Lane-Emden Equation

The Lane-Emden equation is what remains of a star once you refuse to talk about heat. Two of the four equations of stellar structure — hydrostatic equilibrium and mass continuity — are purely mechanical: they balance gravity against a pressure gradient and bookkeep the enclosed mass. The other two carry the thermodynamics: energy transport and generation, opacity, nuclear physics. Close the mechanical pair with a polytropic relation $P=K\rho^{1+1/n}$ , tying pressure to density alone, and the thermodynamic pair drops out entirely — the whole density structure of a self-gravitating sphere collapses into one dimensionless, second-order ODE indexed by a single number, the polytropic index $n$ . That reduction is more powerful than it looks. Solve it once for a given $n$ and you have characterized every star of that class, because the physical scales enter only through two dials, $K$ and the central density $\rho_c$ . The profile $\theta(\xi)$ holds all the shape: where it first hits zero fixes the surface, its slope there fixes the mass. Jonathan Homer Lane wrote the balance down in 1870 to model the Sun; Robert Emden tabulated its solutions in 1907 — decades before anyone knew what powered a star. Polytropes are not a museum piece. A fully convective star is $n=1.5$ because convection enforces $P\propto\rho^{5/3}$ ; a non-relativistic degenerate electron gas is also $n=1.5$ ; an ultra-relativistic degenerate gas is $n=3$ and, through it, the Chandrasekhar mass; Eddington's radiation-supported model is again $n=3$ ; and the isothermal limit $n\to\infty$ governs the cloud cores where stars are born. One equation threads through all of it.

Key takeaways

The Lane-Emden equation is the dimensionless form of hydrostatic equilibrium for a self-gravitating polytrope $P=K\rho^{1+1/n}$ ; index $n$ is the only parameter.
It reads $\frac{1}{\xi^2}\frac{\mathrm d}{\mathrm d\xi}\!\left(\xi^2\frac{\mathrm d\theta}{\mathrm d\xi}\right)=-\theta^{\,n}$ , integrated outward from $\theta(0)=1,\ \theta'(0)=0$ ( $\theta$ scaled density, $\xi$ scaled radius).
Closed forms exist only for $n=0$ ( $\xi_1=\sqrt6$ ), $n=1$ ( $\theta=\sin\xi/\xi,\ \xi_1=\pi$ ), $n=5$ ( $\xi_1=\infty$ ); all else is numerical.
The first zero $\xi_1$ sets the radius and $(-\xi^2\theta')_{\xi_1}$ the mass, so one curve holds the whole mass-radius relation of the class.
The key case: $n=3$ makes mass independent of central density, so degenerate stars have a maximum mass — the Chandrasekhar limit $M_{\rm Ch}\approx1.44\,M_\odot$ for $\mu_e=2$ .
Key idea: a polytropic equation of state collapses hydrostatic equilibrium plus mass conservation into one dimensionless ODE whose single solution per index $n$ describes all stars of that class; physical mass and radius enter only through scaling constants.

The core insight

A polytropic equation of state collapses hydrostatic equilibrium plus mass conservation into a single dimensionless ODE whose one solution per index $n$ describes all stars of that class; physical mass and radius enter only through scaling constants.

Where this lives today

Polytropes remain the working scaffolding of stellar and star-formation astrophysics. Every stellar-evolution code initializes and sanity-checks against them: $n=1.5$ for a fully convective pre-main-sequence star on the Hayashi track, $n=3$ for a radiative main-sequence core. The Chandrasekhar mass — the $n=3$ result — is the trigger scale for Type Ia supernovae, whose standardizable luminosity became the ruler that revealed dark energy, and $M_{\rm Ch}(\mu_e)$ feeds directly into those models. In star formation, the isothermal ( $n\to\infty$ ) limit gives the Bonnor-Ebert sphere, whose critical mass sets when a cloud core collapses — the benchmark for prestellar cores from ALMA and Herschel. And with exact solutions at $n=0,1,5$ , it endures as the standard stress-test for self-gravitating-hydrodynamics codes.

The physical picture

A star hangs together because two tendencies cancel at every radius: gravity pulling matter inward, and a pressure rising inward that pushes back out. Write that balance on a thin shell, add that enclosed mass is the running integral of density, and you have the two mechanical structure equations — not yet closed: two equations, three unknowns ( $P$ , $\rho$ , $m$ ). The polytrope supplies the missing relation, $P=K\rho^{1+1/n}$ with $K$ everywhere the same. Now pressure is known the instant density is, temperature never enters, and the mechanics close on themselves.

