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Astronomy 112: The Physics of Stars
Class 9 Notes: Polytropes
With our discussion of nuclear reaction rates last time, we have mostly completed our survey
of the microphysical properties of stellar matter – its pressure, how energy flows through it,
and how it generates energy from nuclear reactions. For the next few weeks we will be using
those microphysical models to begin to make our first models of stars.
I. The Stellar Structure Equations
To begin we will collect the various equations we have developed thus far to describe
the behavior of material in stars. As always, we consider a shell of material of mass
dm and thickness dr, which is at a distance r from the center of the star and has
a mass m interior to it. The shell has density ρ, pressure P , temperature T , and
opacity κ. The radiation flux passing through it is F , and the shell generates energy
via nuclear reactions at a rate per unit mass q. We will write down the equations
describing this shell, under the assumption that the star is in both hydrostatic and
thermal equilibrium, so we can drop all time derivatives describing change in position,
energy, etc. Since we are assuming equilibrium, for now we will also assume that the
composition is fixed, so that we know X, Y , Z, and any other quantities we need that
describe the chemical makeup of the gas.
The first equation is just the definition of the density for the shell, which says that
ρ = dm/dV . Writing this in Eulerian or Lagrangian form (i.e. with either radius or
mass as the independent variable), we have
dm
= 4πr2 ρ
dr
dr
1
=
.
dm
4πr2 ρ
The second equation is the equation of hydrostatic balance, which we can also write
in either Eulerian or Lagrangian form:
Gm
dP
Gm
dP
=− 2 ρ
=−
.
dr
r
dm
4πr4
This just equates the change in gradient in pressure with the force of gravity. The
third equation is the equation describing how the temperature changes with position
within a star as a result of radiative diffusion:
dT
3 κρ F
dT
3 κ
F
=−
=−
.
3
2
3
dr
4ac T 4πr
dm
4ac T (4πr2 )2
Finally, we have the equation of energy conservation, which for material in equilibrium
just equates the change in energy flux across a shell to the rate at which nuclear
reactions generate power within it:
dF
= 4πr2 ρq
dr
dF
= q.
dm
1
These are four coupled, non-linear ordinary differential equations. As we discussed a
few weeks ago, by themselves they are not a complete system, because by themselves
they contain more unknowns than equations. The unknowns appearing are ρ, T , P ,
F , κ, and q. If we adopt Eulerian coordinates, then m is also an unknown, and r is
the independent variable. For Lagrangian coordinates, r is also an unknown, and m
is the independent variable. Thus we have seven unknowns, but only four equations.
We therefore need three more equations, and that is what we have spent the last few
weeks providing.
The pressure depends on density and temperature via
P =
R
1
ρT + Pe + aT 4 .
µI
3
The first term is the ion pressure, which we have written assuming that ions are
non-degenerate, which they are except in neutron stars. The last term is the radiation
pressure. The middle term, the electron pressure, takes a form that depends on whether
the electrons are degenerate or not, but which is a known function of ρ and T .
The opacity and nuclear energy generate rate, we have seen, are in general quite complicated functions. However, we have also seen that they can be approximated reasonably
well as powerlaws:
κ = κ0 ρa T b
q = q 0 ρm T n .
Regardless of whether we make the powerlaw approximation or not, we now know how
to compute κ and q from ρ and T .
Thus we have written down three more equations involving the unknowns. We are
therefore up to seven equations for our seven unknowns, which is sufficient to fully
specify the system. The only thing missing is boundary conditions, since differential
equations produce constants of integration that must be determined by boundary conditions. Since there are four differential equations, we need four boundary conditions.
Three of them are obvious. In Lagrangian coordinates, we have r = 0 and F = 0 at
m = 0, and P = 0 at m = M . In words, the first condition says that the innermost
mass element must reside at radius r = 0, and it must have zero flux (F = 0) entering
it from below. The third condition says that the pressure falls to zero at the boundary
of the star, m = M .
The fourth condition is slightly more complicated. The simplest approach is to set
T = 0 at the star’s surface. This is actually not a terrible approximation, since the
temperature at the surface is very low compared to that in the interior. A better
approach is to specify the relation between flux and temperature at the stellar surface
as F = 4πr2 σT 4 . An even better approach is to make a detailed model of a stellar
atmosphere and figure out how the flux through it depends on its temperature and
pressure, and use that as a boundary condition for the stellar model.
The set of seven equations and four boundary conditions we have written down now
fully specifies the structure of a star. Solving those equations, however, is another
matter entirely. There is no general method for solving sets of coupled non-linear
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differential-algrebraic equations subject to boundary conditions specified at two points.
