The Lane-Emden Equation for Polytropes

Derivation

The Lane-Emden Equation for Polytropes

D-334 Home PU-308 Threads force · energy · matter Depends on Equation of Stellar Hydrostatic Equilibrium, poisson-equation-gravity

Statement

For a self-gravitating, spherically symmetric fluid in hydrostatic equilibrium obeying the polytropic equation of state \( P = K\rho^{1+1/n} \), the structure equations reduce exactly to the dimensionless Lane\text{-}Emden equation \[ \frac{1}{\xi^2}\frac{d}{d\xi}\!\left(\xi^2\frac{d\theta}{d\xi}\right) = -\,\theta^{\,n}, \] where the density profile is \( \rho = \rho_c\,\theta^{\,n} \), the radial coordinate is scaled as \( r=\alpha\xi \) with \( \alpha^2 = (n+1)P_c/(4\pi G\rho_c^{2}) \), and the physical boundary conditions \( \theta(0)=1 \), \( \theta'(0)=0 \) hold.

Why it matters

The Lane-Emden equation is the single ordinary differential equation that governs the entire mechanical structure of any polytrope. Because the polytropic index \( n \) captures a wide range of physical situations — an ideal degenerate non-relativistic gas (\( n=3/2 \)), a fully convective star or the ultra-relativistic degenerate core (\( n=3 \)), an isothermal-like limit, and the incompressible sphere (\( n=0 \)) — a single dimensionless solution \( \theta(\xi) \) can be re-scaled to describe stars, planetary interiors, and star-cluster envelopes without re-solving the physics each time.

It is the historical and pedagogical gateway to stellar structure: the Chandrasekhar mass, the mass-radius relation of white dwarfs, and the qualitative fact that \( n=3 \) polytropes have a mass independent of central density all fall out of the same scaled solution. It also demonstrates cleanly how choosing a scale (\( \rho_c \), \( \alpha \)) removes every dimensional constant from a problem.

Assumptions

Spherical symmetry.If dropped, \( m(r) \) and \( \nabla^2\Phi \) acquire angular dependence; the ODE becomes a partial differential equation and rotation/tidal distortion must be modelled (e.g. Chandrasekhar-Milne expansions).

Hydrostatic equilibrium — no net acceleration, so pressure gradient exactly balances gravity.If dropped, a \( \rho\,\partial^2 r/\partial t^2 \) term appears and the star is dynamically collapsing or pulsating, not static.

Polytropic equation of state \( P=K\rho^{1+1/n} \) with \( K \), \( n \) constant throughout.If dropped, \( P \) is no longer a single-valued function of \( \rho \); temperature, composition, or radiation entropy enter and the pressure-density relation cannot be inverted to close the system.

Newtonian gravity.If dropped, strong-field compact objects require the Tolman-Oppenheimer-Volkoff equation; the \( -Gm\rho/r^2 \) source gains pressure and metric corrections.

Self-gravity only — the potential \( \Phi \) is sourced entirely by the fluid's own density via Poisson's equation, with no external field.If dropped (e.g. a companion, a central point mass, or an imposed dark-matter halo), an inhomogeneous term breaks the clean \( -\theta^n \) source and the scaling no longer closes.

Derivation

\[ \frac{dP}{dr} = -\frac{G\,m(r)\,\rho}{r^{2}}, \qquad \frac{dm}{dr}=4\pi r^{2}\rho \]

Prior result: hydrostatic equilibrium for a spherical shell, together with the mass-continuity relation defining the enclosed mass \( m(r) \). A

\[ \frac{r^{2}}{\rho}\frac{dP}{dr} = -G\,m(r) \]

Multiply the hydrostatic equation by \( r^2/\rho \) to isolate the enclosed mass; this collects everything density-dependent on the left. A

\[ \frac{d}{dr}\!\left(\frac{r^{2}}{\rho}\frac{dP}{dr}\right) = -G\frac{dm}{dr} = -4\pi G\,r^{2}\rho \]