Figure

A self-gravitating shell balances inward gravity against the outward pressure gradient. The polytrope P = Kργ closes the mechanics without temperature, and scaling out the star's size and central density turns the balance into one dimensionless ODE.

The index $n$ is a stiffness dial. A small $n$ is a stiff, nearly incompressible gas — at $n=0$ the density is uniform and the star has a crisp edge. A large $n$ is a soft gas that piles most of its mass into a dense core wrapped in a puffy envelope. Push $n$ to $5$ and the envelope becomes infinitely extended: finite mass, no finite radius; beyond $n=5$ no bounded polytrope exists. The second figure shows the whole family.

Figure

Density templates θ(ξ) for polytropic index n = 0, 1, 3, 5. A stiff gas (small n) is sharply bounded; a soft gas (large n) is centrally concentrated with an extended envelope. The surface is the first zero ξ₁: √6 for n=0, π for n=1, 6.897 for n=3, while n=5 never reaches zero — infinite radius, finite mass.

The core equations

Start from the closure. A polytrope locks pressure and density by a power law with globally constant $K$ ; the exponent is written $\gamma=1+1/n$ so the index $n$ is the natural label. Nothing here is temperature — that is the point.

P = K\rho^{\,\gamma}, \qquad \gamma = 1 + \frac{1}{n} \qquad (1)

where: $P$ — pressure; $\rho$ — mass density; $K$ — polytropic constant, fixed throughout the star; $\gamma$ — polytropic exponent; $n$ — polytropic index. Reading it: with $K$ and $n$ fixed the equation of state is a curve, not a surface — one density, one pressure, no thermodynamic freedom. The two mechanical structure equations fold into one. Differentiating $r^2\rho^{-1}\,\mathrm dP/\mathrm dr=-Gm$ and using $\mathrm dm/\mathrm dr=4\pi r^2\rho$ gives the Poisson form below — the equation the polytrope is poured into.

\frac{1}{r^2}\frac{\mathrm d}{\mathrm d r}\!\left(\frac{r^2}{\rho}\frac{\mathrm d P}{\mathrm d r}\right) = -4\pi G \rho \qquad (2)

where: $r$ — radial coordinate; $G$ — Newton's constant; $\rho$ — local density; $P$ — local pressure. Reading it: the left side is a spherical Laplacian of a pressure potential, the right its source — hydrostatic equilibrium is Poisson's equation. Inserting the polytrope and rescaling length and density strips out every dimensional constant, leaving the Lane-Emden equation — a nonlinear ODE for one function $\theta(\xi)$ , derived in full below.

\frac{1}{\xi^2}\frac{\mathrm d}{\mathrm d\xi}\!\left(\xi^2\frac{\mathrm d\theta}{\mathrm d\xi}\right) = -\theta^{\,n} \qquad (3)

where: $\xi$ — dimensionless radius, $r=\alpha\xi$ ; $\theta$ — dimensionless density, $\rho=\rho_c\theta^{\,n}$ ; $n$ — index. Reading it: one template curve solves it per $n$ ; the star's identity lives in the scale factors, not the shape. The variable $\theta$ is not the density but its $(1/n)$ -th root; density and pressure recover as powers of it, so $\theta$ runs from $1$ at the center to $0$ at the surface.

\rho = \rho_c\,\theta^{\,n}, \qquad P = P_c\,\theta^{\,n+1}, \qquad P_c = K\rho_c^{\,1+1/n} \qquad (4)

where: $\rho_c$ — central density; $P_c$ — central pressure; $\theta$ — the solution. Reading it: since $P\propto\theta^{n+1}$ and $\rho\propto\theta^{n}$ , both vanish at the surface where $\theta$ first reaches zero. The length unit $\alpha$ is the one combination of $K$ , $\rho_c$ and $G$ that makes (3) dimensionless — the physical size of one $\xi$ unit, and where central density re-enters.

\alpha^2 = \frac{(n+1)P_c}{4\pi G\rho_c^{2}} = \frac{(n+1)K}{4\pi G}\,\rho_c^{\frac{1}{n}-1} \qquad (5)

where: $\alpha$ — radial scale factor, $R=\alpha\xi_1$ ; $\rho_c$ — central density; $K$ — polytropic constant; $n$ — index. Reading it: turn the central-density dial and the star scales by $\alpha$ while its shape never changes. Regularity at the origin fixes both initial data: $\theta(0)=1$ defines $\rho_c$ , and the vanishing slope encodes a center with no enclosed mass and no pressure gradient.

