10. The Interiors of Stars10. The Interiors of StarsGoalsGoals:1. Develop the basic equations describing

equilibrium conditions applying in stellar interiors.

2. Estimate the physical conditions that must exist at the centre of a typical star, the Sun.

3. Outline potential sources of energy generation in stars and investigate the basics of nuclear reactions as a means of providing a self-sustained energy source.

4. Develop the equations of energy transport in stars.

Fact or Fiction?Consider an inhabited planet completely enshrouded by clouds, upon which astronomy is not a scientific discipline since it is not possible to see into space from the planet’s surface. Yet, one can construct abstract mathematical models of massive spheres of hot gas in equilibrium and deduce their properties from what is known about the physics of matter. Would it come as a surprise to scientists on such a planet if the constant canopy of surrounding clouds one day parted and stars came into view?

Models of StarsThe parameters used for studying and modeling stellar interiors include:

r = radial distance from the centre of the star

M(r) = mass interior to r

T(r) = temperature at r

P(r) = pressure at r

L(r) = luminosity at r

ε(r) = energy generation at r

κ(r) = opacity at r

ρ(r) = density at r

In modern models mass M is used as the dependent variable rather than radial distance r, but it is more informative to initiate the study of stellar interiors using the geometrical variable r.

At the “natural” boundaries of the star the corresponding values are:

At the centre: At the surface:

r = 0 r = R

M(r) = 0 M(r) = M*

T(r) = Tc T(r) = 0 (or Teff)

P(r) = Pc P(r) = 0

L(r) = Lc L(r) = L*

ρ(r) = ρc ρ(r) = 0

Rotation and magnetic fields are usually ignored in most models (i.e. spherical symmetry is imposed), as well as any temporal changes (i.e. radial pulsation).

Let us examine the equations of stellar structure.

Hydrostatic EquilibriumFor balance at any point in the interior of a star, the weight of a block of matter of unit cross-sectional area and thickness dr must be balanced by the buoyancy force of the gas pressure, i.e.:

Mass of block = density × volume = ρ drWeight of the block = mass × local gravity = ρg drBut local gravity, g = GM(r)/r2

Buoyant Force = pressure difference (top – bottom) = –dPSo:

2

2

)(

or

)(

r

rMG

dr

dP

r

drrMGdP

Example:Obtain a crude estimate for the pressure at the centre of the Sun. Assume 1 M = 1.9891 × 1033 gm, 1 R = 6.9598 × 1010 cm and:

Solution (instructor):Convert the equation of hydrostatic equilibrium into a difference equation, evaluate it at r = ½R, assume that the mean density and M = ½M* applies there as well. Then:

Set dP = Ps – Pc = 0 – Pc = –Pc and dr = rs – rc = R* – 0 = R*

3310

34

33

3Sun3

4gm/cm4086.1

cm109598.6

gm109891.1SunSun

R

M

5*

2*

3*3

42*

**2

*

*21

2 2

2

2

)(

R

GM

RR

MGM

R

MG

r

rGM

dr

dP

4*

2*

5*

2*

* 2and

2 R

GMP

R

GM

R

Pc

c

So, for the Sun:

Pc ≈ GM2/ R

4

= 6.6726 × 10–8 × (1.989 × 1033)2/ (6.9598 × 1010)4 = 5.63 × 1015 dynes/cm2

The textbook method of solution gives a value of:

Pc = ~2.7 × 1015 dynes/cm2

A more rigorous solution involving integration with the standard solar model gives a more reliable estimate of

Pc = 2.5 × 1017 dynes/cm2

two orders of magnitude larger!

Since 1 atmosphere = 1.013 × 106 dynes/cm2, the pressure at the centre of the Sun is equivalent to 2.5 × 1011 atmospheres!

Conservation of MassThe mass of a star must increase uniformly from the interior to the surface, there cannot be any holes! For each element of thickness dr the volume must increase by the mass of the shell encompassed, i.e. by the spherical area × thickness dr. The resulting equation is:

2

2

4)(

or

4)(

rdr

rdM

drrrdM

Ideal Gas LawThe ideal gas law is introduced in PHYS1211:

PV = NkT,

where P is the gas pressure, V its volume, N the number of particles in the volume, T is the temperature (on the absolute scale), and k = 1.3807 × 10–16 ergs/K is the Boltzmann constant.

It is also expressed using the gas constant R as:

PV = nRT, where n is the number of moles of gas.

The relationship can be derived from first principles through the kinetic theory of gases, which specifies that:

(i) gas consists of small particles (molecules, atoms) that are smaller than the distances separating them,(ii) particles are in constant motion and make perfectly elastic collisions with container walls,(iii) the motions of particles are random, i.e. ⅓ are moving in any specific direction.