Differentiate both sides in \( r \) and substitute mass continuity, eliminating \( m(r) \) entirely. This is equivalent to inserting the radial Poisson equation \( \frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\Phi}{dr})=4\pi G\rho \) with \( \frac{dP}{dr}=-\rho\frac{d\Phi}{dr} \). B

\[ \frac{1}{r^{2}}\frac{d}{dr}\!\left(\frac{r^{2}}{\rho}\frac{dP}{dr}\right) = -4\pi G\,\rho \]

Divide by \( r^2 \) to write the balance in manifestly Laplacian (divergence) form. No physics added; this is the second-order structure equation. B

\[ P = K\rho^{1+1/n} = K\rho^{\frac{n+1}{n}} \]

Impose the polytropic equation of state, closing the system by making \( P \) a fixed function of \( \rho \) alone. A

\[ \rho = \rho_c\,\theta^{\,n}, \qquad P = K\rho_c^{\frac{n+1}{n}}\theta^{\,n+1} = P_c\,\theta^{\,n+1} \]

Define the dimensionless variable \( \theta \) by \( \rho=\rho_c\theta^n \), with \( \rho_c \) the central density so that \( \theta(0)=1 \). The EOS then makes \( P\propto\theta^{n+1} \) with \( P_c=K\rho_c^{(n+1)/n} \). B

\[ \frac{1}{\rho}\frac{dP}{dr} = \frac{1}{\rho_c\theta^{n}}\cdot K\rho_c^{\frac{n+1}{n}}(n+1)\theta^{n}\frac{d\theta}{dr} = K(n+1)\rho_c^{1/n}\frac{d\theta}{dr} \]

Differentiate \( P=P_c\theta^{n+1} \) using the chain rule and divide by \( \rho=\rho_c\theta^n \); the factor \( \theta^n \) cancels exactly — the algebraic reason the polytrope linearises so cleanly. B

\[ K(n+1)\rho_c^{1/n}\,\frac{1}{r^{2}}\frac{d}{dr}\!\left(r^{2}\frac{d\theta}{dr}\right) = -4\pi G\,\rho_c\,\theta^{\,n} \]

Insert Step 7 into Step 4. The constant coefficient \( K(n+1)\rho_c^{1/n} \) pulls outside the derivative, leaving a Laplacian acting on \( \theta \) sourced by \( \theta^n \). B

\[ \frac{1}{r^{2}}\frac{d}{dr}\!\left(r^{2}\frac{d\theta}{dr}\right) = -\underbrace{\frac{4\pi G\,\rho_c^{\,1-1/n}}{K(n+1)}}_{\displaystyle 1/\alpha^{2}}\;\theta^{\,n} \]

Divide through by \( K(n+1)\rho_c^{1/n} \). The prefactor has dimensions of (length)\(^{-2}\); define the scale length \( \alpha \) by \( \alpha^{2}=\dfrac{(n+1)K\rho_c^{1/n-1}}{4\pi G}=\dfrac{(n+1)P_c}{4\pi G\rho_c^{2}} \). C

\[ r=\alpha\xi \;\Rightarrow\; \frac{d}{dr}=\frac{1}{\alpha}\frac{d}{d\xi}, \qquad \frac{1}{\alpha^{2}}\,\frac{1}{\xi^{2}}\frac{d}{d\xi}\!\left(\xi^{2}\frac{d\theta}{d\xi}\right) = -\frac{1}{\alpha^{2}}\,\theta^{\,n} \]

Non-dimensionalise the radius as \( r=\alpha\xi \). Every factor of \( \alpha \) cancels between the two sides, removing all dimensional constants. C

\[ \boxed{\;\frac{1}{\xi^{2}}\frac{d}{d\xi}\!\left(\xi^{2}\frac{d\theta}{d\xi}\right) = -\theta^{\,n}\;}\qquad \theta(0)=1,\ \ \theta'(0)=0 \]

Cancel the common \( 1/\alpha^2 \). Central conditions: \( \theta(0)=1 \) fixes \( \rho=\rho_c \) at the centre, and \( \theta'(0)=0 \) enforces zero pressure gradient (regularity, finite central gravity). A

