Download 4. Polytropes

4. Polytropes
4.1Definitions
An equation of state of the form
P= where γ and κ are constants, is called polytropic.
It is cutomary to define the corresponding polytropic index, denoted by n, as:
= 1
1
n
Note that n does not have to be integer. In the case that γ=5/3 we get n=3/2.
There are several types of stars that can be approximated well by a polytrope:
–
–
–
–
convective cores of stars
fully convective low-mass stars
white dwarfs
neutron stars
4.2The Lane-Emden Equation
=
dM
GM
=4 r2
and
2
Recall
1 dP
d
hence:
r 2 dP
= G M
dr
and
d r 2 dP
dr dr
dr
r
= 4 G r
=G dM
dr
2
(*)
Now we add the equation of state for a polytrope:
r :
d r2
dr 1
1
d
n
dr
= 4 r
2
P= 1
1
n
, giving us a single equation for
G
We can try to simplify this equation:
Define a new variable θ, defined as = c θ is called the Lane-Emden function. θ=1 in the centre of the star and 0 at the surface.
n
So, in terms of θ we can write P as:
1
P = c
1
n
1
1 n 1 = P c n 1 , where P c = c n
Hence:
r 2 dP
dr
=
r2 P c
c n
n 1 n
d
2 d
=
r
dr
dr
n 1
Pc
c
So the derivative with this with respect to r :
=
d r 2 dP
dr dr
Pc d 2 d r
dr
c dr
n 1
=
4 G c r 2 n using formula (*)
So we have:
d
d
r2
dr
dr
=
4 G c
r2 n
n 1 P c
2
Now let's introduce a dimensionless radius ξ, so that r = , where α has the
dimension of length. With ξ we have:
4 G 2c 2
= n1 P 2 n (**)
c
d
d
2
d
d
2
Now choose =
n1 P c
4 G c
2
So that equation (**) becomes:
d
1 d
2
2
d
d
=
n
This is the Lane-Emden equation.
There are exact solutions of this equation for only 3 examples:
n=0: = 1
n=1:
=
2
6
sin =
n=5:
1
1
2
3
The solutions look like:
Draw graph. First zero ( 1 ) correponding to the radius of the star corresponds to
=
6 n= 0 ,
= n= 1
,
Also, at = 0 :
= n= 5
4.3Scaling Relations
1. Central Density
RS
Consider
M S = 4 r r dr
2
0
Set r = , = c
This gives:
MS = 4 n
1
2
d c
n
3
0
but:
d
1 d
2
2
d
d
M S = 4 c
3
=
n
, so
1
dd
2
0
d
d
d
hence
M S = 4 3 c 12
d
d
=
1
which implies for the central density:
c =
3MS
4 R3S
a n , where
1
a n =
3
d
d
=
1
and this means that we can write the relation for c :
= 1 ; dd = 0
c
= a
=
, where
n
MS
4
R3S
3
2. Central Pressure
from the definition of α we have:
4G c Pc =
n 1
2
2
2
= 4nG1
an 3 M S
R 2S
4 RS
12
3
reshuffle:
G M 2S
1
Pc =
4 n 1 R 4S
1
2
d
d
=
1
hence
P c = cn
cn =
2
G MS
R
with
4
S
1
2
4 n 1
d
d
=
1
3. Central Temperature
Tc=
Pc
R
=b
c
2
3
4 cn G M S R S
=R
3 an
G MS
n
R 4S
MS
where bn =
R RS
4 cn
3 an
4.4 A Mass-Radius relation
Recall P = so that P c = 1
1
n
1 1
n
c
This means that given
c
and P c we can calculate κ:
=
Pc
1
c
1
n
=c
4
G MS
n
3 an M S
4
RS
1 1
n
4 RS
3
simplifying this gives:
=d
1 G MS
n
1 RS
1
n
=d
3
n
1
MS
n
1
n
R
3
1
n
S
and d n =
cn 4 1
1
3 an
1
n
1
n
Since κ is a constant, this leads to the Mass-Radius relation:
MS R
n 3
n 1
S
5. Summary
MS
R
,
3
S
GM
and T R R S
G M 2S
Pc R
4
S
S
4.5White dwarfs
1. In the non-relativistic case we have already seen (degenerate):
Pe = = 20 mh m 3m
e
H
H
2
5
3
, where
2
3
1 X
2
5
3
For a polytrope with index n= 2 we have such a relation between P and ρ with
3
= P =c
c
5
3
c
G M 2S
3
2
3 a3
R
4
S
2
5
3
MS
4 R 3S
These constants a n , b n and cn in practise can be lookd up in tables. Doing this
we get:
1
3
S
2
= 0.424 G M RS = 20 mh m 3
mH
e
H
2
3
1 X
2
5
3
putting in the numbers and X=0 (White dwarfs = Helium star):
RS
= 0.8710
R Sun
1
3
MS
M Sun
2
so in case of a solar mass white dwarf, the radius would be about Earth-size. Its
density (see exercises week 1) would be about ~109 kg m 3 . One matchbox of such
material weighs 1 ton.