Consider a box of N particles of mass m, where ⅓N are moving in the x-direction. The averageforce on the right-hand wall(shaded) is given by the rateof change of momentum:

Favg = Δ(mv)/Δt

A particle takes time Δt tocomplete a trip from one wallof the box back to the samewall, i.e. Δt = 2l/v, where l is the dimension of the box.

Δ(mv) = mv(before) – mv(after) = mv – (–mv) = 2mv

The resulting pressure (force/unit area) on the shaded wall is given by the total force exerted by all particles in the box on the end wall, i.e.:

3

2

31

2231

2avg

31

22

3 l

mvN

l

vmv

l

N

tl

mvN

l

FNP

But l 3 = V, the volume of the box, so:

Since Nm = ρV (density × volume):

But the average kinetic energy of a particle ½mv2 = 3⁄2 kT, so:

The average kinetic energy of a gas molecule at room temperature is:

231

2

31 or NmvPV

V

mvNP

2312

31 or vPVvPV

V

Nm

m

kTP

NkTkTNNmvPV

since,or

3312

31

eVeV0388.0K300eV/erg106022.1

K/erg103807.1KE 25

112

16

23

23

kT

Pressure Equation of StateThe ideal gas law is, once again:

PV = NkT, where

P = gas pressureV = volume of the gasN = number of particles in the gasT = temperature (K)k = 1.3807 × 10–16 ergs/K is the Boltzmann constant.

The number density for the gas can be written as n = N/V, so:

P = nkT,

and since the number density can also be written as n = ρ/μmH, where ρ is the actual density (gm/cm3), mH is the mass of a hydrogen atom, and μ is the mean molecular weight. So the gas pressure can be written as:

Hg m

kTkTnP

Mean Molecular WeightAs the name implies, the mean molecular weight is the average mass of a gas particle, but in units of the mass of a hydrogen atom, i.e.:

where is the average mass of a gas particle.

Consider possible examples:

Hydrogen: neutral,

ionized,

molecular,

Hm

m

1H

H

H m

m

m

m

212

1

H

H

H m

m

m

m

22

H

H

H m

m

m

m

m

Helium: neutral,

ionized,

Heavy element:neutral,

ionized,

where Aj is the atomic number.

In stellar interiors the value of μ for a fully-ionized gas is desired. How does one take into consideration the contributions from all elements?

44

H

H

H m

m

m

m

343

4

H

H

H m

m

m

m

jH

Hj

H

Am

mA

m

m2

2

21

2

Hj

Hj

H mA

mA

m

m

Let X = fractional abundance by mass of hydrogen,Y = fractional abundance by mass of helium, andZ = fractional abundance by mass of all heavy elements,

where X + Y + Z ≡ 1, by definition.

In stellar interiors the value of μ for a fully-ionized gas is desired. How does one take into consideration the contributions from all elements?

In a cubic centimeter of gas of density ρ there is Xρ of hydrogen, Yρ of helium, and Zρ of heavy elements by mass. Each of the elements contributes different numbers of electrons to the mix, under the assumption of complete ionization: 1 for hydrogen, 2 for helium, and 3, 4, 5… for the heavy elements. The number of particles per cubic centimeter is:

Xρ/mH × 2 = 2Xρ/mH for hydrogen

Yρ/4mH × 3 = 3Yρ/4mH for helium, and

Zρ/2AjmH × (Aj + 1) ≈ Zρ/2mH for the heavy elements.

The total number of particles per cubic centimeter is therefore given by:

So:

or: for fully ionized gas,

and: for neutral gas,

or, since X + Y + Z ≡ 1:

When Z is negligible:

Hm

ZYX

21

432particlesofNo.

ZYXm

ZYXmmn

H

HH 21

43

21

43 2

1

2

1

ZYXi21

432

1

ZYXjA

n1

41

1

26

4

YXi

35

4

Xi

Example:What is the mean molecular weight for gas in the Sun, where X = 0.75, Y = 0.23, and Z = 0.02?

Solution (instructor):Consider the case for gas that is fully ionized, which gives:

i.e., slightly larger than ½, the value for a pure hydrogen gas.

5944.06825.1

1

0100.01725.05000.1

1

02.023.075.02

1

2

1

21

43

21

43

ZYXi

Example:Obtain an estimate for the temperature at the centre of the Sun. Assume, for simplicity, μi = 0.5944, i.e. fully-ionized gas.

