---
title: White Dwarfs and the Chandrasekhar Limit
module: Degenerate Fermi Gas
moduleNumber: 9
lessonNumber: 3
order: 903
summary: >
  A white dwarf is held up against its own gravity by the degeneracy pressure of
  its electrons. Balancing that pressure against gravity gives a mass-radius
  relation $R\propto M^{-1/3}$: heavier white dwarfs are smaller and denser. As the
  density rises the electrons turn relativistic, the pressure softens from
  $n^{5/3}$ to $n^{4/3}$, and the star can no longer support itself above a
  critical mass. This lesson derives that Chandrasekhar mass, about $1.4\,M_\odot$,
  and what lies beyond it.
topics: [Degenerate Fermi Gas]
sources:
  - book: Pathria & Beale
    ref: "Ch. 8 — Ideal Fermi Systems; §8.4–8.5"
  - book: Schroeder
    ref: "Ch. 7 — Quantum Statistics; §7.3 (the white dwarf)"
  - book: Kardar (Statistical Physics of Particles)
    ref: "Ch. 7 — Quantum Statistical Mechanics; §7.3"
  - book: Carroll & Ostlie
    ref: "Ch. 16 — The Degenerate Remnants of Stars"
draft: false
---

A white dwarf is the exposed core left when a low-mass star exhausts its nuclear
fuel. With no fusion to supply pressure, gravity compresses the core until the
electrons are squeezed into a degenerate gas, and their zero-temperature pressure
halts the collapse. The star is then a cold ball of electron-degenerate matter,
roughly the mass of the Sun packed into the volume of the Earth. The balance
between degeneracy pressure and gravity fixes its size, and the relativistic
softening of that pressure sets an upper limit on its mass.

## Degeneracy pressure against gravity

The matter is fully ionized: bare nuclei sit in a sea of free electrons. The
electrons are degenerate because the density is enormous, while the nuclei stay
classical because their large mass gives them a Fermi temperature far below the
stellar temperature. The pressure is therefore the electron degeneracy pressure of
the previous lessons, evaluated at the local electron density $n_e$.

Mass is carried by the nuclei. With $\mu_e$ nucleons per electron (so $\mu_e\approx2$
for helium, carbon, or oxygen, where each electron is balanced by two nucleons),
the mass density is

$$
\rho = \mu_e\, m_u\, n_e ,
$$

where $m_u$ is the atomic mass unit. The nonrelativistic degeneracy pressure from
the zero-temperature Fermi gas is

$$
P = \frac{(3\pi^2)^{2/3}}{5}\,\frac{\hbar^2}{m_e}\,n_e^{5/3}
= K_{\mathrm{nr}}\,\rho^{5/3},\qquad
K_{\mathrm{nr}} = \frac{(3\pi^2)^{2/3}}{5}\,\frac{\hbar^2}{m_e\,(\mu_e m_u)^{5/3}} .
$$

The star is in **hydrostatic equilibrium**: at every radius the outward pressure
gradient balances the inward pull of the mass interior to that radius,

$$
\frac{\d P}{\d r} = -\frac{G\,m(r)\,\rho(r)}{r^2},\qquad
\frac{\d m}{\d r} = 4\pi r^2 \rho(r).
$$

$$
% caption: In hydrostatic equilibrium the outward pressure difference across a shell balances the inward gravitational pull of the enclosed mass $m(r)$; the star is cold, so the pressure is the electron degeneracy pressure.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\fill[black!4] (0,0) circle (2.3);
\draw[thick] (0,0) circle (2.3);
\draw[thick] (0,0) circle (1.45);
\fill[black!10] (0,0) circle (1.45);
\node[black] at (0,-0.6) {$m(r)$};
\draw[->,black,thick] (3.25,0)--(2.4,0);
\node[black] at (2.95,0.55) {gravity};
\draw[->,acc,thick] (0,1.45)--(0,2.15);
\node[acc,right] at (0.05,1.85) {pressure};
\draw[black,dashed,->] (0,0)--(-1.63,1.63);
\node[black] at (-1.35,0.95) {$r$};
\end{tikzpicture}
$$

## The nonrelativistic mass-radius relation

An order-of-magnitude balance gives the mass-radius relation without solving the
differential equation. The gravitational pressure scale is
$P_{\mathrm{grav}}\sim GM^2/R^4$, obtained by integrating $\d P/\d r$ across the
star with $\rho\sim M/R^3$ and $m\sim M$. The degeneracy pressure scale is
$P\sim K_{\mathrm{nr}}\rho^{5/3}\sim K_{\mathrm{nr}}(M/R^3)^{5/3}$. Setting the two
equal,

$$
\frac{\hbar^2}{m_e (\mu_e m_u)^{5/3}}\,\frac{M^{5/3}}{R^5}
\;\sim\; \frac{G M^2}{R^4}
\quad\Longrightarrow\quad
R \;\sim\; \frac{\hbar^2}{G\,m_e\,(\mu_e m_u)^{5/3}}\;M^{-1/3}.
$$