Result

\[ \frac{1}{\xi^{2}}\frac{d}{d\xi}\!\left(\xi^{2}\frac{d\theta}{d\xi}\right) = -\theta^{\,n}, \quad \theta(0)=1,\ \theta'(0)=0, \quad \rho=\rho_c\theta^{n},\ \ \alpha^{2}=\frac{(n+1)P_c}{4\pi G\rho_c^{2}} \]

Reading. All the dimensional structure of a polytrope collapses onto one universal function \( \theta(\xi) \) depending only on the index \( n \). The physical density is \( \rho_c\theta^n \), the physical radius is \( \alpha\xi \), and the star's surface is the first zero \( \theta(\xi_1)=0 \), giving radius \( R=\alpha\xi_1 \). Two numbers per index — \( \xi_1 \) and \( \xi_1^2|\theta'(\xi_1)| \) — set the mass and radius of every polytrope of that index.

Units check. \( \theta \), \( \xi \) are dimensionless by construction. \( \alpha^2=(n+1)P_c/(4\pi G\rho_c^2) \): with \( [P]=\mathrm{Pa}=\mathrm{kg\,m^{-1}s^{-2}} \), \( [G]=\mathrm{m^3\,kg^{-1}s^{-2}} \), \( [\rho^2]=\mathrm{kg^2\,m^{-6}} \), we get \( \dfrac{\mathrm{kg\,m^{-1}s^{-2}}}{\mathrm{m^3kg^{-1}s^{-2}}\cdot\mathrm{kg^2m^{-6}}}=\mathrm{m^2} \). So \( \alpha \) is a length and \( r=\alpha\xi \) is consistent. Both sides of the ODE are dimensionless.

Limiting cases

\( n=0 \) (incompressible): \( \theta=1-\xi^2/6 \), surface at \( \xi_1=\sqrt{6}\approx 2.449 \); uniform density \( \rho=\rho_c \).
\( n=1 \): the equation is linear; \( \theta=\dfrac{\sin\xi}{\xi} \), \( \xi_1=\pi \). Radius \( R=\pi\alpha \) is independent of \( \rho_c \).
\( n=5 \) (Schuster-Plummer): \( \theta=\left(1+\xi^2/3\right)^{-1/2} \); \( \theta\to0 \) only as \( \xi\to\infty \) — infinite radius but finite mass.
\( n=3/2 \): non-relativistic degenerate gas; \( \xi_1\approx 3.654 \) — white dwarfs, brown dwarfs, fully convective giant-planet cores.
\( n=3 \): ultra-relativistic degenerate / radiation-dominated; mass \( \propto\xi_1^2|\theta'(\xi_1)| \) is independent of \( \rho_c \) — the seed of the Chandrasekhar limit.
\( n\to\infty \): approaches the isothermal sphere \( P\propto\rho \); the equation limits to the isothermal Lane-Emden (Emden) equation with no finite surface.

Breaks when

\( n\ge 5 \): the solution no longer reaches \( \theta=0 \) at finite \( \xi \). For \( n=5 \) the radius is infinite (mass still finite); for \( n>5 \) both radius and mass diverge, so no bounded star exists.
Non-barotropic pressure: when \( P \) depends on temperature or composition as well as density (real radiative envelopes, burning shells, molecular-weight gradients), \( P=K\rho^{1+1/n} \) fails, the \( \theta^n \) cancellation in Step 7 breaks, and the structure equations no longer decouple from the energy equations.
Strong gravity: for compact objects where \( GM/(Rc^2) \) is not small, Newtonian hydrostatic equilibrium is wrong; the relativistic TOV equation replaces \( dP/dr=-Gm\rho/r^2 \) and there is no simple Lane-Emden reduction.
Broken spherical symmetry: rapid rotation, magnetic fields, or a tidal companion make \( \Phi=\Phi(r,\theta_{\text{ang}}) \); the ODE becomes a 2-D/3-D PDE and \( \alpha\xi \) scaling no longer separates.

