Emden–Chandrasekhar equation

${\displaystyle \frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d\psi }{d\xi }\right)=e^{-\psi }}$

where ${\displaystyle \xi }$ is the dimensionless radius and ${\displaystyle \psi }$ is the related to the density of the gas sphere as ${\displaystyle \rho ={\rho }_c{e}^{-\psi }}$, where ${\displaystyle {\rho }_c}$ is the density of the gas at the centre. The equation has no known explicit solution. If a polytropic fluid is used instead of an isothermal fluid, one obtains the Lane–Emden equation. The isothermal assumption is usually modeled to describe the core of a star. The equation is solved with the initial conditions,

${\displaystyle \psi =0,\frac{d\psi }{d\xi }=0\text{at}\xi =0.}$

The equation appears in other branches of physics as well, for example the same equation appears in the Frank-Kamenetskii explosion theory for a spherical vessel. The relativistic version of this spherically symmetric isothermal model was studied by Subrahmanyan Chandrasekhar in 1972.

Derivation

For an isothermal gaseous star, the pressure ${\displaystyle p}$ is due to the kinetic pressure and radiation pressure

${\displaystyle p=\rho \frac{k_B}{WH}T+\frac{4\sigma }{3c}{T}^4}$

where

• ${\displaystyle \rho }$ is the density
• ${\displaystyle k_B}$ is the Boltzmann constant
• ${\displaystyle W}$ is the mean molecular weight
• ${\displaystyle H}$ is the mass of the proton
• ${\displaystyle T}$ is the temperature of the star
• ${\displaystyle \sigma }$ is the Stefan–Boltzmann constant
• ${\displaystyle c}$ is the speed of light

The equation for equilibrium of the star requires a balance between the pressure force and gravitational force

${\displaystyle \frac{1}{r^2}\frac{d}{dr}\left(\frac{r^2}{\rho }\frac{dp}{dr}\right)=-4\pi G\rho }$

where ${\displaystyle r}$ is the radius measured from the center and ${\displaystyle G}$ is the gravitational constant. The equation is re-written as

${\displaystyle \frac{k_{B}T}{WH}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\ln \rho }{dr}\right)=-4\pi G\rho }$

Introducing the transformation

${\displaystyle \psi =\ln \frac{{\rho }_c}{\rho },\xi =r{\left(\frac{4\pi G{\rho }_{c}WH}{k_{B}T}\right)}^{1/2}}$

where ${\displaystyle {\rho }_c}$ is the central density of the star, leads to

${\displaystyle \frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d\psi }{d\xi }\right)=e^{-\psi }}$

The boundary conditions are

${\displaystyle \psi =0,\frac{d\psi }{d\xi }=0\text{at}\xi =0}$

For ${\displaystyle \xi \ll 1}$, the solution goes like

${\displaystyle \psi =\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}+\cdots }$