Solution (instructor):We can use the perfect gas law if we know the density at the centre of the Sun. Approximate the value using estimates for conditions at r = ½R (as before), where we assume that ρ = . Assume also that P = ½Pc at that point, that k ≈ 4⁄3 × 10–16 erg/K, mH ≈ 3⁄2 × 10–24 gm, and μi = 0.5944. Then:

K102.1

10210

10710610

5944.0

so

73316

34

310152423

Sun

3Sun3

4c2

1

Mk

RPm

k

PmT

m

TkP

HH

H

If the average temperature = ½Tc , then Tc = 2.4 × 107 K, fairly close to the actual value of 1.6 × 107 K.

The central density is found from:

whereas the actual value is ~82 gm/cm3.

Since the mass of a hydrogen atom is 1.6726 × 10–24 gm, a density of 1.8 gm/cm3 corresponds to a number density of n = ρ/μmH = 1.81 × 1024 /cm3. The volume occupied by a particle is 4⁄3πr3 = 1/n, so the average particle radius is r = 5.1 × 10–9 cm = 0.51 Å, compared with the Bohr radius of 0.523 Å.

With an actual density for the Sun of 82 gm/cm3, the average particle radius is r = 1.4 × 10–9 cm = 0.14 Å, compared with the Bohr radius of 0.523 Å.

Such highly compressed matter cannot exist as bound atoms, and is referred to as pressure-ionized gas.

3716

1524

gm/cm8.1104.2103607.1

106106726.15944.0

c

cHc kT

Pm

Example (from Mechanics):Calculate the gravitational self-energy (energy of assembly piece-wise from infinity) of a uniform sphere of mass M and radius R.

Solution:Think of assembling the sphere a shell at a time (r = 0 to r = R). For a shell of radius r the incremental potential energy is dV = φdm, where dm is the mass of the shell and φ is the gravitational potential for the mass already assembled, which is a sphere. Since the mass being assembled forms a spherical shell, we have:

So the potential self-energy of the mass is:

3

3

334

33

3

22

334

2

3

4

3

4where,

and3

44

R

rM

R

Mrrm

r

Gm

R

drMrdrr

R

Mdrrdm

dVV

where:

So the potential self-energy of the sphere is:

R

GMR

R

GM

r

R

GMdrr

R

GMV

R R

25

6

2

0 0

5

6

24

6

2

5

3

5

3

5

33

drR

rGM

drR

M

R

GMr

R

drMrr

r

G

r

dmmGdmdV

6

42

3343

4

3

23

3

4

3

3

4

Stellar Energy SourcesAs noted the gravitational potential energy required to contract a star to its present size is given by:

But, of the potential energy lost by a star, according to the Virial Theorem (ASTR 2100), one half is transformed into an increase in the kinetic energy of the gas (heat) and the remainder is radiated into space. The radiation lost by a star upon contraction to the main sequence is therefore given by:

For the Sun, at its present mass (1.9891 1033 gm) and radius (6.9598 1010 cm), the amount of energy radiated through contraction is:

R

GMV

2

5

3

R

GME

2

10

3

ergs10138.1

109598.6

109891.110672.6

10

3 4810

2338

E

The present luminosity of the Sun is L = 3.851 1033 ergs/s, so if it had been shining at the same luminosity for the entire duration of its contraction (clearly erroneous) then the time scale for contraction is given by the following:

where tKH is the Kelvin-Helmholtz time scale (more properly the Helmholtz-Kelvin time scale, although it originated with the Scottish physicist John James Waterston, 1811-1883, whose papers on the subject were rejected by the Royal Society of London!). Here:

or ~107 years. The actual value should be smaller because the Sun’s luminosity was greater during the contraction phase, but the main point is that the estimate is much shorter than the estimated age of the solar system of ~4.6 109 years.

KHSunor

nContractioforScaleTimeLuminosityPresentLostEnergy

tLE

yr109.4s/yr103.1557

s10955.2

ergs/s103.851

ergs10138.1Sun 6

7

14

33

48

SunKH

L

Et

Nuclear Energy SourcesConsider the rest masses of the fundamental nuclear particles:

Proton: 1.672623 × 10–24 gmNeutron: 1.674929 × 10–24 gmElectron: 9.109390 × 10–28 gm

Atomic mass unit, 1 u = 1.660540 × 10–24 gm = 931.49432 MeV,for E = mc2.

The original nucleon symbolism was:

where A = mass number = number of nucleonsZ = number of protons (usually omitted)X = chemical symbol of the element as specified by Z.

i.e., is redundant, since indicates the same thing.

XAZ

H11 H1

Typical masses:

1H = 1.007825 u = 938.78326 MeV2H = 2.014102 u = 1876.12457 MeV4He = 4.002603 u = 3728.40196 MeV5Li = 5.0125 u = 4669.115279 MeV8Be = 8.005305 u = 7456.89614 MeV

The major reaction in astronomy converts 4 hydrogen nuclei (protons) into a helium nucleus (4He).

