Question

A star that is too heavy to stabilize as a white dwarf can collapse further to form a neutron star: a star made entirely of neutrons, supported against gravitational collapse by degenerate neutron pressure. Repeat the steps of the previous problem for a neutron star, to determine the following: the mass radius relation; the radius, density, Fermi energy, and Fermi temperature of a one-solar-mass neutron star; and the critical mass above which a neutron star becomes relativistic and hence unstable to further collapse.

Short answer

The mass radius relation is

$R\propto \frac{1}{M^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

The radius is found to be $12410\mathrm{m}.$

The density of the white dwarf star is $2.5\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3$

The Fermi energy and temperature are $9\times {10}^{-12}\mathrm{J},6.5\times {10}^{11}\mathrm{K}$

The critical mass is given by,$\mathrm{M}=1.38\times {10}^{31}\mathrm{kg}.$

Step-by-step solution

Given information 

We have to calculate the, mass-radius relation; radius of star, density, Fermi energy and Fermi temperature of a one-solar-mass neutron star; and the critical mass above which a neutron star becomes relativistic and hence unstable to further collapse.

Simplify

we now that, the total energy of the white dwarf is given by,

$$\begin{gathered} \mathrm{U}={\mathrm{U}}_{\mathrm{kinetic}}+{\mathrm{U}}_{\mathrm{grav}} \\ \mathrm{U}=(0.0086)\frac{{\mathrm{h}}^2{\mathrm{M}}^{{\displaystyle \frac{5}{3}}}}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\mathrm{R}}^2}-\frac{3}{5}\frac{{\mathrm{GM}}^2}{\mathrm{R}} \end{gathered}$$

Let us consider,

localid="1650022143628" $$\begin{gathered} \mathrm{\alpha }=(0.0086)\frac{{\mathrm{h}}^2{\mathrm{M}}^{{\displaystyle \frac{5}{3}}}}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ \mathrm{\beta }=\frac{3}{5}{\mathrm{GM}}^2 \end{gathered}$$

Then, we can write, the energy as,

localid="1650022013208" $$\begin{gathered} \mathrm{U}=\frac{\mathrm{\alpha }}{{\mathrm{R}}^2}-\frac{\mathrm{\beta }}{\mathrm{R}} \\ \frac{\mathrm{dU}}{\mathrm{dR}}=\frac{-2\mathrm{\alpha }}{{\mathrm{R}}^3}+\frac{\mathrm{\beta }}{{\mathrm{R}}^2}=0 \\ \mathrm{R}=0, \\ \mathrm{R}=\frac{2\mathrm{\alpha }}{\mathrm{\beta }} \end{gathered}$$

Substitute the value of $\mathrm{\alpha },\mathrm{\beta }$then,

localid="1650022149205" $$\begin{gathered} \mathrm{R}=2\times \frac{(0.0086)\frac{{\mathrm{h}}^2{\mathrm{M}}^{{\displaystyle \frac{5}{3}}}}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{\frac{3}{5}{\mathrm{GM}}^2} \\ \mathrm{R}=\left(\frac{(0.0933){\mathrm{h}}^2}{{\mathrm{m}}_{\mathrm{p}}^{{\displaystyle \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\mathrm{G}}\right)\frac{1}{{\mathrm{M}}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \end{gathered}$$

This is the drive relation-ship between the Radius and Mass.

After sustaining all the value,

localid="1650022153954" $\mathrm{R}=12410\mathrm{m}$

Step 3:Simplify

The density of the white dwarf is given by,

$$\begin{gathered} \mathrm{\rho }=\frac{\mathrm{M}}{\mathrm{V}} \\ \mathrm{\rho }=\frac{\mathrm{M}}{{\displaystyle \frac{4}{3}(3.14)\left({\mathrm{R}}^3\right)}} \\ \mathrm{\rho }=2.5\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

The Fermi energy is found out as ,

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{\mathrm{h}}^2}{8{\mathrm{m}}_{\mathrm{e}}}{\left(\frac{9\mathrm{M}}{8{\mathrm{m}}_{\mathrm{p}}{\mathrm{\pi }}^2{\mathrm{R}}^3}\right)}^{\frac{2}{3}} \\ {\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}^2}{8\times 9.1\times {10}^{-31}\mathrm{kg}}{\left(\frac{9\times 2\times {10}^{30}\mathrm{kg}}{8\times (1.67\times {10}^{-27}\mathrm{kg}){\mathrm{\pi }}^2\times {(12410\mathrm{m})}^3}\right)}^{\frac{2}{3}} \\ {\mathrm{\epsilon }}_{\mathrm{f}}=9\times {10}^{-12}\mathrm{J} \end{gathered}$$

The Fermi temperature is

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{B}}\mathrm{T} \\ \mathrm{T}=\frac{{\mathrm{\epsilon }}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{B}}} \\ \mathrm{T}=\frac{9\times {10}^{-12}\mathrm{j}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}} \\ \mathrm{T}=6.5\times {10}^{11}\mathrm{K} \end{gathered}$$

Simplify

For this the kinetic energy and the potential should needs to balance each other at some point.

$$\begin{gathered} {\mathrm{U}}_{\mathrm{kinetic}}={\mathrm{U}}_{\mathrm{grav}} \\ (0.023){\left(\frac{\mathrm{M}}{{\mathrm{m}}_{\mathrm{p}}}\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{\mathrm{hc}}{\mathrm{R}}=\frac{3{\mathrm{GM}}^2}{5\mathrm{R}} \\ \frac{(0.023)(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})(3\times {10}^8\mathrm{m}/\mathrm{s})}{{(1.67\times {10}^{-27}\mathrm{kg})}^{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}=\frac{3{\mathrm{GM}}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{5} \\ \mathrm{M}=1.38\times {10}^{31}\mathrm{kg} \end{gathered}$$