Download Applicability of Polytropes The four equations of stellar structure

Applicability of Polytropes
The four equations of stellar structure divide naturally into two
groups: one describing the mechanical structure of the star using (2.1.3) and (2.2.9), and the other giving the thermal structure via (2.3.3) and (2.4.4). However, the only contact between
the mechanical variables and thermal equations is through the
temperature dependence of the equation of state. Under certain
circumstances, however, the pressure can become independent of
temperature, and only depend on density, i.e.,
P = Kργ = Kρ1+1/n
(15.1.1)
where n is the polytropic index. When the equation of state
can be written in this form, the calculations of stellar structure
simplify enormously.
Decoupling occurs when:
• Stars become supported by electron degeneracy. When this
occurs, the pressure is related to the density by
h2
Pe =
20me ma µe
µ
−5/3
(K = 1.0036 × 1013 µe
Pe =
µ
hc
8ma µe
¶µ
3
πma µe
ρ5/3 = Kρ5/3
(7.3.7)
cgs), or
3
πma µe
−4/3
¶2/3
¶1/3
ρ4/3 = Kρ4/3
(7.3.13)
(K = 1.2435 × 1015 µe
cgs) depending on whether the Fermi
momentum is relativistic or not. For future reference, note here
that the constant K depends only on µ and atomic physics.
• Energy transport comes only from convection, and radiation
pressure is negligible. In this case, P ≈ Pgas , and
P =
ρ
kT
µma
=⇒ T ∝
µ
P
ρ
¶
When combined with the definition of ∇ad ,
µ
∂ ln T
∂ ln P
¶
ad
= ∇ad =⇒ P ∝ T 1/∇ad
this yields
P 1−(1/∇ad ) ∝ ρ−1/∇ad
or
P = Kρ1/(1−∇ad )
(15.1.2)
where K is not a function of atomic physics, but instead depends
on the star’s boundary conditions. Note that for a completely
ionized gas, ∇ad = 2/5, which recovers P ∝ ρ5/3 .
• Regions in which the ratio of the gas pressure to the radiation
pressure is constant. If we let β = Pgas /P , then for an ideal gas
plus radiation pressure equation of state
Pgas
ρ
=
kT
β
µma β
(15.1.3)
Prad
aT 4
P =
=
1−β
3(1 − β)
(15.1.4)
P =
and
If we then use (15.1.4) to substitute for temperature, then equation (15.1.3) becomes
ρk
P =
µma β
µ
3P (1 − β)
a
¶1/4
µ
¶1/4
ρk
3(1
−
β)
P 3/4 =
µma β
a
¶4/3 µ
µ ¶1/3 µ
¶1/3
1−β
3
k
4/3
4/3
ρ
=
Kρ
P =
a
µma
β4
(15.1.5)
where the value of K depends on the importance of radiation
pressure.
This last condition comes about naturally if the ratio κL/M is
constant. To see this, consider that if the star is convective, P ∝
ρ1/(1−∇ad ) as above, but if the star is radiative, then ∇ = ∇rad ,
and
1
3P κL
P dT
∇rad =
=
(3.1.6)
16πcG aT 4 M
T dP
If we then substitute using
Prad
aT 4
=
3
and dPrad
4aT 3
=
dT
3
then
P dPrad
1
P κL
P 3 dPrad
=
=
16πcG Prad M
dP 4aT 4
dP 4 Prad
or
1 κL
dPrad
=
dP
4πcG M
(15.1.6)
Thus, if κL/M is constant, and the surface pressure is ∼ 0, then
Z
0
Prad (r)
dPrad = C
Z
0
P (r)
dP
=⇒
Prad (r)
=C
P (r)
(15.1.7)
This is the condition for the Eddington “standard” model.
(Actually, setting κL/M constant is not a terrible assumption.
Kramer law opacities decrease with temperature as T −3.5 , while
the proton-proton chain gives ²n = L/M ∝ T 3.5 for low mass
stars. Thus, κL/M is approximately constant, and (15.1.7) is
an adequate approximation.)
Polytropic Calculations
To calculate the structure of a polytropic star, begin by assuming
hydrostatic equilibrium, and multiplying both sides by r2 /ρ, i.e.,
r2 dP
r2 GM
ρ = −GM
=−
ρ dr
ρ r2
If we then take the derivative (with respect to r) of both sides,
and divide by r2 , then we get
1 d
r2 dr
µ
2
r dP
ρ dr
¶
=−
G dM
G
4πr2 ρ = −4πGρ
=
−
2
2
r dr
r
(15.2.1)
(Note that this is nothing more than the Poisson equation — if
you substitute for pressure using
dΦ
1 dP
GM
= 2 =−
dr
r
ρ dr
then you recover ∇2 Φ = 4πGρ in spherical coordinates.) If you
then substitute in the polytropic relation (15.1.1), then
µ
¶
dP
1
dρ
=K 1+
ρ1/n
dr
n
dr
and Poisson’s equation becomes
1 d
r2 dr
½
µ
¶
¾
dρ
r2
1
K 1+
ρ1/n
= −4πGρ
ρ
n
dr
(15.2.2)
To simplify this expression, let’s put it in dimensionless form.
First, define a variable θ, such that
1/n
θ(r) = (ρ/ρc )
ρ = ρc θ n
=⇒
(15.2.3)
Poisson’s equation then becomes
½
¾
½
¾
dθ
1 1 d
θ1−n nρc θn−1
K 1+
r2 ρ−1+1/n
= −4πGρc θn
c
2
n r dr
dr
or
(n
1+1/n
+ 1) Kρc
4πGρ2c
1 d
r2 dr
µ
r2
dθ
dr
¶
= −θn
Then define a dimensionless radius, ξ = r/rn , where
rn =
½
(n + 1)K
4πG
¾1/2
ρc(1−n)/2n =
½
(n + 1)Pc
4πGρ2c
¾1/2
(15.2.4)
where we have used (15.1.1) to substitute in the central pressure.
