Download The Mass-Radius Relation for Polytropes The mass within any point

The Mass-Radius Relation for Polytropes
The mass within any point of a polytropic star is
M=
R
Z
4πr2 ρ dr = 4πrn3 ρc
0
ξ0
Z
ξ 2 θn dξ
(16.1.1)
0
This can be simplified by substituting the left side of the LaneEmden equation for θn
µ
¶
µ ¶
Z ξ0
d
dθ
dθ
2
M(ξ 0 ) = 4πrn3 ρc
−
ξ2
dξ = 4πrn3 ρc ξ 0
dξ
dξ
dξ ξ0
0
·
¸ 32
·
¸ 32
(n + 1)K
= 4π
4πG
(n + 1)K
= 4π
4πG
3−3n
2n +1
ρc
(3−n)
2n
ξ0
ρc
ξ0
µ
2
µ
2
dθ
−
dξ
dθ
−
dξ
¶
¶
ξ0
(16.1.2)
ξ0
Now compare this to the radius
·
(n + 1)K
r(ξ 0 ) = rn ξ 0 =
4πG
¸ 12
(1−n)
2n
ρc
ξ0
(16.1.3)
If you solve for ρc in the mass equation (16.1.2)
ρc =

· M ¸ ·
 4π
4πG
(n + 1)K
¸ 23 "
−ξ 0
2
µ
dθ
dξ
 2n
¶ #−1  3−n
ξ0

and eliminate it from the radius equation (16.1.3), you get
r = (4π)
1
n−3
·
(n + 1)K
G
n
·
¸ 3−n
−ξ
0 2 dθ
dξ
¸ n−1
3−n
M
1−n
3−n
(16.1.4)
In other words, the radius of a polytropic star is related to its
mass by
(1−n)/(3−n)
R ∝ MT
Note what this means: for a polytropic index of 3/2 (the γ = 5/3
case), R ∝ M−1/3 . Thus, for a set of stars with the same K and
n (i.e., white dwarfs, or a fully convective star that is undergoing
mass transfer), the stellar radius is inversely proportional to the
mass.
Equation (16.1.4) can also be re-written to give the polytropic
constant, K, in terms of the star’s total mass and radius. The
expression is
K=
"
4π
ξ n+1
µ
−
dθ
dξ
¶1−n #1/n
ξ1
G
1−1/n −1+3/n
R
MT
n+1
(16.1.5)
The Chandrasekhar Mass Limit
For n = 3, (γ = 4/3), the total mass of a polytropic star is
independent of its central density. This is important, for if you
substitute in the equation of state for a completely relativistic
electron gas (K = 1.2435 × 1015 µ−4/3 cgs) into (16.1.2), and do
the math, the total mass of the star becomes
MCh =
5.82
M¯
µ2e
(16.1.6)
This is the Chandrasekhar mass limit. For M < MCh , a star
can adjust its structure and back away from the completely relativistic limiting case. However, since (16.1.4) already represents
the limit to electron pressure, degenerate stars cannot exceed
MCh . If we assume such a star has no hydrogen, then, by (5.1.6),
µe = 2, and MCh = 1.456M¯ .
The Central Temperature and Density
With a little algebra, equations (16.1.2) and (16.1.3) can be manipulated to yield the conditions in the center of a polytropic
star. By combining the two equations, while eliminating K, the
expression for central density as a function of MT and R becomes
¶−1
µ
dθ
MT
1
ρc =
ξ1 −
4π
dξ ξ1 R3
(16.2.1)
The central pressure then follows, by combining this with the
equation for K given in (16.1.5)
Pc =
1
=
4π(n + 1)
Kρ1+1/n
c
µ
dθ
−
dξ
¶−2
GM2T
R4
ξ1
(16.2.2)
Finally, the central temperature can be found from (16.2.1),
(16.2.2), and the ideal gas law
µ ma
µ ma Pc
=
Tc =
k ρc
(n + 1)k
µ
dθ
−ξ
dξ
¶−1
ξ1
GMT
R
(16.2.3)
Numerically, these expressions evaluate to
ρc = 0.47 ξ1
µ
dθ
−
dξ
14
Pc =
8.97 × 10
(n + 1)
2.29 × 107
Tc =
(n + 1)
¶−1 µ
µ
µ
ξ1
−
dθ
dξ
MT
M¯
¶µ
¶−2 µ
dθ
−ξ
dξ
ξ1
MT
M¯
¶−1 µ
ξ1
R
R¯
¶−3
¶2 µ
MT
M¯
¶µ
g cm−3
¶−4
dyne cm−2
¶−1
K
R
R¯
R
R¯
The Hayashi Track
We can locate polytropic-like stars in the HR diagram by matching fully convective models to boundary conditions for cool, convective stars. To do this, we first assume that the star is entirely
convective, except for a thin radiative region at the surface, and
that the star is cool, so that H − opacity dominates. From our
boundary condition analysis
Pt = 22/3 Pp
(9.26)
Now recall that the pressure at the photosphere is
Pp =
2 GMT
3 R2 κ
(9.12)
and that κ itself is a function of pressure
κ = κ00 P α T β
So,
Pp =
1
2 GMT
β
3 R2 κ00 Ppα Teff
Ppα+1 =
=⇒
2 GMT
β
3 R2 κ00 Teff
4
or, if we substitute for R using LT = 4πR2 σTeff
,
Pp =
½
2 4πσ GMT 4−β
Teff
3 κ00 LT
¾1/(α+1)
(16.3.1)
The pressure at the top of the convective polytrope is therefore
Pt = c1
µ
MT
LT
¶1/(α+1)
(4−β)/(α+1)
Teff
(16.3.2)
where
c1 = 22/3
µ
2 4πσG
3 κ00
¶1/(α+1)
(16.3.3)
Now, consider the polytropic equation
P = Kρ1+1/n =
µ
k
µ ma
¶n+1
K −n T n+1
(16.3.4)
where the constant K is given by (16.1.5), i.e.,
µ
¶ µ
¶n
dθ
n
+
1
1
−ξ n+1
MT1−n Rn−3
K −n =
4π
dξ ξ1
G
=
1
4π
µ
−ξ n+1
dθ
dξ
¶ µ
ξ1
n+1
G
¶n µ
LT
4πσ
¶ (n−3)
2
6−2n
MT1−n Teff
At the top of the convective layer,
µ ¶2/9
8
Tt =
Teff
5
(9.24)
so if we substitute this in (16.3.4), the pressure becomes
(n−3)/2
Pt = c2 MT1−n LT
7−n
Teff
(16.3.5)
where
c2 =
µ
k
µ ma
¶n+1 µ
dθ
−ξ n+1
dξ
¶ µ
ξ1
n+1
G
3−n
¶n µ ¶ 2n+2
9
(4πσ) 2
8
5
4π
If we now set (16.3.2) equal to (16.3.5), the relation between MT ,
LT , and Teff for fully convective stars becomes
β−4
7−n+ α+1
Teff
=
µ
c1
c2
¶
1
n−1+ α+1
MT
3−n
1
2 − α+1
(16.3.6)
LT
Or, for the case of fully convective (n = 3/2) stars with H −
opacity (α = 1/2, β = 17/2),
Teff ∼ 2600 µ13/51
µ
MT
M¯
¶7/51 µ
LT
L¯
¶1/102
K
This is the Hayashi track (although, due to our boundary condition assumptions, the constant is a factor of 1.5 too small). Note
that the temperature is nearly independent of luminosity, and is
only weakly dependent on mass. All fully convective stars will
therefore fall in the same cool area of the HR diagram.