There is every reason to believe that such equations cannot, in general, be solved in
closed form.
Today the standard approach is to hand the problem to a computer. A computer can
integrate the equations and find solutions to any desired level of accuracy, and this
problem is sufficiently simple that the calculations will run on an ordinary desktop
machine in a matter of seconds. However, in the days when people first approached
these problems, there were no such things as computers. Instead, people were forced
to come up with analytic approximations, and it turns out that one can understand a
great deal about the behavior of stars using such approximations. (It turns out that
a significant fraction of being a good physicist consists in the ability to come up with
good approximations for intractable differential equations.)
II. Polytropes
A. Definition and Motivation
The first two stellar structure equations, describing the definition of density and
hydrostatic equilibrium, are linked to the second two only via the relationship
between pressure and temperature. If we can write the pressure in terms of
the density alone, without reference to the temperature, then we can separate
these two equations from the others and solve them by themselves. Solving two
differential equations (plus one algebraic equation relating P and ρ is much easier
than solving seven equations.
As a first step in this strategy, we can combine the first two first-order ODEs into
a single second-order ODE. To do so, we start with the equation of hydrostatic
equilibrium and multiply by r2 /ρ to obtain
r2 dP
= −Gm.
ρ dr
Next we differentiate both sides:
d
dr
r2 dP
ρ dr
!
= −G
dm
.
dr
Finally, we substitute for dm/dr using the definition of density, dm/dr = 4πr2 ρ.
Doing so we obtain
!
1 d r2 dP
= −4πGρ.
r2 dr ρ dr
This is just another form of the equation of hydrostatic balance, this time with
the definition of density explicitly substituted in.
Thus far everything we have done is exact. Now we make our approximation. We
approximate that the pressure and density are related by a powerlaw
P = KP ργ P .
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Equations of state of this sort are called polytropes. For historical reasons, it is
common to define
1
1
γP = 1 + or n =
n
γP − 1
where we call n the polytropic index.
Before going any further, it is important to consider whether an equation of state
like this is at all sensible. Why should a star ever obey such an equation of state?
The answer to this question becomes clearer if we recall that, for an adiabatic
gas, the equation of state reads
P = Ka ργ a ,
where Ka is the adiabatic constant and adiabatic index. It is important to understand the difference between this relation and the polytropic relation. The
polytropic relation describes how the pressure changes with density inside as one
moves through a star, while the adiabatic equation of state describes how a given
gas shell would respond to being compressed. The constant Ka depends on the
entropy of the gas in a given shell, so different shells in a star can have different
values of Ka . If different shells have different Ka values, then as I move through
the star the pressure will not vary as P ∝ ργ , because different shells will have different constants of proportionality. Thus a star can be described by a polytropic
relation only if Ka is the same for every shell.
While this condition might seem far-fetched, it is actually satisfied under a wide
range of circumstances. One circumstance when it is satisfied is if a star is dominated by the pressure of degenerate electrons, since in that case we proved a few
classes back that Pe = K10 (ρ/µe )5/3 for a non-relativistic gas, or Pe = K20 (ρ/µe )4/3 .
The proportionality constants K10 and K20 depend only on constants of nature like
h, c, the electron mass, and the proton mass, and thus do not vary within a star.
For such stars, γP = 5/3 or 4/3, corresponding to n = 1.5 or n = 3, for the
non-relativistic and highly-relativistic cases, respectively.
Another situation where Ka is constant is if a star is convective. As we will discuss
in a week or so, under some circumstances the material in a star can be subject
to an instability that causes it to move around in such a way as to enforce that
the entropy is constant. In a region undergoing convection, Ka does not vary
from shell to shell, and a polytropic equation of state is applicable. Significant
fractions of the mass in the Sun and similar stars are subject to convection, and
for those parts of stars, a polytropic equation of state applies well, and is a good
approximation for the star as a whole. In low mass stars where the gas is nonrelativistic and radiation pressure is not significant, γa = 5/3, so constant Ka
means that γP = 5/3 as well, and n = 1.5.
Because they apply in a broad range of situations, polytropic models turn out to
be extremely useful.
B. The Lane-Emden Equation
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Having motivated the choice P = KP ργP = KP ρ(n+1)/n for an equation of state,
we can proceed to substitute it into the equation of hydrostatic balance. Doing
so, after a little bit of simplification we find
(n + 1)KP 1 d
4πGn r2 dr
r2
dρ
ρ(n−1)/n dr
!
= −ρ.
