Everyone quotes 1.4 solar masses. Almost nobody says where it comes from — and the honest answer has no Sun in it, and no stars either.
When someone says a quantity has a limit, you normally picture a wall. Push harder and you hit it. The Chandrasekhar limit is not like that, and the difference is the whole content.
Take a cold star — no fusion, nothing burning, just matter held up against its own gravity by the fact that electrons refuse to occupy the same state. Squeeze it. The electrons get squeezed into higher momentum states, they push back harder, and the star supports more weight. That is a white dwarf, and it is why a thing the mass of the Sun can sit stably in a volume the size of the Earth.
So: how much can it hold? Squeeze harder, get more support, hold more mass. Where does that stop?
It stops because of a number that turns out to be zero.
Write the pressure of the degenerate electrons as $P = K\rho^{1+1/n}$ — a polytrope, with $n$ an index that depends on how relativistic the electrons are. Solving hydrostatic equilibrium for a polytrope is a classical exercise, and it hands you the star's mass:
$$M \;=\; 4\pi\,\rho_c^{\,\frac{3-n}{2n}}\left[\frac{(n+1)K}{4\pi G}\right]^{3/2}\!\omega_n$$$\rho_c$ is the central density — how hard you squeezed. $\omega_n$ is a pure number you get by solving one ordinary differential equation, Lane–Emden.
Now look at the exponent on $\rho_c$. For slow electrons the gas is $n = 3/2$, the exponent is $+1/2$, and mass grows as you squeeze. Fine. But push the electrons to relativistic speeds — and you must, because squeezing is exactly what does that — and the gas becomes $n = 3$. Then:
$$\frac{3-n}{2n}\;=\;\frac{3-3}{6}\;=\;0$$The mass stops depending on the central density. Not "stops increasing much." Stops depending. Over nine decades of central density, from $10^4$ to $10^{12}$ grams per cubic centimetre, the computed mass holds constant to one part in $10^{12}$:
That constant column is the Chandrasekhar mass. It is not a wall the star runs into. It is a number that stopped caring what you do.
Which is why you cannot rescue a star by compressing it further: compression is the only lever, and at $n=3$ the lever has come free in your hand.
The ultra-relativistic Fermi gas gives $P = \frac{hc}{8}\left(\frac{3}{\pi}\right)^{1/3} n_e^{4/3}$, and with $n_e = \rho/(\mu_e m_u)$ — $\mu_e$ nucleons per electron, which is 2 for helium, carbon and oxygen — the Lane–Emden solution at $n=3$ gives $\xi_1 = 6.896849$, $\omega_3 = 2.018236$, and
$$M_{\rm ch} \;=\; 4\pi\,\omega_3\left(\frac{K}{\pi G}\right)^{3/2} \;=\; 2.8957\times 10^{30}\ \text{kg} \;=\; 1.45630\ M_\odot$$Now substitute $K$ and collect. Every $h$, every $c$, every $G$ lands in one place:
where $M_{\rm Pl} = \sqrt{\hbar c/G}$ and $3.09797 = \tfrac{1}{2}\omega_3\sqrt{3\pi}$ is a pure number produced by one ODE.
∎Read what that says. The maximum mass of a cold star is the Planck mass cubed, divided by the nucleon mass squared.
There is no Sun in that sentence. No star, no astronomy, no observation. It is a statement about gravity and about the proton, and it would be true in a universe that never formed a single star. The famous 1.4 is a coincidence of units: we chose to measure stellar masses in Suns, and the Sun happens to weigh roughly what gravity and the proton allow.
Count it in nucleons and the point gets sharper still:
$$N \;=\; \frac{M_{\rm ch}}{m_u} \;=\; \frac{3.098}{\mu_e^2}\left(\frac{M_{\rm Pl}}{m_u}\right)^{3} \;=\; 1.74\times 10^{57}$$People say a star is about $10^{57}$ nucleons as though it were an observation. It is not. It is $\alpha_G^{-3/2}$, where $\alpha_G = (m_u/M_{\rm Pl})^2 = 5.9\times10^{-39}$ is gravity's coupling constant for nucleons. The size of a star is the weakness of gravity, raised to the three-halves.
