The White Dwarf Mass–Radius Relation

White dwarfs behave in a way that surprises everyone the first time: the more massive a white dwarf, the smaller it is. It follows from combining degeneracy pressure with what hydrostatic equilibrium demands. A typical white dwarf holds about 0.6 solar masses in a ball about the size of Earth (that is how a Sun-like star ends).

Why it works

Gravity needs a central pressure $P_c \sim GM^2/R^4$. Degeneracy supplies $P \propto n_e^{5/3} \propto \rho^{5/3} \propto (M/R^3)^{5/3} = M^{5/3}/R^5$. Setting them proportional: $$\frac{M^{5/3}}{R^5} \propto \frac{M^2}{R^4} \quad\Rightarrow\quad R \propto M^{-1/3}.$$ Adding mass makes gravity's demand grow faster than the supply at the old size, so the star must shrink until the denser electrons push back hard enough. Another way to say it: $MR^3$ is the same for all such white dwarfs, so volume goes as $1/M$.

Key idea

For white dwarfs, $R \propto M^{-1/3}$: heavier ones are smaller and much denser ($\bar\rho \propto M/R^3 \propto M^2$). A $0.6\,M_\odot$ white dwarf has $R \approx 8{,}700$ km and a mean density near $4 \times 10^{8}$ kg/m³.

Radius against mass for white dwarfs. Solid: the simple $R \propto M^{-1/3}$ law. Dashed: a fit that includes relativistic electrons (Nauenberg 1972), which drops to zero at about $1.44\,M_\odot$. Dots: two measured white dwarfs. Drag the mass to read off sizes and densities.

Worked examples

Worked example: A heavier white dwarf

A $0.6\,M_\odot$ white dwarf has $R = 8700$ km. Estimate the radius of a $1.0\,M_\odot$ white dwarf with $R \propto M^{-1/3}$.

$R_2 = R_1(M_2/M_1)^{-1/3} = 8700 \times (1.67)^{-1/3}$.
$(1.67)^{-1/3} = 0.843$.

Answer: About 7,300 km. Real ones are smaller still (Sirius B: about 5,800 km), because near this mass relativity softens the electrons' pressure.

Worked example: A teaspoon of white dwarf

Find the mean density of a $0.6\,M_\odot$, 8,700 km white dwarf and the mass of 5 cm³ of it.

$V = \tfrac43\pi(8.7 \times 10^{6}\text{ m})^3 = 2.76 \times 10^{21}$ m³.
$\bar\rho = 1.19 \times 10^{30}/2.76 \times 10^{21} = 4.3 \times 10^{8}$ kg/m³.

Answer: A teaspoon would hold about 2,200 kg, the mass of a large car.

Worked example: The exponent for a general law

If $P \propto \rho^{\gamma}$, what power $x$ appears in $R \propto M^{x}$?

$M^{\gamma}R^{-3\gamma} \propto M^2R^{-4}$, so $R^{4 - 3\gamma} \propto M^{2 - \gamma}$.
$x = (2 - \gamma)/(4 - 3\gamma)$. For $\gamma = 5/3$: $x = (1/3)/(-1) = -1/3$.

Answer: $x = -1/3$ for nonrelativistic electrons. At $\gamma = 4/3$ the denominator is zero, a warning sign that the next topic explains.

Common mistake

Thinking more massive white dwarfs must be bigger, as for planets made of ordinary rock. Ordinary matter resists compression almost regardless of the load, so adding mass adds volume. Degenerate matter is soft by comparison: extra mass squeezes all of it. Check: Sirius B ($1.0\,M_\odot$) is smaller than Earth, while 40 Eridani B ($0.57\,M_\odot$) is larger.

The White Dwarf Mass–Radius Relation

Worked examples

Practice

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Sources and further reading