But 4 1H = 1.007825 u × 4 = 4.031280 uand 1 4He = 4.002603 u = 4.002603 uDifference = 0.028677 u = 0.0071 of 4 1H

The energy released = mc2 = 0.028677 u × 1.660540 × 10–24 gm c2

= 26.71 MeV.

The lifetime of a star via nuclear reactions depends upon how much of its hydrogen content is converted to energy via nuclear reactions.

For the Sun we can estimate:

At L = 3.851 × 1033 ergs/s,

Also, if tnuclear = Enuclear/L* ≈ XM*c2/M*4

then tnuclear = Xc2/M*3 ≈ 1010 yrs/(M*/M)3 .

1 M, tnuclear = 1010 years 2 M, tnuclear = 109 years (A-star)4.6 M, tnuclear = 108 years 10 M, tnuclear = 107 years (B-star)21.5 M, tnuclear = 106 years (O-star)0.5 M, tnuclear = 1011 years > 1/H0 (estimated age of the universe)

The lifetime of the Sun and stars via nuclear reactions is consistent with the nuclear ages of meteorites, as well as the oldest rocks on the Earth and the Moon.

ergs103.10071.010.0 512Sunnuclear cME

yr10s/yr103.1557

s1038.3

ergs/s103.851

ergs103.1Sun 10

7

17

33

51

Sun

nuclearnuclear

L

Et

Properties of nuclear particles:Particle Baryon Lepton Spin Chargeproton ±½ћ +1neutron ±½ћ 0electron ±½ћ –1positron ±½ћ +1neutrino ±½ћ 0muon ±½ћ ±1Other baryons, the hyperons:xi (Ξ) ±½ћ 0, sigma (Σ) ±½ћ 0, ±1lamda (Λ) ±½ћ 0Middle family, the mesons:κ-mesons 0 or ±1ћ 0, ±1π-mesons 0 or ±1ћ 0, ±1photons 0 0

Nuclear reactions conserve: (i) number of nucleons, (ii) number of leptons, (iii) electronic/nuclear charge, (iv) particle spin.

Example:What is the missing particle in the following reaction?37Cl + νe ↔ 37Ar + ?The collision of an electron neutrino with a chlorine-37 nucleus produces an argon-37 nucleus plus a to-be-identified particle.

Solution:37Cl is element 17 with 17 protons, 20 neutrons. 37Ar is element 18 with 18 protons, 19 neutrons.Nucleons in = 17 + 20 = 37. Nucleons out = 18 + 19 = 37. Leptons in = 1. Leptons out = 0, so missing particle is a lepton. Charge in = +17 (17 protons). Charge out = +18 (18 protons), so missing particle has a charge of –1. Spin in = 19ћ (17 protons, 20 neutrons, 1 neutrino). Spin out = 18½ћ (18 protons, 19 neutrons), so missing particle has spin ½ћ. The only lepton with spin ½ћ and charge of –1 is the electron.

i.e., 37Cl + νe ↔ 37Ar + e–

The proton-proton reaction:The first step in the proton-proton cycle is stymied by the fact that the colliding particles are small and both positively charged. Although the nuclear strong force takes over at small separations of the particles, at larger distances they are blocked from interacting by mutual Coulomb repulsion.

How is the Coulomb barrier overcome? Consider the potential energy of a Coulomb barrier:

From statistical mechanics we know that kinetic energy and thermal energy are related through the reduced mass of a particle:

so:

which is much higher than the Sun’s central temperature.

But, by the Heisenberg uncertainty principle, the uncertainty in a particle’s momentum, Δpx, is related to the uncertainty in its position, Δx, via:

ΔpxΔx ≥ ½ћ = h/4π

r

eZZ 221EnergyPotential

classical232

21 kTvm

K10114.1

cm101erg/K10381.13

esu10803.4112

3

2 101316

210221

classical

rk

eZZT

Reaction Rates:The rates of individual nuclear reactions depend upon a variety of factors:1. Atomic nuclei must have sufficient energy to penetrate the Coulomb barrier of the target nucleus, only nuclei with specific energies specified by the high velocity “tail” of the Maxwell-Boltzmann distribution can react with the target nuclei.2. The cross-section for the reaction, σ(E), must be substantial.

