Neutrinos and counting particle species

Guth's slides (2–5) review Lecture 16: $a \propto \sqrt t$, $H = 1/(2t)$, the $2ct$ horizon and $\rho = 3/(32\pi Gt^2)$ for a flat radiation-dominated universe, and the formulas for $u$, $p$, $n$ and $s$. Photons have $g = 2$ because they are massless spin-1 particles with only the two extreme spin states, $\pm 1$ along the motion — the quantum version of splitting light into left- and right-circular (or $x$ and $y$) polarisation.

Entropy, slide 5 adds, is conserved in the early universe for every process the course discusses, except inflation. Guth explains aloud (transcript p. 4) that if inflation is right, nearly all of today's entropy was produced in the phase transition at its end.

Asked why entropy only increases, Guth says that, among physicists, you might get ten different answers. It does not follow from the other known laws, which are essentially time-reversal invariant. The usual story is that the universe began, for unknown reasons, in a state of peculiarly low entropy and has been heading towards equilibrium ever since. Guth mentions that he was writing a paper with a slightly different idea; the arrow of time, he says, remains something of a mystery.

Neutrinos carry a significant share of the universe's entropy even today. They come in three flavours ($\nu_e$, $\nu_\mu$, $\nu_\tau$); oscillation between flavours shows that the mass differences are non-zero, so at most one could be massless. For cosmology they can be treated as massless — though that needs justifying, because mass changes the number of spin states.

Helicity. In the massless approximation a neutrino has one spin state: it is left-handed, with its spin opposite to its momentum. Antineutrinos are right-handed. This is a Lorentz-invariant statement only for a massless particle. If a neutrino had mass, it would move slower than light, and a fast enough rocket could overtake it: in the rocket's frame the momentum would reverse while the spin would not, turning a left-handed neutrino into a right-handed one.

A student asks how a spin-½ particle can have only one spin state. For massless particles this is consistent: the photon, with spin 1, lacks the $m = 0$ state. Only for massive particles do Lorentz transformations mix all spin states, so that having one forces having all.

Fermions and bosons. Neutrinos have spin ½. By the spin–statistics theorem of quantum field theory, half-integer spin means fermion (obeying the exclusion principle: no two in the same state), integer spin means boson (photons; bosons even prefer sharing states). The black-body formulas come from counting states, each weighted by $e^{-E/kT}$. Fermions allow fewer states, so their factors are smaller:

The 7/8 is bigger than the 3/4, so fermions carry slightly more energy per particle — as you'd expect, since the exclusion principle pushes particles into higher states.

Each species' $g$ is a product of factors: 7/8 for fermions, the number of species, 2 if particle and antiparticle differ, and the number of spin states. For the three neutrinos:

(A photon is its own antiparticle, so photons get no factor 2 for that.) Go further back, to where $kT \gg m_ec^2$, and electron–positron pairs join the black-body radiation too — one species, particle and antiparticle, two spin states:

Between the electron threshold ($m_ec^2 \approx 0.5$ MeV) and the next one, the muon ($m_\mu c^2 \approx 106$ MeV), all three are present at the same temperature:

This number governs a long stretch of the universe's history.

Today $kT$ is far below $m_ec^2$, so the electron–positron pairs are long gone, leaving photons and neutrinos. But we can't simply add their formulas, because they no longer share a temperature. (Working it out is an 8.286 homework problem; Guth gives the outline.)

As $kT$ fell below $m_ec^2$, the pairs left the thermal mix. Following their energy is hard, because it depends on the pressure, which is complicated near a threshold. Their entropy is easy: it is conserved. By then the photons were still interacting strongly, but the neutrinos had essentially decoupled. So the pairs handed nearly all their entropy to the photons and none to the neutrinos, heating the photons relative to the neutrinos. The result:

So the universe should be filled with a thermal bath of neutrinos at about 1.95 K, a little colder than the 2.725 K photons. [Guth recalls the value as "about 2.4" K; $(4/11)^{1/3} \times 2.725$ K $= 1.95$ K.] Neutrinos interact so weakly that nobody has come close to detecting this background — a challenge, Guth tells the students, for their careers. If it turned out not to be there, that would be a major shift in cosmology.

With the neutrino temperature in hand, the radiation density of today's universe follows. Energy density goes as $T^4$, so the neutrino term carries $(4/11)^{4/3}$:

— the number Guth pulled "out of a hat" in Lecture 15 to date matter–radiation equality.

At least two neutrinos do have mass. Then the counting above seems wrong: a massive particle must have both spin states, so where are the right-handed neutrinos, which no one has ever seen? There are two theories (both named after people, so don't read physics into the names), and we don't yet know which is right.

• Dirac mass, the same kind of mass the electron has. Right-handed neutrinos would be a new species. Because of the peculiar way neutrinos interact, they could interact vastly more weakly than the left-handed ones — so weakly that they are not seen in the laboratory and were essentially never produced even in the early universe.
• Majorana mass, possible only for neutral particles, and not known for any other particle. The right-handed partner of the left-handed neutrino would be a particle we already know: the right-handed antineutrino. The neutrino would be its own antiparticle — like the photon.

Either way, the counting above stands. For Majorana neutrinos the factor 2 counts two spin states instead of particle and antiparticle, with the same result. What distinguishes the options is lepton number: in every interaction seen, electron number (electrons and electron-neutrinos minus their antiparticles), and likewise muon and tau number, is conserved. A Majorana mass requires lepton number not to be exactly conserved. A student points out that oscillations already violate the separate electron, muon and tau numbers; Guth agrees, but total lepton number might still be conserved, which is believed not to be exactly true but not ruled out.

• Each effectively massless species adds to black-body radiation; fermions count with $\frac78$ in $g$ (energy, pressure, entropy) and $\frac34$ in $g^*$ (number).
• Massless neutrinos are left-handed (antineutrinos right-handed): $g_\nu = 21/4$, $g^*_\nu = 9/2$ for three flavours. Electrons and positrons: $g = 7/2$, $g^* = 3$.
• For $m_ec^2 \ll kT \ll m_\mu c^2$, $g = 10\frac34$.
• When the $e^\pm$ pairs disappeared, their entropy went to the photons, not the already decoupled neutrinos: $T_\nu = (4/11)^{1/3}T_\gamma \approx 1.95$ K, and today $u_r = 7.01 \times 10^{-14}$ J/m³.
• Neutrino masses may be Dirac (new, very weakly interacting right-handed neutrinos) or Majorana (neutrino = antineutrino, lepton number not conserved); the counting is the same either way.