The other end of the same formula. And it comes out 38% too big — which is the most useful thing about it.
§8.8 asked how heavy a cold star can be, and got a number with no stars in it: $3.098\,M_{\rm Pl}^3/(\mu_e m_u)^2$. So ask the other one.
How light can a star be? Not how light can a ball of gas be — there is no floor on that, Jupiter exists — but how little mass will still ignite hydrogen and hold a star together against its own cooling.
The same machinery answers it. It gives the wrong number, and the way it is wrong turns out to say something the right number would not.
Take a ball of hydrogen and let it contract. Two things happen at once.
The centre heats up. Gravity does work, and for an ideal gas the central temperature scales as $M/R$ — shrink the ball, heat the core.
But the electrons are also being squeezed, and degeneracy pressure grows faster — as $\rho^{5/3}$, which is $R^{-5}$ — and it does not care about temperature at all. It is happy to hold the ball up cold.
So write both into hydrostatic equilibrium and solve for the central temperature at radius $R$:
$$\frac{k T_c}{\mu m_u} \;=\; \underbrace{\frac{W}{D_c}\frac{GM}{R}}_{\text{rises as }R^{-1}} \;-\; \underbrace{\frac{K_1 D_c^{2/3} M^{2/3}}{R^2}}_{\text{rises as }R^{-2}}$$One term rises as the ball shrinks. The other rises faster. So $T_c$ climbs, turns over, and comes back down. It has a maximum, and the maximum is in closed form:
$$R_* = \frac{2K_1 D_c^{5/3}}{WG}M^{-1/3}, \qquad T_{\max} = \frac{\mu m_u W^2G^2}{4kK_1D_c^{8/3}}\,M^{4/3}$$Both checked against numerical maximisation, four masses, to two parts in $10^3$. And the exponent $4/3$ is exact.
That is the whole argument. An object whose $T_{\max}$ falls short of ignition never gets hot enough — not slowly, not eventually, not ever. Degeneracy takes the support away from gravity before the fire starts. It is not a slow star. It is a thing that was never going to be one, and it spends the rest of its existence cooling.
Set $T_{\max} = T_{\rm ign}$ and solve. With the usual $3\times10^6$ K for the p–p chain and $X = 0.70$:
$$M_{\min} = 0.1035\,M_\odot = 108\,M_{\rm Jup}$$Now substitute $K_1$ and collect, exactly as in §8.8. Everything lands in the same place it did there:
against, from the other end of the range,
$M_{\rm ch} \;=\; 3.098\;\dfrac{M_{\rm Pl}^3}{(\mu_e m_u)^2}$
∎The same object — $M_{\rm Pl}^3/m_u^2$, the Planck mass cubed over the nucleon mass squared — times a dimensionless number, at both ends. The entire main sequence lives between two multiples of one quantity.
But look at what the two dimensionless numbers are.
Chandrasekhar's is $3.098 = \tfrac12\omega_3\sqrt{3\pi}$. A pure number. It comes out of one ordinary differential equation and it will be the same in a thousand years.
This one is $16.32\,\varepsilon^{3/4}$, where $\varepsilon = kT_{\rm ign}/m_ec^2 = 5.06\times 10^{-4}$ — a temperature divided by the electron rest mass. That is nuclear physics, not mathematics. And it is small, $\varepsilon^{3/4} = 0.0034$, which is precisely why stars span a factor of twenty in mass instead of a factor of one.
Chabrier, Baraffe, Phillips and Debras (2023) put the hydrogen-burning minimum mass at $0.075\,M_\odot$. This derivation says 0.1035. The predicted radius, 1.31 Jupiter radii, misses by about the same fraction — real brown dwarfs sit near one.
Four reasons, and they all push the same way:
Ignition is not a threshold. The real criterion is that nuclear luminosity balance surface luminosity over the object's life, and that is satisfied below $3\times10^6$ K. A weaker condition means a smaller minimum mass. This is the largest of the four and the one this model is least equipped to fix, having no atmosphere and therefore no luminosity.
The equation of state is too stiff. Ideal gas plus ideal degeneracy omits the Coulomb attraction between ions and the electron sea, and the exchange term. Both lower the pressure; a softer gas contracts further and gets hotter, and ignites at lower mass.
Partial degeneracy is not a sum. Adding the two pressures is not the finite-temperature Fermi–Dirac answer, and it overestimates support in exactly the regime that decides the question.
$n=3/2$ was assumed. An object in the transition region is not a polytrope of any single index, and both structural constants came from one.
A model whose errors happened to cancel would be the suspicious one. This one is wrong in a direction it can name, by an amount it can bound.
Invert the calculation and the ignition temperature that would give $0.075\,M_\odot$ is $1.98\times10^6$ K. That is physically sensible — p–p burning switches on gradually — and it is still a fitted number. The chapter does not adopt it. The 38% stands.
Here is what the two numbers have done.
The hydrogen-burning limit, as the equation of state improved: 0.073 (Saumon–Chabrier–van Horn, 1995), 0.074 (CMS19, 2019), 0.075 (Chabrier–Debras, 2023). It moved because the physics input moved. It will move again.
The Chandrasekhar mass: Chandrasekhar published 0.91 $M_\odot$ in 1931. Today we say 1.456. That looks like a bigger revision — until you notice why. He used $\mu_e = 2.5$; we use 2. Feed $\mu_e = 2.5$ into the modern formula and it returns 0.932. The number moved because an input was wrong. The formula never moved at all, and cannot: it is one ODE and three constants.
So the top of the white dwarf range is a theorem about a Fermi gas, and the bottom of the stellar range is a line drawn through a continuum by a criterion we chose. Both get quoted to three significant figures. Only one of them has earned it.
That distinction is not visible from either number alone. It is visible from the pair, which is the reason to derive the one you already know you will get wrong.
Six gaps in the script. The one that would most change the answer: no atmosphere, so no luminosity — which means reason one above is stated and not modelled, and it is the biggest of the four. The whole result also scales as $T_{\rm ign}^{3/4}$, so the softest number in the calculation controls it.
And the deuterium limit is not attempted. The same machinery at $T_{\rm ign}\approx 5\times10^5$ K would give the planet/brown-dwarf boundary at 13 Jupiter masses. It was left out because the systematic above is not pinned well enough to make a second uncalibrated prediction worth printing — which is a reason, not an excuse, and it is the next thing to fix.
Producing script: book8/ch8-8b-brown-dwarfs-verify.py — 7 sections, 6 gaps. CITED: Chabrier, Baraffe, Phillips & Debras 2023 (A&A 671 A119) for the HBMM and its history. Pedagogical derivations of the minimum stellar mass along these lines are published — e.g. Pinochet, arXiv:1909.08575 — and nothing here is claimed as new physics; what is offered is the side-by-side. Cross-references: §8.8, the upper limit, Book VII · 1103 and 26390.