Failure modes

Confusing the two \( n \)'s: writing \( P=K\rho^n \) instead of \( P=K\rho^{1+1/n} \). The polytropic index \( n \) and the adiabatic exponent \( \gamma=1+1/n \) are different numbers; \( n=3/2 \) means \( \gamma=5/3 \), not \( \gamma=3/2 \).
Dropping the \( (n+1) \) in \( \alpha \): forgetting that \( dP/d\theta \) brings down \( n+1 \) in Step 7, giving a scale length wrong by \( \sqrt{n+1} \).
Wrong central condition: imposing \( \theta'(0)= \) something nonzero, which produces a cusp/point mass at the origin instead of a regular, finite-gravity centre.
Treating \( K \) as fixed by \( \rho_c \): for a given microphysical EOS \( K \) is a constant; \( \rho_c \) is the free parameter. Solving for one when you meant the other scrambles the mass-radius relation.
Using \( \xi_1 \) as the physical radius: forgetting the factor \( \alpha \); \( \xi_1 \) is dimensionless, \( R=\alpha\xi_1 \).
Assuming a surface exists for \( n\ge5 \): looking for a first zero of \( \theta \) that never occurs.

Discussion

The deep content of the reduction is scale invariance. A polytrope has no intrinsic length: given a solution, one can rescale \( \rho_c \) and read off a completely different star with the same shape function \( \theta(\xi) \). This is why a single table of \( \theta(\xi) \) per index suffices for an entire family of objects, and why the mass, radius, and central pressure are related by pure powers of \( \rho_c \). Physically, the star is the fixed point where the outward push of a \( \gamma=1+1/n \) fluid exactly balances self-gravity at every radius simultaneously.

The role of the index is to encode the stiffness of matter. Small \( n \) (stiff, large \( \gamma \)) gives centrally concentrated but compact configurations; large \( n \) (soft) gives extended, diffuse envelopes. The threshold \( n=5 \) (\( \gamma=6/5 \)) is exactly where the configuration stops being spatially bounded — a statement about how weakly pressure must respond to compression before gravity wins at large radius. The related index \( n=3 \) (\( \gamma=4/3 \)) is where mass becomes independent of central density, the mechanical origin of a maximum mass for relativistic degenerate matter.

A rigorous view treats the derivation as the reduction of a semilinear elliptic PDE, \( \nabla^2\Phi=4\pi G\rho \) with \( \rho \) a power of \( (\Phi_s-\Phi) \), to a boundary-value problem by exploiting spherical symmetry and the one-parameter scaling group \( \rho\to\lambda^{2n/(1-n)}\rho \), \( r\to\lambda r \) under which the equation is invariant. For \( n=5 \) this scaling group becomes conformal and generates the exact Schuster-Plummer solution; the Emden-Fowler class \( \theta''+ (2/\xi)\theta'+\theta^n=0 \) is a canonical testbed in nonlinear ODE theory, with the existence and uniqueness of a regular solution guaranteed by a contraction-mapping argument on the equivalent integral equation near \( \xi=0 \).

Common misconceptions. The Lane-Emden equation does not assume the star is isothermal or uniform — those are only the \( n\to\infty \) and \( n=0 \) endpoints. It also does not by itself determine \( K \) or \( n \); those come from the microphysics (degeneracy, ideal gas, radiation), and the equation only takes them as input. Finally, \( \theta \) is not the density: the density is \( \rho_c\theta^n \), and \( \theta \) can be thought of as a normalised temperature-like or enthalpy-like variable that happens to obey a Poisson-type equation.

