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The Emden-Fowler transformation is a change of variables for scaling-invariant radial equations, including the Lane-Emden equation. The Emden-Fowler transformation replaces a radial variable $r$ by logarithmic time $s=\log r$ and rescales the dependent variable by a power of $r$. It converts many scale-invariant radial equations into autonomous ordinary differential equations.

A Generalized Lane-Emden equation is a semilinear partial differential equation whose highest-order term is the Laplacian:For a radial function in $d$ dimensions it becomes $u^{\prime \prime }+(d-1)u^{\prime }/r+u^p=0$. The case $d=3$ gives the classical Lane-Emden equation.

The classical Lane-Emden equation is the dimensionless equationfor a spherically symmetric polytrope in three dimensions. The Generalized Lane-Emden equation extends it to arbitrary dimensions and to functions without radial symmetry.

For $p=K{\rho }^{1+1/n}$, the Lane-Emden mass formula gives $M\propto {\rho }_c^{(3-n)/(2n)}$ at fixed $K$, while $M/R\propto {\rho }_c^{1/n}$. For $3<n<5$, the Lane-Emden equation has a finite-radius solution, yet $M\to \infty$ as ${\rho }_c\to 0$. The relativistic Tolman–Oppenheimer–Volkoff equation approaches this limit since $p_c/{\rho }_c\to 0$. Monotone pressure and density alone therefore do not impose a finite stellar maximum mass.

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.

To turn this density model into a CNO cycle burning model, use gas-dominated pressure and uniform mean molecular weight. The ideal gas law and the polytropic equation of state then imply $T/T_c=\theta$. Without this thermal assumption a polytropic equation of state for total pressure does not alone fix the temperature profile. Since the stellar energy-generation rate is power per unit mass, its contribution to luminosity is $ϵdm$, giving the luminosity integral of an index-five polytrope:Put $u=\xi /\sqrt{3+{\xi }^2}$. Then $\xi =\sqrt{3}u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt{3}{\int }_0^1{u}^2(1-u^2{)}^{8}du$. Evaluation givesThe PDF's power of $\alpha$ in the proposed luminosity scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies ${\alpha }^3$; replacing it by ${\alpha }^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.

The polytrope of index five is unlikely to model a real burning stellar core well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly temperature-sensitive CNO cycle heating concentrates the luminosity centrally and tends to produce a convection zone there. A well-mixed gas-pressure-dominated convective core is closer to an adiabatic stellar polytrope of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.

Efficient convection keeps the interior at nearly uniform specific entropy. For the fully ionized monatomic gas, the heat capacity ratio is $5/3$, so the adiabatic process gives $P=K_{\rho }{\rho }^{5/3}$, with $K_{\rho }$ spatially constant. The ideal gas relation $P=\mathcal{R}\rho T/\mu$ then impliesThe constant is spatial, not temporal: loss of specific entropy during contraction changes it. This is a stellar polytrope with polytropic index $n=3/2$. We neglect radiation pressure and take the thin photosphere as an atmospheric matching layer rather than a significant fraction of the stellar mass.

For a fixed dimensionless profile of the Lane-Emden equation, the stellar homology relations have the formwhere $C_{\rho },C_P$ are positive dimensionless structure constants. More explicitly, write $\rho (r)={\rho }_{c}f(x)$ and $m(r)=Mh(x)$ with $x=r/R$ and fixed dimensionless profiles. Mass conservation gives $M=4\pi {\rho }_c{R}^3{\int }_0^{1}f(x)x^{2}dx$. Integrating the hydrostatic pressure support equation, with negligible surface pressure on the interior scale, gives $P_c=(GM{\rho }_c/R){\int }_0^{1}h(x)f(x)x^{-2}dx$. Both integrals are fixed positive numbers, proving the two central scalings. Substituting these relations into $K=P_c/T_c^{5/2}$ givesThus the same polytropic index fixes $K_0$ throughout this homologous sequence, while the dimensional stellar polytrope constant changes.

At the photosphere, whose optical depth is $2/3$, identify the temperature with the effective temperature $T_e$. The ideal gas relation and the interior adiabat give ${\rho }_{\mathrm{ph}}=\mu K{T}_e^{3/2}/\mathcal{R}$. With the prescribed opacity and stellar surface boundary condition,Eliminating $K$ therefore provesUsing the Stefan–Boltzmann law, with $\sigma =ac/4$, gives the luminosity

The gravitational energy of a stellar polytrope is $\Omega =-6G{M}^2/(7R)$ for $n=3/2$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, and hence the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. With no appreciable stellar nuclear fusion or accretion, energy conservation givesCombining the two expressions for $L$ yieldsMultiplication by $-7R^{-9/2}/2$ and integration proves the contraction law with density-linear fourth-power stellar opacity:When the accumulated contraction term dominates the initial-radius term, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Substitution into the central ideal gas relation giveswhich has the stated $t^{2/7}$ behavior when the time origin is negligible. This late-time approximation still requires the assumed fully convective star, fixed composition and opacity law to remain applicable.

