After the Main Sequence

Section 6 of 8

White Dwarfs

Part 5: White Dwarfs Are Degenerate Stellar Remnants

In HR diagrams, white dwarfs are hot but faint. That means they must have small radii. Spectra and binary masses show they can contain roughly half a solar mass or more inside an Earth-sized volume. Observationally, that is an extraordinary combination: high mass, tiny radius, and no active fusion.

White dwarf

The exposed, degenerate carbon-oxygen (or helium) core left after a low-mass star sheds its envelope. It is roughly Earth-sized, holds $\sim 0.5\text{–}1.4\,M_\odot$ , and is supported against gravity by temperature-independent electron degeneracy pressure rather than fusion — so it simply cools over billions of years.

A white dwarf is supported by electron degeneracy pressure, not by ordinary thermal gas pressure: the pressure comes from quantum state packing, not from thermal agitation. The key scaling for a non-relativistic degenerate electron gas is $P_{\text{deg}} \propto \rho^{5/3}$ , assuming the electrons are degenerate, still non-relativistic, and the dominant pressure source. Because this pressure comes from quantum state filling, it is largely independent of temperature. That is why a white dwarf can cool without losing its pressure support. Electron degeneracy pressure supports the white dwarf mechanically against gravity, but it is not a continuing energy source — the white dwarf shines only because it is still hot and slowly cooling. So a white dwarf remains standing not because it is still generating fusion energy, but because quantum mechanics supplies a pressure that does not disappear as the star cools.

Misconception — a white dwarf collapses once it cools

If thermal pressure holds ordinary stars up, surely a white dwarf — with no fusion and no heat source — must eventually cool, lose its pressure, and collapse.

Why more massive white dwarfs are smaller

Combine the degeneracy-pressure scaling with a characteristic gravitational pressure scaling. For a star of mass $M$ and radius $R$ , $\rho \sim M/R^3$ , so

P_{\text{deg}} \propto \rho^{5/3} \propto \left(\frac{M}{R^3}\right)^{5/3} = \frac{M^{5/3}}{R^5}.

A characteristic self-gravitational pressure scale is $P_{\text{grav}} \sim GM^2/R^4$ (a scaling, not an exact local formula). Equilibrium requires these to scale together:

\frac{M^{5/3}}{R^5} \propto \frac{GM^2}{R^4} \;\Longrightarrow\; M^{5/3} \propto G M^2 R \;\Longrightarrow\; R \propto \frac{M^{5/3}}{M^2} = M^{-1/3}.

So the white-dwarf mass-radius relation is:

Adding mass strengthens gravity, so equilibrium requires higher density; in a degenerate object higher pressure means packing electrons into a smaller volume, so the radius decreases.

More massive white dwarfs are smaller because stronger gravity demands higher density, and higher density means larger degeneracy pressure.

Density example with units

Worked Example 2Average Density of a White Dwarf

Problem

Estimate the mean density of a white dwarf with $M = 0.6\,M_\odot$ and $R = R_\oplus = 6.4 \times 10^8~\text{cm}$ .

StepConvert the mass

$M = 0.6\,(2.0 \times 10^{33}~\text{g}) = 1.2 \times 10^{33}~\text{g}$

StepCompute the volume

$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (6.4 \times 10^8~\text{cm})^3 \approx 1.10 \times 10^{27}~\text{cm}^3$

StepDivide mass by volume

$\bar{\rho} = \frac{1.2 \times 10^{33}~\text{g}}{1.10 \times 10^{27}~\text{cm}^3} \approx 1.09 \times 10^6~\text{g}\,\text{cm}^{-3}$

Dimensional check

$\text{g} / \text{cm}^3 = \text{g}\,\text{cm}^{-3}$ , a density ✓.

Result

$\bar{\rho} \approx 1.1 \times 10^6~\text{g}\,\text{cm}^{-3}.$ A million times denser than water — a teaspoon would weigh several tonnes. This extreme density is what makes the degenerate electron gas the dominant pressure source.

An object that is hot but faint on the HR diagram must have a very small radius. That rules out an ordinary gas-supported star of comparable temperature. An Earth-sized, hot, faint stellar remnant with no fusion source therefore points to a white dwarf supported by electron degeneracy pressure. The inverse mass-radius relation is a signature of quantum pressure, not ordinary gas support.

Quick check

A white dwarf has far more mass than Earth but a radius of order Earth’s. Use $R_{\text{WD}} \propto M^{-1/3}$ to explain why increasing the mass makes the remnant smaller rather than larger. Your answer must mention both gravity and degeneracy pressure.