As a second-order equation, this ODE requires two boundary conditions. At the
surface r = R, the pressure P (R) = 0, and since P ∝ ργP , one of our boundary
conditions could be ρ(R) = 0. However, usually one instead sets the density at
the center, ρ(0) = ρc , and then R is the radius at which ρ first goes to zero. For
the second boundary condition, recall that the original equation of hydrostatic
equilibrium read dP/dr = −ρGm/r2 . Unless the density becomes infinite in the
center, m must vary as m ∝ ρc r3 near the center of the star, so m/r2 must vary
as r. Thus dP/dr goes to zero at r = 0, and, for our polytropic equation of state,
dρ/dr must therefore approach zero in the center as well. This is our second
boundary condition: dρ/dr = 0 at r = 0.
Note that our equation now involves three constants: KP and n, which come from
the equation of state, and R, which sets the total radius of the star. Since these
are the only constants that appear (other than physical ones), these must fix the
solution. In other words, for a given polytrope with a given choice of KP , n, and
R, there is a single unique density profile ρ(r) which is in hydrostatic equilibrium.
In fact, it is even simpler than that, as becomes clear if we make a change of
variables. Let ρc be the density in the center of the star, and let us define the
new variable Θ by
ρ
= Θn .
ρc
Note that Θ is a dimensionless number, and that it runs from 1 at the center of
the star to 0 at the edge of the star. With this change of variable, we can re-write
the equation of hydrostatic balance as
"
#
(n + 1)KP
(n−1)/n
4πGρc
1 d
2 dΘ
r
r2 dr
dr
!
= −Θn .
The quantity in square brackets has units of length squared, and this suggests a
second change of variables. We let
"
2
α =
(n + 1)KP
#
(n−1)/n
4πGρc
and we then set
r
α
so that ξ is also a dimensionless number. With this substitution, the equation
becomes
!
1 d
2 dΘ
ξ
= −Θn .
ξ 2 dξ
dξ
ξ=
5
The two central boundary conditions are now Θ = 1 and dΘ/dξ = 0 at ξ = 0.
The equation we have just derived is called the Lane-Emden Equation. Clearly
the only constant in the equation is n.
Before moving on to solve the equation, it is worth mentioning the procedure used
to derive it, which is a very general and powerful one. This procedure is called
non-dimensionalization. The basic idea is to take an equation relating physical
quantities, like the hydrostatic balance equation we started with, and re-write all
the physical quantities as dimensionless numbers times their characteristic values.
For example, we re-wrote the density as ρ = ρc Θn . We re-wrote the length as
r = αξ. The advantage of this approach is that it allows us to factor out all the
dimensional quantities and leave behind only a pure mathematical equation, and
in the process we often discover that the underlying problem does not depend on
the dimensional quantities. In this case, it was far from obvious that the structure of a polytrope didn’t depend on R, ρc , KP , etc. After all, those quantities
appear in the hydrostatic balance equation. However, the non-dimensionalization
procedure shows that they just act as multipliers on an underlying solution whose
behavior depends only on n. Tricks like this come up all the time in the study of
differential equations, and are well worth remembering if you plan to think about
them at any point in the future.
With that aside out of the way, consider the Lane-Emden equation itself. To get a
sense of how it behaves, we can solve it for some chosen values of n. First consider
n = 0, corresponding to the limit γP → ∞. In this case the equation is
d
dΘ
ξ2
dξ
dξ
!
= −ξ 2 ,
which we can integrate to obtain
ξ2
ξ3
dΘ
= − + C,
dξ
3
where C is a constant of integration. Bringing the ξ 2 to the other side and
integrating again gives
ξ2 C
Θ = − − + D.
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ξ
Applying the boundary conditions that Θ = 1 and dΘ/dξ = 0 at ξ = 0, we
immediately see that we must choose C = 0 and D = 1; so the solution is
therefore
ξ2
Θ=1− .
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This √
is a function that decreases monotonically at ξ > 0, and reaches 0 at ξ =
ξ1 = 6. Analytic solutions also exist for n = 1 and n = 5.
Numerically integrating the equation for other values of n shows that, for n < 5,
this is generic behavior: Θ decreases monotonically with ξ and reaches zero at
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some finite value ξ1 . As we will see in a moment, it is particularly useful to know
ξ1 and −ξ12 (dΘ/dξ)ξ1 , and these can be obtained trivially from the numerical
solution. Some reference values are
ξ1
−ξ12 (dΘ/dξ)ξ1
3.14
3.14
3.65
2.71
4.35
2.41
5.36
2.19
6.90
2.02
n
1.0
1.5
2.0
2.5
3.0
The radius at which ξ reaches zero is clearly the radius of the star, so
R = αξ1 .
Similarly, given a solution Θ(ξ), we can also compute the mass of the star:
Z R
M =
4πr2 ρ dr
0
= 4πα3 ρc
3
Z ξ1
ξ 2 Θn dξ
0
= −4πα ρc
Z ξ1
0
= −4πα3 ρc ξ12
d
dΘ
ξ2
dξ
dξ
!
dΘ
.
dξ ξ1
!
dξ
In the third step we used the equation of hydrostatic balance to replace ξ 2 Θn .
From a polytropic model, we can derive a bunch of other useful numbers and
relationships. As one example, it is often convenient to know how centrally concentrated a star is, i.e. how much larger its central density is than its mean density.
We define this quantity as
Dn ≡
ρc
ρ
= ρc
4πR3
3M