A stellar polytrope replaces a detailed equation of state and thermal structure by $P=K{\rho }^{1+1/n}$ for $n>0$, with spatially constant $K$ and polytropic index $n$. Together with spherical hydrostatic equilibrium and mass conservation, it gives a tractable family of stellar density profiles. Introduce the Lane-Emden variables for a stellar polytropeCombining $dP/dr=-Gm\rho /r^2$ with $dm/dr=4\pi r^2\rho$ eliminates $m$ to give $r^{-2}d[r^2{\rho }^{-1}(dP/dr)]/dr=-4\pi G\rho$. Substitution of the pressure law and the dimensionless variables gives the Lane-Emden equationA regular finite-radius model for $0\le n<5$ ends at the first zero ${\xi }_1$. Its radius and mass areThese formulas supply central-to-mean density ratios and stellar homology scalings without solving a full evolutionary model. Eliminating ${\rho }_c$, for $0<n<5$ and $n\ne 3$, gives the polytropic mass-radius relation

An efficient-convection monatomic gas has $n=3/2$. The same index describes a cold, nonrelativistic electron degeneracy pressure law; at fixed composition it gives $R\propto M^{-1/3}$ for a white dwarf. Ultrarelativistic electron degeneracy pressure has $n=3$, for which the mass is independent of ${\rho }_c$ and scales as $(K/G{)}^{3/2}$. This is the origin of the Chandrasekhar limit. A radiation-dominated, approximately constant-entropy interior also approaches $P\propto {\rho }^{4/3}$ and $n=3$. The polytrope of index zero is an incompressible comparison model, useful for illustrating the role of central concentration.

The gravitational energy of a stellar polytrope, $\Omega =-3G{M}^2/[(5-n)R]$, and the stellar virial theorem make polytropes useful for contraction-energy estimates. They also provide structural models for perturbation and stability calculations. A polytrope is a structural approximation, not a complete stellar evolution calculation. One must still specify stellar nuclear fusion, opacity, energy transport, composition and boundary conditions to determine the luminosity and history. Moreover, the structural exponent $1+1/n$ need not equal the adiabatic exponent relevant to a rapid perturbation. Real stars can have different effective indices in their cores and envelopes, as well as ionization zones or varying electron degeneracy pressure.

Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient pressure while retaining its specific entropy and stellar composition, so an adiabatic process gives $d\log {\rho }_{\mathrm{parcel}}=d\log P/\gamma$. Its excess mass density over the new surroundings is, to first order,Restoring buoyancy requires this bracket to be positive. With $\gamma =5/3$, the Schwarzschild criterion is thereforeEquality is neutral stability. For uniform mean molecular weight, the ideal gas law gives $d\log \rho =d\log P-d\log T$. Since hydrostatic equilibrium pressure decreases outward, the inequality is equivalent to $\nabla =d\log T/d\log P<{\nabla }_{\mathrm{ad}}=2/5$. Dividing the radiative diffusion in a star and hydrostatic equilibrium stellar structure equations givesThis is the stellar radiative temperature gradient test for stellar convective stability.

In the thin upper stellar atmosphere take $m\simeq M$, $L_r\simeq L$ and use gas pressure dominance. The prescribed opacity becomes $\kappa ={\kappa }_0\mu P{T}^{12}/\mathcal{R}$. Dividing the two structure equations then yieldsAt zero optical depth, the atmospheric boundary has $T_s^4=T_e^4/2$ and negligible gas pressure. Integrating from that boundary, with $T_s^{-8}=4T_e^{-8}$, givesTo locate the onset, put $y=T_e^8/T^8$. Logarithmic derivative gives $d\log P/d\log T=4y/(4-y)$, henceThis starts at zero at the outer boundary and increases inward. It first reaches $2/5$ atThe corresponding optical depth is ${\tau }_b=\frac{4}{3}(\sqrt{13/20}-\frac{1}{2})>0$, so the onset lies inside the stellar atmosphere. This is grey-atmosphere convection onset with thirteenth-power opacity.

