Question

Certain cold stars (called white dwarfs) are stabilized against gravitational collapse by the degeneracy pressure of their electrons (Equation 5.57). Assuming constant density, the radius R of such an object can be calculated as follows:

$$\begin{gathered} \mathbf{P}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}\frac{{\mathbf{E}}_{\mathbf{tot}}}{\mathbf{V}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{k}}_{\mathbf{F}}^{\mathbf{5}}}{\mathbf{10}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{m}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{3}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{5}\mathbf{m}}{\mathbf{p}}^{\mathbf{5}\mathbf{/}\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{57}\mathbf{)} \end{gathered}$$

(a) Write the total electron energy (Equation 5.56) in terms of the radius, the number of nucleons (protons and neutrons) N, the number of electrons per nucleon d, and the mass of the electron m. Beware: In this problem we are recycling the letters N and d for a slightly different purpose than in the text.

$$\begin{gathered} {\mathbf{E}}_{\mathbf{tot}}\mathbf{=}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}\mathbf{V}}{\mathbf{2}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{m}}{\mathbf{\int }}_{\mathbf{0}}^{{\mathbf{k}}_{\mathbf{F}}}{\mathbf{K}}^{\mathbf{4}}\mathbf{dk}\mathbf{=}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{k}}_{\mathbf{F}}^{\mathbf{5}}\mathbf{V}}{\mathbf{10}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{m}}\mathbf{=}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{\left(}\mathbf{3}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{Nd}\mathbf{\right)}}^{\mathbf{5}\mathbf{/}\mathbf{3}}}{\mathbf{10}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{m}}{\mathbf{V}}^{\mathbf{-}\mathbf{2}\mathbf{/}\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{56}\mathbf{)} \end{gathered}$$

(b) Look up, or calculate, the gravitational energy of a uniformly dense sphere. Express your answer in terms of G (the constant of universal gravitation), R, N, and M (the mass of a nucleon). Note that the gravitational energy is negative.

(c) Find the radius for which the total energy, (a) plus (b), is a minimum.

$\mathbf{R}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{9}\mathbf{\pi }}{\mathbf{4}}\mathbf{\right)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{d}}^{\mathbf{5}\mathbf{/}\mathbf{3}}}{{\mathbf{GmM}}^{\mathbf{2}}{\mathbf{N}}^{\mathbf{1}\mathbf{/}\mathbf{3}}}$

(Note that the radius decreases as the total mass increases!) Put in the actual numbers, for everything except , using d=1/2 (actually, decreases a bit as the atomic number increases, but this is close enough for our purposes). Answer:

(d) Determine the radius, in kilometers, of a white dwarf with the mass of the sun.

(e) Determine the Fermi energy, in electron volts, for the white dwarf in (d), and compare it with the rest energy of an electron. Note that this system is getting dangerously relativistic (seeProblem 5.36).

Short answer

Answer is not given in the question.

Step-by-step solution

:(a)The total number of electron energy in terms of radius

$$\begin{gathered} \mathrm{V}=\frac{4}{3}{\mathrm{\pi }}^3 \\ \mathrm{So} \\ \mathrm{E}=\frac{{\sout{\mathrm{h}}}^2{\left(3{\mathrm{\pi }}^2\mathrm{Nq}\right)}^{5/3}}{10{\mathrm{\pi }}^2\mathrm{m}}{\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)}^{-2/3} \end{gathered}$$

(b) gravitational energy of a uniformly dense sphere

Imagine building up a sphere by layers. When it has reached mass m, and radius r, the work necessary to bring in the next increment dm is: dW=-(Gm/r)dm . In terms of the mass density $\mathrm{p},\mathrm{m}=\frac{4}{3}{\mathrm{\pi r}}^3,$where dr is the resulting increase in radius.

Thus:

And the total energy of a sphere of radius R is therefore.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{grav}}=-\frac{16{\mathrm{\pi }}^2}{3}{\mathrm{p}}^2\mathrm{G}{\int }_0^{\mathrm{R}}{\mathrm{r}}^4\mathrm{dr} \\ =-\frac{16{\mathrm{\pi }}^2{\mathrm{p}}^2{\mathrm{R}}^5}{15}\mathrm{G}. \end{gathered}$$

But

So

$$\begin{gathered} {\mathrm{E}}_{\mathrm{grav}}=-\frac{16{\mathrm{\pi }}^2{\mathrm{R}}^5}{15}\mathrm{G}\frac{9{\mathrm{N}}^2{\mathrm{M}}^2}{16{\mathrm{\pi }}^2{\mathrm{R}}^6} \\ =-\frac{3}{5}\mathrm{G}\frac{{\mathrm{N}}^2{\mathrm{M}}^2}{\mathrm{R}} \end{gathered}$$

