In astrophysics, the Emden–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 gravitational force, named after Robert Emden and Subrahmanyan Chandrasekhar.[1] [2] The equation was first introduced by Robert Emden in 1907.[3] The equation[4] readswhere
\xi
\psi
\rho=\rhoce-\psi
\rhoc
\psi=0,
d\psi | |
d\xi |
=0 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.[5]
For an isothermal gaseous star, the pressure
p
p=\rho
kB | |
WH |
T+
4\sigma | |
3c |
T4
where
\rho
kB
W
H
T
\sigma
c
The equation for equilibrium of the star requires a balance between the pressure force and gravitational force
1 | |
r2 |
d | \left( | |
dr |
r2 | |
\rho |
dp | |
dr |
\right)=-4\piG\rho
where
r
G
kBT | |
WH |
1 | |
r2 |
d | |
dr |
| ||||
\left(r |
\right)=-4\piG\rho
\psi=ln
\rhoc | |
\rho |
, \xi=r\left(
4\piG\rhocWH | |
kBT |
\right)1/2
where
\rhoc
1 | |
\xi2 |
d | |
d\xi |
\left(\xi2
d\psi | |
d\xi |
\right)=e-\psi
The boundary conditions are
\psi=0,
d\psi | |
d\xi |
=0 at \xi=0
For
\xi\ll1
\psi=
\xi2 | |
6 |
-
\xi4 | |
120 |
+
\xi6 | |
1890 |
+ …
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\gg1
\rho | |
\rhoc |
=e-\psi=
2 | \left[1+ | |
\xi2 |
A | \cos\left( | |
\xi1/2 |
\sqrt7 | |
2 |
ln\xi+\delta\right)+O(\xi-1)\right]
where
A
\delta
Introducing the transformation
x=1/\xi
x4
d2\psi | |
dx2 |
=e-\psi
The equation has a singular solution given by
-\psis | |
e |
=2x2, or -\psis=2lnx+ln2
Therefore, a new variable can be introduced as
-\psi=2lnx+z
z
d2z | - | |
dt2 |
dz | |
dt |
+ez-2=0, where t=lnx
This equation can be reduced to first order by introducing
y= | dz |
dt |
=\xi
d\psi | |
d\xi |
-2
then we have
y | dy |
dz |
-y+ez-2=0
There is another reduction due to Edward Arthur Milne. Let us define
u=
\xie-\psi | |
d\psi/d\xi |
, v=\xi
d\psi | |
d\xi |
then
u | |
v |
dv | |
du |
=-
u-1 | |
u+v-3 |
\psi(\xi)
\psi(A\xi)-2lnA
A
d\psi/d\xi=0
\xi=0