Let the interior of the fully convective star have $P=K_T{T}^{5/2}$, with $K_T$ spatially constant. At its boundary $T_b/T_e$ is fixed, so the atmospheric pressure formula givesThe Stefan–Boltzmann law, $L=4\pi R^2\sigma T_e^4$ with $\sigma =ac/4$, consequently impliesThe ideal gas equation transforms the interior relation intoFor completeness, the mass-radius scaling follows by taking $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$ with ${\alpha }^2=5K_{\rho }{\rho }_c^{-1/3}/(8\pi G)$. Hydrostatic equilibrium and mass conservation reduce to the Lane-Emden equation $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2{\theta }^{3/2}$, with $\theta (0)=1$, ${\theta }^{\prime }(0)=0$. Its fixed dimensionless profile gives $R\propto \alpha$ and $M\propto {\alpha }^3{\rho }_c$. Eliminating ${\rho }_c$ gives $R\propto K_{\rho }{G}^{-1}{M}^{-1/3}$, or $K_{\rho }\propto G{M}^{1/3}R$. Thus $K_T^2\propto M^{-1}{R}^{-3}$ at fixed stellar composition. Equating this with the atmospheric expression gives $L^{17/4}\propto M^2{R}^{19/2}$, and henceThis Hayashi relation with thirteenth-power opacity concerns the interior of a fully convective star beneath a thin stellar atmosphere. The coefficient $K_T$ may vary between stars; spatially constant specific entropy does not mean equal specific entropy across the whole sequence.

A fully ionized helium nucleus supplies two Electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Filling all Electron momentum states up to the Fermi momentum givesFor a nonrelativistic Electron, the speed is $p/m_e$. The three mean squared Cartesian components of momentum are equal by rotational symmetry and sum to $p^2$, so each is $p^2/3$. This is the isotropic tensor integral. Thus the momentum flux, or electron degeneracy pressure, isSubstituting the expression for $p_0$ and simplifying the powers of two givesHere $h$ is the Planck constant. This is the pure-helium specialization of nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture.

For a cold helium white dwarf, neglect ion thermal pressure and use this electron degeneracy pressure as the support against Newtonian gravity. Combining hydrostatic equilibrium with mass conservation givesSet $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$, where ${\alpha }^2=5K{\rho }_c^{-1/3}/(8\pi G)$. Direct substitution gives the Lane-Emden equationIts dimensionless solution is independent of ${\rho }_c$. Let ${\xi }_1$ be its first zero and ${\omega }_1=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)={\int }_0^{{\xi }_1}{\xi }^2{\theta }^{3/2}d\xi$. Integrating mass conservation givesEliminating ${\rho }_c$ yields the polytropic mass-radius relationThus a more massive nonrelativistic white dwarf is smaller at fixed stellar composition. The assumptions require $p_0\ll m_{e}c$ and $k_{B}T\ll p_0^2/(2m_e)$ in the region providing most support; the law is not extended into the relativistic or thermally supported regimes.

Efficient convection makes the interior follow a nearly adiabatic process and maintains uniform specific entropy. For the fully ionized monatomic ideal gas, the specific-heat ratio is $5/3$, so the adiabatic temperature gradient gives $d\log T/d\log P=2/5$. Integrating at a fixed time givesThe coefficient is spatially constant because the composition and specific entropy are uniform. It can change as the star loses energy. Equivalently the gas is an adiabatic stellar polytrope with $P=K_{\rho }{\rho }^{5/3}$ and polytropic index $n=3/2$; $K_{\rho }$ and the pressure-temperature coefficient $K$ are different quantities.

This fixes the dimensionless Lane-Emden equation profile throughout a stellar homology sequence. Writing $\rho (r)={\rho }_{c}f(r/R)$ in the mass integral gives ${\rho }_c=C_{\rho }M/R^3$, with a fixed positive shape constant $C_{\rho }$. Integrating the stellar hydrostatic equation through the same profile, with negligible surface pressure compared with central pressure, gives $P_c=C_{P}G{M}^2/R^4$. HenceThese are homology scalings of central stellar pressure. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yieldsThus $K_0$ is constant for the fixed stellar polytrope family.

At the approximate photosphere, take $T=T_e$ and use the ideal gas relation $\rho =\mu P/(\mathcal{R}T)$. The prescribed opacity givesThe stellar surface boundary condition and surface gravity of a star therefore implyThis is the temperature-four special case of convective contraction with power-law opacity.