(c) The radius for total energy

$$\begin{gathered} {\mathrm{E}}_{\mathrm{tot}}=\frac{\mathrm{A}}{{\mathrm{R}}^2}-\frac{\mathrm{B}}{\mathrm{R}},\mathrm{where}\mathrm{A}\equiv \frac{2{\sout{\mathrm{h}}}^2}{15\mathrm{\pi m}}{\left(\frac{9}{4}\mathrm{\pi Nq}\right)}^{5/3}\mathrm{and}\mathrm{B}\equiv \frac{3}{5}{\mathrm{GN}}^2{\mathrm{M}}^2 \\ \frac{{\mathrm{dE}}_{\mathrm{tot}}}{\mathrm{dR}}=-\frac{2\mathrm{A}}{{\mathrm{R}}^3}+\frac{\mathrm{B}}{{\mathrm{R}}^2}=0\Rightarrow 2\mathrm{A}=\mathrm{BR}, \\ \mathrm{So} \\ \mathrm{R}=\frac{2\mathrm{A}}{\mathrm{B}} \\ =\frac{4{\sout{\mathrm{h}}}^2}{15\mathrm{\pi m}}{\left(\frac{9}{4}\mathrm{\pi Nq}\right)}^{5/3}\frac{5}{3{\mathrm{GN}}^2{\mathrm{M}}^2}\mathrm{dR}. \\ \mathrm{R}=\left[\left(\frac{4}{9\mathrm{\pi }}\right){\left(\frac{9\mathrm{\pi }}{4}\right)}^{5/3}\right]\left(\frac{{\mathrm{N}}^{5/3}}{{\mathrm{N}}^2}\right)\frac{{\sout{\mathrm{h}}}^2}{{\mathrm{GmM}}^2}{\mathrm{q}}^{5/3} \\ ={\left(\frac{9\mathrm{\pi }}{4}\right)}^{2/3}\frac{{\sout{\mathrm{h}}}^2}{{\mathrm{GmM}}^2}\frac{{\mathrm{q}}^{5/3}}{{\mathrm{N}}^{1/3}}. \\ \mathrm{R}={\left(\frac{9\mathrm{\pi }}{4}\right)}^{2/3}\frac{{\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2{\left(1/2\right)}^{5/3}}{\left(6.673\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(9.109\times {10}^{-31}\mathrm{kg}\right){\left(1.674\times {10}^{-27}\mathrm{kg}\right)}^2} \end{gathered}$$

Step 4:(d)Determining the radius in km of a white dwarf with the mass of sun Mass of sun

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{sun}:1.989\times {10}^{30\mathrm{kg}} \\ \mathrm{So} \\ \mathrm{N}=\frac{1.989\times {10}^{30}}{1.674\times {10}^{-27}} \\ =1.188\times {10}^{57} \\ {\mathrm{N}}^{-1/3}=9.44\times {10}^{-20} \\ \mathrm{R}=\left(7.58\times {10}^{25}\right)\left(9.44\times {10}^{-20}\right)\mathrm{m}\left(\mathrm{Slightly}\mathrm{larger}\mathrm{than}\mathrm{the}\mathrm{earth}\right). \\ =7.16\times {10}^6\mathrm{m} \end{gathered}$$

$$\begin{gathered} \mathrm{From}\mathrm{Eq}.5.54:{\mathrm{E}}_{\mathrm{F}}=\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}{\left(3{\mathrm{p\pi }}^2\right)}^{2/3}\left(5.54\right) \\ {\mathrm{E}}_{\mathrm{F}}=\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}{\left(3{\mathrm{\pi }}^2\frac{\mathrm{Nq}}{4/3{\mathrm{\pi R}}^3}\right)}^{2/3} \\ =\frac{{\sout{\mathrm{h}}}^2}{2{\mathrm{mR}}^2}{\left(\frac{9\mathrm{\pi }}{4}\mathrm{Nq}\right)}^{2/3} \\ \mathrm{Numerically}: \\ {\mathrm{E}}_{\mathrm{F}}=\frac{{\left(1.055\times {10}^{-34}\mathrm{j}.\mathrm{s}\right)}^2}{2\left(9.109\times {10}^{-31}\mathrm{kg}\right){\left(7.16\times {10}^6\mathrm{m}\right)}^2}{\left[\frac{9\mathrm{\pi }}{4}\left(1.188\times {10}^{57}\right)\frac{1}{2}\right]}^{2/3} \\ =3.102\times {10}^{-14}\mathrm{J} \\ \mathrm{or},\mathrm{in}\mathrm{electron}\mathrm{volts}{\mathrm{E}}_{\mathrm{F}}=\frac{3.102\times {10}^{-14}}{1.602\times {10}^{-19}}\mathrm{eV} \\ =1.94\times {10}^5\mathrm{eV}. \\ {\mathrm{E}}_{\mathrm{rest}}={\mathrm{mc}}^2=5.11\times {10}^5\mathrm{eV},\mathrm{So}\mathrm{the}\mathrm{F}\mathrm{ermi}\mathrm{energy}(\mathrm{Which}\mathrm{is}\mathrm{the}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{most} \\ \mathrm{energetic}\mathrm{electrons})\mathrm{is}\mathrm{comparable}\mathrm{to}\mathrm{the}\mathrm{rest}\mathrm{energy},\mathrm{so}\mathrm{they}\mathrm{are}\mathrm{getting}\mathrm{fairly}\mathrm{relativistic} \end{gathered}$$