2. In the relativistic case we have a different polytrope:
(as M S rises P e rises as well, until the electrons become relativistic).
P
= 4
3
with
=
hc
8 mH
1 X
2
1
3
3
mH
4
3
This means that we have n=3.
As before:
=
Pc
=
4
3
c
=c
2
G MS
3
4
RS
2
3
S
G M c3
3 a3 M S
4
3
4 R 3S
4
3 a3
4
3
This does not depend on RS , which means that the mass is uniquely determined:
MS
=
hc
8 Gm H
3
2
If X=0 (pure Helium)
3
mH
1
2
1 X
2
2
2
c3
3
2
3 a3
4
M S = M Crit = 1.44 M Sun
If M M Crit there will be no stable, degenerate, relativistic solution, so the star will
undergo further collapse, and become a neutron star, or a Black Hole.
Since R S m e 1 a white dwarf is about a factor 1000 larger than a neutron star.
4.6A practical case: the Sun
In general we need to integrate the Lane-Emden equation numerically. This goes as
follows:
We can write ther Lane-Emden equation as:
d
2
d
2
= 2 d
d
n
We would like to integrate by starting in the centre and going outward in steps of
. We can write:
i =
i 1
=
i d
d
=
i
In the same way we can write for the first derivative:
d
d
i 1
=
d
d
d2
d 2
i
Now use the Lane-Emden equation:
d
d
i 1
=
d
d
i
In the centre we know that
integrate outward until
Now, for the Sun
2 d
d
0
=1
n
i
d
d
and
=0
, so we start from those and
0
reaches 0.
and should be converted to density and radius:
r = , so at the photosphere of the Sun we haveL
=
R Sun
1
We also had:
c =
3MS
4 R3S
a n with
1
a n =
3
d
d
=
1
We now input M Sun and RSun from observations, and then have a solution for :
= c
n
3
2
We had: M = 4 c d
d
, through which we have the mass profile.
P and T can be calculated as before.
In the figure profiles are plotted calculated in this way for a n=3 polytrope model. A
comparison is made with the Standard Solar Model (Bahcall 1998; Physics Letters
B, 433, 1), which is the most up-to-date solution to the equations of stellar structure
currently available. It can be seen that the polytrope model does remarkably well,
considering how simple the physics is - we have used only the mass and radius of the
Sun and an assumption about the relationship between internal pressure and density as
a function of radius. The agreement is particularly good at the core of the star, where
the polytrope gives a central density of 7.65 x 104 kg m-3, a central pressure of 1.25 x
1016 N m-2 and a central temperature of 1.18 x 107 K, in comparison with the SSM
values of 1.52 x 105 kg m-3, 2.34 x 1016 N m-2 and 1.57 x 10 7 K. Only in the outer
regions of the Sun, where convection takes place, do the two solutions deviate
significantly from one another.