Using the Stefan–Boltzmann law, the luminosity isFor $n=3/2$, the supplied gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, so the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. During Kelvin-Helmholtz contraction, with constant mass and no significant nuclear source, $L=-dE/dt$. ThusMultiplying by $-(7/2)R^{-9/2}$ and integrating givesThe sign makes the radius decrease as time increases.

When the accumulated contraction term dominates the initial-radius term, this yields $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Combining it with the central ideal gas temperature gives the late-time central-temperature scalingReplacing $t-t_0$ by $t$ gives the stated late-time form. The asymptotic step also requires a time long enough to forget the initial radius, not merely a choice of zero for the age.

Assume $K>0$ and a positive central mass density. For this polytrope of index one, the spherical hydrostatic pressure support equation becomesDifferentiate the second equation and use mass conservation. With $k^2=2\pi G/K$, this givesThe regular central solution, with $\rho (0)={\rho }_c$ and ${\rho }^{\prime }(0)=0$, isThis is also the index-one solution of the Lane-Emden equation. The mass density remains positive up to its first zero, so the free surface is at $kR=\pi$, givingTaking a later zero would include a region of negative mass density and would not describe a physical star.

Integrating the mass givesConsequentlyThe radius is independent of the central mass density, whereas the mass is proportional to it; this is the special polytropic mass-radius relation at index one.

For a spherical stellar polytrope in hydrostatic equilibrium, combine $d{m}_r/dr=4\pi r^2\rho$ with $dP/dr=-G{m}_r\rho /r^2$ to eliminate the enclosed mass:For $n>0$, put $P=K{\rho }^{1+1/n}$, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=a\xi$. Since ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/(a{\rho }_c)$, chooseThe Lane-Emden equation is thenA regular center requires $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$, with positive chosen central mass density and pressure. Locally $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. The stellar surface is the first positive zero ${\xi }_1$, where the idealized external pressure is zero; retain the positive solution before it. Thus $R=a{\xi }_1$ and the Lane-Emden mass formula isThe surface condition selects where to stop a centrally regular solution, rather than replacing its central regularity conditions.

For a polytrope of index zero, the density is constant and $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2$. Regularity givesThe pressure is $P_c{\theta }_0$ and $a^2=P_c/(4\pi G{\rho }_c^2)$, so $R^2=3P_c/(2\pi G{\rho }_c^2)$. Index zero is the structural incompressible limit: the expression $K{\rho }^{1+1/n}$ is not itself defined at $n=0$.

For a polytrope of index one, set $u=\xi \theta$. The equation becomes $u^{\prime \prime }+u=0$, while central regularity requires $u(0)=0$, $u^{\prime }(0)=1$. HenceThe mass is $4{\pi }^2{a}^3{\rho }_c$, so it can change with central mass density while the radius stays fixed.

The moment of inertia of a polytropic star about any axis through its center follows by integrating $r^2{\sin }^2\vartheta$ over spherical shells:For constant mass density this gives $I_0=2M{R}^2/5$. For index one, integration by parts gives ${\int }_0^{\pi }{\xi }^3\sin \xi d\xi =\pi ({\pi }^2-6)$, and thereforeThese are axial moments of inertia, not the scalar second mass moment $\int r^{2}dm$.

For a finite-radius centrally regular stellar polytrope with $0<n<5$ and $n\ne 1$, eliminate ${\rho }_c$ from $R=a{\xi }_1$ and $M=4\pi a^3{\rho }_c{\omega }_n$. The resulting polytropic mass-radius relation isHere ${\xi }_1,{\omega }_n$ are dimensionless functions of $n$. At $n=1$ no such single-valued mass-as-a-power-of-radius relation exists at fixed $K$: the radius is fixed instead. At $n=3$ the exponent is zero and $M=4\pi {\omega }_3(K/(\pi G){)}^{3/2}$ is independent of central mass density. For the incompressible case, $M=(4\pi {\rho }_c/3)R^3$ at fixed mass density. Regular $n\ge 5$ solutions have no finite zero-pressure surface, so the finite-radius formula does not apply to them.

The useful high-density core bound from TOV compactness follows without choosing the unknown high-density equation of state. Let $a$ be the radius at which the decreasing energy density first reaches ${\rho }_0$, and put $\mu =m(a)$, $p_0=p({\rho }_0)$. The Tolman–Oppenheimer–Volkoff equation gives $p^{\prime }<0$ away from a nonempty centre; since $dp/d\rho >0$, energy density decreases outward. HenceFor the local stellar compactness, if $z\ge 0$, $1-z+\sqrt{1+z}\le 2$, since its derivative is $-1+(2\sqrt{1+z}{)}^{-1}<0$. Applying the supplied compactness inequality at the core boundary therefore gives $\mu /a<4/9$. Combining the two bounds yieldsThe pressure-dependent form further restricts the allowed core-boundary data:In particular, positivity of its right side gives $6\pi a^2{p}_0<3$, so $a^2<1/(2\pi p_0)$ when $p_0>0$. These restrictions depend only on known quantities, not on the equation of state above ${\rho }_0$.

