Question

We can extend the theory of a free electron gas (Section 5.3.1) to the relativistic domain by replacing the classical kinetic energy, $\mathbf{E}\mathbf{=}{\mathbf{p}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{m}\mathbf{,}$,with the relativistic formula, $\mathbf{E}\mathbf{=}\sqrt{{\mathbf{p}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{4}}}\mathbf{-}{\mathbf{mc}}^{\mathbf{2}}$. Momentum is related to the wave vector in the usual way: $\mathbf{p}\mathbf{=}\sout{\mathbf{h}}\mathbf{k}$. In particular, in the extreme relativistic limit, $\mathbf{E}\mathbf{\approx }\mathbf{pc}\mathbf{=}\sout{\mathbf{h}}\mathbf{ck}\mathbf{.}$

(a) Replace ${\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{k}}^{\mathbf{2}}$n Equation 5.55 by the ultra-relativistic expression, $\sout{\mathbf{h}}\mathbf{ck}$, and calculate${\mathbf{E}}_{\mathbf{tot}}$in this regime.

$\mathbf{dE}\mathbf{=}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{k}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{V}}{{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{k}}^{\mathbf{2}}\mathbf{dk}$ (5.55).

(b) Repeat parts (a) and (b) of Problem 5.35 for the ultra-relativistic electron gas. Notice that in this case there is no stable minimum, regardless of R; if the total energy is positive, degeneracy forces exceed gravitational forces, and the star will expand, whereas if the total is negative, gravitational forces win out, and the star will collapse. Find the critical number of nucleons, Nc , such that gravitational collapse occurs for $\mathbf{N}\mathbf{>}\mathbf{N\_}\left\{C\right\}$is called the Chandrasekhar limit.

(c) At extremely high density, inverse beta decay${\mathbf{e}}^{\mathbf{-}}\mathbf{+}{\mathbf{p}}^{\mathbf{+}}\mathbf{\to }\mathbf{n}\mathbf{+}\mathbf{v}$,converts virtually all of the protons and electrons into neutrons (liberating neutrinos, which carry off energy, in the process). Eventually neutron degeneracy pressure stabilizes the collapse, just as electron degeneracy does for the white dwarf (see Problem 5.35). Calculate the radius of a neutron star with the mass of the sun. Also calculate the (neutron) Fermi energy, and compare it to the rest energy of a neutron. Is it reasonable to treat a neutron star non relativistic ally?

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{value}\mathrm{of}{\mathrm{E}}_{\mathrm{tot}}\mathrm{is}\mathrm{E}=\frac{\sout{\mathrm{h}}\mathrm{c}}{3\mathrm{\pi R}}{\left(\frac{9\mathrm{\pi Nq}}{4}\right)}^{4/3} \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{critical}\mathrm{number}\mathrm{of}\mathrm{nucleons}\mathrm{is}{\mathrm{N}}_{\mathrm{c}}=2.05.{10}^{57} \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{Fermi}\mathrm{energy}\mathrm{is}{\mathrm{E}}_{\mathrm{F}}=55.6\mathrm{MeV} \end{gathered}$$

Step-by-step solution

Definition of Chemical potential

The chemical energy per mole of a substance is its "chemical potential." Gibbs free energy is defined here as chemical energy, and the substance can either be a single, pure substance or a system of several substances.

Calculate Etot

(a)

Now energy in ultra-relativistic regime is:

$$\begin{gathered} \mathrm{dE}=\frac{\mathrm{V}}{{\mathrm{\pi }}^2}\sout{\mathrm{h}}{\mathrm{ckk}}^2\mathrm{dk} \\ =\frac{\mathrm{V}\sout{\mathrm{h}}\mathrm{c}}{{\mathrm{\pi }}^2}{\mathrm{k}}^3\mathrm{dk} \\ \mathrm{E}=\frac{\mathrm{V}\sout{\mathrm{h}}\mathrm{c}}{{\mathrm{\pi }}^2}{\int }_0^{{\mathrm{k}}_{\mathrm{p}}}{\mathrm{k}}^3\mathrm{dk} \\ =\frac{\mathrm{V}\sout{\mathrm{h}}\mathrm{c}}{{\mathrm{\pi }}^2}\frac{{\mathrm{k}}_{\mathrm{F}}^4}{4} \\ {\mathrm{k}}_{\mathrm{F}}={\left(3{\mathrm{\pi }}^2\frac{\mathrm{Nq}}{\mathrm{V}}\right)}^{1/3} \\ \mathrm{E}=\frac{\mathrm{V}\sout{\mathrm{h}}\mathrm{c}}{4{\mathrm{\pi }}^2}{\left(3{\mathrm{\pi }}^2\frac{\mathrm{Nq}}{\mathrm{V}}\right)}^{4/3} \\ =\frac{\sout{\mathrm{h}}\mathrm{c}}{4}{\left(3\mathrm{Nq}\right)}^{4/3}{\left(\frac{{\mathrm{\pi }}^2}{\mathrm{V}}\right)}^{1/3} \\ \mathrm{V}=\frac{4{\mathrm{\pi R}}^3}{3} \\ \mathrm{E}=\frac{\sout{\mathrm{h}}\mathrm{c}}{3\mathrm{\pi R}}{\left(\frac{9\mathrm{\pi Nq}}{4}\right)}^{4/3} \end{gathered}$$

Finding the critical number of neutrons.

