(2)$ for Dirac (Majorana) neutrinos in the framework of the Standard Model. The number density of the neutrino sea is revealed to be related to the photon number density by entropy conservation (entropy conservation is the key of this important cosmological result!) in the adiabatic expansion of the Universe:

$n_\nu=\dfrac{3}{11}n_\gamma$

From this, we can derive the relationship of the cosmic relic neutrino background (neutrinos coming from the Big Bang radiation when they lost the thermal equilibrium with photons!) or $C\nu B$ and the cosmic microwave background (CMB):

$T_{C\nu B}=\left(\dfrac{3}{11}\right)^{1/3}T_{CMB}$

From the CMB radiation measurements we can obtain the value

$n_\nu=411(photons)cm^{-3}$

for a perfect Planck blackbody spectrum with temperature

$T_{CMB}=2.725\pm0.001 K\approx 2.35\cdot 10^{-4}eV$

This CMB temperature implies that the $C\nu B$ temperature should be about

$T_{C\nu B}^{theo}=1.95K\approx 0.17meV$

Remark: if you do change the number of neutrino degrees of freedom you also change the temperature of the $C\nu B$ and the quantity of neutrino “hot dark matter” present in the Universe!

Moreover, the neutrino density $\Omega_\nu$ is related to the total neutrino density and the critical density as follows:

$\Omega_\nu=\dfrac{\rho_\nu}{\rho_c}$

and where the critical density is about

$\rho_c=\dfrac{3H_0^2}{8\pi G_N}$

When neutrinos “decouple” from the primordial plasma and they loose the thermal equilibrium, we have $m_\nu>>T$, and then we get

$\Omega_\nu h^2=10^{-2}m_{tot}eV$

with $h$ the reduced Hubble constant. Recent analysis provide $h\approx 67-71\cdot 10^{-2}$ (PLANCK/WMAP).

There is another useful requirement for the neutrino density in Cosmology. It comes from the requirements of the BBN (Big Bang Nucleosynthesis). I talked about this in my Cosmology thread. Galactic structure and large scale observations also increase evidence that the matter density is:

$\Omega_Mh^2\approx 0.05-0.2$

These values are obtained through the use of the luminosity-density relations, galactic rotation curves and the observation of large scale flows. Here, the $\Omega_M$ is the total mass density of the Universe as a fraction of the critical density $\rho_c$. This $\Omega_M$ includes radiation (photons), bayrons and non-baryonic “cold dark matter” (CDM) and “hot dark matter” (HDM). The two first components in the decomposition of $\Omega_M$

$\Omega_M=\Omega_r+\Omega_b+\Omega_{nb}+\Omega_{HDM}+\Omega_{CDM}$

are rather well known. The photon density is

$\Omega_rh^2=\Omega_\gamma h^2=2.471\cdot 10^{-5}$

The deuterium abundance can be extracted from the BBN predictions and compared with the deuterium abundances in the stellar medium (i.e. at stars!). It shows that:

$0.017\leq\Omega_Bh^2\leq 0.021$

The HDM component is formed by relativistic long-lived particles with masses less than about $1keV$. In the SM framework, the only HDM component are the neutrinos!

The simulations of structure formation made with (super)computers fit the observations ONLY when one has about 20% of HDM plus 80% of CDM. A stunning surprise certainly! Some of the best fits correspond to neutrinos with a total mass about 4.7eV, well above the current limit of neutrino mass bounds. We can evade this apparent contradiction if we suppose that there are some sterile neutrinos out there. However, the last cosmological data by PLANCK have decreased the enthusiasm by this alternative. The apparent conflict between theoretical cosmology and observational cosmology can be caused by both unprecise measurements or our misunderstanding of fundamental particle physics. Anyway observations of distant objects (with high redshift) favor a large cosmological constant instead of Hot Dark Matter hypothesis. Therefore, we are forced to conclude that the HDM of $\Omega_M$ does not exceed even $0.2$. Requiring that $\Omega_\nu <\Omega_M$, we get that $\Omega_\nu h^2\leq 0.1$. Using the relationship with the total mass density, we can deduce that the total neutrino mass (or HDM in the SM) is about

$m_\nu\leq 8-10 eV$ or less!

