Estimate the radius of a white dwarf whose mass is equal to...

Estimate the radius of a white dwarf whose mass is equal to that of the Sun by the following method, assuming there are $N$ nucleons and $\frac{1}{2} N$ electrons (why $\left.\frac{1}{2} ?\right):(a)$ Use Fermi-Dirac statistics (Section $40-6$ ) to show that the total energy of all the electrons $$ E_{\mathrm{e}}=\frac{3}{5}\left(\frac{1}{2} N\right) \frac{h^{2}}{8 m_{\mathrm{e}}}\left(\frac{3}{\pi} \frac{N}{2 V}\right)^{\frac{2}{3}} $$ [Hint: See Eqs. $40-12$ and $40-13$; we assume electrons fill energy levels from 0 up to the Fermi energy.] (b) The nucleons contribute to the total energy mainly via the gravitational force (note that the Fermi energy for nucleons is negligible compared to that for electrons - why?). Use a gravitational form of Gauss's law to show that the total gravitational potential energy of a uniform sphere of radius $R$ is $$ -\frac{3}{5} \frac{G M^{2}}{R} $$ by considering the potential energy of a spherical shell of radius $r$ due only to the mass inside it (why?) and integrate from $r=0$ to $r=R .$ (See also Appendix D.) $(c)$ Write the total energy as a sum of these two terms, and set $d E / d R=0$ to find the equilibrium radius, and evaluate it for a mass equal to the Sun's $\left(2.0 \times 10^{30} \mathrm{~kg}\right)$.

Estimate the radius of a white dwarf whose mass is equal to that of the Sun by the following method, assuming there are $N$ nucleons and $\frac{1}{2} N$ electrons (why $\left.\frac{1}{2} ?\right):(a)$ Use Fermi-Dirac statistics (Section $40-6$ ) to show that the total energy of all the electrons
$$
E_{\mathrm{e}}=\frac{3}{5}\left(\frac{1}{2} N\right) \frac{h^{2}}{8 m_{\mathrm{e}}}\left(\frac{3}{\pi} \frac{N}{2 V}\right)^{\frac{2}{3}}
$$
[Hint: See Eqs. $40-12$ and $40-13$; we assume electrons fill energy levels from 0 up to the Fermi energy.] (b) The nucleons contribute to the total energy mainly via the gravitational force (note that the Fermi energy for nucleons is negligible compared to that for electrons - why?). Use a gravitational form of Gauss's law to show that the total gravitational potential energy of a uniform sphere of radius $R$ is
$$
-\frac{3}{5} \frac{G M^{2}}{R}
$$
by considering the potential energy of a spherical shell of radius $r$ due only to the mass inside it (why?) and integrate from $r=0$ to $r=R .$ (See also Appendix D.) $(c)$ Write the total energy as a sum of these two terms, and set $d E / d R=0$ to find the equilibrium radius, and evaluate it for a mass equal to the Sun's $\left(2.0 \times 10^{30} \mathrm{~kg}\right)$.

Key Concepts

Energy Equilibrium Condition

The equilibrium configuration of a white dwarf is found by minimizing the total energy of the system, which is the sum of the electron kinetic energy (from degeneracy pressure) and the gravitational potential energy. By setting the derivative of the total energy with respect to the radius to zero, one can derive the equilibrium radius, a critical step in estimating the size of the white dwarf.

Gravitational Potential Energy of a Uniform Sphere

This concept involves calculating the total gravitational binding energy of a uniformly distributed mass, derived by integrating the gravitational potential due to each spherical shell of the star. The resulting expression, which scales as -GM²/R, is essential in balancing the kinetic energy of the electrons against the pull of gravity in a white dwarf.

Fermi-Dirac Statistics

This concept explains how fermions, such as electrons, occupy energy levels according to a specific statistical distribution at low temperatures. It is fundamental in understanding the behavior of electrons in a white dwarf, where, due to the Pauli exclusion principle, electrons fill up available quantum states up to a maximum energy known as the Fermi energy.

Degenerate Electron Gas

In the very dense environment of a white dwarf, electrons form a degenerate gas, meaning that their pressure is not determined by temperature but primarily by quantum mechanical effects. This degeneracy pressure is what supports the star against gravitational collapse, making it a key concept in determining the star's equilibrium radius.

Fermi Energy

The Fermi energy represents the highest occupied energy state in a system of fermions at zero temperature. In the context of white dwarfs, it sets the scale of the electron kinetic energy, which is directly related to the electron degeneracy pressure that counteracts gravitational forces.

Transcript

00:01 So for part a, we use equation 40 -12 and 40 -13.

00:08 And we find that if all electron levels are filled up to the fermi energy, the average electron energy is 3 -5s ef.

00:17 So e -e is equal to n -e times 3 -5s ef, that's equal to 1 1⁄2n 3 -5s h squared, over 8m .e times 3 over pi times n e over v to the two -thirds power.

00:46 And so that equals three -fifths, one -half n times h squared over 8m .e times 3 pi in 2v over, i mean times the two -third power.

01:05 So that's your answer for part a.

01:08 For part b, we have, we use that density to calculate the masses and then integrate over the full radius of the star to find the total gravitational energy of the star.

01:26 So m is equal to 4 thirds pi r cubed, which is equal to m over 4 thirds pi r cubed, times 4 thirds pi r cubed, is equal to m r cubed over r cubed.

01:49 So the derivative of m is equal to p4 pi r squared d r, and that's equal to m over four thirds pi r cubed times four pi r squared d r, and that's equal to three mr squared over r cubed d r.

02:17 So d -u is equal to negative g, m -d -m over r.

02:22 That's equal to negative g times m r -r -cubed over r -cubed times 3m -r squared over r -cubed times dr -tover r -cubed times dr over r.

02:45 And so that's equal to negative 3 -g -m squared over r.

02:53 To the 6th power r4 dr.

03:02 So then we have to take the integral, so u is equal to r and 0, so that would be negative 3gm squared over r to the 6th times r to the 4th, and that's equal to negative.

03:20 3gm squared over r to the 6th.

03:24 That would be r0.

03:30 R to the fourth, dr, and that's equal to negative 3gm squared over r to the sixth, r to the fifth, over five.

03:42 And so lastly, that equals negative three -fifths, gm squared over r.

03:52 And lastly, for part c, we have e total is equal to e plus u, and that's three -fifths times 1 1 half n times h squared over 8 m e times 3 over pi times n over 2 v to the 2 3 3 3 3 .5s gm squared over r so you solve that out and you get 9 2 thirds 3h squared m to the 5 thirds over 320 pi to the 4 thirds m .e...

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