The intended stellar-mass argument now attaches the known low-density envelope to each admissible pair $(a,\mu )$ with pressure $p_0$. Its Tolman–Oppenheimer–Volkoff equation involves only the known equation of state. For a physical low-density envelope with a uniformly bounded mass on this bounded set of core data, taking the supremum of the envelope-plus-core masses gives an upper mass limit independent of any high-density continuation. Stars with ${\rho }_c\le {\rho }_0$ are also determined entirely by the known equation of state. Unknown dense-core physics cannot evade the displayed core bounds.

A qualification is necessary for a claim about the entire star: the printed monotonicity assumptions alone do not guarantee a finite maximum total mass for an arbitrary known low-density equation of state. An explicit low-density polytropic mass divergence occurs for $p=K{\rho }^{5/4}$ with fixed $K>0$. In the low-density Newtonian limit, this is a stellar polytrope of index $n=4$, whose Lane-Emden equation has a finite first zero. With $\rho ={\rho }_c{\vartheta }^4$ and $r=\ell \xi$, hydrostatic balance fixeswhere ${\mu }_4>0$ is the dimensionless Lane-Emden surface mass constant. Thus $M\to \infty$ as ${\rho }_c\to 0$, although $M/R\propto {\rho }_c^{1/4}\to 0$. The first Lane-Emden zero is simple. The continuous dependence of an ODE solution on parameters therefore preserves this finite surface for sufficiently small relativistic parameter. The relativistic correction is controlled by $p_c/{\rho }_c=K{\rho }_c^{1/4}\to 0$, so the corresponding regular finite-radius Tolman–Oppenheimer–Volkoff equation solutions have the same limiting scalings. They never sample $\rho >{\rho }_0$, and satisfy the local positivity and monotonicity assumptions in their nonvacuum interiors. This shows exactly why a suitable bounded-envelope hypothesis, or an appropriate restriction to physically stable stellar models, must accompany the intended total-mass conclusion. The compactness inequality supplies a bound on the unknown core, not a bound on the mass of every conceivable dilute envelope.

For a stellar polytrope of index $n>0$, write $P=K{\rho }^{1+1/n}$ and combine hydrostatic equilibrium with mass conservation to obtainIntroduce Lane-Emden variables for a stellar polytrope, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=\alpha \xi$, whereSince ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/{\rho }_c\alpha$, the mechanical equation reduces to the Lane-Emden equationA constant-density interior corresponds to the formal polytrope of index zero, with $\rho ={\rho }_c{\theta }^0={\rho }_c$ where $\theta >0$. At $n=0$, regularity first gives ${\xi }^2{\theta }^{\prime }=-{\xi }^3/3$, and a second integration givesHere $P=P_c\theta =P_c(1-r^2/R^2)$ and ${\alpha }^2=P_c/(4\pi G{\rho }_c^2)$, reproducing $P_c=2\pi G{\rho }_c^2{R}^2/3$. The mass density jumps from its constant interior value to zero at the surface, while pressure vanishes continuously. The relation $P=K{\rho }^{1+1/n}$ is singular at $n=0$; the regular dimensionless pressure/mass density formulation defines this incompressible structural limit. It does not mean that the gas's perturbative stellar adiabatic exponent is infinite: the hydrostatic ideal-gas toy model and its adiabatic response are distinct choices.

For a fixed polytropic index $n>0$ and fixed equation-of-state constant $K$, let ${\xi }_1$ be the first zero of the regular Lane-Emden equation solution. A finite-radius model requires such a zero; for the usual nonnegative indices this holds for $n<5$. The surface radius and mass follow from $r=\alpha \xi$ and $\rho ={\rho }_c{\theta }^n$:The last equality integrates the Lane-Emden equation; define the positive Lane-Emden surface mass constant ${\omega }_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. Substituting $\alpha =C_1{\rho }_c^{(1-n)/(2n)}$ givesThe central-density exponents cancel when the required powers are taken. Thus the polytropic mass-radius relation isIt is independent of ${\rho }_c$, with $n,K$ and composition held fixed. For $n=1$ the radius is fixed; for $n=3$ the mass is fixed. These limiting powers need no division by a vanishing exponent.