A white dwarf star (see Figure 7.12) is essentially a degenerate electron gas, with a bunch of nuclei mixed in to balance the charge and to provide the gravitational attraction that holds the star together. In this problem you will derive a relation between the mass and the radius of a white dwarf star, modeling the star as a uniform-density sphere. White dwarf stars tend to be extremely hot by our standards; nevertheless, it is an excellent approximation in this problem to set T=0.

(a) Use dimensional analysis to argue that the gravitational potential energy of a uniform-density sphere (mass M, radius R) must equal

${\mathrm{U}}_{\mathrm{grav}}=-\left(\mathrm{constant}\right)\frac{{\mathrm{GM}}^2}{\mathrm{R}}$

where (constant) is some numerical constant. Be sure to explain the minus sign. The constant turns out to equal 3/5; you can derive it by calculating the (negative) work needed to assemble the sphere, shell by shell, from the inside out.

(b) Assuming that the star contains one proton and one neutron for each electron, and that the electrons are nonrelativistic, show that the total (kinetic) energy of the degenerate electrons equals

${\mathrm{U}}_{\mathrm{kinetic}}=(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}$

Figure 7.12. The double star system Sirius A and B. Sirius A (greatly overexposed in the photo) is the brightest star in our night sky. Its companion, Sirius B, is hotter but very faint, indicating that it must be extremely small-a white dwarf. From the orbital motion of the pair we know that Sirius B has about the same mass as our sun. (UCO /Lick Observatory photo.)

( c) The equilibrium radius of the white dwarf is that which minimizes the total energy ${\mathrm{U}}_{\mathrm{gravity}}+{\mathrm{U}}_{\mathrm{kinetic}}$· Sketch the total energy as a function of R, and find a formula for the equilibrium radius in terms of the mass. As the mass increases, does the radius increase or decrease? Does this make sense?

( d) Evaluate the equilibrium radius for $\mathrm{M}=2\times {10}^{30}\mathrm{kg}$, the mass of the sun. Also evaluate the density. How does the density compare to that of water?

( e) Calculate the Fermi energy and the Fermi temperature, for the case considered in part (d). Discuss whether the approximation T = 0 is valid.

(f) Suppose instead that the electrons in the white dwarf star are highly relativistic. Using the result of the previous problem, show that the total kinetic energy of the electrons is now proportional to 1 / R instead of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{R}}^2$}\right.$• Argue that there is no stable equilibrium radius for such a star.

(g) The transition from the nonrelativistic regime to the ultra relativistic regime occurs approximately where the average kinetic energy of an electron is equal to its rest energy, ${\mathrm{mc}}^2$Is the nonrelativistic approximation valid for a one-solar-mass white dwarf? Above what mass would you expect a white dwarf to become relativistic and hence unstable?

we have to drive the relationship between the radius and mass of the white dwarf star.

Using dimensional analysis, we know that the dimensions of the

$$\begin{gathered} \mathrm{G}=\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2 \\ \mathrm{M}=\mathrm{kg} \\ \mathrm{R}=\mathrm{m} \end{gathered}$$

Then,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{grav}}=-\left(\mathrm{constant}\right)\frac{{\mathrm{GM}}^2}{\mathrm{R}} \\ {\mathrm{U}}_{\mathrm{grav}}=\frac{({\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2})\left({\mathrm{M}}^2\right)}{\mathrm{L}} \\ {\mathrm{U}}_{\mathrm{grav}}={\mathrm{ML}}^2{\mathrm{T}}^{-2} \end{gathered}$$

Since the work done by gravitational force which is attractive then sign of gravitational potential is negative.

We have to drive the giving expression by assuming the star contains one proton and one neutron for one electron.

