Neutrinos are very weakly interacting, electrically neutral particles that are involved in nuclear interactions where protons and changed into neutrons or vice versa, and in other reactions as well. An example of a weak nuclear interaction involving a neutrino is the free neutron decay

neutron --> proton + electron + anti-neutrinoThis decay has a mean life of 887 seconds or a half life of 10.25 minutes. The cross section for a typical interaction involving a neutrino is 5*10

Because the Universe was once so hot and dense that even neutrinos interacted many times during the Hubble time 1/H, there once was a thermal background of neutrinos in equilibrium with the thermal background of photons that is the CMBR. But since neutrino interactions are so weak, this thermal equilibrium only survived until 1 second after the Big Bang. But the neutrino background is still present today, with about 56 electron neutrinos, 56 electron anti-neutrinos, 56 muon neutrinos, etc., per cubic centimeter, for a total of 337 neutrinos per cubic centimeter in the Universe. The photons of the CMBR are slightly more numerous with 411 photons/cc.

Because the number of neutrinos in the Universe is so large, even a very small neutrino mass can have drastic consequences for cosmology. So experiments to measure the neutrino mass are obviously very significant. There are three ways to detect a neutrino mass:

- The energy spectrum of the observable electrons in a radioactive beta decay is modified if the electron neutrino has a non-zero mass. The unseen neutrinos are emitted uniformly in momentum, but for a massive neutrino the change in energy for momenta up to about 0.5*m*c is small, so a relatively large number of electrons are emitted at close to the maximum energy. This leads to a large number of electrons being emitted at just below the maximum possible energy, and thus an abrupt decline to zero at the maximum possible energy. For zero rest mass, the number of electrons per unit energy declines at a constant slope to zero at the maximum energy. What is actually seen is a smooth transition from zero to a constant slope instead of the "cliff" expected for non-zero mass, so only an upper limit of several eV is obtained for the mass of the electron neutrino. Radioactive decays producing muon and tau neutrinos are very hard to observe, so this method gives very weak limits on their masses.
- Time-of-flight data can be used to detect the fact that massive
neutrinos travel slightly slower than the speed of light.
The speed is given by
v/c = sqrt[1-(mc

which leads to an arrival time difference of^{2}/E)^{2}] = 1 - 0.5*(mc^{2}/E)^{2}+ ...Dt = (L/c)*0.5*(mc

Since Dt < 10 seconds with (L/c) = 5*10^{2}/E)^{2}^{12}seconds for 10 MeV neutrinos from SN1987A in the LMC, the mass is less than 20 eV. All the detected neutrinos from SN1987A were electron neutrinos, so this limit only applies to one of the three types. The Sudbury Neutrino Observatory (SNO) will be able to detect all three types of neutrinos, and if we are lucky enough to have a nearby supernova SNO may be able to improve the limits on the muon and tau neutrinos. - Neutrino oscillations: quantum mechanics says that the wavefunction
for any particle oscillates at a frequency of mc
^{2}/h cycles per second,*in the rest frame of the particle.*Relativity says that a particle sees a time interval in its rest frame that is smaller by the factor (mc^{2}/E) than the lab frame time interval due to time dilation. Thus the effective lab frame oscillation frequency of a particle is [mc^{2}]^{2}/[Eh] and the*beat frequency*between two particles of different masses isDf = [c

which is 2.7 kHz for m^{4}/Eh]*[m_{2}^{2}-m_{1}^{2}]_{1}= 0, m_{2}c^{2}= 0.1 eV, and E = 1 GeV. Since it takes 40 msec for neutrinos to travel through the Earth, many cycles of the beat frequency can occur and the neutrinos become a mixture of types when traveling through the Earth.