The restriction from (1) implies that energetic nuclei capable of reacting with the target nuclei are described by the MB distribution, i.e.,

where KE = ½μmv2 is the kinetic energy of the particle. The restriction from (2) produces:

dEeETk

ndEn Tk

E

E

2

12

3

21

12

area/timeparticles/incidentofnumber

menucleus/tireactions/ofnumberE

Denote dNE as the number of particles of velocity vE = (2E/μm)½ that can strike a target nucleus in time dt, i.e.:

So:

For nx targets/unit volume, the total reaction rate per unit volume per unit time is:

.and,,particlesincidentof

fluxtotaltheoffractionsome.,.,But

.energyofparticlesincidentofnumbertheiswhere

,

00

dEnndEnn

eidEnn

ndEn

En

dtdEnvEdn

E

E

E

E

iiE

Ei

i

i

iEE

dEnn

nvE

dt

dtdEnvE

dt

dNE

iE

iEE E

intervaltime

nucleusreactions/#

0

, dEn

nvEnnr EEixxi

But we still need to evaluate σ(E).

The radius of an atomic nucleus can be estimated as r ~ λ (= h/p), the de Broglie wavelength.

The size of the Coulomb barrier VC is also important, so:

m

m

pvKE

Ep

hE

2since,

1~~

22

21

2

22

EVCeE22~

so,KEand

22

2But

221

221

221

2

221

v

hv

eZZ

pvph

eZZ

v

reZZ

E

V

m

m

C

Presumably the function describes the main type of variability, although an additional term that is a slowly varying function of E cannot be excluded. The reaction rate therefore becomes:

Where the MB relation has been used as a substitute for nE. The resulting functional dependence of the reaction rate resembles a Gaussian, and is referred to as the Gamov peak.

E

eEei

h

eZZ

h

eZZ

E

V

bE

mmC

21

21

21

21

21

~.,.

2

KE2

2 221

221

0

,

21

21

23

2dEeeES

nn

kTr kTEEb

m

ixxi

In actual cases there may be resonant peaks superposed because of resonances with excited energy levels in the nucleus (analogous to excited energy levels for atoms).

21

Ebe

Examples of potential resonant cross-sections in the reaction rate functional dependence.

Electrons can partially shield the positive charges of nuclei, resulting in lower effective Coulomb barriers to reactions, i.e.:

where Ves(r) < 0, is the contribution from electron screening.

When electron screening is ignored, the integration results in a function that can be approximated as:

where r0 is a constant,Xi is the mass fraction of impacting particles for the reaction,Xx is the mass fraction of target particles for the reaction, andα' and β are exponents established by the integration using a power-law expansion for the reaction rate equations, which have messy integrals.

rVr

eZZV es

221

eff

TXXrr xixi'

0,

The energy released by nuclear reactions per gram of stellar material is given by:

where α = α' – 1. The units are ergs/gm/s.

Energy generated through nuclear reactions is responsible for a star’s luminosity, through the equation of continuity:

where ε = εnuclear + εgravity, where the latter term is not always negligible.

TXX

r

xixi

ixxi

0,

0, or

24 rdr

dL

Nuclear Reaction Chains:All must conserve momentum, energy, spin, charge, etc.:The Proton-Proton Reaction.

PPI: (69%)

½ = 7.9 × 109 yr

½ = 4.4 × 10–8 yr

½ = 2.4 × 105 yr

PPII: (30.85%)

½ = 9.7 × 105 yr

½ = 3.9 × 10–1 yr

½ = 1.8 × 10–5 yr

e eHHH 211

HeHH 312

H2HeHeHe 1433

BeHeHe 743

e LieBe 77

HeHeHLi 4417

PPIII: (0.15%)

½ = 9.7 × 105 yr

½ = 6.6 × 101 yr

instantaneous

½ = 3.0 × 10–8 yr

Net Reaction:

26.73 MeV of energy

Energies of Neutrino Products:

From 1H: Maximum Energy = 0.42 MeV

From 7Be: Maximum Energy = 0.86 MeV

From 8B: Maximum Energy = 14.0 MeV

BeHeHe 743

*BHBe 817

e eBe*B 88

HeHeBe 448

22e2HeH4 41 e

The nuclear energy generation rate for the process can be written as:

where T6 = units of temperature in 106 K,fPP = fPP(X, Y, ρ, T) ≈ 1 is the electron screening factor,φPP = φPP(X, Y, T) ≈ 1 is a correction factor to account for the various branches of the PP chain, andcPP ≈ 1 is a correction factor for higher order terms.

In simple form the relationship is written as:

for temperatures near 1.5 107 K. In other words, the energy generation rate for the proton-proton chain varies as the local density and the temperature to the fourth power, ε ~ ρT4.

31

632 80.33

6PPPPPP26

PP 1038.2

T

eTcfX

46PPPPPP

2PP,0PP TcfX

The CNO Bi-Cycle.