dΘ
4π
=
ρc (αξ1 )3 −4πα3 ρc ξ12
3
dξ

= −
3
ξ1
dΘ
dξ
! −1

ξ1
! −1

ξ1
A second useful relationship is between mass and radius. We start by expressing
the central density ρc in terms of the other constants in the problem and our
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length scale α:
"
(n + 1)KP
ρc =
4πGα2
#n/(n−1)
.
Next we substitute this into the equation for the mass:
"
M = −4πα
3
(n + 1)KP
4πGα2
#n/(n−1)
ξ12
dΘ
dξ
!
.
ξ1
Finally, we substitute α = R/ξ1 . Making the substitution and re-arranging, we
arrive at
"
#n−1
!3−n
GM
R
[(n + 1)KP ]n
=
−ξ12 (dΘ/dξ)ξ1
ξ1
4πG
Thus mass and radius are related by M ∼ R(n−3)/(n−1) .
A third useful expression is for the central pressure. From the equation of state
we have Pc = KP ρ(n+1)/n
. We can then use the mass-radius relation to solve for
c
KP and then substitute it into the equation of state, which gives
GM
(4πG)1/n
Pc =
2
n+1
−ξ1 (dΘ/dξ)ξ1
"
#(n−1)/n
R
ξ1
!(3−n)/n
ρ(n+1)/n
.
c
Then, using the centrally concentrated measure, Dn , to elliminate R, we get
Pc = (4π)1/3 Bn GM 2/3 ρ4/3
c
where


2
Bn = − 
(n + 1) ξ1

! 2/3 −1
dθ  

dξ
ξ1
which is relatively independent of n. As we will see, all realistic models have
n = 1.5 to n = 3. Thus we expect a nearly universal relation among the central
pressure, central density and mass of stars.
C. The Chandrasekhar Mass and Relativistic Gasses
Consider what this analysis of polytropes implies for stars where the pressure is
dominated by electron degeneracy pressure. White dwarf stars are examples of
such stars. To see why, consider what their position on the HR diagram and their
masses tells us about them. Observations of white dwarfs in binary systems imply
that they have masses comparable to the mass of the Sun. On the other hand,
their extremely low luminosities, combined with surface temperatures that are
not very different from that of the Sun, implies that they must have very small
radii: r ∼ rE = 6 × 108 cm. The corresponding mean density is ρ ∼ 2 × 106 g
cm−3 . In contrast, we showed a few classes ago that electrons become degenerate
at a density above ρ/µe ≈ 750(T /107 K)3/2 g cm−3 . Thus, unless white dwarf
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interiors are hotter than ∼ 109 K, which they do not appear to be, the gas must
be degenerate.
As long as the gas is non-relativistic, this implies that its pressure is P = K10 (ρ/µe )5/3 ,
where K10 is a constant that depends only on fundamental constants. Thus a white
dwarf is a polytrope, and γa = γP = 5/3 implies that n = 3/2. For a polytrope
of index n, we have just shown that mass and radius are related by
R ∝ M (n−1)/(n−3) = M −1/3 .
Thus more massive white dwarfs have smaller radii, with the radius falling as
M −1/3 . It is important to realize that this statement is not just true of a particular
white dwarf, it is true of all white dwarfs. The constant of proportionality between
M and R depends only on KP and G, and for a degenerate gas KP = K10 /µ5/3
e .
For a given composition (fixed µe ), this is a universal constant of nature, since K10
depends only on fundamental constants. Thus all white dwarfs in the universe
follow a common mass-radius relation that is imposed by quantum physics.
However, this cannot continue forever, with ever more massive white dwarfs having smaller and smaller radii. The thing that breaks is that, as M increases and
R shrinks, eventually the electrons in the white dwarf are forced by the Pauli
exclusion principle to occupy higher and higher momenta. As a result, the gas
becomes relativistic. The mean density varies as
ρ∝
M
∝ M 2,
R3
and we showed a few weeks that a degenerate gas becomes relativistic when its
density exceeds
ρ
≈ 3 × 106 g cm−3 .
µe
We got a density of 2 × 106 g cm−3 for a mass of M and a radius of rE , so it
doesn’t take much of an increase in mass to push the star above the relativistic
limit.
For a relativistic electron gas, we have shown that P = K20 (ρ/µe )4/3 , where, as
with K10 , the quantity K20 depends only on fundamental constants. Since such a
gas has γa = 4/3 and fixed Ka , it can be described by a polytropic equation of
state with index n = 1/(γP − 1) = 3.
The mass-radius relation has an interesting behavior near n = 3. We showed that
it can be written
"
GM
2
−ξ1 (dΘ/dξ)ξ1
#n−1
R
ξ1
!3−n
=
[(n + 1)KP ]n
,
4πG
or equivalently
M =−
1
(n+1)/(n−1)
ξ
1/(n−1) 1
(4π)
9
dΘ
dξ
!
ξ1
[(n + 1)KP ]n/(n−1) (n−3)/(n−1)
R
.
Gn/(n−1)
Notice that, for n = 3, the R dependence disappears, and we find
4
dΘ
M = − √ ξ12
π
dξ
!
ξ1
KP
G
3/2
A relativistic electron gas has
KP =
K20
1/3
4/3
µe
=
3
π
hc
,
8(µe mH )4/3
and plugging this into the mass we have just calculated gives
3
=
32
M = Mch
1/2