More massive white dwarfs are **smaller**: adding mass strengthens gravity, which
compresses the star until the stiffer degeneracy pressure at higher density
restores balance. The product $R\,M^{1/3}$ is a constant fixed only by fundamental
constants and the composition $\mu_e$. Numerically, a solar-mass white dwarf has a
radius near $7\times10^3\,\mathrm{km}$, comparable to the Earth, at a mean density
of order $10^{9}\,\mathrm{kg\,m^{-3}}$.[^wdscale]

> **Worked example.** Estimate the radius of a solar-mass carbon-oxygen white dwarf
> ($\mu_e=2$). The dimensional balance gives
>
> $$
> R \sim \frac{\hbar^2}{G\,m_e\,(\mu_e m_u)^{5/3}\,M^{1/3}}
> = \frac{(1.055\times10^{-34})^2}
> {(6.67\times10^{-11})(9.11\times10^{-31})(3.32\times10^{-27})^{5/3}(2\times10^{30})^{1/3}}
> \approx 2\times10^{6}\,\mathrm{m}.
> $$
>
> The dimensional estimate omits the $O(1)$ Lane-Emden coefficient, which is about
> $2.8$; including it raises the radius to $\sim5\times10^3\,\mathrm{km}$, matching
> the measured radius of Sirius B ($M\approx1.0\,M_\odot$, $R\approx5.9\times10^3\,\mathrm{km}$).

[^wdscale]: The full treatment solves the Lane-Emden equation for a polytrope
$P\propto\rho^{1+1/n}$ with $n=\tfrac32$; the scaling here reproduces its
mass-radius exponent. Pathria & Beale, §8.4; Carroll & Ostlie, Ch. 16.

$$
% caption: The nonrelativistic degeneracy balance gives $R\propto M^{-1/3}$: a heavier white dwarf is more compact, so the curve falls as the mass grows.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(6,0) node[right,black]{$M$};
\draw[->,black] (0,0)--(0,4) node[above,black]{$R$};
\draw[acc,very thick] plot[domain=0.55:5.4,samples=80] (\x,{2.4/(\x)^0.3333});
\node[acc,align=center] at (3.6,2.7) {heavier is smaller};
\end{tikzpicture}
$$

## Relativistic softening of the pressure

The mass-radius relation predicts unbounded density as $M$ grows, but the
electrons cannot supply the required pressure indefinitely. Compression raises the
Fermi momentum $p_F=\hbar(3\pi^2 n_e)^{1/3}$, and once $p_F\gtrsim m_e c$ the
electrons at the Fermi surface become relativistic. The dispersion crosses over
from $\epsilon=p^2/2m_e$ to $\epsilon\approx pc$, and the pressure changes its
density scaling.

The onset density follows from $p_F=m_ec$, or
$n_e=(m_ec/\hbar)^3/3\pi^2\approx6\times10^{35}\,\mathrm{m^{-3}}$, corresponding to
a mass density $\rho=\mu_e m_u n_e\approx2\times10^{9}\,\mathrm{kg\,m^{-3}}$. This
is reached in the core of a white dwarf whose mass approaches a solar mass, so the
relativistic correction is not academic: it governs the most massive white dwarfs,
precisely those near the stability limit.

In the ultrarelativistic limit the Fermi energy is $\epsilon_F=\hbar c\,k_F$, the
ground-state energy is $U=\tfrac34 N\epsilon_F$, and the pressure of an
ultrarelativistic gas obeys $PV=\tfrac13 U$. This gives

$$
P = \frac{1}{4}\,n_e\,\epsilon_F
= \frac{\hbar c}{4}\,(3\pi^2)^{1/3}\,n_e^{4/3}
= K_{\mathrm{ur}}\,\rho^{4/3},\qquad
K_{\mathrm{ur}} = \frac{\hbar c}{4}\,\frac{(3\pi^2)^{1/3}}{(\mu_e m_u)^{4/3}} .
$$

The exponent has softened from $\tfrac53$ to $\tfrac43$. On logarithmic axes the
pressure-density relation is a line whose slope decreases from $\tfrac53$ to
$\tfrac43$ as the gas turns relativistic.

$$
% caption: The degeneracy pressure follows $P\propto n^{5/3}$ while electrons are nonrelativistic and softens to $P\propto n^{4/3}$ once $p_F$ exceeds $m_e c$; the slope on log axes drops from $\tfrac53$ to $\tfrac43$.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(6,0) node[right,black]{$\log n$};
\draw[->,black] (0,0)--(0,4.4) node[above,black]{$\log P$};
\draw[acc,very thick] (0.5,0.4)--(3.0,3.05);
\draw[black,very thick,dashed] (3.0,3.05)--(5.4,4.0);
\filldraw[black] (3.0,3.05) circle (2.0pt);
\node[acc] at (1.35,2.4) {slope $\tfrac{5}{3}$};
\node[black] at (4.5,3.15) {slope $\tfrac{4}{3}$};
\node[black,align=center] at (3.1,2.3) {relativistic\\onset};
\end{tikzpicture}
$$

