Transcript Lecture 9

Lecture 9
Stellar structure equations
The Eddington Approximation
dFrad
 0  4  I  S 
d v
dPrad 1
 Frad
d v
c
3 4 
2
I 
Te  v  
4 
3
Approximation #5: Local thermodynamic equilibrium
In this case the source function is equal to the blackbody function.
T 4
I B

3 
2
 T 4  Te4  v  
4 
3
The Eddington Approximation
• Note that T=Te when v=2/3
 Thus the photons we see (that
give us the luminosity we use to
define the effective
temperature) originate at an
optical depth of 2/3, not 0.
3 4
2
T  Te  v  
4 
3
4
Te
Limb darkening
• The solar disk is
darker at the edge
(limb) than at the
centre
• The light rays that
we see from the edge
of the Sun must
originate from higher
in the atmosphere
(since otherwise they
would have to travel
through a greater
optical depth to
reach us).
Limb darkening
 4
Te 2  3 cos  
4
I   
2  3 cos 

I  (  0)
5
I 
We can compare this with
observations, and the
agreement is pretty good.
This doesn’t prove our
numerous assumptions are
correct, but does show that
they do not produce a result
that is in strong conflict with
the data.
Break
Hydrostatic equilibrium
The force of gravity is always directed toward the
centre of the star. Why does it not collapse?
 The opposing force is the gas pressure. As the star
collapses, the pressure increases, pushing the gas back out.
• How must pressure vary
with depth to remain in
equilibrium?
Hydrostatic equilibrium
Consider a small cylinder
at distance r from the
centre of a spherical
star.
Pressure acts on both
the top and bottom of
the cylinder.
By symmetry the
pressure on the sides
cancels out
FP,t
A
dm dr
FP,b
Hydrostatic equilibrium
GM r  dP
d 2r


 2
2
r
dr
dt
If we now assume the gas is static, the acceleration
must be zero. This gives us the equation of hydrostatic
equilibrium (HSE).
dP
GM r 

2
dr
r
• It is the pressure gradient that supports the star
against gravity
• The derivative is always negative. Pressure must get
stronger toward the centre
Mass Conservation
The second fundamental equation of stellar structure is
a simple one relating the enclosed mass to the density.
Consider a shell of mass dMr and thickness dr, in a spherically symmetric
star
dM r  dV
  4r 2 dr
Rearranging we get the equation of mass conservation
dM r
 4r 2 
dr
Example
Make a crude estimate of the central solar pressure,
assuming the density is constant.
3M
-3


1410
kg
m
4R 3
This is a big underestimate because the density increases
strongly toward the centre.
The accepted value is
2.5 1016 Pa  2.5 1011 atm
Pressure equation of state
We now need to assume something about the source of pressure in
the star.
 We require an equation of state to relate the pressure to
macroscopic properties of the gas (i.e. temperature and density)
Consider the ideal gas approximation:
 Gas is composed of point particles, each of mass m, that interact
only through perfectly elastic collisions
P  nkT
Note the particle mass does not enter into this equation
 The momentum of each collision depends on mass, but lighter particles are moving
faster in a way that exactly cancels out
Thus, tiny electrons contribute as much to the pressure as massive protons
Mean molecular weight
We want to relate the particle number density to the
mass density of the gas.
The two quantities are related by the average particle
mass:
Define the mean molecular
weight:
m
mH
n

m
i.e. this is the
average mass of a
free particle, in
units of the mass of
hydrogen
So we can express the ideal gas law in terms of density:

kT
P
mH
Mean molecular weight
The mean molecular weight is an important quantity,
because the pressure support against gravity depends on
the number of free particles
m

mH
 Sudden changes in the ionization state or chemical
composition of the star can lead to sudden changes in the
pressure.
In general, the value of  requires solving the Saha equation to
determine the ionization state of every atom.
 We can derive two useful expressions for the cases of fully neutral or
fully ionized gasses:
Define: X,Y,Z are the mass fractions of H, He and metals, respectively.
Neutral:
For solar
abundances,
1
1
1
X Y
Z
n
4
A n
1
A

n
1
15.5
Ionized:
1
3
1
 2X  Y  Z
i
4
2
Example
By how much does the pressure increase following
complete ionization, for a neutral gas with the
following composition (typical of young stars):
X  0.70
Y  0.28
Z  0.02
Radiation pressure
We earlier derived an expression for the
radiation pressure of a blackbody:
Prad
The equation of state then becomes P 
4 4

T
3c
kT 4 4

T
mH 3c
In a standard solar model, the central density and
temperature are
Tc  2.0 107 K
 c  1.5 105 kg m 3
Calculate the gas pressure and radiation pressure. Assume
complete ionization, so =0.62.