½ = 1.3 × 107 yr

½ = 2.8 × 10–5 yr

½ = 2.7 × 106 yr

½ = 3.2 × 108 yr

½ = 5.6 × 10–6 yr

½ = 1.1 × 105 yr

99.96% of the time, or 0.04% of the time

½ = 3.0 × 10–6 yr

NHC 13112

e eCN 1313

NHC 14113

OHN 14114

e eNO 1515

HeCHN 412115

OHN 16115

FHO 17116

e eOF 1717

HeNHO 414117

In the CNO cycle, discovered by Hans Bethe in 1938, isotopes of carbon (C), nitrogen (N), and oxygen (O) act as catalysts for the reaction. The element fluorine (F) is also involved. Although the elements are not destroyed in the reactions, they proceed at such different rates that the isotopes of nitrogen (N) increase in abundance while those of carbon (C) and oxygen (O) decrease.

The proton-proton chain dominates for cool stars like the Sun, the CNO bi-cycle for stars hotter than the Sun, which have higher core temperatures.

The Triple-Alpha Process.

½ = 1.3 × 107 yr

½ < 8.2 × 10–24 yr

Summary:

Reaction ρ-dependence X-dependence T-dependence

PP Chain ρ1 X2 T64

CNO Bi-Cycle ρ1 XXCNO T619.9

Triple-Alpha ρ2 Y3 T841.0

Note the higher temperature dependence of the CNO cycle. It is dependent upon the CNO abundances, but dominates over the PP chain for stars somewhat more massive than the Sun where the central temperatures are higher.

*BeHeHe 844

C*BeHe 1284

The triple-alpha reaction involves an unstable isotope in the production of 8Be. The reaction proceeds because, under the high density conditions at the centers of evolved stars, a third alpha particle (4He nucleus) can collide with 8Be before it has time to decay. The resulting production of energy has an extremely strong temperature dependence.

The Helium Flash.When stellar core material is electron degenerate, the local pressure does not depend upon T. Thus, when He-burning is initiated, the high T-dependence of the 3α process means the energy is generated, raises T locally, thereby increasing the reaction rate, but does not produce a pressure or density decrease to moderate the reaction. The result is known as a helium flash. It only occurs in red giants for stars roughly as massive as the Sun or less, and in more advanced stages of other stars, typically in the white dwarf stage. It is best pictured using the pressure equation:

When H-burning or He-burning occurs, the result is a gradual increase in the mean molecular weight μ. If T and ρ remain unchanged, P decreases and unbalances the previously-existing hydrostatic equilibrium. The core of the star collapses so that both P and ρ increase. That enhances the energy generation rate, making the star more luminous. If Teff does not change, the star becomes larger, since L = 4πR2σTeff

4.

Hg m

kTkTnP

Other Reactions.More advanced reactions involve fusion of 12C to 16O, 23Mg (endothermic) or 20Ne, 23Na, 24Mg (exothermic), as well as fusion of 16O to 24Mg (endothermic) or 28Si, 31P, 31S, 32S, (exothermic). Various reactions are possible. Consider the binding energy per nucleon:

A

mmZAZm

A

mc

A

Eb nucleusnp2 )(

At low atomic weights the most stable nuclei are 1H, 2H, 3He, 6Li, 4He, 12C, 16O, 24Mg, 40Ca, and 56Fe. At high atomic weights the most stable nuclei are 86Kr, 107Ag, 127I, 174Yb, 208Pb, and 238U. Such unusually stable nuclei are called magic nuclei. The maximum binding energy per nucleus occurs at the iron peak, and all other fusion reactions producing heavier nuclei are endothermic.

The problem of explaining the existence of heavy elements in the universe can be restricted to explaining the existence of nuclear reactions that are endothermic in stars. The solution is advanced stages of evolution in massive stars (proton and alpha capture) and supernova explosions (neutron capture).

Energy Transport and Thermodynamics:

So far we have developed the following equations of stellar structure:

Equation of Continuity:

Hydrostatic Equilibrium:

Energy Generation:

But what about the temperature gradient, dT/dr?

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r

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dr

dP

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rdr

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Radiative Transport:Energy can be transported through a star by radiation, convection, or conduction. Conduction is unimportant in most gaseous stellar interiors, but the other two processes are important. In stellar atmospheres radiative energy transport is described by:

But Prad = ⅓aT4, so:

Thus:

But Frad = Lr/4πr2, giving, for radiative energy transport:

radrad F

cdr

dP

rad

3rad

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4F

cdr

dTaT

dr

dP

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Tacdr

dT

2r

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r

L

Tacdr

dT

Convective Transport:Convection is a three-dimensional process that must be approximated by one-dimensional equations for simple stellar interior models. It is a process that is difficult to model correctly, but begins with certain assumptions.