In this form, the Poisson equation becomes
µ
¶
dθ
1 d
ξ2
= −θn
(15.2.5)
2
ξ dξ
dξ
This is called the Lane-Emden equation; its solution gives the
run of dimensionless density θ as a function of the dimensionless
radius, ξ. Since it is a second order differential equation, you
need two boundary conditions. The first is at the center: from
spherical symmetry, the pressure gradient at the center (θ = 1)
must be zero. The second condition comes from the surface, ξ1 ,
where the density should go to zero. So our boundary conditions
are
dθ
(ξ = 0) = 0 (the center)
dξ
and
θ(ξ = ξ1 ) = 0
(the surface)
Unfortunately, the Lane-Emden equation does not have an analytic solution for arbitrary values of n. In fact, there are only
four analytic solutions. The first is for n = 0, which implies
ρ(r) = ρc , or a constant density sphere. For these models,
ξ2
θ(ξ) = 1 −
6
(15.2.6)
√
with ξ1 = 6 to satify the boundary condition of θ(ξ1 ) = 0. (This
can be trivially checked via substition.) The second solution is
for the n = 1 case, where
θ(ξ) =
sin ξ
ξ
(15.2.7)
Note here that there are an infinite number of values of ξ1 for
which θ(ξ1 ) = 0. However, in practice, ξ1 = π, since all other
values would give an unrealistic zero density somewhere in the
middle of the star. A third solution exists for n = 5,
¢−1/2
θ(ξ) = 1 + ξ /3
¡
2
For this model, θ only goes to zero when ξ1 −→ ∞, thus stars
with n = 5 have infinite radius. All models with n > 5 have both
infinite radius and infinite mass, but the n = ∞ solution is noteworthy. From (15.1.1), n = ∞ corresponds to P = Kρ, which
is the isothermal case. If we go back to the Poisson equation
(15.2.2) and let n = ∞, then the equation becomes
K d
r2 dr
½
d ln ρ
r2
dr
¾
= −4πGρ
(15.2.8)
If we let Ψ = − ln ρ and ξ = (4πG/K)1/2 , then this translates to
µ
¶
1 d
dΨ
2
−Ψ
ξ
=
e
(15.2.9)
ξ 2 dξ
dξ
This is the equation for an isothermal sphere. It has many applications in astrophysics (especially extragalactic astrophysics).
The remaining solutions of the Lane-Emden equation, including
that for n = 3/2 (γ = 5/3) and n = 3 (γ = 4/3), must be computed numerically. For the inner part (ξ < 1) of the polytrope,
this can be done by writing θ out as a power law, i.e.,
θ = 1 + aξ + bξ 2 + cξ 3 + dξ 4 + . . .
(15.2.10)
and substituting into the Lane-Emden equation. When this is
done, the left side of the equation becomes
f (ξ) = 2aξ −1 + 6b + 12cξ + 20dξ 2 + . . .
(15.2.11)
Meanwhile, the right side of the equation can be evaluated by
expanding θn as a Taylor series about ξ = 0. In other words,
f 00 (0)ξ 2
f (ξ) = θ = f (0) + f (0)ξ +
+ ...
2!
n
0
where, through (15.2.10),
f (0) = 1
f 0 (0) = n θn−1
dθ
= n (1) a
dξ
2
00
n−2 dθ dθ
n−1 d θ
f (0) = n (n − 1)θ
+θ
dξ dξ
dξ 2
©
ª
= n (n − 1)(1)a2 + (1)2b
½
¾
Thus,
n(n − 1)a2 + 2b 2
−θ = −1 − n aξ −
ξ + ...
2
n
(15.2.12)
The coefficients of (15.2.10) can then be found by equating the
terms of (15.2.11) and (15.2.12). In other words, since (15.2.12)
does not contain a ξ −1 term, a = 0, and
6b = −1
=⇒
12c = −n a
=⇒
20d = − n(n−1)a
2
2
+2b
=⇒
b = − 61
c=0
d=
n
120
Note that all the odd power terms are zero, due to the fact that
a = 0. This leaves
1
n 4 n(8n − 5) 6
θ(ξ) = 1 − ξ 2 +
ξ −
ξ + ...
6
120
15120
The outer regions of the polytrope can now be computed numerically by translating the second-order Lane-Emden equation into
two first-order equations with known starting point boundary
conditions. In other words, if we write Lane-Emden as
1 d
ξ 2 dξ
µ
then
z=
dθ
ξ2
dξ
dθ
dξ
¶
= −θ
and
n
=⇒
d2 θ
2 dθ
n
=
−θ
−
dξ 2
ξ dξ
dz
2
= −θn − z
dξ
ξ
(15.2.13)
This can be solved in a straightforward manner by starting from
the polynomial expansion and using Runge-Kutta integration.
Some key values resulting from this integration are tabulated
below.
n
ξ1
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.25
3.5
4.0
4.5
4.9
5.0
2.4494
2.7528
3.14159
3.65375
4.35287
5.35528
6.89685
8.01894
9.53581
14.97155
31.83646
169.47
∞
−ξ12
³
dθ
dξ
´
4.8988
3.7871
3.14159
2.71406
2.41105
2.18720
2.01824
1.94980
1.89056
1.79723
1.73780
1.7355
1.73205
ξ1
ρc /ρ̄
1.0000
1.8361
3.28987
5.99071
11.40254
23.40646
54.1825
88.153
152.884
622.408
6189.47
934800.
∞