(b)

We calculate total energy:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{tot}}=\frac{\mathrm{A}}{\mathrm{R}}-\frac{\mathrm{B}}{\mathrm{R}}, \\ \mathrm{A}=\frac{\sout{\mathrm{h}}\mathrm{c}}{3\mathrm{\pi }}{\left(\frac{9\mathrm{\pi Nq}}{4}\right)}^{4/3},\mathrm{B}=\frac{3{\mathrm{GN}}^2{\mathrm{M}}^2}{5} \\ \frac{{\mathrm{dE}}_{\mathrm{tot}}}{\mathrm{dR}}=0=\frac{1}{{\mathrm{R}}^2}\left(\mathrm{B}-\mathrm{A}\right)\Rightarrow \mathrm{A}=\mathrm{B} \end{gathered}$$

We have no condition on the radius of a star. But we can calculate critical number of nucleons.

$$\begin{gathered} \frac{\sout{\mathrm{h}}\mathrm{c}}{3\mathrm{\pi }}{\left(\frac{9{\mathrm{\pi N}}_{\mathrm{c}}\mathrm{q}}{4}\right)}^{4/3}=\frac{3{\mathrm{GN}}_{\mathrm{C}}^2{\mathrm{M}}^2}{5} \\ {\mathrm{N}}_{\mathrm{C}}^{2/3}=\frac{\sout{\mathrm{h}}\mathrm{c}}{3\mathrm{\pi }}\frac{5}{3{\mathrm{GM}}^2}{\left(\frac{9\mathrm{\pi q}}{4}\right)}^{4/3} \\ {\mathrm{N}}_{\mathrm{C}}={\left(\frac{5\sout{\mathrm{h}}\mathrm{c}}{9{\mathrm{\pi GM}}^2}\right)}^{3/2}{\left(\frac{9\mathrm{\pi q}}{4}\right)}^2 \\ {\mathrm{N}}_{\mathrm{C}}=2.05.{10}^{57} \end{gathered}$$

Stellar mass is equal to:

$$\begin{gathered} \frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{sun}}}=\frac{\mathrm{M}.{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{M}}_{\mathrm{sun}}} \\ =1.72\mathrm{M} \\ =1.72{\mathrm{M}}_{\mathrm{sun}} \end{gathered}$$

Converting all the protons and electrons into neutrons

(c)

From previous task we have:

$\mathrm{R}={\left(\frac{9\mathrm{\pi }}{4}\right)}^{2/3}\frac{{\sout{\mathrm{h}}}^2{\mathrm{q}}^{5/3}}{{\mathrm{mM}}^2{\mathrm{GN}}^{1/3}},\mathrm{q}=1,\mathrm{M}={\mathrm{m}}_{\mathrm{e}},\mathrm{m}={\mathrm{m}}_{\mathrm{N}}$

Because now we have neutron degeneracy instead of electron degeneracy.

$$\begin{gathered} \mathrm{R}={\left(\frac{9\mathrm{\pi }}{4}\right)}^{2/3}\frac{{\sout{\mathrm{h}}}^2}{{\mathrm{m}}_{\mathrm{n}}{\mathrm{m}}_{\mathrm{e}}^2{\mathrm{GN}}^{1/3}} \\ \mathrm{R}=1.31.{10}^{23{\mathrm{N}}^{-1/3}\mathrm{m}} \end{gathered}$$

For Sun we have R=12,4 km.

Fermi energy is equal to:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{F}}=\frac{{\sout{\mathrm{h}}}^2}{2{\mathrm{m}}_{\mathrm{n}}}{\left(3{\mathrm{\pi }}^2\frac{3\mathrm{Nq}}{4{\mathrm{\pi R}}^3}\right)}^{2/3}=8.9203635.{10}^{-12}\mathrm{J} \\ ={\mathrm{E}}_{\mathrm{F}}=55.6\mathrm{MeV} \end{gathered}$$

Rest energy of a neutron is 940MeV which is more than Fermi energy, so we can say that neutrons are non-relativistic.