The Quantum Limit

Section 3 of 6

The Chandrasekhar Limit

Part 3: The Chandrasekhar Limit

Why Massive White Dwarfs Fail

As a white dwarf’s mass increases, gravity is stronger, requiring higher density to generate enough degeneracy pressure. Higher density means electrons are confined to smaller volumes, giving them larger momenta ( $p_F \sim \hbar n_e^{1/3}$ ). At some point, the fastest electrons approach the speed of light — they become relativistic. For relativistic particles, the energy-momentum relation changes:

\text{Non-relativistic: } E = \frac{p^2}{2m_e} \quad \xrightarrow{\text{relativistic}} \quad E = pc.

This changes the degeneracy pressure scaling:

Regime	Pressure	Scaling with $\rho$
Non-relativistic ( $p_F \ll m_e c$ )	$P \sim \hbar^2 n_e^{5/3}/m_e$	$\rho^{5/3}$
Relativistic ( $p_F \sim m_e c$ )	$P \sim \hbar c\,n_e^{4/3}$	$\rho^{4/3}$

The relativistic pressure scales more slowly with density ( $\rho^{4/3}$ instead of $\rho^{5/3}$ ). This is the key: gravity’s demand grows as $\rho^{4/3}$ , and now the pressure supply also grows as $\rho^{4/3}$ . The two scalings match — so increasing the density no longer helps. Either there is an equilibrium or there isn’t, and it depends only on the mass.

Misconception — degeneracy pressure switches off at the Chandrasekhar mass

It is tempting to picture the Chandrasekhar limit as a switch: below $1.4\,M_\odot$ degeneracy pressure is “on,” and at the limit it suddenly turns “off,” letting gravity win.

Deriving the Chandrasekhar Mass — Reading the Limit at Its Cleanest

This is the crest of the move we have been practicing all module. The Chandrasekhar mass is Reading the Limit done as purely as it ever gets — write the balance, cancel the size, read off the constants.

① Write the balance. Set the gravitational pressure required (from hydrostatic equilibrium, Module 3) against the relativistic degeneracy pressure supplied, with $n_e \sim M/(m_p R^3)$ :

P_\text{grav} \sim \frac{GM^2}{R^4}, \qquad P_\text{deg,rel} \sim \hbar c\,n_e^{4/3} \sim \hbar c\,\frac{M^{4/3}}{m_p^{4/3} R^4}.

② Solve — the size cancels. Set $P_\text{grav} = P_\text{deg,rel}$ . Both sides carry exactly $R^4$ — because in the relativistic regime pressure and gravity grow at the same rate, $\rho^{4/3}$ — so the radius cancels completely (the same disappearing act as the Eddington ceiling in Reading 1, and for the same reason):

\frac{GM^2}{R^4} \sim \frac{\hbar c\,M^{4/3}}{m_p^{4/3}\,R^4} \;\Longrightarrow\; M^{2/3} \sim \frac{\hbar c}{G\,m_p^{4/3}} \;\Longrightarrow\; M_\text{Ch} \sim \left(\frac{\hbar c}{G}\right)^{3/2}\frac{1}{m_p^2}.

③ Read off the constants. The mass is fixed — no radius, no temperature, just $\hbar$ , $c$ , $G$ , $m_p$ . And this combination should look familiar: it is the same natural mass scale $M_0 = (\hbar c/G)^{3/2}/m_p^2 \approx 1.8\,M_\odot$ that set the minimum stellar mass in Reading 1. The floor and the ceiling of degenerate stardom are built from the same four constants — the floor is $M_0$ times a small thermal factor, the ceiling is $M_0$ times an order-unity number. That is the through-line of the whole module in one equation.

The Chandrasekhar mass is therefore independent of the white dwarf’s size:

The dimensional estimate gives $\sim 1.9\,M_\odot$ ; the exact relativistic, composition-corrected value (with $Y_e = 0.5$ for C/O) is $\approx 1.44\,M_\odot$ .

Chandrasekhar mass

The maximum mass of a stable white dwarf, $M_\text{Ch} \sim (\hbar c/G)^{3/2}/m_p^2 \approx 1.44\,M_\odot$ (for $Y_e = 0.5$ ). It is set purely by fundamental constants: above it, relativistic electron degeneracy pressure can no longer balance gravity at any radius, so the core collapses to a neutron star or black hole.

Putting Numbers on the Limit

Worked Example 1The Chandrasekhar Mass from Four Constants

Problem

Evaluate $M_\text{Ch} \sim (\hbar c/G)^{3/2}/m_p^2$ in CGS and convert to solar masses. Use $\hbar = 1.055\times10^{-27}$ , $c = 3\times10^{10}$ , $G = 6.674\times10^{-8}$ , $m_p = 1.67\times10^{-24}~\text{g}$ , $M_\odot = 2.0\times10^{33}~\text{g}$ .

StepAssemble ℏc/G

$\frac{\hbar c}{G} = \frac{(1.055\times10^{-27})(3\times10^{10})}{6.674\times10^{-8}} \approx 4.75\times10^{-10}.$

StepRaise to the 3/2 power and divide by the proton mass squared

$M_\text{Ch} \sim (4.75\times10^{-10})^{3/2}\,\frac{1}{(1.67\times10^{-24})^2} = (1.03\times10^{-14})(3.58\times10^{47}) \approx 3.7\times10^{33}~\text{g}.$

Dimensional check

The constants close to a mass. With $\text{erg} = \text{g}\,\text{cm}^2\text{s}^{-2}$ , $[\hbar c] = (\text{erg·s})(\text{cm·s}^{-1}) = \text{g}\,\text{cm}^3\text{s}^{-2}$ , so $[\hbar c/G] = \dfrac{\text{g}\,\text{cm}^3\text{s}^{-2}}{\text{cm}^3\text{g}^{-1}\text{s}^{-2}} = \text{g}^2$ . Then $[(\hbar c/G)^{3/2}] = \text{g}^3$ , and dividing by $[m_p^2] = \text{g}^2$ leaves $\text{g}$ ✓.

Result

$\frac{M_\text{Ch}}{M_\odot} \approx \frac{3.7\times10^{33}}{2.0\times10^{33}} \approx 1.85.$ The bare scaling gives $\sim 1.9\,M_\odot$ ; the exact relativistic, composition-corrected value ( $Y_e = 0.5$ for carbon-oxygen) is $1.44\,M_\odot$ . The dimensional argument lands within $\sim 30\%$ — typical for this kind of estimate, and more than enough to understand why the wall sits near a solar mass.

Quick check

In the Chandrasekhar mass derivation, the radius $R$ canceled out of the equation. Why is this physically significant?