Download Lecture 16

Introduction to stellar structure
Paul Hickson
October 17, 2016
www.windows2universe.org
Hydrostatic equilibrium
Stars are gaseous spheres that exist in a state of hydrostatic
equilibrium. The gravitational force that pulls the gas inwards is
balanced by pressure (force per unit area), which opposes gravity.
The pressure results from thermal motion of the gas molecules and
also the from momentum of photons carrying energy outward from
the core of the star.
In equilibrium, the inward
gravitational force on the
material in any small region
must be balanced by the net
outward pressure force. Thus
gρ dA dr “ rP ´pP `dP qsdA
where ρprq is the density and
gprq is the gravitational
acceleration at radius r.
Hydrostatic equilibrium
Simpifying this, we get the equation of hydrostatic equilibrium,
dP
“ ´ρg.
dr
The gravitational acceleration, at radius r, is determined by the
mass M prq that is contained within this radius,
g“
GM prq
.
r2
Since ρ is generally a function of r, we must obtain M prq by
integrating outward from the centre, over shells of thickness dr
and surface area 4πr2 ,
żr
ż
2
M prq “
ρprqp4πr qdr “ 4π ρprqr2 dr.
0
Example: estimating central pressure
In order to accurately model the interior of a star, we need many
more equations, describing temperature, pressure, energy flow, etc.
Solving all these equations is not easy but can be done numerically.
However, we can easily get a rough estimate of the pressure at the
centre of a star by making some approximations in the equation of
hydrostatic equilibrium.
Let the pressure at the centre be P p0q. The pressure acting on the
outside of a star is zero, since the star is in empty space, therefore
P pRq “ 0.
Let’s now replace dP {dr with the average pressure gradient,
dP
P pRq ´ P p0q
P p0q
»
“´
.
dr
R
R
Example: estimating central pressure
Next, we approximate the actual density inside the star (which
varies with distance r from the centre) with the average density,
3M
M
“
.
ρ»
V
4πR3
Finally, we approximate the gravitational acceleration within the
star (which also depends on r) with a constant, the gravitational
acceleration at the surface,
GM
g» 2 .
R
Putting these approximations into the equation of hydrostatic
equilibrium, we get
P p0q
GM 3M
´
»´ 2
R
R 4πR3
Therefore,
3GM 2
.
P p0q »
4πR4
Central pressure of the Sun
Putting in numbers for the Sun,
3 ˆ 6.6734 ˆ 10´11 ˆ p1.9886 ˆ 1030 q2
“ 2.68 ˆ 1014 Pa
4πp6.955 ˆ 108 q4
“ 2.65 ˆ 109 atm.
P p0q »
More accurate
calculations,
using all the
equations of
stellar structure,
give a value
about 100 times
higher.
ifa.hawaii.edu
Equation of state
So far we have one equation with two unknowns, P prq and ρprq.
To find either of these, we need a second equation relating these
quantities. That is provided by the equation of state.
The form of the equation of state depends on the nature of the
matter. In the Sun, and many other stars, the gas pressure is given
by the ideal gas law
ρkT
P “
µmH
here T is the temperature, mH is the mass of the hydrogen atom,
and k is Boltzmann’s constant.
µ is the mean molecular weight which is defined as the mean mass
of a gas particle, in units of mH .
It should be evident from the definition of µ that the denominator
in the ideal gas law is just the mean mass of a gas particle.
Mean molecular weight
Let’s estimate the value of µ for the Sun. The reciprocal of µ is
the mean number of particles per hydrogen mass.
Let X be the fraction of hydrogen, by mass, Y be the fraction of
helium, and Z be the fraction of everything else.
Since this covers everything, the fractions must add up to one,
X ` Y ` Z “ 1.
Now, the Sun is very hot, so the gas is ionized. From one hydrogen
atom (mass mH ), we get two particles (a proton and an electron).
So two particles per mH .
From helium (mass 4mH ) we get three particles, a helium nucleus
and two electrons.
From heavier elements, we get about 1{2 particle per mH , because
the the nucleus contains about as many neutrons as protons, so for
every 2mH we get one electron.
Mean molecular weight
Putting this all together, we have
1
3
1
“ 2X ` Y ` Z.
µ
4
2
This simplifies to
µ“
4
6X ` Y ` 2
For the Sun, one finds that X » 0.747, Y » 0.236 and Z » 0.017.
Therefore,
µd » 0.6.
Central temperature of the Sun
We can use our previous result for the central pressure, with the
ideal gas law, to estimate the central temperature of the Sun.
Rearranging the ideal gas law, we find
µmH P
T “
kρ
Lets evaluate this at the centre of the Sun. For P we use our
estimate of P p0q and for ρ we again use the mean density,
µmH 4πR3 3GM 2
,
k 3M 4πR4
µmH GM
“
.
kR
T p0q »
Putting in numbers for the Sun, we get Tc » 1.4 ˆ 107 K.
This is quite close to the prediction of the current solar model,
which is 1.57 ˆ 107 K.
Radiation pressure
In high-mass stars, the flux of photons propagating through the
star from the nuclear reactions in the core is very intense and one
must include radiation pressure in the equation of state.
Photons carry momentum p “ h{λ and exert a pressure given by
Prad “
4σSB T 4
.
3c
The total pressure is therefore
P “ Pgas ` Prad “
ρkT
4σSB T 4
.
`
µmH
3c
Energy transport
Energy, generated in the core, is transported to the surface by a
combination of radiation and convection, depending on the mass
of the star.
www.sun.org
The equations of stellar structure
In addition to hydrostatic equilibrium and an equation of state, one
needs several additional equations.
These describe the rate of energy generation by nuclear reactions,
and transport of energy by convection.
Those equations will not be discussed in this course. It is sufficient
to know that they exist, and allow a complete solution to the
structure of stars.
Normally the equations of stellar structure are expressed as a set of
coupled differential equations. They are integrated numerically
with specified boundary conditions.
The solutions are called stellar models.