\theta(0)=1, \qquad \left.\frac{\mathrm d\theta}{\mathrm d\xi}\right|_{\xi=0}=0 \qquad (6)

where: $\theta(0)=1$ — density equals $\rho_c$ at the center; $\theta'(0)=0$ — zero central pressure gradient, from $m\to0$ . Reading it: these are the only conditions needed — a polytrope specifies no surface pressure; $\theta$ finds its own zero. Once the curve is in hand, $\xi_1$ and the boundary derivative deliver the two observables. The mass is a pure surface term, since the Lane-Emden integrand is a total derivative.

R = \alpha\,\xi_1, \qquad M = 4\pi\alpha^3\rho_c\left(-\xi^2\frac{\mathrm d\theta}{\mathrm d\xi}\right)_{\!\xi_1} \qquad (7)

where: $\xi_1$ — first zero of $\theta$ ; $R$ — radius; $M$ — total mass; $(-\xi^2\theta')_{\xi_1}$ — dimensionless mass factor, tabulated once per $n$ . Reading it: two template numbers plus the dials $\alpha$ and $\rho_c$ reconstruct any star of the class. Eliminating $\alpha$ between (5) and (7) gives the mass as an explicit function of central density — the relation whose exponent is the hinge of the theory. When $(3-n)/2n$ vanishes at $n=3$ , mass stops depending on compression and a maximum appears.

M = -4\pi\left[\frac{(n+1)K}{4\pi G}\right]^{3/2}\rho_c^{\frac{3-n}{2n}}\left(\xi^2\frac{\mathrm d\theta}{\mathrm d\xi}\right)_{\!\xi_1} \qquad (8)

where: $\rho_c$ — central density; $n$ — index; $K$ — polytropic constant. Reading it: for $n<3$ mass rises with central density, for $n>3$ it falls, for $n=3$ it is frozen — the sign of $3-n$ decides whether squeezing adds mass, sheds it, or neither. Finally, the central pressure of any polytrope is a fixed multiple $W_n$ of $GM^2/R^4$ , with $W_n$ set by the boundary slope — turning mass and radius into an interior pressure.

P_c = \frac{1}{4\pi(n+1)\,[\theta'(\xi_1)]^{2}}\,\frac{GM^2}{R^4} \equiv W_n\,\frac{GM^2}{R^4} \qquad (9)

where: $W_n$ — dimensionless structure constant ( $W_0=3/8\pi\approx0.119$ , $W_{1.5}\approx0.770$ , $W_3\approx11.05$ ); $M,R$ — mass and radius. Reading it: the deeper the central concentration (larger $n$ ), the larger $W_n$ — an $n=3$ star crushes its core to ~100 times the central pressure of an $n=0$ ball of equal $M$ and $R$ .

Derivation AReducing the two structure equations to a single dimensionless ODE

The goal is to show how hydrostatic equilibrium and mass continuity — two coupled first-order equations in $P$ , $\rho$ and $m$ — become the one nonlinear second-order equation (3) once a polytropic closure is imposed. We fold the mechanics into Poisson form, insert $P=K\rho^{1+1/n}$ so pressure becomes a power of density, define a density variable $\theta$ that turns those powers into a clean Laplacian, and choose a length unit $\alpha$ that annihilates the dimensional constants. The reward is an equation that knows nothing about which star it describes.

Step 1. Combine $\mathrm dP/\mathrm dr=-Gm\rho/r^2$ with $\mathrm dm/\mathrm dr=4\pi r^2\rho$ : multiply the first by $r^2/\rho$ , differentiate, and substitute the second to get $\frac{1}{r^2}\frac{\mathrm d}{\mathrm dr}\!\big(\frac{r^2}{\rho}\frac{\mathrm dP}{\mathrm dr}\big)=-4\pi G\rho$ . Why: this eliminates the enclosed mass $m$ , leaving a self-contained relation between $P$ and $\rho$ .

Step 2. Insert the polytrope. Since $P=K\rho^{1+1/n}$ , $\frac1\rho\frac{\mathrm dP}{\mathrm dr}=(n+1)K\frac{\mathrm d}{\mathrm dr}\rho^{1/n}$ , so the factor $\rho^{-1}\,\mathrm dP/\mathrm dr$ becomes the gradient of a single power of density. Why: the density-weighted pressure gradient collapses to $\nabla\rho^{1/n}$ , which will become a plain Laplacian.