The Fermi energy is given by,

${\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}{\left(\frac{3\mathrm{N}}{\mathrm{\pi V}}\right)}^{\frac{2}{3}}$

and the kinetic energy is given by,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{kinetic}}=\frac{3}{5}{\mathrm{N\epsilon }}_{\mathrm{f}} \\ {\mathrm{U}}_{\mathrm{kinetic}}=\frac{3}{5}\mathrm{N}\left(\frac{{\mathrm{h}}^2}{8\mathrm{m}}{\left(\frac{3\mathrm{N}}{\mathrm{\pi V}}\right)}^{\frac{2}{3}}\right) \end{gathered}$$

Since the star constrains one proton ns one neutron for one electron the,

$\mathrm{N}=\frac{\mathrm{M}}{2{\mathrm{m}}_{\mathrm{p}}}$

and the volume of the electrons in spherical star,

$\mathrm{V}=\frac{4}{3}{\mathrm{\pi R}}^3$

Put this values in above equation of kinetic energy ,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{kinetic}}=\frac{3}{5}\mathrm{N}\left(\frac{{\mathrm{h}}^2}{8\mathrm{m}}{\left(\frac{3\mathrm{N}}{\mathrm{\pi V}}\right)}^{\frac{2}{3}}\right) \\ {\mathrm{U}}_{\mathrm{kinetic}}=\frac{3}{5}\left(\frac{\mathrm{M}}{2{\mathrm{m}}_{\mathrm{p}}}\right)\left(\frac{{\mathrm{h}}^2}{8\mathrm{m}}\right){\left(\frac{3\mathrm{M}}{2\mathrm{m\pi }{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\right)}^{\frac{2}{3}} \\ {\mathrm{U}}_{\mathrm{kinetic}}=\left(\frac{3^{\frac{7}{3}}}{4.4^{\frac{2}{3}}{\mathrm{\pi }}^{{\displaystyle \frac{4}{3}}}{2}^{\frac{5}{3}}}\right)\frac{{\mathrm{h}}^2{\mathrm{M}}^{\frac{5}{3}}}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}^{\frac{5}{3}}{\mathrm{R}}^2} \\ {\mathrm{U}}_{\mathrm{kinetic}}=(0.00880855)\frac{{\mathrm{h}}^2{\mathrm{M}}^{\frac{5}{3}}}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}^{\frac{5}{3}}{\mathrm{R}}^2} \end{gathered}$$

We have to minimize the total energy and to plot the graph between the energy and radius.

Since the total energy of the white dwarf is ,

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

then,

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

After substitute this values we found the graph

As the mass increases the equilibrium radius will reduced.

We have to find the equilibrium radius and density.

Since,

$$\begin{gathered} \mathrm{R}=\frac{2\mathrm{\alpha }}{\mathrm{\beta }} \\ \mathrm{R}=9\times {10}^{16}{\mathrm{M}}^{\frac{-1}{3}} \end{gathered}$$

then put the value of M,

$\mathrm{R}=7.14509\times {10}^6\mathrm{m}$

Then density will be,

localid="1650019901945" $$\begin{gathered} \mathrm{\rho }=\frac{\mathrm{M}}{{\displaystyle \frac{4}{3}}(3.14)\left({\mathrm{R}}^3\right)} \\ \mathrm{\rho }=1.31\times {10}^9\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

We have to calculate the Fermi energy and temperature.

Since, the Fermi energy can be written as,

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{\mathrm{h}}^2}{8{\mathrm{m}}_{\mathrm{e}}}{\left(\frac{9\mathrm{M}}{8{\mathrm{m}}_{\mathrm{p}}{\mathrm{\pi }}^2{\mathrm{R}}^3}\right)}^{\frac{2}{3}} \\ {\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}^2}{8\times 9.1\times {10}^{-31}\mathrm{kg}}{\left(\frac{9\times 2\times {10}^{30}\mathrm{kg}}{8\times (1.67\times {10}^{-27}\mathrm{kg}){\mathrm{\pi }}^2\times {(7.1\times {10}^6\mathrm{m})}^3}\right)}^{\frac{2}{3}} \end{gathered}$$

$$\begin{gathered} {\mathrm{ϵ}}_{\mathrm{f}}=3.12437\times {10}^{-14}\mathrm{J} \\ \mathrm{T}=2.3\times {10}^9\mathrm{K} \end{gathered}$$

We have to find the kinetic energy relation with the radius.

