Q. 7.49 - An Introduction to Thermal Physics

Question

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 7.21). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (ea.ch 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 Figure 7.21. When the temperature was greater than the electron mass times $\raisebox{1ex}{$c^2$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$, 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. the previous problem. Recall from special relativity that the energy of a massive particle is $ϵ=\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}$.

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

$u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}{e^{\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}+1}dx; where u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}{e^{\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}+1}dx$

(b) Show that $u\left(T\right)$ goes to zero when $kT\ll m{c}^2$ , and explain why this is a

reasonable result.

( c) Evaluate $u\left(T\right)$ in the limit $kT\gg m{c}^2$, and compare to the result of the

the previous problem for the neutrino radiation.

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

(e) Use the method of Problem 7.46, part (d), to show that the free energy

density of the electron-positron radiation is

$\frac{F}{V}=-\frac{16\pi (kT{)}^4}{(hc{)}^3}f\left(T\right); where f\left(T\right)={\int }_0^{\infty }{x}^2\ln \left(1+e^{-\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}\right)dx$

Evaluate $f\left(T\right)$ in both limits, and use a computer to calculate and plot $f\left(T\right)$ at intermediate

temperatures.

(f) Write the entropy of the electron-positron radiation in terms of the functions

$u\left(T\right)$ and $f\left(T\right)$. Evaluate the entropy explicitly in the high-T limit.

Step-by-Step Solution

Answer

(a). The energy density of electrons is $U=\frac{16\pi (kT{)}^{4}V}{(hc{)}^3}u\left(T\right)$

(b). By observing the graph, it can be concluded that the result is reasonable.

(c). The result of the neutrino radiation

$U=\frac{14{\pi }^{5}V(kT{)}^4}{15(hc{)}^3}$

(d).  The plot at intermediate temperatures is

(e). Since, $\frac{F}{V}=-\frac{16\pi (kT{)}^4}{(hc{)}^3}f\left(T\right)$. Hence, Proved.

(f). The entropy of the electron-positron radiation $F=-\frac{16\pi V(kT{)}^4}{(hc{)}^3}f\left(T\right)$.

Part(a) Step 1: Given information

We have to prove that energy density of electron and positrons is given by $u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}{e^{\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}+1}dx$

Part(a) Step 2: Solution

Probability of any single state state to be occupied${\overline{n}}_{\mathrm{FD}}=\frac{1}{e^{ϵ/kT}+1}$

Total energy equals$U=2·2\sum _{n_x}\sum _{n_y}\sum _{n_z}ϵ{\overline{n}}_{\mathrm{FD}}$

Energy of massive particle$ϵ=\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}$

$U=4\sum _{n_x,n_y,n_z}\frac{\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}}{e\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}/kT+1}$

Changing sum to integral $U=4{\int }_0^{\pi /2}d\Phi {\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{\infty }\frac{n^2\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}}{e^{\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}/kT}+1}dn$

$U=2\pi {\int }_0^{\infty }\frac{n^2\sqrt{(hcn/2L{)}^2+{\left(m{c}^2\right)}^2}}{e^{\sqrt{(hcn/2L{)}^2+{\left(m{c}^2\right)}^2}/kT}+1}dn$

$U=2\pi {\left(\frac{2LkT}{hc}\right)}^3{\int }_0^{\infty }\frac{x^2\sqrt{(xkT{)}^2+{\left(m{c}^2\right)}^2}}{e^{\sqrt{(xkT{)}^2+{\left(m{c}^2\right)}^2}/kT}+1}dx$

$U=\frac{16\pi (kT{)}^{4}V}{(hc{)}^3}u\left(T\right)$ where$u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}}{e\sqrt{x^2+{\left(m{c}^2/kT\right)}^2}+1}dx$

Part(b) Step 1: Given information:

We have to Evaluate :$u(T\to 0)=0$

Part(b) Step 2: Simplify

$u\left(T\right)={\int }_0^{\infty }\frac{x^3}{e^x+1}dx$

${\int }_0^{\infty }\frac{x^3}{e^x+1}dx=\frac{7{\pi }^4}{120}$

$U=\frac{14{\pi }^{5}V(kT{)}^4}{15(hc{)}^3}$

Part(c) Step 1: Given information

We have been given that $u\left(T\right)={\int }_0^{\infty }\frac{x^2\sqrt{x^2+(1/t{)}^2}}{e^{\sqrt{x^2+(1/t{)}^2}}+1}dx$

Part(c) Step 2:Simplify

To plot the function,we get graph

Part(d) Step 1: given information

we have been given that $F=-kT\ln \left(Z\right)$

Part(d) Step 2: Simplify

$F=-4\sum _{n_x}\sum _{n_y}\sum _{n_z}kT\ln \left(1+e^{-ϵ/kT}\right)=-4\sum _{n_x,n_y,n_z}kT\ln \left(1+e^{-ϵ/kT}\right)$

$ϵ=\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}$

$F=-\frac{16\pi V(kT{)}^4}{(hc{)}^3}f\left(T\right)$

Part(e) Step 1:Given information

Se have been given that $F=-kT\ln \left(1+e^{-ϵ/kT}\right)$

Part(e) Step 2:Simplify

The spherical coordinator is

$F=-4kT{\int }_0^{\pi /2}d\Phi {\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{\infty }{n}^2\ln \left(1+e^{-ϵ/kT}\right)dn$

$ϵ=\sqrt{(pc{)}^2+{\left(m{c}^2\right)}^2}$

Prt(f) Step 1:Given information

We have been that $F=-\frac{16\pi V(kT{)}^4}{(hc{)}^3}f\left(T\right)$

Part(f) Step 2:Simplify

The Helmotz free energy us 

$S=\frac{16\pi V(kT{)}^3}{(hc{)}^3}\left(u\right(T)+f(T\left)\right)k$

$S=\frac{56{\pi }^{5}V(kT{)}^3}{45(hc{)}^3}k$