Step 3. Define $\rho=\rho_c\theta^{\,n}$ , so $\rho^{1/n}=\rho_c^{1/n}\theta$ with $\theta=1$ at the center. The equation becomes $(n+1)K\rho_c^{1/n}\,\frac{1}{r^2}\frac{\mathrm d}{\mathrm dr}\!\big(r^2\frac{\mathrm d\theta}{\mathrm dr}\big)=-4\pi G\rho_c\theta^{\,n}$ . Why: the left side is now the plain spherical Laplacian of $\theta$ , with all the nonlinearity in the source $\theta^{\,n}$ .

Step 4. Choose the length unit that clears the constants: $\alpha^2=(n+1)K\rho_c^{1/n-1}/(4\pi G)$ , $r=\alpha\xi$ . Every dimensional prefactor cancels, leaving $\frac{1}{\xi^2}\frac{\mathrm d}{\mathrm d\xi}\!\big(\xi^2\frac{\mathrm d\theta}{\mathrm d\xi}\big)=-\theta^{\,n}$ . Why: $\alpha$ is the unique length-dimensioned combination of $K$ , $\rho_c$ , $G$ , so a dimensionless equation forces this scale — and with it the star's physical size.

Go deeper · Derivation B · 1 minThe closed-form solution at index one

Derivation BThe closed-form solution at index one

The aim is to exhibit the one physically interesting index where the Lane-Emden equation is exactly solvable in elementary functions, and to extract its structural numbers $\xi_1=\pi$ and $(-\xi^2\theta')_{\xi_1}=\pi$ . At $n=1$ the source term $\theta^{\,n}$ becomes simply $\theta$ , so the equation is linear; the substitution $\theta=\chi/\xi$ then converts the spherical Laplacian into a harmonic-oscillator equation whose solution reads off at once — the cleanest proof that the equation really contains a finite-radius star.

Step 1. Set $n=1$ : the equation becomes $\frac{1}{\xi^2}\frac{\mathrm d}{\mathrm d\xi}\!\big(\xi^2\frac{\mathrm d\theta}{\mathrm d\xi}\big)=-\theta$ , a linear second-order ODE. Why: linearity is what makes $n=1$ special — elementary solutions exist, which fails for every other physical index.

Step 2. Substitute $\theta=\chi(\xi)/\xi$ ; the derivatives conspire and the equation reduces to $\mathrm d^2\chi/\mathrm d\xi^2=-\chi$ . Why: $\theta=\chi/\xi$ is the universal trick for the 3D radial Laplacian — it trades a spherical operator for a flat one.

Step 3. The solution is $\chi=A\sin\xi+B\cos\xi$ , so $\theta=(A\sin\xi+B\cos\xi)/\xi$ . Regularity at the origin forces $B=0$ , and $\theta(0)=1$ (with $\sin\xi/\xi\to1$ ) forces $A=1$ . So $\theta=\sin\xi/\xi$ . Why: the two boundary conditions select one member of the family — the spherical Bessel function $j_0$ .

Step 4. The surface is the first zero: $\sin\xi=0$ at $\xi_1=\pi$ . With $\theta'=\cos\xi/\xi-\sin\xi/\xi^2$ , $\theta'(\pi)=-1/\pi$ , so the mass factor $(-\xi^2\theta')_{\pi}=\pi$ . Why: these are the two numbers every application needs; for $n=1$ the radius $R=\alpha\pi$ is famously independent of mass, since $\alpha$ does not depend on $\rho_c$ .

Go deeper · Derivation C · 1 minWhy index three fixes a maximum mass

Derivation CWhy index three fixes a maximum mass

This derivation traces the most consequential fact about polytropes: at $n=3$ the mass becomes independent of central density, and if $K$ is itself fixed by microphysics — as for an ultra-relativistic degenerate electron gas — then only one mass is consistent with equilibrium at all. That is the Chandrasekhar limit. We read the exponent off the mass formula, watch it vanish at $n=3$ , pin $K$ from relativistic degeneracy, and assemble the value near $1.4\,M_\odot$ .

