For a brief time in the early universe, the temperature

Chapter 7, Problem 49P

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Problem 49P

For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of “background radiation,” in addition to the photons and neutrinos (see Figure). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (each with the same mass), and each has two independent polarization states. During the time period of interest the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in the previous problem. Recall from special relativity that the energy of a massive particle is

(a) Show that the energy density of electrons and positrons at temperature T is given by

(b) Show that u(T) goes to zero when kT ≪ mc2 and explain why this is a reasonable result.

(c) Evaluate u(T) in the limit kT ≫ mc2, and compare to the result of the previous problem for the neutrino radiation.

(d) Use a computer to calculate and plot u(T) at intermediate temperatures.

(e) Use the method of Problem, part (d), to show that the free energy density of the electron-positrou radiation is

Evaluate f(T) in both limits, and use a computer to calculate and plot f(T) at intermediate temperatures.

(f) Write the entropy of, the electron-positron radiation in terms of the functions u(T) and f(T). Evaluate the entropy explicitly in the high-T limit.

Figure: When the temperature was greater than the electron mass times c2/k the universe was filled with three types of radiation: electrons and positrons (solid arrows); neutrinos (dashed); and photons (wavy). Bathed in this radiation were a few protons and neutrons, roughly one for every billion radiation particles.

Problem:

Sometimes it is useful to know the free energy of a photon gas.

d) A more interesting way to calculate F is to apply the formula F = −KT In Z separately to each mode (that is, each effective oscillator), then sum over all modes Carry out this calculation, to obtain

Integrate by parts, and check that your answer agrees with part (a).