## The Chandrasekhar mass

The softening removes the stabilizing feedback. Repeat the equilibrium balance
with the ultrarelativistic pressure:

$$
K_{\mathrm{ur}}\left(\frac{M}{R^3}\right)^{4/3}\!\!\frac{1}{R}
\;\sim\; \frac{GM^2}{R^4}
\quad\Longrightarrow\quad
\hbar c\,\frac{M^{4/3}}{(\mu_e m_u)^{4/3}} \;\sim\; G M^2 .
$$

The radius $R$ cancels from both sides. Instead of fixing $R$ for each $M$, the
balance fixes a single mass at which degeneracy pressure and gravity scale
together at every radius. Solving for that mass,

$$
M_{\mathrm{Ch}} \;\sim\; \left(\frac{\hbar c}{G}\right)^{3/2}\frac{1}{(\mu_e m_u)^2}.
$$

The full polytrope calculation supplies the numerical coefficient,

$$
M_{\mathrm{Ch}} = \frac{\sqrt{3\pi}}{2}\,\omega_3\left(\frac{\hbar c}{G}\right)^{3/2}
\frac{1}{(\mu_e m_u)^2}
\approx 1.44\left(\frac{2}{\mu_e}\right)^{2} M_\odot ,
$$

with $\omega_3\approx2.018$ the constant of the $n=3$ Lane-Emden solution.[^chandra]
The Chandrasekhar mass is built entirely from $\hbar$, $c$, $G$, and the nucleon
mass: the combination $(\hbar c/G)^{3/2}/m_u^2\approx1.9\,M_\odot$ is a fundamental
stellar mass scale, and the composition factor $\mu_e^{-2}$ and the $O(1)$
coefficient bring it to $1.4\,M_\odot$. A white dwarf below this mass sits stably
on the mass-radius curve; as $M\to M_{\mathrm{Ch}}$ the equilibrium radius shrinks
toward zero and no cold configuration exists above it.

[^chandra]: Pathria & Beale, §8.5, carries out the polytrope calculation and
obtains this coefficient; Carroll & Ostlie, Ch. 16, gives the astrophysical
$1.4\,M_\odot$ value. The Planck-scale combination uses NIST constants,
<https://physics.nist.gov/cuu/Constants/>.

$$
% caption: Including the relativistic softening, the mass-radius curve bends down and the equilibrium radius falls to zero as the mass approaches the Chandrasekhar limit; no stable white dwarf exists beyond it.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(6,0) node[right,black]{$M$};
\draw[->,black] (0,0)--(0,4) node[above,black]{$R$};
\draw[acc,very thick] plot[domain=0.5:4.7,samples=90] (\x,{1.25*sqrt(4.8-\x)/(\x)^0.3333});
\draw[black,thick,dashed] (4.85,0)--(4.85,3.4);
\node[black,align=center] at (5.3,2.4) {$M_{\mathrm{Ch}}$};
\node[acc] at (2.0,2.6) {stable};
\end{tikzpicture}
$$

## Beyond the limit

A white dwarf pushed over $M_{\mathrm{Ch}}$, by accretion from a companion or by
the merger of two dwarfs, cannot regain equilibrium. Two outcomes follow. If the
carbon and oxygen ignite under the rising density, runaway fusion unbinds the star
as a type Ia supernova. If instead the density climbs first, the electrons are
forced onto the nuclei by **inverse beta decay** (electron capture),
$p+e^-\to n+\nu_e$, which removes the very electrons that supply the pressure. The
loss of pressure accelerates the collapse until the matter reaches nuclear density
and neutron degeneracy takes over. That neutron-supported remnant is the subject of
the next lesson.

## Summary

- White-dwarf matter is a cold electron-degenerate gas; the electrons supply the
  pressure and the nuclei ($\mu_e$ nucleons per electron) supply the mass,
  $\rho=\mu_e m_u n_e$.
- Balancing nonrelativistic degeneracy pressure $P\propto\rho^{5/3}$ against
  gravity gives $R\propto M^{-1/3}$: heavier white dwarfs are smaller and denser,
  a solar mass in an Earth-sized volume.
- At high density the electrons turn relativistic and the pressure softens to
  $P\propto\rho^{4/3}$, removing the radius from the equilibrium balance and
  fixing a maximum mass.
- The Chandrasekhar mass $M_{\mathrm{Ch}}\approx1.44(2/\mu_e)^2 M_\odot$ is set by
  $(\hbar c/G)^{3/2}/(\mu_e m_u)^2$; beyond it the star collapses toward a neutron
  star or a supernova.