Since the chemical potential is given by,

$\mathrm{\mu }=\frac{\mathrm{hc}}{2}{\left(\frac{3\mathrm{N}}{\mathrm{\pi V}}\right)}^{\frac{1}{3}}$

and then the we found total energy which is the energy occupied by the electron and we also consider the factor of 2 due to half spin of electron in one energy level.

Then, the total energy is given by,

$\mathrm{U}=2\sum _{{\mathrm{n}}_{\mathrm{x}}}\sum _{{\mathrm{n}}_{\mathrm{y}}}\sum _{{\mathrm{n}}_{\mathrm{z}}}\mathrm{\epsilon }\left(\mathrm{n}\right)$

Then, the spherical coordinates system,

the volume term will be,

$\mathrm{dV}={\mathrm{n}}^2\mathrm{sin\theta }\mathrm{dn}\mathrm{d\theta }\mathrm{d\varphi }$

Then, the energy will becomes,

$\mathrm{U}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_0^{{\mathrm{n}}_{\max }}\mathrm{\epsilon }\left(\mathrm{n}\right){\mathrm{n}}^2\mathrm{sin\theta }\mathrm{dn}\mathrm{d\theta }\mathrm{d\varphi }$

since,

$$\begin{gathered} ϵ=\frac{hc}{2L}\sqrt{n_x^2+n_y^2+n_z^2} \\ ϵ=\frac{hcn}{2L} \end{gathered}$$

Then,

$$\begin{gathered} \mathrm{U}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_0^{{\mathrm{n}}_{\max }}\frac{\mathrm{hcn}}{2\mathrm{L}}{\mathrm{n}}^2\mathrm{sin\theta }\mathrm{dn}\mathrm{d\theta }\mathrm{d\varphi } \\ U=\frac{hc}{L}\left[\frac{\pi }{2}\right]\left[1\right]\left[\frac{n_{\max }^4}{4}\right] \\ U=\frac{hc\pi n_{\max }^4}{8L} \end{gathered}$$

since,

${\mathrm{n}}_{\max }={\left(\frac{3\mathrm{N}}{\mathrm{\pi }}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Then,

$$\begin{gathered} U=\frac{3hcN}{8}{\left(\frac{3N}{\pi V}\right)}^{1/3} \\ U=\frac{3N}{4}\frac{hc}{2}{\left(\frac{3N}{\pi V}\right)}^{1/3} \\ U\mathit{\propto }\frac{\mathit{1}}{\mathit{R}} \end{gathered}$$

Since it is inversely proportional to R then, we an say that it is not have stable equilibrium point.

We have given, the transition is from non relativistic to the ultra relativistic of the electron has been occur.

we have to find the mass for which white dwarf becomes relativistic and unstable.

Since average kinetic energy of the electron is given by,

$$\begin{gathered} \frac{\mathrm{U}}{\mathrm{N}}=0.6{\epsilon }_f=1.87462\times {10}^{-14}J \\ and, \\ {m}_e{c}^2=8.178105\times {10}^{14}J \end{gathered}$$

then, we can say that the nonrelativistic approach will be same as relativistic only for one solar mass white dwarf.

Then,

To find the most stable mass the kinetic energy should be equal to the gravitation potential energy.

localid="1650020379881" $$\begin{gathered} {\mathrm{U}}_{\mathrm{kinetic}}={\mathrm{U}}_{\mathrm{grav}} \\ (0.091){\left(\frac{\mathrm{M}}{{\mathrm{m}}_{\mathrm{p}}}\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\mathrm{hc}=\frac{3{\mathrm{GM}}^2}{5} \\ \frac{(0.091)(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})(3\times {10}^8\mathrm{m}/\mathrm{s})}{{\mathrm{m}}_{\mathrm{p}}^{{\displaystyle \frac{4}{3}}}}=\frac{3{\mathrm{GM}}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{5} \\ \mathrm{M}=3.44\times {10}^{30}\mathrm{kg}. \end{gathered}$$