White Dwarfs

The sun is a little less than halfway through its lifetime and in about 5 million years it will become a white dwarf. Currently it is on the main sequence when stars undergo nuclear fusion and convert Hydrogen into Helium. When there is no more hydrogen in the sun's core to be converted, the force of gravity will cause it to collapse until his is balanced out by some outward force due to pressure. One possible outward force is the force due to degeneracy pressure. White dwarfs are very dense so the fermions in the sun's core do not have space between them. They cannot occupy the same quantum state as their neighbors so they must constantly be in motion to avoid other fermions. The Broglie wavelength $\lambda$ is when the space between the particles is on the order of the particles themselves. $\lambda$ is related to the momentum p of the particle by the formula: $\lambda= \frac{h}{p}$ where h is the Planck constant, $h = 6.6 \times 10^{27} erg s$

(a) For a stellar core of a given temperature, which particles reach this critical density first: electrons or protons?

$K_e$ = kinetic energy of an electron

$m_e$= mass of an electron

$v_e=$ velocity of an electron

$p_e=m_e \times v_e=$ momentum of an electron

$K_p$= kinetic energy of a proton

$m_p$= mass of a proton

$v_p$= velocity of a proton

$p_p=m_p \times v_p=$ momentum of a proton

Since the core is at a certain temperature, we can say that the protons and electrons have the same kinetic energy.

$K_e=K_p$

$\frac{1}{2}m_ev_e^2=\frac{1}{2}m_pv_p^2$

$m_ev_e^2=m_pv_p^2$

$\frac{v_e^2}{v_p^2}=\frac{m_p}{m_e}$

$\frac{v_e}{v_p}=(\frac{m_p}{m_e})^{1/2}$

Multiply by $\frac{m_e}{m_p}$ on each side: $\frac{v_e}{v_p}\times \frac{m_e}{m_p}=(\frac{m_p}{m_e})^{1/2} \times \frac{m_e}{m_p}$

replace mv with p

$\frac{p_e}{p_p}=\frac{m_p}{m_e})^{1/2} \times \frac{m_e}{m_p}$

$\frac{p_e}{p_p}=\frac{m_e}{m_p})^{1/2}$

$\lambda=\frac{h}{p}$

Therefore $lambda  _e > \lambda _p$ so the electrons will reach critical density first.

(b) If a typical white dwarf has roughly half the mass of the Sun ($M_o= 2\times 10^{33} g$) and the radius of the Earth ($R_e=  6.4 \times 10^8 cm$), what is the typical density of a white dwarf in grams per cubic centimeter? Estimate the volume of white dwarf material that weighs as much (has as
much mass) as a car.

$M_{WD}= 1\times 10^{33} g$

$R_{WD}=  6.4 \times 10^8 cm$

$\rho = M/V$

$\rho = \frac{M_{WD}}{\frac{4}{3}\pi R_{WD}^3}$

$\rho = \frac{1\times 10^{33} g}{\frac{4}{3}\pi 6.4 \times 10^8 cm^3}$

$\rho = 1.25 \times 10^6 g/cm^3$

$V=\frac{M_{car}}{\rho}$

$V= \frac{1\times 10^6 g}{1.25 \times 10^6 g/cm^3}$

$V= 0.8 cm^3$

So white dwarfs are very dense.

(c) A white dwarf can be considered a gravitationally bound system of massive particles. Express
the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual
notation using its speed $v$.

$K=\frac{mv^2}{2} = \frac{m}{2}\times (\frac{p}{m})^2 = \frac{p^2}{2m}$

(d) What is the relationship between the total kinetic energy of the electrons that are supplying
the pressure in a white dwarf, and the total gravitational energy of the WD?

Virial Theorem!

$K=\frac{-1}{2} U_g$

(e) According to the Heisenberg uncertainty Principle, one cannot know both the momentum and
position of an election such that $∆p∆x > \frac{h}{4π}$  Use this to express the relationship between the kinetic energy of electrons and their number density $n_e$ (Hint: what is the relationship between
an object’s kinetic energy and its momentum? From here, assume p is approximately ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming volume (V)= (\delta x)^3.

N= number of electrons

$∆p∆x > \frac{h}{4π}$

$K= \frac{\delta p^2}{2m}= \delta p = (2mK)^{1/2}$

$m_e = \frac{N}{V}= \frac{1}{V/N}= \frac{1}{V}= \frac{1}{(\delta x)^3}$

$\delta x = (\frac{1}{n_e})^{1/3}$

$\delta p >\frac{h}{4 \pi \delta x}= \frac{h}{4 \pi}\times (n_e)^{1/3}$

$(2mK)^{1/2}> \frac{h}{4 \pi}\times (n_e)^{1/3}$

$K^{1/2}>\frac{h}{4 \pi} \times \frac{(n_e)^{1/3}}{(2m)^{1/2}}$

$K>\frac{h^2}{32 \pi^2} \times \frac{(n_e)^{2/3}}{m}$

(f) Substitute back into your Virial energy statement. What is the relationship between $n_e$ and
the mass M and radius R of a WD?

$K=-1/2U_g > \frac{h^2}{32 \pi ^2} \times \frac{(n_e)^{2/3}}{m_e}$

$\frac{3GM^2}{5R}> \frac{h^2 (n_e)^{2/3}}{16 \pi ^2 m_e}$

Therefore, as mass increases, $n_e$ increases, and as radius increases, $n_e$ decreases.

(g) Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

Above we got the inequality: $\frac{3GM^2}{5R}> \frac{h^2 (n_e)^{2/3}}{16 \pi ^2 m_e}$

When we drop constants, which include whole numbers, $G, h, \pi$, and $m_e$

we get: $\frac{M^2}{R} \sim (n_e)^{\frac{2}{3}}

(h) What would happen to the radius of a white dwarf if you add mass to it?

From the virial theorem we know:

$K_e=\frac{U_ge}{2}= \frac{-GMM_e}{2R}$

$\frac{dP}{dR}= -gp \sim -\frac{GMM_e}{R^5}$

We can approximate dP/dR as P/R

$P/R \sim \frac{-GM^2}{R^5}$

$P \sim \frac{-GM^2}{R^4}$

Pressure is also related to $\rho$: $P_e \sim \rho^{5/3}$

$P \sim (\frac{M}{R^3})^{\frac{3}{5}}= \frac{M^{\frac{5}{3}}}{R^5}$

$P \sim \frac{GM^2}{R^4} \sim \frac{M^{\frac{5}{3}}}{R^5}$

$\frac{M^2}{M^{\frac{5}{3}}} \sim \frac{1}{R}$

$R \sim M^{- \frac{1}{3}}$

Figures generated using Biorender.