Mass limits, in this case lower limits, for heavy or superheavy neutrinos ($M_N\sim 1GeV$ or higher) can also be obtained along the same reasoning. The puzzle gets very different if the neutrinos were “unstable” particles. One gets then joint bounds on mass and timelife, and from them, we deduce limits that can overcome the previously seen limits (above).

There is another interesting limit to the density of neutrinos (or weakly interacting dark matter in general) that comes from the amount of accumulated “density” in the halos of astronomical objects. This is called the Tremaine-Gunn limit. Up to numerical prefactors, and with the simplest case where the halo is a singular isothermal sphere with $\rho\propto r^{-2}$, the reader can easily check that

$\rho=\dfrac{\sigma^2}{2\pi G_Nr^2}$

Imposing the phase space bound at radius r then gives the lower bound

$m_\nu>(2\pi)^{-5/8}\left(G_Nh_P^3\sigma r^2\right)^{-1/4}$

This bound yields $m_\nu\geq 33eV$. This is the Tremaine-Gunn bound. It is based on the idea that neutrinos form an important part of the galactic bulges and it uses the phase-space restriction from the Fermi-Dirac distribution to get the lower limit on the neutrino mass. I urge you to consult the literature or google to gather more information about this tool and its reliability.

Remark: The singular isothermal sphere is probably a good model where the rotation curve produced by the dark matter halo is flat, but certainly breaks down at small radius. Because the neutrino mass bound is stronger for smaller $\sigma r^2$, the uncertainty in the halo core radius (interior to which the mass density saturates) limits the reliability of this neutrino mass bound. However, some authors take it seriously! As Feynman used to say, everything depends on the prejudges you have!

The abundance of additional weakly interacting light particles, such as a light sterile neutrino $\nu_s$ or additional relativistic degrees of freedom uncharged under the Standard Model can change the number of relativistic degrees of freedom $g_\nu$. Sometimes you will hear about the number $N_{eff}$. Planck data, recently released, have decreased the hopes than we would be finding some additional relativistic degree of freedom that could mimic neutrinos. It is also constrained by the BBN and the deuterium abundances we measured from astrophysical objects. Any sterile neutrino or extra relativistic degree of freedom would enter into equilibrium with the active neutrinos via neutrino oscillations! A limit on the mass differences and mixing angle with another active neutrino of the type

$\Delta m^2\sin^2 2\theta\leq 3\cdot 10^{-6}eV^2$ should be accomplished in principle. From here, it can be deduced that the effective number of neutrino species allowed by neutrino oscillations is in fact a litle higher the the 3 light neutrinos we know from the Z-width bound:

$N_\nu (eff)<3.5-4.5$

PLANCK data suggest indeed that $N_\nu (eff)< 3.3$. However, systematical uncertainties in the derivation of the BBN make it too unreliable to be taken too seriously and it can eventually be avoided with care.

# LOG#109. Basic Cosmology (IV).

Today, a new post in this fascinating Cosmology thread…

## The Big Bang Nucleosynthesis in a nutshell

When the Universe was young, about $T\sim 1MeV$, the following events happen

1st. Some particles remained in thermodynamical equilibrium with the primordial plasma (photons, positrons and electrons, i.e., $\gamma$, $e^+$, $e^-$).

2nd. Some relativistic particles decoupled, the neutrinos! And the neutrinos are a very important particle there (the $\nu$ “power” is even more mysterious since the discovery of the neutrino flavor oscillations).

3rd. Some non relativistic particles, the baryons, experienced a strong and subtle asymmetry. Even if we do not understand the physics behind this initial baryon asymmetry, it is important to explain the current Universe. The initial baryon asymmetry is estimated to be

$\dfrac{n_b-n_{\bar{b}}}{s}\sim 10^{10}$

At $T\sim 1MeV$, the baryon number was very large compared with the number of antibaryons. The reason or explanation of this fact is not well understood, but there are some interesting ideas for this asymmetric baryogenesis coming from leptogenesis. I will not discuss this fascinating topic today. Moreover, the fraction or ratio between the baryon density and the photon density is known to be the quantity

$\eta_b\equiv \dfrac{n_b}{n_\gamma}\approx 5\mbox{.}5\cdot 10^{-10}\left(\dfrac{\Omega_b h^2}{0\mbox{.}020}\right)$

In that moment, the question is: how the baryons end up? The answer is pretty simple. There are two main ideas:

1) The thermal equilibrium is kept thorugh out the whole phase in the early Universe. It means that the nuclear state will be the one with the lowest energy per baryon. That is, the most stable nucleus is iron (Fe), but, in fact,  most of the baryons end up in hydrogen (H) or Helium (He).

