Introduction to Astrophysics

Tutorial 2: Polytropic Models

Iair Arcavi

1 Summary of the Equations of Stellar Structure

We have arrived at a set of dierential equations which can be used to describe the structure of a star
in equilibrium. The equations can be written either as function of the radius coordinate r or the mass
coordinate m:

Hydrostatic Equilibrium: dP
dr = −ρ Gm
r2
dP
dm
Gm
= − 4πr 4

Continuity: dm
dr = 4πr2 ρ dr
dm = 1
4πr 2 ρ

Radiative Transfer: dT
dr
3 κρ F
= − 4ac T 3 4πr 2
dT
dm
3 κ
= − 4ac F
T 3 (4πr 2 )2
(if energy is transferred only
by radiative diusion)

Thermal Equilibrium: dF
dr = 4πr2 ρq dF
dm =q

These structure equations are supplemented by:

P = PI + Pe + Prad
R 1
= ρT + Pe + aT 4
µI 3
a b
κ = κ0 ρ T
q = q0 ρm T n
Solving these equations is very dicult, since they are non-linear, coupled and have two-point boundary
conditions. However, we can gain insight into the structure of stars by analyzing the equations without
solving them, and by using simple models. One such model is called the Polytropic Model.

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2 The Polytropic Model
Note that the rst pair of equations is connected to the second pair by the equation of state. However,
if the pressure is independent of temperature, then the rst pair of equations is separated from the
second pair. Multiplying the equation of hydrostatic equilibrium by r2 /ρ and dierentiating with
respect to r, we nd:
r2 dP
 
d dm
= −G
dr ρ dr dr
Using the continuity equation on the right hand side gives:
r2 dP
 
1 d
= −4πGρ
r2 dr ρ dr

Now enters the model. We consider equations of state of the form:

P = Kργ

where K and γ are constants. This is known as a polytropic equation of state, and we have seen that
it pops up with degenerate gases and adiabatic processes.
We dene the polytropic index n as:
1
γ =1+
n
so a degenerate gas has index n = 1.5 for the non-relativistic case and n = 3 for the ultra relativistic
case.
Using this model in the structure equation we constructed gives:
r2 dρ
 
(n + 1) K 1 d
n−1 = −ρ
4πGn r2 dr ρ n dr
With the boundary conditions:
ρ (R) = 0

dρ 
= 0
dr r=0

(where the second condition comes from the equation of hydrostatic equilibrium).
The solution ρ (r) is called a polytrope, and is uniquely dened by K , n and R.
It is convenient to dene dimensionless variables θ and ξ by:
ρ = ρc θ n
r = αξ

where ρc is the central density and s

(n + 1) K
α= n−1
4πGρc n

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Our equation then becomes the Lane-Emden equation:
 
1 d dθ
ξ2 = −θn
ξ 2 dξ dξ

with boundary conditions:

θ|ξ=0 = 1

dθ 
= 0
dξ  ξ=0

For n < 5, θ (ξ) decreases monotonically, and the radius of the star can be found by looking for the
rst zero of θ (usually numerically), called ξ1 (then R = αξ1 ), since at the surface ρ = 0 so θ = 0.
There are only three cases where an analytical solution to the Lane-Emden equation exists:
√
1. n = 0, θ0 = 1 − ξ6 , ξ1 = 6, γ = ∞
2

2. n = 1, θ1 = sin ξ
ξ , ξ1 = π , γ = 2
3. n = 5, θ5 = √1+ξ
1
2 /3
, ξ1 = ∞, γ = 6
5

In case 1, we have P = Pc θ, and this is applicable for an incompressible uid (since n = 0 means
constant density). Case 3 is not physical because the density never reaches zero.
The total mass of a star can be found by:
ˆR
M = 4πr2 ρdr
0
ˆξ1
3
= 4πα ρc ξ 2 θn dξ
0

Substituting from the Lane-Emden equations we have:

ˆξ1  
3 d dθ
M = −4πα ρc ξ2 dξ
dξ dξ
0

which gives:

3 dθ 
M = −4πα ρc ξ12
dξ ξ1

The mean density ρ̄ = 4

M
3 can be shown to be related to the central density by:
3 πR

ρc = Dn ρ̄

3
with: " #−1
3 dθ 
Dn = −
ξ1 dξ ξ1
This enables us to nd a relation between the mass and the radius:
 n−1  3−n n
GM R [(n + 1) K]
=
Mn Rn 4πG

with:

dθ 
Mn = −ξ12
dξ ξ1
Rn = ξ1

It is interesting to see what happens in the special cases n = 1 and n = 3. For n = 3, the mass is
independent of the radius:
 3/2
K
M = 4πM3
πG
while for n = 1 we can nd a radius independent of mass:
 1/2
K
R = R1
2πG
Generally, for n between these values, we note that:
1
R3−n ∝
M n−1

The more massive the star, the smaller (and hence denser) it gets!
Finally, we would like to nd an expression for the central1 pressure, which can be done by substituting
the central density into the equation of state P = Kρ1+ n :
1   n−1   3−n
(4πG) n GM n
R n n+1
Pc = ρc n
n+1 Mn Rn
which can be written as:
1/3
Pc = (4π) Bn GM 2/3 ρ4/3
c

where Bn includes all the n-dependent coecients. It turns out that Bn varies quite slowly with n,
implying that the above expression is almost universal, and therefore will be useful later.

3 The Chandrasekhar Mass

Stars which are dominated by degeneracy pressure, can be described by a polytropic equation of state
with index n = 1.5 and K = K1 (from the previous tutorial). White dwarfs are an example of such

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stars (having roughly the mass of the sun but the radius of the earth, giving them a mean density of
ρ̄ ≈ 105 g/cm3 ). For such stars the mass-radius relation is:

R ∝ M −1/3

and so the average density increases with the mass as:

ρ̄ ∝ M R−3
∝ M2

For increasing mass, then, the star becomes denser, and so ultimately its electrons become relativistic.
In this case, n = 3, for which the mass-radius relation gives a mass that is independent of the radius:
 3/2
K
M = 4πM3
πG

Plugging in K = K2 from the relativistic degenerate equation of state, gives the Chandrasekhar Mass:
√ !3/2
M3 1.5 hc
MCh = 4/3
µ−2
e
4π GmH

Substituting all the constants, this can be written as:

5.83
MCh = M
µ2e

This is the maximal mass a stable conguration can have. For a white dwarf composed mostly of He
and heavier elements, for which we can take µe = 2, we nd MCh = 1.46M .

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