Worked examples

\[ \textbf{Incompressible sphere } (n=0)\text{: find the surface and radius.} \]

Set \( n=0 \). The source term is \( -\theta^0=-1 \), so the equation becomes linear and solvable in closed form. A

\[ \frac{1}{\xi^2}\frac{d}{d\xi}\!\left(\xi^2\theta'\right)=-1 \;\Rightarrow\; \frac{d}{d\xi}\!\left(\xi^2\theta'\right)=-\xi^2 \;\Rightarrow\; \xi^2\theta'=-\frac{\xi^3}{3}+C_1 \]

Integrate once. Regularity \( \theta'(0)=0 \) forces \( C_1=0 \), so \( \theta'=-\xi/3 \). A

\[ \theta = -\frac{\xi^2}{6}+C_2, \qquad \theta(0)=1\Rightarrow C_2=1 \;\Rightarrow\; \theta=1-\frac{\xi^2}{6} \]

Integrate again and apply \( \theta(0)=1 \). A

\[ \theta(\xi_1)=0 \Rightarrow \xi_1=\sqrt{6}\approx 2.449, \qquad \xi_1^2|\theta'(\xi_1)|=6\cdot\frac{\sqrt6}{3}=2\sqrt6\approx 4.899 \]

The surface is the first zero. The combination \( \xi_1^2|\theta'| \) sets the mass. A

\[ \text{Take }\rho_c=\rho=5.0\times10^{3}\,\mathrm{kg\,m^{-3}},\ P_c=1.0\times10^{11}\,\mathrm{Pa}. \quad \alpha^2=\frac{(n+1)P_c}{4\pi G\rho_c^2} \]

Insert numbers only now. \( n+1=1 \). B

\[ \alpha^2=\frac{(1)(1.0\times10^{11})}{4\pi(6.674\times10^{-11})(5.0\times10^{3})^2}=\frac{1.0\times10^{11}}{2.096\times10^{-2}}=4.77\times10^{12}\,\mathrm{m^2} \]

\( 4\pi G\rho_c^2=4\pi(6.674\times10^{-11})(2.5\times10^{7})=2.096\times10^{-2} \). B

\[ \alpha=2.18\times10^{6}\,\mathrm{m}, \qquad R=\alpha\xi_1=2.18\times10^{6}\times2.449=5.35\times10^{6}\,\mathrm{m} \]

Reading. A uniform-density sphere with this central pressure and density has radius \( \approx5.3\times10^3 \) km (comparable to a terrestrial-planet interior). Units check. \( \sqrt{\mathrm{m^2}}=\mathrm{m} \); \( \xi_1 \) dimensionless, so \( R \) in metres.

\[ \textbf{Linear polytrope } (n=1)\text{: radius and mass of a toy star.} \]

For \( n=1 \), \( \theta^n=\theta \) makes the ODE linear; substitute \( \theta=u/\xi \). B

\[ \theta''+\frac{2}{\xi}\theta'+\theta=0 \;\xrightarrow{\ \theta=u/\xi\ }\; u''+u=0 \;\Rightarrow\; u=A\sin\xi+B\cos\xi \]

The substitution turns it into a simple harmonic equation for \( u(\xi) \). B

\[ \theta=\frac{A\sin\xi+B\cos\xi}{\xi};\quad \theta(0)=1\Rightarrow A=1,\ \ \text{finiteness at }0\Rightarrow B=0 \;\Rightarrow\; \theta=\frac{\sin\xi}{\xi} \]

\( \cos\xi/\xi \) diverges at the origin, so \( B=0 \); \( \sin\xi/\xi\to1 \) gives \( \theta(0)=1 \) and automatically \( \theta'(0)=0 \). B

\[ \theta(\xi_1)=0\Rightarrow \xi_1=\pi, \qquad \theta'(\xi)=\frac{\xi\cos\xi-\sin\xi}{\xi^2},\ \ \theta'(\pi)=-\frac{1}{\pi},\ \ \xi_1^2|\theta'(\xi_1)|=\pi \]

First zero of \( \sin\xi/\xi \) is \( \pi \); evaluate the derivative there. B

\[ \text{For }n=1,\ \alpha^2=\frac{2K\rho_c^{0}}{4\pi G}=\frac{K}{2\pi G}\ \text{(independent of }\rho_c\text{).}\quad \text{Take } K=4.0\times10^{4}\,\mathrm{m^5\,kg^{-1}s^{-2}} \]

