Emden–Chandrasekhar equation

For an isothermal gaseous star, the pressure p is due to the kinetic pressure and radiation pressure :p = \rho\frac{k_B}{W H} T + \frac{4\sigma}{3c} T^4 where • \rho is the density • k_B is the Boltzmann constant • W is the mean molecular weight • H is the mass of the proton • T is the temperature of the star • \sigma is the Stefan–Boltzmann constant • c is the speed of light The equation for equilibrium of the star requires a balance between the pressure force and gravitational force :\frac{1}{r^2} \frac{d}{dr} \left(\frac{r^2}{\rho}\frac{dp}{dr}\right)= - 4\pi G \rho where r is the radius measured from the center and G is the gravitational constant. The equation is re-written 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 Introducing the transformation :\psi = \ln \frac{\rho_c}{\rho}, \quad \xi = r \left(\frac{4\pi G \rho_c W H}{k_B T}\right)^{1/2} where \rho_c is the central density of the star, leads to :\frac{1}{\xi^2} \frac{d}{d\xi}\left(\xi^2 \frac{d\psi}{d\xi}\right)= e^{-\psi} The boundary conditions are :\psi =0, \quad \frac{d\psi}{d\xi} =0 \quad \text{at} \quad \xi=0 For \xi\ll 1, the solution goes like :\psi = \frac{\xi^2}{6} - \frac{\xi^4}{120} + \frac{\xi^6}{1890} + \cdots ==Limitations of the model==

Assuming isothermal sphere has some disadvantages. Though the density obtained as solution of this isothermal gas sphere decreases from the centre, it decreases too slowly to give a well-defined surface and finite mass for the sphere. It can be shown that, as \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] where A and \delta are constants which will be obtained with numerical solution. This behavior of density gives rise to increase in mass with increase in radius. Thus, the model is usually valid to describe the core of the star, where the temperature is approximately constant. ==Singular solution==

Introducing the transformation x=1/\xi transforms the equation to :x^4 \frac{d^2\psi}{dx^2} = e^{-\psi} The equation has a singular solution given by :e^{-\psi_s} = 2x^2, \quad \text{or} \quad -\psi_s = 2 \ln x+ \ln 2 Therefore, a new variable can be introduced as -\psi = 2 \ln x + z, where the equation for z can be derived, :\frac{d^2z}{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 :y=\frac{dz}{dt} = \xi \frac{d\psi}{d\xi} - 2 then we have :y\frac{dy}{dz} - y + e^z- 2 = 0 ==Reduction==

There is another reduction due to Edward Arthur Milne. Let us define :u = \frac{\xi e^{-\psi}}{d\psi/d\xi}, \quad v = \xi \frac{d\psi}{d\xi} then :\frac{u}{v}\frac{dv}{du} = -\frac{u-1}{u+v-3} ==Properties==

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