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2004/12/01 | 莱恩-埃姆登方程[Lane-Emden Equation]
类别(Ω〖物理〗)
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发表于 14:27
A
second-order ordinary differential equation
that applies to
polytropic profiles
in
density
, defined as those which have the form
(1)
where
P
is pressure,
K
is a constant,
is the density, and
(2)
with
n
is an integer (Chandrasekhar 1960).
The Lane-Emden equation is also applicable to
magnetohydrodynamic fluids
under the action of
force-free magnetic fields
.This implies that mass configurations obtained for neutral fluids through the Lane-Emden equation also exist for conducting fluids in
force-free magnetic fields
.Such fields are believed to exist in several astrophysical situations (Krishan 1999).
Poisson's equation
for gravity states
(3)
where
is the
gravitational potential
,
G
is the
gravitational constant
, and
is the mass
density
. For spherical symmetry,
in spherical coordinates is
(4)
Now, plug (1) into the
hydrostatic law
(5)
where
g
is the
gravitational acceleration
, giving
(6)
Plugging (6) into (4),
(7)
This is actually a form of the Lane-Emden equation
(8)
(9)
(10)
Subject to the boundary conditions
(11)
(12)
These boundary conditions establish the allowed values of
R
and
M
. By appropriate change of variables
(13)
(14)
equation (10) can be transformed to the Lane-Emden equation
(15)
(16)
and the boundary conditions become
(17)
(18)
The cases
n
= 0 and 1 can be solved analytically; the others must be obtained numerically. The mass of a spherical body of radius
R
is given by the integral
(19)
so the central density is
(20)
giving
(21)
The central pressure is then given by
(22)
and the
moment of inertia
by
(23)
so
(24)
From the
ideal gas law
,
(25)
where
is the mean molecular mass, and
k
is
Boltzmann's constant
. Therefore, the central temperature is
(25)
Finally, the
gravitational potential energy
is
(27)
For
n
= 0(
), the
Lane-Emden equation
is
The boundary condition
then gives
and
so
so
is parabolic. The first zero
is found by solving
For
n
= 1 (
)in
space, the differential equation becomes
which is the
spherical Bessel differential equation
with
k
= 1 and
n
= 0, so the solution is
(28)
Applying the boundary condition
gives
In
space,
]
(29)
The solution is
(30)
The boundary condition (11) requires
B
= 0. Applying (12) to (30) with
B
= 0,
(31)
for
n
= 1, 2, .... It is physically impossible for
to equal 0 anywhere but at the planet's boundary. Therefore, it must be true that the first place where (30) is 0 is also the boundary. This implies that (31) must equal
(32)
which implies the surprising result that
R
is independent of
M
!
(33)
Therefore, the solution (30) is constrained by boundary conditions to
(34)
A can be expressed in terms of
M
using the condition
(35)
Let
(36)
(37)
The integral (35) then becomes
(38)
(39)
(40)
(41)
(42)
(43)
(44)
Let
(45)
then
(46)
To find
, solve (33) for
K
(47)
(48)
To find
I
,
(49)
(50)
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