$h$ and $c$ have been exact by definition since 2019. $G$ is the worst-measured constant in physics, to about two parts in $10^5$. And the Sun's mass is not measured at all — what is measured is $GM_\odot$, to one part in $10^{10}$, and the mass is that divided by $G$.
So $M_{\rm ch}$ scales as $G^{-3/2}$ and $M_\odot$ as $G^{-1}$, and their ratio as $G^{-1/2}$. The number 1.456 is known to about $10^{-5}$ — better than either mass on its own. Quoting a stellar limit in solar masses cancels most of our ignorance of gravity. That is a good reason to keep doing it, and a bad reason to think we know what a solar mass is.
The polytrope was a simplification — real electrons are neither fully slow nor fully relativistic. So drop it: take Chandrasekhar's exact equation of state for a cold Fermi gas and integrate hydrostatic equilibrium outward from the centre, for a range of central densities. The mass climbs toward the limit and never crosses it, and heavier dwarfs come out smaller:
Now the test. Sirius B is the nearest white dwarf and one of the best-measured objects in the sky. Bond and colleagues, using Hubble astrometry, give it $1.018 \pm 0.011\,M_\odot$ and a radius of $5634 \pm 34$ km.
predicted radius 5583.5 km
measured radius 5634 ± 34 km
error −0.9 %
Input: $\mu_e = 2$ and the CODATA constants. Nothing else. A calculation from 1935 and a measurement from 2017, agreeing to within one percent.
∎Run the same prediction on the heaviest white dwarf known. Caiazzo and colleagues found ZTF J1901+1458 at about $1.35\,M_\odot$ with a radius of 2140 km — barely larger than the Moon.
A quarter off. And that is not arithmetic going wrong; it is every assumption in the derivation expiring at once. At $1.35\,M_\odot$ the star is within 8% of the limit, and there:
general relativity is no longer a correction one may drop; inverse beta decay starts removing the very electrons the argument is about; Coulomb attraction between the ions and the electron sea lowers the pressure below the ideal-gas value; and Caiazzo's star turns once every 6.9 minutes and is strongly magnetised, so some of what holds it up is not pressure at all.
Every one of those makes the real star smaller than the ideal one. That is the direction of the miss. The model fails where it says it will, by the amount and in the sense it predicts. A model that failed somewhere else would be the worrying one.
This calculation gives 1.4563. Textbooks say 1.4, or 1.44. The difference is not rounding: the ideal figure is an upper bound on an upper bound, and Coulomb corrections, inverse beta decay and general relativity all push it down, to roughly 1.38–1.40 for a real carbon–oxygen star.
So 1.4 is the corrected number and 1.456 is the clean one, and a citation that does not say which is citing an ambiguity. The dependence on composition is exact and steep — $M_{\rm ch} \propto (2/\mu_e)^2$ — so an iron core would cap out near $1.26\,M_\odot$. Sirius B's interior composition has never been observed.
The script records six gaps. The two that would change a number:
Everything is Newtonian. Section 6 integrates Newtonian hydrostatic equilibrium, not the Tolman–Oppenheimer–Volkoff equation — and then section 7 leans on that near the limit, where it is wrong. The relativistic calculation is the obvious next thing and is not done here.
Two different surfaces. The measured radius is a photospheric radius. The computed one is where the degenerate gas runs out. Those are not the same surface, and the one-percent agreement at Sirius B does not know that. Getting it right needs an envelope model, which this has none of.
Producing script: book8/ch8-8-chandrasekhar-verify.py — 9 sections, 6 gaps, runs in 20s. Observational values are CITED: Bond et al. 2017 (ApJ 840:70) for Sirius B, Caiazzo et al. 2021 (Nature 595:39) for ZTF J1901+1458. Cross-references: Book VII · 1103 and 26390 (the same shape — a constant that looks found is forced), §8.2 Nebulae, Book VII · what a child can enter (where Chandrasekhar and Mário Schenberg appear, so far without their mathematics).