At the time of weak decoupling, about 1 second after the Big Bang,
the neutrinos and the photon-electron-positron plasma had the
same temperature, which I will call T_{n}. All these particles
were relativistic since k*T_{n} > 1 MeV, where k is the Boltzmann
constant. The energy of a relativistic plasma occupying a
volume a^{3} ("a" will be the scale factor of the Universe) is

Q = (2\sigma/c)(gwhere \sigma is the Stefan-Boltzmann constant, c is the speed of light, and g_{b}+(7/8)g_{f})a^{3}T^{4}

Thus the photon-electron-positron plasma has

Q = (4\sigma/c)(1 + 7/4) aAs the Universe expands and cools adiabatically the entropy in the volume a^{3}T^{4}

dS = dQ/Tor

S = (4\sigma/c)(1 + 7/4) agiving^{3}\int 4 T^{2}dT

S = (4/3)(4\sigma/c)(1 + 7/4) aIn the absence of annihilation conservation of entropy gives aT = constant.^{3}T^{3}

During the period from 1 second after the Big Bang until
3 minutes after the Big Bang the temperature falls to
well below the rest mass of the electron. Thus the
electron-positron plasma annihilates and transfers
its energy and entropy to the photons. This leaves
the photons with a temperature T_{p} that is larger
than the neutrino temperature T_{n}. Energy is
not conserved in an adiabatically expanding gas
because the pressure of the gas does external work,
but the entropy is conserved. Thus the entropy
before from photons and electron-positron pairs,

(4/3)(4\sigma/c)(1 + 7/4) ais equal to the entropy afterward just from photons,^{3}T_{n}^{3},

(4/3)(4\sigma/c) aso^{3}T_{p}^{3},

TThe photon temperature now is T_{n}/T_{p}= (4/11)^{1/3}.

If neutrinos are massless then we can compute their equivalent mass density using

rho = [Q/awhere N^{3}]/c^{2}= N_{n}(2\sigma/c)(7/8)g_{f}T_{n}^{4}

Even though this density is negligible now, it was significant during the time that helium was formed during Big Bang Nucleosynthesis. The increased density due to the neutrino background during helium synthesis caused the universe to expand faster, and this reduced the time required for the temperature to fall to the point where deuterium could survive. As a result the helium abundance is a few percent larger than it would have been without the neutrino background.

If the neutrinos are not massless, then they could have a larger mass density now consisting of their number density times their rest mass. Each neutrino species has a number density of

n = (3/4)(4\pi)\Gamma(3)\zeta(3) (kTwhere \Gamma is the gamma function (\Gamma(n+1) = n! so \Gamma(3) = 2), \zeta is the zeta function_{n}/hc)^{3}

\zeta(s) = 1 + 1/2With T^{s}+ 1/3^{s}+ ... and \zeta(3) = 1.202...

rho = 112*(mwhere m_{n-e}+ m_{n-mu}+ m_{n-tau})

rho = 2*10compared to the critical density of 8*10^{-31}(m_{n-e}+ m_{n-mu}+ m_{n-tau}) gm/cc

Thus a neutrino rest mass of 40 eV for one type would give the critical density in neutrinos, and a rest mass of 10 eV for one type or a sum of rest masses of 10 eV would be a significant factor in the formation of large scale structures in the Universe such as clusters and superclusters of galaxies.

For these masses the neutrinos are traveling slowly now but their thermal velocities were large in the past. The typical momentum of a relativistic particle in a thermal distribution is p = 3kT/c, and the product of the scale factor and the momentum, ap, is a constant. Thus neutrinos with rest mass m will be moving at redshift z with a typical velocity

v = pc/sqrt[pand the distance traveled, measured now (the comoving distance), is^{2}+ (mc)^{2}] = 3(1+z)k(1.947 K)/mc - ...

D = \int (1+z) v dt = 2(c/Hwhich gives D = 100 Mpc for mc_{o}) sqrt(3*k*(1.947 K))/mc^{2}) + ...

Tutorial:
Part 1 |
Part 2 |
Part 3 |
Part 4

FAQ |
Age |
Distances |
Bibliography |
Relativity

© 1997-1998 Edward L. Wright. Last modified 21-Sep-1998