91影视

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,

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

and the kinetic energy is given by,

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

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,

$\mathrm{V}=\frac{4}{3}{\mathrm{蟺搁}}^3$

Put this values in above equation of kinetic energy ,

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

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 ,

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

then,

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

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,

$$\begin{gathered} \mathrm{R}=\frac{2\mathrm{伪}}{\mathrm{尾}} \\ \mathrm{R}=9脳{10}^{16}{\mathrm{M}}^{\frac{-1}{3}} \end{gathered}$$

then put the value of M,

$\mathrm{R}=7.14509脳{10}^6\mathrm{m}$

Then density will be,

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

We have to calculate the Fermi energy and temperature.

Since, the Fermi energy can be written as,

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

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

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

Since the chemical potential is given by,

$\mathrm{渭}=\frac{\mathrm{hc}}{2}{\left(\frac{3\mathrm{N}}{\mathrm{蟺痴}}\right)}^{\frac{1}{3}}$

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,

$\mathrm{U}=2\underset{{\mathrm{n}}_{\mathrm{x}}}{鈭�}\underset{{\mathrm{n}}_{\mathrm{y}}}{鈭�}\underset{{\mathrm{n}}_{\mathrm{z}}}{鈭�}\mathrm{蔚}\left(\mathrm{n}\right)$

Then, the spherical coordinates system,

the volume term will be,

$\mathrm{dV}={\mathrm{n}}^2\mathrm{蝉颈苍胃}\mathrm{dn}\mathrm{诲胃}\mathrm{诲蠒}$

Then, the energy will becomes,

$\mathrm{U}={鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_0^{{\mathrm{n}}_{\max }}\mathrm{蔚}\left(\mathrm{n}\right){\mathrm{n}}^2\mathrm{蝉颈苍胃}\mathrm{dn}\mathrm{诲胃}\mathrm{诲蠒}$

since,

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

Then,

$$\begin{gathered} \mathrm{U}={鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_0^{{\mathrm{n}}_{\max }}\frac{\mathrm{hcn}}{2\mathrm{L}}{\mathrm{n}}^2\mathrm{蝉颈苍胃}\mathrm{dn}\mathrm{诲胃}\mathrm{诲蠒} \\ U=\frac{hc}{L}\left[\frac{蟺}{2}\right]\left[1\right]\left[\frac{n_{\max }^4}{4}\right] \\ U=\frac{hc蟺n_{\max }^4}{8L} \end{gathered}$$

since,

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

Then,

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

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,

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

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