Lecture 9Lecture 9

Stellar structure equations

The Eddington ApproximationThe Eddington Approximation

3

2

4

3 4veTI

Approximation #5: Local thermodynamic equilibriumIn this case the source function is equal to the blackbody function.

SId

dF

v

rad

40

3

2

4

3 44

4

veTT

TBI

radv

rad Fcd

dP 1

The Eddington ApproximationThe Eddington Approximation

3

2

4

3 44veTT

• 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.

Te

Limb darkeningLimb 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 darkeningLimb darkening

5

cos32

)0(

cos324

4

I

I

TI e

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.

BreakBreak

Hydrostatic equilibriumHydrostatic 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 equilibriumHydrostatic 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

drA

dm

FP,b

FP,t

Hydrostatic equilibriumHydrostatic equilibrium

If we now assume the gas is static, the acceleration must be zero. This gives us the equation of hydrostatic equilibrium (HSE).

2

2

2 dt

rd

dr

dP

r

GM r

2r

GM

dr

dP 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 ConservationMass Conservation

24 rdr

dM r

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

drr

dVdM r

24

Rearranging we get the equation of mass conservation

ExampleExample

Make a crude estimate of the central solar pressure, assuming the density is constant.

33

m kg 14104

3 -

R

M

This is a big underestimate because the density increases strongly toward the centre.

The accepted value is atmPa 1116 105.2105.2

Pressure equation of statePressure 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

nkTP 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 weightMean molecular weight

We want to relate the particle number density to the mass density of the gas.

mn

Hm

m

Hm

kTP

The two quantities are related by the average particle mass:

Define the mean molecular weight: 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:

Mean molecular weightMean molecular weight

The mean molecular weight is an important quantity, because the pressure support against gravity depends on the number of free particles

Sudden changes in the ionization state or chemical composition of the star can lead to sudden changes in the pressure.

Hm

m

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:

Neutral:Ionized:

Define: X,Y,Z are the mass fractions of H, He and metals, respectively.

ZA

YXnn

1

4

11

5.15

11

nAFor solar abundances,

ZYXi 2

1

4

32

1

ExampleExample

By how much does the pressure increase following complete ionization, for a neutral gas with the following composition (typical of young stars):

02.0

28.0

70.0

Z

Y

X

Radiation pressureRadiation pressure

We earlier derived an expression for the radiation pressure of a blackbody:

4

3

4Tc

Prad

The equation of state then becomes 4

3

4Tcm

kTP

H

In a standard solar model, the central density and temperature are

35

7

m kg105.1

100.2

c

c KT

Calculate the gas pressure and radiation pressure. Assume complete ionization, so =0.62.