When a star has nearly bumped up its intimal fuel, it may become a white dwarf. It is crushed under its own enormous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle, "squeezing" the potential well necessarily increases the energies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is minimum total energy and corresponding equilibrium radius.

(a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67) is
${\mathit{U}}_{\mathbf{c}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{i}\mathbf{m}\mathbf{n}\mathbf{s}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}{\mathbf{\left(}\frac{\mathbf{3}{\mathbf{\pi }}^{\mathbf{2}}\sout{{\mathbf{h}}^{\mathbf{3}}}}{{\mathbf{m}}_{\mathbf{e}}^{\raisebox{1ex}{$\mathbf{3}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{V}}\mathbf{\right)}}^{\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}{\mathit{N}}^{\raisebox{1ex}{$\mathbf{5}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}$

Note that as the volume Vis decreased, the energy does increase. For a neutral star. the number of electrons, N, equals the number of protons. If protons account for half of the white dwarf's mass M (neutrons accounting for the other half). Show that the minimum electron energy may be written

${\mathit{U}}_{\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{c}\mathbf{t}\mathbf{r}\mathbf{o}\mathbf{n}\mathbf{s}}\mathbf{=}\frac{\mathbf{9}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{80}{\mathbf{m}}_{\mathbf{e}}}{\left(\frac{3{\pi }^2{M}^5}{m_p^5}\right)}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}\frac{\mathbf{1}}{{\mathbf{R}}^{\mathbf{2}}}$

Where, R is the star's radius?

(b) The gravitational potential energy of a sphere of mass Mand radius Ris given by

${\mathit{U}}_{\mathbf{g}\mathbf{r}\mathbf{a}\mathbf{y}}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{5}}\frac{\mathbf{G}{\mathbf{M}}^{\mathbf{2}}}{\mathbf{R}}$

Taking both factors into account, show that the minimum total energy occurs when

$\mathit{R}\mathbf{=}\frac{\mathbf{3}\sout{{\mathbf{h}}^{\mathbf{2}}}}{\mathbf{8}\mathbf{G}}{\mathbf{\left(}\frac{\mathbf{3}{\mathbf{\pi }}^{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{e}}^{\mathbf{3}}{\mathbf{m}}_{\mathbf{p}}^{\mathbf{5}}\mathbf{M}}\mathbf{\right)}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

(c) Evaluate this radius for a star whose mass is equal to that of our Sun 2x10³⁰kg.

(d) White dwarfs are comparable to the size of Earth. Does the value in part (c) agree?

The minimum energy that is allowed by the exclusion principle is given by,

$U_{\mathrm{electrons}}=\frac{3}{10}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{h}}^3}{{\mathrm{m}}_{\mathrm{e}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{V}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Formula used:

The expression for the gravitational energy of a sphere of mass M and radius Ris given by,

${\mathit{U}}_{\mathbf{g}}\mathbf{=}\frac{\mathbf{3}\mathbf{G}{\mathbf{M}}^{\mathbf{2}}}{\mathbf{5}\mathbf{R}}$

The expression for the volume of the sphere is given by.

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

The minimum energy that is allowed by the exclusion principle is

$\begin{array}{rcl}{U}_{electrons}& =& \frac{3}{10}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{h}}^3}{m_e^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}V}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{3{\mathrm{h}}^2}{10m_e}{\left(\frac{3{\mathrm{\pi }}^2}{V}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{3{\mathrm{h}}^2}{10m_e}{\left(\frac{3{\mathrm{\pi }}^2}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{3{\mathrm{h}}^2}{10m_e}{\left(\frac{9\mathrm{\pi }}{{\displaystyle 4R^3}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}$

On further solving,

$$\begin{gathered} U_{electron}=\frac{3h^2}{10m_e}{\left(\frac{81{\mathrm{\pi }}^2}{16}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{N^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{R^2} \\ {U}_{electron}=\frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{\pi }}^2{\mathrm{N}}^5}{16}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$

Since star is assumed electrically neutral thus the number of protons N that makes the half of dwarf's mass M is expressed as:

$\begin{array}{rcl}\frac{M}{2}& =& N{m}_P\\ N& =& \frac{M}{2m_P}\end{array}$

Put the value of N in equation (I),

role="math" localid="1659165527416" $\begin{array}{rcl}{U}_{electron}& =& \frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{\pi }}^2{\left({\displaystyle \frac{\mathrm{M}}{2{\mathrm{m}}_{\mathrm{p}}}}\right)}^5}{16}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{\pi }}^2\left({\displaystyle \frac{{\mathrm{M}}^5}{32{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}}\right)}{16}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{\pi }}^2{\mathrm{M}}^5}{512{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{10m_e{R}^2}\left(\frac{3}{8}\right){\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)}^{\frac{1}{3}}\end{array}$

On further solving,

role="math" localid="1659165629679" $U_{electron}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{9h^2}{80m_e{R}^2}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$

Therefore, the minimum electron energy is equal to $U_{electron}=\begin{array}{rcl}& & \frac{9h^2}{80m_e{R}^2}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$.

Formula used:

The expression for the total energy of the star is given by,

${\mathit{U}}_{\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{a}\mathbf{l}}\mathbf{=}{\mathit{U}}_{\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{c}\mathbf{t}\mathbf{r}\mathbf{o}\mathbf{n}}\mathbf{+}{\mathit{U}}_{\mathbf{g}}$

The expression for the gravitational energy of a sphere of mass Mand radius Ris given by,

${\mathit{U}}_{\mathbf{g}}\mathbf{=}\frac{\mathbf{3}\mathbf{G}{\mathbf{M}}^{\mathbf{2}}}{\mathbf{5}\mathbf{R}}$

The total energy of the star is calculated as:

$\begin{array}{rcl}{U}_{Total}& =& U_{electron}+U_g\\ & =& \frac{9h^2}{80m_{\mathcal{l}}{R}^2}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}+\frac{3G{M}^2}{5R}\end{array}$

The minimum total energy is calculated by taking its derivative with respect radius equal to zero.

$\begin{array}{rcl}\frac{d}{dR}\left(U_{Total}\right)& =& 0\\ \frac{d}{dR}\left[\frac{9h^2}{80m_e{R}^2}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}-\frac{3G{M}^2}{5R}\right]& =& 0\\ \left(-2\right)\frac{9h^2}{80m_e{R}^3}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}-\left(-1\right)\frac{3G{M}^2}{5R^2}& =& 0\\ \frac{-9h^2}{80m_e{R}^3}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}+\frac{3G{M}^2}{5R^2}& =& 0\end{array}$

On further solving,

$\begin{array}{rcl}\frac{G{M}^2}{R^2}& =& \frac{3h^2}{8m_e{R}^3}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}\\ G{M}^2& =& \frac{3h^2}{8m_e{R}^3}{\left(\frac{3{\mathrm{\pi }}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{8G}{\left(\frac{3{\mathrm{\pi }}^2}{m_e^3{m}_p^{5}M}\right)}^{\frac{1}{3}}\end{array}$

Therefore, the minimum total energy occurs when radius of star role="math" localid="1659167163703" $R\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{3h^2}{8G}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{\pi }}^2}{m_e^3{m}_p^{5}M}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$.

The mass of sun is m = 2x10³⁰kg.

Formula used:

The expression for the radius of the stars given by,

$R\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{3h^2}{8G}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{\pi }}^2}{m_e^3{m}_p^{5}M}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$

The radius of the star is calculated as:

$\begin{array}{rcl}R& =& \frac{3h^2}{8G}\left(\frac{3{\mathrm{\pi }}^2}{m_e^3{m}_p^{5}M}\right){}^{\frac{1}{3}}\\ & =& \frac{3{\left(1.055\times {10}^{-34}J·s\right)}^2}{8\left(6.67\times {10}^{-11}N.m^2/kg\right)}\left(\frac{3{\left(3.14\right)}^2}{{\left(9.1\times {10}^{-31}Kg\right)}^3{\left(1.67\times {10}^{-27}kg\right)}^5\left(2\times {10}^{30}kg\right)}\right)\frac{1}{3}\\ & =& 6.26\times {10}^{-59}{\left(0.00151\times {10}^{198}\right)}^{\frac{1}{3}}\\ & =& 6.26\times {10}^{-59}\times 0.11T4\times {10}^{66}\\ R& =& 0.713\times {10}^{7}m\\ & =& 7.13\times {10}^{6}m\\ & =& 7\times {10}^{6}m\\ & & \end{array}$

Therefore, the radius of a star with mass as that of our sun is7x10⁶m .