advanced35 min studyLesson 894 in Physics

Polytropic models and the Lane–Emden equation

Reduce a spherical barotropic model to a dimensionless boundary problem and interpret its mass–radius scaling without confusing equilibrium with stability.

Before you begin

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The question for today

What you’ll understand

Derive the dimensionless Lane–Emden equation and regular central conditions.
Recover physical radius and mass from the first zero and surface derivative.
Distinguish a polytropic equilibrium exponent from an adiabatic perturbation exponent.

A polytropic model replaces the full thermal problem by a specified relation between pressure and density. This lets us isolate how self-gravity responds to the stiffness of a barotropic pressure law. It can approximate a cold degenerate object or an isentropic ideal-gas region, but it is not automatically a complete nuclear-burning stellar model. The polytropic constant must either be fixed by microscopic physics or declared as a free structural parameter. Keeping those interpretations separate prevents a scaling relation from being mistaken for a universal evolutionary track.

Choose a pressure law and regular center

Write pressure as K times density to the power one plus one over the polytropic index n, taking a positive constant K and positive n. Combine mass continuity with hydrostatic balance to eliminate the enclosed mass. Rescale density by its central value and radius by a length alpha chosen to absorb the dimensional coefficients. The resulting dimensionless equation depends only on n. Physical units reside in alpha, K, and central density rather than in the dimensionless solution.

At the center the dimensionless density variable theta equals one and its first derivative vanishes by regular spherical symmetry. The equation appears singular at zero radius, but a regular series starts with one minus radius squared divided by six. This expansion supplies a controlled numerical starting point away from the coordinate singularity. A finite-radius model ends at the first positive zero of theta, provided that zero exists. For the usual nonnegative indices below five the regular solution has such a finite first zero.

P=K\rho^{1+1/n},\quad \rho=\rho_c\theta^n,\quad r=\alpha\xi,\quad \alpha^2=\frac{(n+1)K}{4\pi G}\rho_c^{1/n-1},\quad \frac1{\xi^2}\frac{d}{d\xi}\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n

θ and ξ are dimensionless, α is m, and ρc is kg m⁻³. K has units Pa divided by (kg m⁻³)^(1+1/n). Central conditions are θ(0)=1 and θ′(0)=0. This model assumes Newtonian spherical hydrostatic balance and constant K.↔ Scroll wide equations horizontally.

The surface derivative contains the total mass

The first zero gives the radius after multiplication by alpha. The derivative there determines how much mass was accumulated inside the model. It is negative for the regular decreasing solution, so the minus sign in the mass expression is necessary for a positive mass. Merely locating the zero is insufficient to infer total mass without the normalization and derivative. A zero-pressure surface is part of this idealized model, rather than an exact atmosphere boundary.

The case n equal to one has an elementary solution: theta is sine of xi divided by xi, with its central value understood by continuity. Its first zero is pi and its surface mass coefficient is also pi. At fixed K, alpha and radius are independent of central density, while mass is proportional to central density. This unusual behavior is a property of the chosen quadratic pressure law. It does not imply that increasing the central density of an arbitrary real star leaves its radius fixed.

R=\alpha\xi_1,\qquad M=4\pi\alpha^3\rho_c[-\xi_1^2\theta^{\prime}(\xi_1)],\qquad n=1:\quad\theta=\frac{\sin\xi}{\xi},\quad\xi_1=\pi

The first positive zero is ξ₁. For n=1, −ξ₁²θ′(ξ₁)=π and M=4π²α³ρc. The sine quotient has the regular limit θ(0)=1. These are exact results for the stated polytrope.↔ Scroll wide equations horizontally.

Scaling says what is held fixed

Eliminating central density gives a relation among mass, radius, and K. For a cold nonrelativistic electron gas of fixed composition, n is three halves and K is set by microscopic constants and electron abundance. The resulting inverse cube-root mass–radius relation describes that restricted degeneracy limit. At n equal to three, the Newtonian model instead has a mass fixed by K and independent of central density. Relativistic cold-electron support motivates the limiting-mass calculation, with important corrections in actual remnants.

An equilibrium polytropic index describes how pressure varies through the background model. The adiabatic exponent describes the pressure response of a moving parcel during a rapid perturbation. They agree only under additional thermodynamic conditions, such as an isentropic ideal gas with a consistent composition. Therefore the existence of a Lane–Emden solution is not by itself proof of dynamical or thermal stability. Nor does the solution determine opacity, burning luminosity, composition evolution, or a uniquely selected age. Those require restoring the missing thermal and evolutionary equations.

R\propto K^{n/(3-n)}M^{(1-n)/(3-n)}\quad(n\ne3),\qquad n=\frac32:\ R\propto K M^{-1/3},\qquad n=3:\ M\propto K^{3/2}

The proportionalities compare fixed-index Newtonian polytropes. Dimensionless factors depend on n. The n=3 relation is written separately because eliminating ρc through the ordinary mass–radius power law is singular there; it is not a general statement about every radiation-dominated or relativistic star.↔ Scroll wide equations horizontally.

Work it through

Recover an n=1 model

For an n=1 toy polytrope, take α=10⁶ m and ρc=10⁴ kg m⁻³. Use the exact sine-quotient solution to find R and M. Then double ρc at the same K and state what changes.

The first zero is π, so $R=\pi\times10^6\ {\rm m}$ . The center is regular even though the quotient notation contains ξ in the denominator.
The surface derivative gives the mass coefficient π, hence $M=4\pi^2\times10^{22}\ {\rm kg}$ , approximately 3.95 × 10²³ kg.
For n=1, α²=K/(2πG) does not contain ρc. Fixed K therefore preserves the radius while the doubled central density doubles the mass.

The result

R ≈ 3.14 × 10⁶ m and M ≈ 3.95 × 10²³ kg; doubling central density doubles mass at unchanged radius in this specific model.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A quadratic pressure law is inspectable

Reset and set BOTH Coefficient a to 0.5 and Position x to 1; predict y=0.5 and slope=1. Then Reset and set BOTH a to 0.5 and x to 2; predict y=2 and slope=2. With x interpreted as scaled density, this is the n=1 pressure law only. It does not solve Lane–Emden, find a stellar surface, compute enclosed mass, or establish stability.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA chosen barotropic pressure law is input to a self-gravitating model, not its complete solution.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The mass normalization contains ρc raised to (3−n)/(2n), which vanishes at n=3. Radius and stability still require their own analysis.

Your turn

Try it yourself

A cold nonrelativistic degenerate toy sequence has fixed composition and K, with R proportional to M to the power −1/3. Compare a model of mass 8M₀ with a reference of mass M₀. Find its radius ratio and mean-density ratio, and explain when the result stops being valid.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the mass–radius power law first, then mean density proportional to mass divided by radius cubed.

Compare with the solution

The radius ratio is $8^{-1/3}=1/2$ . Mean density therefore rises by $8/(1/2)^3=64$ . The comparison assumes both models remain cold and nonrelativistic with the same composition. Increasing density raises the Fermi momentum and can eventually invalidate the nonrelativistic equation of state, so extending this scaling to arbitrary mass would be inconsistent.

Keep the idea

Take it with you

Lane–Emden isolates spherical barotropic structure with a regular center and specified K.
The first zero and its derivative determine physical radius and mass.
A polytropic background exponent neither supplies thermal evolution nor guarantees perturbative stability.

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