1  2 dΘ
−ξ1
π
dξ
! 
hc

G
ξ1
!3/2
1
5.83
= 2 M .
2
(mH µe )
µe
Everything on the right-hand side except µe is a constant, which means that for
a highly relativistic electron gas, there is only a single possible mass which can
be in hydrostatic equilibrium. If we have a gas that is depleted of hydrogen, so
X = 0, then µe ≈ 2, and we have Mch = 1.46 M .
This quantity is called the Chandresekhar mass, after Subrahmanyan Chandresekhar, who first derived it. He did the calculation while on his first trip
out of India, to start graduate school at Cambridge... at age 20... which goes to
prove that the rest of us are idiots....
To understand the significance of this mass, consider what happens when a star
has a mass near it. If the mass is slightly smaller than Mch , the star can respond
by puffing out a little and adopting a larger radius. This reduces the density and
makes the gas slightly less relativistic, so that n decreases slightly and the star
can be in equilibrium.
On the other hand, if M > Mch , the gas is already relativistic, and there is no
adjustment possible. If a Chandresekhar mass white dwarf gets just a little bit
more massive, there simply is no equilibrium state that the star can possibly
reach. Instead, it is forced to collapse on a dynamical timescale. The result is
invariably a massive explosion. For a white dwarf composed mainly of carbon and
oxygen, the carbon and oxygen undergo rapid nuclear burning to elements near
the iron peak, and the result is essentially that a nuclear bomb goes off with a
solar mass worth of fuel. The resulting explosion, known as a type Ia supernova,
can easily outshine an entire galaxy.
D. Very Massive Stars
The Chandrasekhar limit is one way that a star can get into trouble if it is
supported by a relativistic gas. It is, however, not the only way. Massive stars
are also close to being n = 3 polytropes, but instead of being supported by a
relativistic gas of electrons, they are supported by a relativistic gas of photons –
i.e. by radiation pressure.
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Massive stars don’t explode while they are on the main sequence, but they still
suffer from instabilities associated with being close to the instability line of n = 3.
In massive stars, this instability tends to manifest as rapid mass loss and ejection
of gas. The mechanism is not fully understood today, but observations make it
clear that massive stars do suffer from instabilities.
Perhaps the most spectacular example is the star η Carinae, a massive star that is
reasonably close to Earth. In 1843, η Carinae (which is visible with the naked eye,
despite being > 2 kpc away), suddenly brightened, briefly becoming the secondbrightest star in the night sky. Subsequently it dimmed again, although it remains
naked-eye visible. In modern times, observations using the Hubble Telescope show
that η Carinae is surrounded by a nebula of gas, called the Homunculus Nebula,
which was presumably ejected in the 1843 eruption.
[Slide 1 – η Carinae and the Homunculus Nebula]
The physical mechanism behind the explosion is still not very well understood,
and it is an active area of research. Nonetheless, it seems clear that it is connected
with the fact that η Carinae is supported primarily by radiation pressure, which
puts it dangerously close to the n = 3 line of instability. We’ll discuss the nature
of this instability more in a week or so.
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