Since \( \rho_c^{1/n-1}=\rho_c^0=1 \) at \( n=1 \), the radius does not depend on central density — a signature feature. C

\[ \alpha^2=\frac{4.0\times10^{4}}{2\pi(6.674\times10^{-11})}=\frac{4.0\times10^{4}}{4.193\times10^{-10}}=9.54\times10^{13}\,\mathrm{m^2},\ \ \alpha=3.09\times10^{7}\,\mathrm{m} \]

Numerical substitution. B

\[ M=4\pi\alpha^3\rho_c\,\xi_1^2|\theta'(\xi_1)| =4\pi\alpha^3\rho_c\,\pi,\quad \text{take }\rho_c=1.0\times10^{9}\,\mathrm{kg\,m^{-3}} \]

Mass follows from \( M=\int_0^R4\pi r^2\rho\,dr=-4\pi\alpha^3\rho_c\,\xi_1^2\theta'(\xi_1) \), using the ODE to rewrite \( \xi^2\theta^n=-\frac{d}{d\xi}(\xi^2\theta') \). C

\[ R=\alpha\pi=9.7\times10^{7}\,\mathrm{m}, \qquad M=4\pi(3.09\times10^{7})^3(1.0\times10^{9})(\pi)=1.16\times10^{33}\,\mathrm{kg} \]

Reading. This toy \( n=1 \) configuration has \( R\approx9.7\times10^4 \) km and \( M\approx0.58\,M_\odot \). Changing \( \rho_c \) rescales \( M\propto\rho_c \) but leaves \( R \) fixed — the hallmark of the \( n=1 \) polytrope. Units check. \( [\alpha^3\rho_c]=\mathrm{m^3\cdot kg\,m^{-3}}=\mathrm{kg} \); \( \xi_1^2|\theta'| \) dimensionless, so \( M \) in kg.