The pressure scale height HP is defined as:

or, if HP = constant,

HP is the distance over which the gas pressure P decreases by a factor of 1/e.Example:Estimate a typical value for the pressure scale height in the Sun.Solution:At the midpoint of the Sun we estimated ρ = 4M/3πR

3.

But dP/dr = –GMρ/r2 ≈ –GM2/2R5 = –Pc/R.So the pressure scale height can be estimated as:HP = –P/dP/dr = –½Pc/(–Pc/R) = ½R

More typical values in the Sun are of order HP ≈ 0.1R.

dr

dP

PHP

11

PHrePP 0

Thermodynamics:According to the first law of thermodynamics, the change in internal energy of a mass element, per unit mass, is the difference between the amount of heat added and the work done by the element on its surroundings:

dU = dQ – dW.

The internal energy of the mass element, U, is:

where nR = k/μmH is the universal gas constant.

nRTm

kT

U

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The change in heat of a gaseous mass element is usually expressed in terms of the specific heat C of the gas, where C is the amount of heat required to raise the temperature of a unit mass by 1K, i.e.:

The amount of work done per unit mass by a gas on its surroundings is dW. The usual expression for work done is:

dW = PdV,

so dU = dQ – PdV.

When the volume does not change, i.e., dV = 0, we have:

volumeconstantat

pressureconstantat

VV

PP

dT

dQC

dT

dQC

dTCdTdT

dQdQdU V

V

But:

If the pressure is held constant instead so that the volume changes as heat is added, then:

But PV = nRT,

so:

and

Define: as the ratio of specific heats.

nRCdTnRnRTddU V 23

23

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dTCdTdT

dVPCdT

dT

dVPdT

dT

dQdU V

PP

PP

VP CnRC

nRdT

dVP

and

nRnRnRnRCC VP 25

23

V

P

C

C

For a monatomic gas:

When a gas undergoes ionization, some of the heat dU that would normally increase the average kinetic energy of the gas is used instead for ionization. The temperature of the gas therefore increases less rapidly so dT is lower than otherwise. In such instances:

so: in such instances.

The effect is particularly pronounced in stellar ionization zones.

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nR

C

C

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dT

dQC

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C

For adiabatic processes there is no net heat flow into or out of a mass element, so dQ = 0 and dU = –dW = –PdV. From PV = nRT we have:

PdV + VdP = nRdT and

dU = CVdT, so:

From the definition of specific heats we have:

so: PdV + VdP = –(γ–1)PdV

PdVC

nRnRdTVdPPdV

C

dVP

C

dUdT

V

VV

and

11

V

P

V

VP

V C

C

C

CC

C

nR

That gives: (1+γ–1)PdV = –VdP

or: γPdV = –VdP

or:

The result is the adiabatic gas law: PVγ = K, where K is a constant.

But the perfect gas law also gives: PV = nRT, so:

P

dP

V

dV

11

1

1

where,

or

K

nRKTKP

TK

nRP

K

TnR

PV

VP

TnRVP

The sound speed in a gas is related to the incompressibility and inertia of the gas, specifically:

For adiabatic sound waves:

For convective energy transport one assumes that the gas bubbles are adiabatic. Also the specific volume V = 1/ρ refers to the volume per unit mass.

So if P = KV–γ , we must also have P = Kργ .

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s dV

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Pv

PV

PV

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The formula for the pressure gradient in stellar interiors therefore becomes:

With the perfect gas law:

so:

If there is no gradient in mean molecular weight in a star (not necessarily a valid assumption!), then:

from the adiabatic expression for P and ρ. The resulting equation gives an expression for the adiabatic temperature gradient.

dr

dP

dr

dK

dr

dK

dr

dP

1

Hg m

kTkTnP

dr

dT

T

P

dr

dP

dr

dP

dr

dP

dr

dP

dr

dT

T

P

dr

dP

dr

dP

Namely:

or:

But:

so:

dr

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P

T

dr

dP

K

K

P

T

dr

dP

P

K

P

T

dr

dP

P

T

dr

dT

ad

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The resulting temperature gradient is:

If: temperature gradient is superadiabatic.

The temperature gradient is a negative quantity since the temperature decreases with increasing radius inside a star.

If: heat will be transported by convection.

Otherwise the heat is transported outwards by radiation. The deep interiors of stars are relatively simple to understand. They have either convective or radiative cores depending upon the temperature gradient. They may also be semiconvective in the region immediately outside a convective core. Stellar atmospheres are more complicated since heat may be transported by both methods.

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actual

The test for whether or not convective or radiative transport describes the temperature gradient inside a star is therefore to test whether or not a displaced bubble of gas rises or falls. It will rise, i.e. convection applies if ρi(bubble) < ρi(surroundings).

