Stellar polytrope

A stellar polytrope obeys $P=K{\rho }^{1+1/n}$ for constant $K$ and polytropic index $n$. Its dimensionless density profile satisfies the Lane-Emden equation.

The regular $n=5$ Lane-Emden equation solution has no finite first zero. Its Lane-Emden mass formula gives $m(\xi )=4\pi {\rho }_c{\alpha }^3{\xi }^3/[3(1+{\xi }^2/3{)}^{3/2}]$, so its infinite-radius total mass is $4\pi \sqrt{3}{\rho }_c{\alpha }^3$. Its enclosed mean mass density is ${\rho }_c(1+{\xi }^2/3{)}^{-3/2}$ and tends to zero. A finite externally confined core instead requires a specified truncation pressure.

For a gas-pressure-supported polytrope of index five with uniform mean molecular weight, $T/T_c=\theta$. A specific stellar energy-generation rate $ϵ={ϵ}_0\rho T^q$ therefore integrates to $4\pi {ϵ}_0{\alpha }^3{\rho }_c^2{T}_c^q{\int }_0^{\infty }{\xi }^2{\theta }^{10+q}d\xi$. Substituting $u=\xi /\sqrt{3+{\xi }^2}$ gives the displayed formula, convergent for $q>-7$. At $q=11$ its dimensionless coefficient is approximately $1.02942$. The scale factor is a volume, ${\alpha }^3$, rather than an area.

For a spherical stellar polytrope, the axial moment of inertia is $I=(8\pi /3){\int }_0^R\rho (r)r^{4}dr$. With Lane-Emden variables for a stellar polytrope, first zero ${\xi }_1$ and Lane-Emden surface mass constant ${\omega }_n$, its dimensionless form isA polytrope of index zero gives $2/5$, while a polytrope of index one gives $(2/3)(1-6/{\pi }^2)$. The latter is smaller because more of the mass lies near the center. The scalar second mass moment $\int r^{2}dm$ is $3I/2$, not the axial moment of inertia.

The regular $n=0$ solution of the Lane-Emden equation is $\theta =1-{\xi }^2/6$, with first zero $\sqrt{6}$. Its interior density is constant and its pressure varies as $\theta$. This is the structural incompressible limit; $P=K{\rho }^{1+1/n}$ itself is singular at $n=0$, and does not assign the perturbative stellar adiabatic exponent of a thermally stratified ideal-gas model.

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.

For a spherical stellar polytrope of finite radius and zero surface pressure, the specific enthalpy is $(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =-GM/R$, hence $(n+1)\int PdV=-G{M}^2/R-2\Omega$. Eliminating the pressure integral with the stellar virial theorem proves the displayed formula. For a gas adiabatic stellar polytrope, $n=1/(\gamma -1)$ and $U=G{M}^2/[(5\gamma -6)R]$; a general polytropic index need not have this relation to the specific-heat ratio.

Writingreduces Newtonian hydrostatic balance toThe stellar surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$.

A stellar polytrope with polytropic index one has $\nabla \Phi =-2K\nabla \rho$ in hydrostatic equilibrium. Combining this with the Poisson equation for Newtonian gravity gives the Helmholtz equationThe spherical solution regular at the origin is $\rho (r)={\rho }_c\sin (kr)/(kr)$. Its first zero gives $R=\pi /k=\sqrt{\pi K/(2G)}$, independent of central mass density, and $\bar{\rho }/{\rho }_c=3/{\pi }^2$.

The interior Helmholtz equation for a polytrope of index one also admits a positive solution vanishing on the faces of a cube:Its central value is ${\rho }_c$ and its average is $\bar{\rho }=8{\rho }_c/{\pi }^3$. Together with $\Phi =C-2K\rho$, it solves the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. A local interior solution need not match the external field of its own mass.

For the cubic polytropic interior, $\nabla \rho$ vanishes at each vertex, so the locally defined potential $\Phi =C-2K\rho$ predicts zero gravitational acceleration there. But at the vertex $(0,0,0)$, the actual self-gravitational acceleration of the positive mass distribution iswhose three components are strictly positive. The fields cannot match continuously. Thus the interior Helmholtz equation and zero density on the faces do not construct an isolated static cubic star. External stresses or an external gravitational field would be needed to realize such a boundary-value construction.

The sine-product cubic polytropic interior has zero mass density gradient at every corner. Its hydrostatic potential $C-2K\rho$ therefore has zero gradient there. In contrast, at a corner of a cube containing a positive mass density, every interior mass element contributes a gravitational-potential gradient pointing into the same coordinate octant. Each component of the gradient is strictly nonzero. This directly proves failure of global isolated gravitational matching, even though the interior hydrostatic and Poisson equations hold.

The Eddington standard model is a chemically homogeneous $n=3$ stellar polytrope with constant gas-pressure fraction $\beta$. It approximates massive radiative stars in which gas and radiation both contribute appreciably to pressure.

The polytropic index $n$ is defined by $P=K{\rho }^{1+1/n}$. A monatomic convective ideal gas has $n=3/2$, while a radiation-supported Eddington model has $n=3$.

An adiabatic stellar polytrope has $P=K{\rho }^{\gamma }$ with constant specific entropy, so $n=1/(\gamma -1)$. For a monatomic perfect gas, $\gamma =5/3$ and $n=3/2$.

Eliminating central density from the Lane-Emden scalings givesfor $n\ne 3$. In particular, $R\propto K{M}^{-1/3}$ for $n=3/2$.

With $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$, the exact rearranged relation iswhereFor $n=3$, the radius dependence disappears and the mass is proportional to $K^{3/2}$.

For a finite regular polytrope with fixed index, the scale ${\alpha }^2\propto K{\rho }_c^{(1-n)/n}$ implies $M^{n-1}{R}^{3-n}\propto K^n$. Thus, for $n\ne 3$, $R\propto K^{n/(3-n)}{M}^{(1-n)/(3-n)}$. For $n=3/2$, $R\propto K{M}^{-1/3}$. Fixed composition makes the cold electron degeneracy pressure coefficient approximately fixed, but a convective ideal gas has $K$ controlled by entropy. A stellar sequence with changing entropy therefore need not follow the fixed-$K$ negative slope.

If $\rho ={\rho }_c{\theta }^n$, $r=\alpha \xi$, and ${\xi }_1$ is the first zero of the Lane-Emden function, then a spherical polytrope has

For a Lane-Emden function whose first zero is ${\xi }_1$,

For a chemically uniform $n=3$ gas-plus-radiation polytrope with constant gas-pressure fraction $\beta$, the mass obeysIt follows from the $n=3$ mass scaling $M\propto K^{3/2}$ and the equation of state.