Step 1. From equation (8), $M\propto\rho_c^{(3-n)/2n}$ ; at $n=3$ the exponent is $(3-3)/6=0$ , so $M$ loses all dependence on central density. Why: for $\gamma=4/3$ the pressure gravity demands and the pressure the gas supplies both scale as $\rho^{4/3}$ , so compression shifts neither — equilibrium holds at one mass only.

Step 2. Fix the microphysics. An ultra-relativistic degenerate electron gas has $P=K\rho^{4/3}$ ( $n=3$ ), with $K=\frac{(3\pi^2)^{1/3}}{4}\frac{\hbar c}{(\mu_e m_H)^{4/3}}$ , $\mu_e$ the nucleons per electron. Why: here $K$ is not a free label but a combination of constants, so the "one mass" becomes a definite number.

Step 3. Assemble: at $n=3$ , $M=4\pi\,(-\xi^2\theta')_{\xi_1}\,[K/(\pi G)]^{3/2}$ , with $\xi_1=6.897$ , $(-\xi^2\theta')_{\xi_1}=2.018$ . Why: the two template numbers are the only structural input; everything else is fundamental constants, so $M$ is essentially built from $\hbar$ , $c$ , $G$ and $m_H$ .

Step 4. For $\mu_e=2$ (helium, carbon, oxygen), $K\approx4.89\times10^{9}$ SI gives $M_{\rm Ch}=2.85\times10^{30}\ \mathrm{kg}\approx1.44\,M_\odot$ , and generally $M_{\rm Ch}\approx1.44\,(2/\mu_e)^2\,M_\odot$ . Why: this is the threshold above which degeneracy pressure fails — the trigger mass for Type Ia supernovae and the end of the white-dwarf sequence. *Direct numerical integration gives $\xi_1=6.8968$ and $(-\xi^2\theta')_{\xi_1}=2.0182$ , yielding $M_{\rm Ch}=1.44\,M_\odot$ for $\mu_e=2$ .*

Limiting cases

The entire character of a polytrope — sharp-edged or diffuse, finite or unbounded, analytic or numerical — is set by the one index $n$ :

Case	Result	Meaning
$n=0$ (incompressible)	$\theta=1-\dfrac{\xi^2}{6},\ \xi_1=\sqrt6$	Uniform-density sphere; $(-\xi^2\theta')_{\xi_1}=2\sqrt6$ ; upper bound $P_c=\tfrac{3}{8\pi}GM^2/R^4$ .
$n=1$	$\theta=\dfrac{\sin\xi}{\xi},\ \xi_1=\pi$	Linear, exactly solvable. Radius $R=\alpha\pi$ independent of mass, since $\alpha$ is $\rho_c$ -independent at $n=1$ .
$n=1.5$	$\gamma=5/3,\ \xi_1\approx3.654$	Non-relativistic degenerate gas and fully convective stars; $(-\xi^2\theta')_{\xi_1}\approx2.714$ . Numerical.
$n=3$	$\gamma=4/3,\ \xi_1\approx6.897$	Ultra-relativistic degeneracy and Eddington's model; mass independent of $\rho_c$ — the Chandrasekhar mass.
$n=5$	$\theta=\left(1+\dfrac{\xi^2}{3}\right)^{-1/2},\ \xi_1=\infty$	Finite mass ( $(-\xi^2\theta')_\infty=\sqrt3$ ), infinite radius — marginal case; the Plummer profile.
$n\to\infty$ (isothermal)	$\dfrac{1}{\xi^2}\dfrac{\mathrm d}{\mathrm d\xi}\!\left(\xi^2\dfrac{\mathrm d\psi}{\mathrm d\xi}\right)=e^{-\psi}$	Isothermal Lane-Emden, $P=K\rho$ . Infinite mass unless truncated — Bonnor-Ebert cores.

Common trapThe polytropic index is just the adiabatic exponent

It is tempting to read $\gamma=1+1/n$ as though the polytropic index were always the gas's adiabatic exponent — but the two coincide only when $K$ is genuinely set by a single adiabat. A fully convective star is an $n=1.5$ polytrope because convection enforces $P\propto\rho^{5/3}$ along one entropy, so $\gamma_{\rm poly}=\gamma_{\rm ad}=5/3$ . But Eddington's standard model is an $n=3$ polytrope for a different reason: radiation and gas pressure hold a fixed ratio $\beta$ , making the total pressure scale as $\rho^{4/3}$ with no reference to any single-particle adiabat. Same $n$ , unrelated physics. A sharper trap: polytropes with $n\ge5$ have no finite radius — $\theta$ never reaches zero, so there is no surface. Only $n<5$ is a bounded star; $n=5$ is the knife-edge with finite mass but infinite extent.