2) Simple nucleosynthesis is based on 2 elementary ideas and simplifications. Firstly, no element heavier than helium $^4He$ are produced at high ratios. Therefore, this means that protons(p), neutrons(n), the deuterium(D), the tritium (T), Helium-3 ($^3He$) and Helium-4 ($^4He$) are the main subproducts of Big Bang Nucleosynthesis in this standard scenario. Sedondly, until the Universe froze below $T=0\mbox{.}1MeV$, no light nuclei could form and there were only a “primordial soup” made of protons and neutrons. The neutron to proton ratio $n/p$ IS an input for the synthesis of D, T, $^3He$, and $^4He$. Indeed, the 0.1 MeV temperature is relatively low considering the typical nuclear binding energy, about some MeV’s. We could have some quantities of “heavy” elements, but this effect is small, at the level of $\eta_b\approx 10^{-9}$, i.e., the effect of large number of photons compared to the number of baryons is comparable to the number of baryons and heavy elements beyond helium isotopes. We can loot at this effec in the deuterium (D) production:

$n+p\longrightarrow D+\gamma$

where the reaction is taken in the thermal equilibrium and $n_\gamma=n_\gamma^0$. In this equilibrium, we get

$\dfrac{n_p}{n_nn_p}=\dfrac{n_D^0}{n_n^0n_p^0}=\dfrac{3}{4}\left(\dfrac{2\pi m_D}{m_nm_pT}\right)^{3/2}e^{(m_n+m_p-m_D)/T}=\dfrac{3}{4}\left(\dfrac{4\pi}{m_pT}\right)^{3/2}e^{B_p/T}$

and where $m_D\approx 2m_p$, $m_n\approx m_p$ and $B_p$ is the binding energy of deuterium (D). Indeed, we get $n_p\sim n_n\sim n_b=\eta_bn_\gamma$ and so $n_\gamma\sim T^3$. Moreover we also get

$\dfrac{n_D}{n_b}\sim \eta_b \left(\dfrac{T}{m_p}\right)^{3/2}e^{B_p/T}$

and the exponential “compensates” $\eta_b$ and $T$, since the smal factor $\eta_b$ must be chosen to be smaller than the binding energy to temperature ratio $B_p/T$.

By the other hand, the nuetron abundance can be also estimated. From a simple proton-neutron conversion, we obtain

$p+e^- \leftrightarrow n+\nu_e$

due to the weak interaction! The proton/neutron equilibrium ratio for temperatures greater than 1MeV becomes

$\dfrac{n_p^0}{n_n^0}=\dfrac{e^{-m_p/T}\int dpp^2e^{-p^2/2m_pT}}{e^{-m_nT}\int dpp^2e^{-p^2/2m_nT}}=e^{Q/T}$

where $Q=m_n-m_p=1\mbox{.}293MeV$. In fact, the exponential will not be maintained below $T\approx 1 MeV$ and we define the neutron fraction as follows. Firstly

$X_n=\dfrac{n_n}{n_p+n_n}$

In equilibrium, this becomes

$X_n (eq)=\dfrac{1}{1+n_p^0/n_n^0}$

Boltzmann equation for the process $n+\nu_e\leftrightarrow p+e^-$ can be easily derived

$a^{-3}\dfrac{d( n_na^3)}{dt}=n_n^0n_\nu^0\langle \sigma v\rangle \left( \dfrac{n_pn_e}{n_p^0n_e^0}-\dfrac{n_nn_\nu}{n_n^0n_\nu^0}\right)=n_\nu^0\langle \sigma v\rangle \left(\dfrac{n_pn_n^0}{n_p^0}-n_n\right)$

and where

$e^{-Q/T}=\dfrac{n_n^0}{n_p^0}$

Therefore, we obtain

$n_n=\left(n_n+n_p\right)X_n$

and from the LHS, we calculate

$a^{-3}\dfrac{d}{dt}\left[a^3(n_n+n_p)X_n\right]=a^{-3}X_n\dfrac{d}{dt}\left(a^3(n_n+n_p)\right)+\dfrac{dX_n}{dt}\left(n_n+n_p\right)$

