Emden–Chandrasekhar equation: Difference between revisions

In astrophysics, the Chandrasekhar equation is a dimensionless form of the Poisson equation for the density distribution of a spherically symmetric isothermal gas sphere subjected to its own gravitaional force, named after the Indian American astrophysicist Subrahmanyan Chandrasekhar.^[1][2] The equation^[3] reads

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

where ${\displaystyle \xi }$ is the dimensionless radius and ${\displaystyle \psi }$ is the related to the density of the gas sphere as ${\displaystyle \rho =\rho _{c}e^{-\psi }}$. The equation has no known explicit solution. If a polytropic fluid is used instead of an isotropic fluid, one obtains the Lane–Emden equation. 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.

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

${\displaystyle p={\frac {k_{B}}{WH}}T+{\frac {1}{3}}\sigma T^{4}}$

where

• ${\displaystyle k_{B}}$ 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

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

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

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

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

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

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

Introducing the transformation ${\displaystyle x=1/\xi }$ transforms the equation to

${\displaystyle x^{4}{\frac {d^{2}\psi }{dx^{2}}}=e^{-\psi }}$

The equation has a singular solution given by

${\displaystyle e^{-\psi _{s}}=2x^{2},\quad {\text{or}}\quad -\psi _{s}=2\ln x+\ln 2}$

Therefore, a new variable can introduced as ${\displaystyle -\psi =2\ln x+z}$, where the equation for ${\displaystyle z}$ can be derived,

${\displaystyle {\frac {d^{2}z}{dt^{2}}}-{\frac {dz}{dt}}+e^{z}-2=0,\quad {\text{where}}\quad t=\ln x}$

This equation can be reduced to first order by introducing

${\displaystyle y={\frac {dz}{dt}}=\xi {\frac {d\psi }{d\xi }}-2}$

then we have

${\displaystyle y{\frac {dy}{dz}}-y+e^{z}-2=0}$

There is another reduction directly from the original equation. Let us define

${\displaystyle u={\frac {\xi e^{-\psi }}{d\psi /d\xi }},\quad v=\xi {\frac {d\psi }{d\xi }}}$

then

${\displaystyle {\frac {u}{v}}{\frac {dv}{du}}={\frac {u-1}{u+v-3}}}$

• If ${\displaystyle \psi (\xi )}$ is a solution to Chandrasekhar equation, then ${\displaystyle \psi (A\xi )-2\ln A}$ is also a solution of the equation, where ${\displaystyle A}$ is an arbitrary constant.
• The solutions of the Chandrasekhar equation which are finite at the origin have necessarily ${\displaystyle d\psi /d\xi =0}$ at ${\displaystyle \xi =0}$

• Lane–Emden equation
• Frank-Kamenetskii theory
• Chandrasekhar's white dwarf equation