Worked examples

1 · Central conditions of the Sun as an n = 3 polytrope (fully worked)

Model the Sun as an $n=3$ polytrope ( $M_\odot=1.989\times10^{30}$ kg, $R_\odot=6.96\times10^8$ m). Estimate its central density, pressure, and temperature, and compare with the real Sun.

Step 1. Mean density: $\bar\rho=\dfrac{3M_\odot}{4\pi R_\odot^3}=\mathbf{1.41\times10^{3}\ kg\,m^{-3}}$ — a shade denser than water, averaged over the star.

Step 2. The central-to-mean ratio is $\rho_c/\bar\rho=\xi_1^3/\big[3(-\xi^2\theta')_{\xi_1}\big]=6.897^3/(3\times2.018)=\mathbf{54.2}$ , so $\rho_c=54.2\times1.41\times10^3=\mathbf{7.6\times10^{4}\ kg\,m^{-3}}$ (≈76 g cm⁻³).

Step 3. Central pressure from (9) with $W_3=11.05$ : $P_c=11.05\,\dfrac{G M_\odot^2}{R_\odot^4}=\mathbf{1.25\times10^{16}\ Pa}$ .

Step 4. Central temperature from the ideal gas $P_c=\rho_c k_B T_c/(\mu m_H)$ with $\mu\approx0.62$ : $T_c=\dfrac{P_c\,\mu m_H}{\rho_c k_B}=\mathbf{1.2\times10^{7}\ K}$ .

Step 5. Compare. The real Sun has $\rho_c\approx1.5\times10^5$ kg m⁻³, $P_c\approx2.5\times10^{16}$ Pa, $T_c\approx1.57\times10^7$ K. The $n=3$ polytrope lands within a factor of two on every quantity using no nuclear physics, opacities, or energy transport — and runs low because the real Sun is more centrally condensed than a single polytrope. A one-parameter mechanical model reproduces the Sun's core to a factor of two.

2 · The Chandrasekhar mass from the n = 3 template

Compute the limiting white-dwarf mass for a carbon-oxygen composition ( $\mu_e=2$ ) from the $n=3$ formula.

Step 1. For $n=3$ , equation (8) reduces to $M=4\pi\,(-\xi^2\theta')_{\xi_1}\,[K/(\pi G)]^{3/2}$ , independent of $\rho_c$ , with $(-\xi^2\theta')_{\xi_1}=2.018$ .

Step 2. The degeneracy constant $K=\dfrac{(3\pi^2)^{1/3}}{4}\dfrac{\hbar c}{(\mu_e m_H)^{4/3}}$ ; with $\hbar c=3.16\times10^{-26}$ J m and $\mu_e m_H=3.35\times10^{-27}$ kg, $K=\mathbf{4.89\times10^{9}}$ (SI).

Step 3. Assemble: $K/(\pi G)=2.33\times10^{19}$ , $[K/(\pi G)]^{3/2}=1.13\times10^{29}$ , so $M_{\rm Ch}=4\pi(2.018)(1.13\times10^{29})=\mathbf{2.85\times10^{30}\ kg}=1.44\,M_\odot$ .

Step 4. Scaling $M_{\rm Ch}\propto\mu_e^{-2}$ : hydrogen ( $\mu_e=1$ ) gives $\sim5.8\,M_\odot$ , iron-group ( $\mu_e\approx2.15$ ) $\sim1.25\,M_\odot$ . $M_{\rm Ch}\approx1.44\,(2/\mu_e)^2\,M_\odot$ — built entirely from $\hbar$ , $c$ , $G$ and the proton mass.

3 · Why more massive white dwarfs are smaller (n = 1.5)

A cool white dwarf below the relativistic regime is an $n=1.5$ polytrope. Show its radius shrinks as mass grows.

Step 1. For non-relativistic degeneracy $\gamma=5/3$ , so $n=1.5$ with $K$ a fixed constant of nature. The mass relation (8) gives $M\propto\rho_c^{(3-n)/2n}=\rho_c^{1/2}$ .

Step 2. The length scale (5) gives $\alpha\propto\rho_c^{(1/n-1)/2}=\rho_c^{-1/6}$ , and since $R=\alpha\xi_1$ with $\xi_1$ fixed, $R\propto\rho_c^{-1/6}$ .