By the other hand, from the RHS

$n_\nu^0\langle \sigma v\rangle \left[(n_n+n_p)(1-X_n)e^{-Q/T}-(n_n+n_p)X_n\right]$

Thus, $\Gamma_{np}\longrightarrow \lambda_{np}$ implies that

$\dfrac{dX_n}{dt}=\lambda_{np}\left[(1-X_n)e^{-Q/T}-X_n\right]$

If we change the variable $t\longrightarrow x=Q/T$, then we write

$\dfrac{dX_n}{dt}=\dfrac{dX_n}{dx}\dfrac{dx}{dt}=-\dfrac{Q\dot{T}}{T^2}=-x\dfrac{\dot{T}}{T}=+x\dfrac{\dot{a}}{a}=xH$

where

$H=\sqrt{\dfrac{\rho_R}{3M_p^2}}=\sqrt{\dfrac{\pi^2g_\star}{90}}\dfrac{T^2}{M_p}=\sqrt{\dfrac{\pi^2g_\star}{90}}\dfrac{Q^2}{M_p}x^{-2}=H(x=1)$

and

$\dfrac{dX_n}{dt}=\dfrac{\lambda_{np}x}{H(x=1)}\left[e^{-x}-X_n(1+e^{-x})\right]$

with

$\lambda_{np}(x)=\dfrac{255}{\tau_nx^5}\left(12+6x+x^2\right)$

and $\tau_n$ is the neutron lifetime, i.e. $\tau_n\approx 886\mbox{.7}s$

The numerical integration of these equations provides the following qualitative sketch for $X_n$:

At $T$ below 0.1MeV, the neutron decays $n\longrightarrow p+e^-+\nu_e$ via weak interaction. It yields

$X_ne^{-t/\tau_n}$ and \$latex $X_n(T_{BBN})=0\mbox{.}15\times 0\mbox{.74}=0\mbox{.}11$

such as the deuterium production is done through the processes

$n+p\longrightarrow D+\gamma$

and it started at about $T\sim 0\mbox{.}07MeV$ and

$t=132s\left(\dfrac{0\mbox{.}1MeV}{T}\right)^2$

## The light element abundances

A good approximation is to consider that light element production happens instantaneously at $T=T_{BBN}$. Of course, the issue is…How could we determine that temperature? If we measure the abundance of deuterium abundance today, and the baryon abundance today (i.e., if we know their current densities), we can use the cosmological equations to deduce the ratio

$\dfrac{n_D}{n_b}\approx \eta_b\left(\dfrac{T}{M_p}\right)^{3/2}e^{B_D/T}\sim 1$

Then, we obtain from these equations and the measured densities that $T_{BBN}\approx 0\mbox{.}07MeV\sim 0\mbox{.}1MeV$

Moreover, since $B(He)>B_D$, it implies that helium-4 ($^4He$) production is favoured by BBN! It means that all neutron are processed inside helium-4 or hydrogen. In fact, the helium-4 abundance is known to be

$X_4=\dfrac{4n(^4He)}{n_b}=2X_n(T_{BBN})\approx 0\mbox{.}22$

We can compare this with an exact solution for the “yield” $Y_p=0\mbox{.}2262+0\mbox{.}0135\ln (\eta_b (10^{-10}))$

The observed helium-4 abundance is in good agreement with the theoretical expectations from the Standard Cosmological Model! What an awesome hit! We can also compare this with the primordial helium abundances from cosmological observations

$0\mbox{.}22\sim 0\mbox{.}25$

Thus, we have learned that the deuterium abundance IS a powerful probe of the baryon density!!!!

Remark: Nowadays, there is a problem with the lithium-7 abundances in stars. The origin of the discrepancy is not known, as far as I know. Then, the primordial lithium abundance is a controversial topic in modern Cosmology, so we understand BBN only as an overall picture, and some details need to be improved in the next years.