The results of the previous problem can be used to explain

Chapter 7, Problem 50P

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Problem 50P

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where KT/c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they “heated” the photon radiation but not the neutrino radiation.

(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have been conserved in the early universe, assuming that no other species of particles interacted with these.

(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by

as the universe expands and cools. Evaluate the constant by assuming that T = Tv, when the temperatures are very high.

(c) Calculate the ratio T/Tv in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K.

(d) Use a computer to plot the ratio T/Tv as a function of T, for kT/mc2 ranging from 0 to 3

Problem:

In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and anti neutrinos , currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless.

(a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of antineutrinos, so t hat their chemical potentials are equal:  Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction

(where γ is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that µ = 0 for both the neutrinos and the antineutrinos.

(b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. (Hint: There are very few differences between this “neutrino gas” and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a Lable or consult Appendix B.

(c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K.

(d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn’t have affected the production of neutrinos in the early universe, when mc2 would have been negligible compared to typical thermal energies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass m. What would mc2 have to be (in eV), in order for the total mass of neutrinos in the universe to be comparable to the to the mass of ordinary matter?