Step 3. Eliminate $\rho_c$ : from Step 1, $\rho_c\propto M^{2}$ , so $R\propto(M^2)^{-1/6}=\mathbf{M^{-1/3}}$ . More mass means a smaller star — the opposite of ordinary matter, because added weight compresses the degenerate gas faster than it adds volume. It runs up to where electrons turn relativistic ( $n\to3$ ) and $R\to0$ at the Chandrasekhar mass. $R\propto M^{-1/3}$ — Sirius B, at $\approx1\,M_\odot$ , is only Earth-sized.

Physics of the portrait

The Lane-Emden equation is nineteenth-century mathematical physics that turned out exactly right about the twentieth century's stars — written to model the Sun long before anyone knew a star had a nuclear furnace.

Origins · 1 minLane-Emden equation and how this was discovered

Lane-Emden equation

the polytrope ODE ∇²θ = −θⁿ · Lane 1870, Emden 1907 · stellar structure before nuclear physics

The equation carries two names because it was built in two acts. Jonathan Homer Lane, an American patent examiner and physicist, published the first quantitative theory of a star as a gaseous sphere in equilibrium in 1870, asking whether a self-gravitating gas on the ideal-gas and convective-adiabatic laws could reproduce the Sun's size and temperature — and estimating a solar central temperature decades before anyone knew the source of solar energy. August Ritter systematized polytropic gas spheres through the 1870s-1880s and introduced the family of indices, and Robert Emden gathered and vastly extended the numerical solutions in his 1907 treatise Gaskugeln ("Gas Spheres"). The climax belongs to Subrahmanyan Chandrasekhar, who in 1930-31 recognized that a relativistic degenerate gas is an $n=3$ polytrope whose density-independent mass implied a hard upper limit for white dwarfs — the result Eddington disputed and that later won the Nobel Prize. What Lane began as a model of the Sun is now the scaffolding for degenerate stars, convective envelopes, and cloud cores.

History

1870

— Jonathan Homer Lane publishes the first theory of the Sun as a gas sphere in equilibrium

1878

— August Ritter systematizes polytropic gas spheres and the family of indices

1907

— Robert Emden's Gaskugeln tabulates the solutions θ(ξ) for many polytropic indices

1930

— Subrahmanyan Chandrasekhar derives the limiting white-dwarf mass from the n=3 polytrope

1939

— Chandrasekhar's An Introduction to the Study of Stellar Structure codifies polytrope theory

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., §10.5 and §16.4 · Chandrasekhar, An Introduction to the Study of Stellar Structure, ch. IV · Kippenhahn, Weigert & Weiss, Stellar Structure and Evolution, 2nd ed., ch. 19 · Hansen, Kawaler & Trimble, Stellar Interiors, 2nd ed., §7.2 · Binney & Tremaine, Galactic Dynamics, 2nd ed., §4.3.3 · Mo, van den Bosch & White, Galaxy Formation and Evolution, §8.2. Key results to keep in view: combining hydrostatic equilibrium and mass continuity with $P=K\rho^{1+1/n}$ yields the boxed ODE; the exact solutions are $n=0$ ( $\xi_1=\sqrt6$ ), $n=1$ ( $\theta=\sin\xi/\xi,\ \xi_1=\pi$ ), $n=5$ ( $\theta=(1+\xi^2/3)^{-1/2}$ ); the numerical values are $\xi_1\approx3.654$ at $n=1.5$ and $\xi_1\approx6.897$ , $(-\xi^2\theta')_{\xi_1}\approx2.018$ at $n=3$ ; the Sun-as- $n=3$ estimates give $\rho_c\approx7.6\times10^4$ kg m⁻³, $P_c\approx1.25\times10^{16}$ Pa, and $M_{\rm Ch}\approx1.44\,M_\odot$ for $\mu_e=2$ . The structure constants $W_0=3/8\pi$ , $W_{1.5}=0.770$ , $W_3=11.05$ follow from $W_n=1/[4\pi(n+1)\theta'(\xi_1)^2]$ .

Prerequisites. Hydrostatic Equilibrium · Mass Continuity Equation · Equation of State for Stellar Matter · Equations of Stellar Structure. Leads to. Convective Energy Transport · Adiabatic Temperature Gradient · Radiation Pressure in Stellar Interiors.