Problems

(A) Verify by direct substitution that \( \theta=\left(1+\xi^2/3\right)^{-1/2} \) satisfies the Lane-Emden equation for \( n=5 \), and confirm \( \theta(0)=1 \), \( \theta'(0)=0 \).
Solution
Let \( \theta=(1+\xi^2/3)^{-1/2} \). Then \( \theta'=-\tfrac12(1+\xi^2/3)^{-3/2}\cdot\tfrac{2\xi}{3}=-\tfrac{\xi}{3}(1+\xi^2/3)^{-3/2} \). At \( \xi=0 \): \( \theta(0)=1 \), \( \theta'(0)=0 \). Now \( \xi^2\theta'=-\tfrac{\xi^3}{3}(1+\xi^2/3)^{-3/2} \). Differentiate: \( \tfrac{d}{d\xi}(\xi^2\theta')=-\xi^2(1+\xi^2/3)^{-3/2}+\tfrac{\xi^3}{3}\cdot\tfrac{3}{2}(1+\xi^2/3)^{-5/2}\cdot\tfrac{2\xi}{3} \) \( =-\xi^2(1+\xi^2/3)^{-5/2}\big[(1+\xi^2/3)-\tfrac{\xi^2}{3}\big]=-\xi^2(1+\xi^2/3)^{-5/2} \). Divide by \( \xi^2 \): LHS \( =-(1+\xi^2/3)^{-5/2}=-\theta^5 \). Since \( n=5 \), this equals \( -\theta^n \). Verified. Note \( \theta\to0 \) only as \( \xi\to\infty \): infinite radius.
(A) A polytrope has adiabatic exponent \( \gamma=5/3 \). State the polytropic index \( n \) and the value of \( \xi_1 \) to look up (given \( \xi_1\approx3.654 \) for this index). If \( \alpha=6.0\times10^{6}\,\mathrm{m} \), find the stellar radius.
Solution
\( \gamma=1+1/n\Rightarrow 1/n=\gamma-1=2/3\Rightarrow n=3/2 \). This is the non-relativistic degenerate index (white dwarfs), \( \xi_1\approx3.654 \). Radius \( R=\alpha\xi_1=6.0\times10^{6}\times3.654=2.19\times10^{7}\,\mathrm{m}\approx2.2\times10^4 \) km.
(B) Starting from \( \alpha^2=(n+1)K\rho_c^{1/n-1}/(4\pi G) \), show that the mass of a polytrope scales as \( M\propto\rho_c^{(3-n)/(2n)} \) and hence that for \( n=3 \) the mass is independent of \( \rho_c \).
Solution
Mass \( M=-4\pi\alpha^3\rho_c\,\xi_1^2\theta'(\xi_1) \); the bracket is a pure number for fixed \( n \), so \( M\propto\alpha^3\rho_c \). Now \( \alpha\propto\rho_c^{(1/n-1)/2}=\rho_c^{(1-n)/(2n)} \), so \( \alpha^3\propto\rho_c^{3(1-n)/(2n)} \). Therefore \( M\propto\rho_c^{3(1-n)/(2n)+1}=\rho_c^{[3(1-n)+2n]/(2n)}=\rho_c^{(3-3n+2n)/(2n)}=\rho_c^{(3-n)/(2n)} \). For \( n=3 \): exponent \( =(3-3)/6=0 \), so \( M \) is independent of \( \rho_c \). This is the origin of the Chandrasekhar mass for the ultra-relativistic (\( \gamma=4/3 \)) limit.
(B) For the \( n=0 \) sphere (\( \theta=1-\xi^2/6 \)), compute the ratio of mean density to central density, \( \bar\rho/\rho_c \), directly and comment.
Solution
For \( n=0 \) the density is uniform: \( \rho=\rho_c\theta^0=\rho_c \) everywhere. Hence \( \bar\rho=\rho_c \) and \( \bar\rho/\rho_c=1 \). Consistency check via the general formula \( \bar\rho/\rho_c=-\tfrac{3}{\xi_1}\theta'(\xi_1) \): with \( \theta'=-\xi/3 \), \( \theta'(\sqrt6)=-\sqrt6/3 \), so \( -\tfrac{3}{\sqrt6}\cdot(-\tfrac{\sqrt6}{3})=1 \). The incompressible limit is the only polytrope whose average and central densities coincide; higher \( n \) are increasingly centrally concentrated (e.g. \( n=3 \) gives \( \bar\rho/\rho_c\approx0.018 \)).
(C) An \( n=1 \) polytrope models a body with \( K=3.0\times10^{4}\,\mathrm{m^5\,kg^{-1}s^{-2}} \). Find its radius (independent of \( \rho_c \)), then find the central density required for a mass \( M=1.0\times10^{30}\,\mathrm{kg} \). Use \( \xi_1=\pi \), \( \xi_1^2|\theta'(\xi_1)|=\pi \).
Solution
\( \alpha^2=K/(2\pi G)=3.0\times10^{4}/(2\pi\cdot6.674\times10^{-11})=3.0\times10^{4}/4.193\times10^{-10}=7.15\times10^{13}\,\mathrm{m^2} \), so \( \alpha=8.46\times10^{6}\,\mathrm{m} \) and \( R=\pi\alpha=2.66\times10^{7}\,\mathrm{m}\approx2.7\times10^4 \) km. Mass: \( M=4\pi\alpha^3\rho_c\,\pi=4\pi^2\alpha^3\rho_c \). Solve for \( \rho_c=M/(4\pi^2\alpha^3) \). \( \alpha^3=(8.46\times10^{6})^3=6.06\times10^{20}\,\mathrm{m^3} \); \( 4\pi^2=39.48 \). \( \rho_c=1.0\times10^{30}/(39.48\times6.06\times10^{20})=1.0\times10^{30}/2.39\times10^{22}=4.18\times10^{7}\,\mathrm{kg\,m^{-3}} \). So \( \rho_c\approx4.2\times10^{7}\,\mathrm{kg\,m^{-3}} \), with the radius fixed at \( 2.7\times10^4 \) km regardless of this choice.