Chandrasekhar equation

Numerical solution of the Chandrasekhar equation

The Chandrasekar equation in astrophysics is the dimensionless form of the Poisson equation for the density distribution of a spherically symmetric isothermal gas sphere under the action of its own gravitational force, named after the American astrophysicist Subramanyan Chandrasekara . ^[1] ^[2] Equation ^[3] has the form

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

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

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

Where $\xi$ ${\ displaystyle \ xi}$ $\ xi$ is a dimensionless radius, $\psi$ ${\ displaystyle \ psi}$ $\ psi$ associated with the density of the gas sphere by the ratio $\rho =\rho _{c}e^{-\psi }$ ${\ displaystyle \ rho = \ rho _ {c} e ^ {- \ psi}}$ ${\ displaystyle \ rho = \ rho _ {c} e ^ {- \ psi}}$ where $\rho _{c}$ ${\ displaystyle \ rho _ {c}}$ ${\ displaystyle \ rho _ {c}}$ represents the density of the gas in the center. The equation has no known explicit solution. If instead of isothermal matter we take polytropic, the written equation will be the Lane-Emden equation . Usually, the isothermal approximation is used to describe the core of a star. In this case, the equation is solved with the initial conditions

\xi =0:\quad \psi =0,\quad {\frac {d\psi }{d\xi }}=0.

{\ displaystyle \ xi = 0: \ quad \ psi = 0, \ quad {\ frac {d \ psi} {d \ xi}} = 0.}

{\ displaystyle \ xi = 0: \ quad \ psi = 0, \ quad {\ frac {d \ psi} {d \ xi}} = 0.}

The equation also arises in other areas of physics, for example, in the developed by D. A. Frank-Kamenetsky .

Content

Derivation of the equation

For an isothermal gas star, the pressure $p$ ${\ displaystyle p}$ consists of kinetic pressure and radiation pressure :

p=\rho {\frac {k_{B}}{WH}}T+{\frac {4\sigma }{3c}}T^{4}

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

,

Where

$\rho$ ${\ displaystyle \ rho}$ - density
$k_{B}$ ${\ displaystyle k_ {B}}$ - Boltzmann constant
$W$ ${\ displaystyle W}$ - average molecular weight
$H$ ${\ displaystyle H}$ Is the mass of the proton ,
$T$ ${\ displaystyle T}$ Is the temperature of the star,
$\sigma$ ${\ displaystyle \ sigma}$ - Stefan-Boltzmann constant ,
$c$ ${\ displaystyle c}$ Is the speed of light.

The equation for the equilibrium state of a star requires a balance between the force of pressure and the force of gravity:

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

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

Where $r$ ${\ displaystyle r}$ equal to the radius measured from the center, $G$ ${\ displaystyle G}$ is the gravitational constant . The equation is rewritten as

{\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 .

{\ 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.}

Exact and asymptotic solutions

Introducing conversions

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

{\ 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 $\rho _{c}$ ${\ displaystyle \ rho _ {c}}$ Is the density of the star in the central part, we obtain the expression

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

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

The boundary conditions are as follows:

\xi =0:\quad \psi =0,\quad {\frac {d\psi }{d\xi }}=0.

{\ displaystyle \ xi = 0: \ quad \ psi = 0, \ quad {\ frac {d \ psi} {d \ xi}} = 0.}

At $\xi \ll 1$ ${\ displaystyle \ xi \ ll 1}$ the solution is close to

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

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

Model Limitations

The assumption that the sphere is isothermal has some drawbacks. Although the density of the isothermal gas sphere obtained by the solution decreases with distance from the center, the decrease is still too slow to obtain a reliably determined surface and the mass of the sphere is finite ^[4] . It can be shown that for $\xi \gg 1$ ${\ displaystyle \ xi \ gg 1}$ ,

{\frac {\rho }{\rho _{c}}}=e^{-\psi }={\frac {2}{\xi ^{2}}}\left[1+{\frac {A}{\xi ^{1/2}}}\cos \left({\frac {\sqrt {7}}{2}}\ln \xi +\delta \right)+O(\xi ^{-1})\right]

{\ displaystyle {\ frac {\ rho} {\ rho _ {c}}} = e ^ {- \ psi} = {\ frac {2} {\ xi ^ {2}}} left [1 + {\ frac {A} {\ xi ^ {1/2}}} \ cos \ left ({\ frac {\ sqrt {7}} {2}} \ ln \ xi + \ delta \ right) + O (\ xi ^ {-1}) \ right]}

,

Where $A$ ${\ displaystyle A}$ and $\delta$ ${\ displaystyle \ delta}$ are constant values that can be obtained by numerical solution. This behavior of the density leads to an increase in mass with increasing radius. Therefore, the model is usually suitable for describing stellar cores, where the temperature is approximately constant.

Special Decision

Conversion $x=1/\xi$ ${\ displaystyle x = 1 / \ xi}$ reduces the equation to

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

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

The equation has a special solution of the form

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

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

Therefore, you can enter a new variable $-\psi =2\ln x+z,$ ${\ displaystyle - \ psi = 2 \ ln x + z,}$ while the equation for $z$ ${\ displaystyle z}$ can be deduced:

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

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

This equation can be reduced to a first order equation by introducing a variable

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

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

then

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

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

Other Equations

The equation can be brought to a different kind. Let be

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

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