Ask the tutor: Ask me to walk through the numerical integration of the $n=1.5$ or $n=3$ polytrope, or to show exactly how the Chandrasekhar mass falls out of the $n=3$ scaling.

Why can we ignore temperature and the energy equations when computing polytrope structure?

Because the closure $P=K\rho^{1+1/n}$ ties pressure directly to density with a globally constant $K$ . Hydrostatic equilibrium then involves only $P$ and $\rho$ , so the mechanical structure closes without ever mentioning temperature, opacity, or nuclear generation. Those can be recovered afterward if you want them (for an ideal gas, $T\propto P/\rho\propto\theta$ ), but they are not needed for the shape of the star. That is why a polytrope is such a shortcut: it isolates the mechanical skeleton from all the microphysics a full stellar model must otherwise solve at once.

Is the polytropic index n the same thing as the gas adiabatic exponent?

Not in general. The relation $\gamma=1+1/n$ holds by definition, but whether that $\gamma$ equals the thermodynamic adiabatic exponent depends on why the star is a polytrope. For a convective star or a degenerate gas, $K$ is set by a single adiabat and the two coincide ( $n=1.5$ , $\gamma=5/3$ ). But Eddington's standard model is $n=3$ because radiation and gas pressure keep a constant ratio, making the total pressure scale as $\rho^{4/3}$ for reasons unrelated to any particle's adiabat. Same $n$ , unrelated physics — so reading $n$ as "the adiabatic exponent" is an error unless you have checked that $K$ is fixed by one adiabat.

Why does n = 1.5 keep appearing for both convective stars and white dwarfs?

Two independent routes to the same exponent $\gamma=5/3$ . A fully convective star is nearly isentropic because convection enforces $P\propto\rho^{5/3}$ along one entropy — a thermodynamic argument. A non-relativistic degenerate electron gas also has $P\propto\rho^{5/3}$ , but for a quantum reason: the Pauli principle and the filling of a Fermi sea. Both give $n=1.5$ , hence the identical template with $\xi_1\approx3.654$ . The coincidence is in the exponent, not the physics — which is exactly what makes the polytrope economical: one curve serves a red dwarf and a cool white dwarf alike.

What exactly goes wrong for polytropic index n greater than or equal to 5?

The solution $\theta(\xi)$ never crosses zero. For $n<5$ the curve hits $\theta=0$ at a finite $\xi_1$ — the surface, where density and pressure vanish and the star has a definite radius. For $n\ge5$ it decays too slowly ever to reach zero, so there is no surface and the configuration extends to infinity. The case $n=5$ is the exact knife-edge: $\theta=(1+\xi^2/3)^{-1/2}$ has infinite radius but finite mass (factor $\sqrt3$ ). For $n>5$ even the mass diverges. So only $n<5$ is a bound star; the isothermal $n\to\infty$ limit must be truncated by an external pressure to be physical.

Why does the Chandrasekhar mass depend on composition but not on central density?

Two facts combine. First, at $n=3$ the exponent $(3-n)/2n$ vanishes, so mass is independent of central density — the same-mass star can be built at any compression. Second, $K$ for a relativistic degenerate gas is fixed by constants and by nucleons-per-electron $\mu_e$ : $K\propto\mu_e^{-4/3}$ , and since $M\propto K^{3/2}$ , $M_{\rm Ch}\propto\mu_e^{-2}$ . Composition enters through how many nucleons each electron supports — hydrogen ( $\mu_e=1$ ) gives a larger limit, carbon-oxygen ( $\mu_e=2$ ) the canonical $1.44\,M_\odot$ . Central density is exactly the variable that drops out, which is why a maximum mass exists at all.

What is the isothermal Lane-Emden equation and where does it matter?

It is the $n\to\infty$ limit, where $P=K\rho^{1+1/n}$ degenerates to $P=K\rho$ . Taking that limit carefully (with $\theta\to e^{-\psi/n}$ ) gives $\frac{1}{\xi^2}\frac{\mathrm d}{\mathrm d\xi}(\xi^2\frac{\mathrm d\psi}{\mathrm d\xi})=e^{-\psi}$ , the isothermal or Emden-Chandrasekhar equation. Its untruncated solution has infinite mass, so alone it is not a star — but truncated by an external confining pressure it becomes the Bonnor-Ebert sphere, the model of a molecular-cloud core on the verge of collapse. That sphere's maximum stable mass sets the critical mass for gravitational instability, and the same machinery reappears in the singular isothermal sphere of halo modeling.