The condition is:

or:

or:

See textbook, also for mixing length model.

1

dT

dP

P

T

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ln

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addr

dT

dr

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Stellar Models:The complete set of differential equations describing the interiors of stars is therefore:

Equation of Continuity:

Hydrostatic Equilibrium:

Energy Generation:

Temperature Gradient:

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rdr

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r

rMG

dr

dP

24 rdr

dL

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Stellar Models:The results for a series of stellar models at the start of hydrogen burning, i.e. with their initial composition unchanged, is depicted in the following figures. In each case solid circles denote models with a solar metallicity (X = 0.73, Y = 0.25, Z = 0.02), whereas open circles denote models of extremely low metallicity (X = 0.749, Y = 0.25, Z = 0.001). The actual metallicity of the Sun is presently under debate as a result of newer models describing turbulence in the solar atmosphere.

Models are constructed numerically using the differential equations as difference equations. They are also rearranged so that the dependent variable is mass rather than radius. Models are then denoted by the number of mass cells they contain. Spherical symmetry is also assumed, although that assumption is relaxed in more recent models.

15

16

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18

19

20

0.0 5.0 10.0 15.0 20.0

M/Msun

log Pc

0

1

2

3

0.0 5.0 10.0 15.0 20.0

M/Msun

log rhoc

5

6

7

8

9

0.0 5.0 10.0 15.0 20.0

M/Msun

log Tc

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10

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12

13

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0.0 5.0 10.0 15.0 20.0

M/Msun

log R

-2

-1

0

1

2

3

4

5

6

7

3.54.04.5

log Teff

log L/Lsun

One of the basic tenets of stellar evolutionary models is the Vogt-Russell Theorem, which states that the mass and chemical composition of a star, and in particular how the chemical composition varies within the star, uniquely determine its radius, luminosity, and internal structure, as well as its subsequent evolution. A consequence of the theorem is that it is possible to uniquely describe all of the parameters for a star simply from its location in the Hertzsprung-Russell Diagram. There is no proof for the theorem, and in fact, it does fail in some special instances.

A prime example of where ambiguities arise occurs when one compares models for two stars, one of which is spherically symmetric and the other of which is flattened as a result of rapid rotation. Both stars can occupy the same location in the Hertzsprung-Russell diagram, at different evolutionary ages and even for different masses.

Insights into Stellar Evolution:Consider the sequence of events that happens as stars evolve:

During hydrogen burning, the primary 41H → 4He reaction converts two protons and two electrons into two neutrons. The reaction chain may suggest that two positrons are produced, but they quickly self-annihilate through collisions with electrons, so the net result is as stated.

The production of α-particles from protons and electrons has two effects: (i) the mean molecular weight μ of the gas increases slightly, and (ii) the gas opacity, which is dominated by electron scattering in stellar cores, decreases because of the depleted abundance of electrons.

Consequence (i) implies that the gaspressure must decrease, since the massdensity of the gas is unaffected.

Consequence (ii) implies that radiation escapes the core of the star more easily.

Hg m

kTP

Since the gas pressure is reduced, the core of the star responds to the pressure imbalance by contracting. But according to the Virial Theorem, a contracting sphere of gas converts half of the decreased potential energy of the system into kinetic energy, i.e. heat, while the remainder escapes as radiation.

The consequence of an increased temperature for the gas at the stellar core is an increase in the nuclear reaction rate, so, along with the radiation increase induced by contraction of the core, the luminosity of the stellar core increases sharply, raising the photon flux from the interior.

The decreased electron scattering opacity of thegas near the stellar core means that radiation escapes more easily into the stellar envelope,where the main sources of opacity (from atomsand ions) exist. The increased photon flux onthe gas transfers more outwards-directedmomentum to the gas particles, resulting inan outwards expansion of the envelope gases.

So, as the core of the star contracts, the envelope expands!

Stellar evolution thus results in an increased luminosity as the star evolves. The resulting change in the effective temperature of the star is more complicated, since increased luminosity can be accommodated by an increase in the stellar radius or an increase in the effective temperature. Since the radius of the star must increase because of envelope expansion, it is not clear what will happen to the star’s effective temperature.

Stellar evolutionary models indicate that, for massive stars, the effective temperature decreases as the star evolves. For low mass stars like the Sun, however, the effective temperature actually increases during the initial stages of hydrogen burning.

The difference presumably originates from the differences in how energy is transmitted outwards in the two types of stars. Massive stars have convective cores and radiative envelopes, whereas low-mass stars like the Sun have radiative cores and convective envelopes. Convection mixes gas so that any changes in chemical composition are transmitted throughout the convective region, which is clearly more important for high mass stars: M > 1¼ M.

Question. Air is mostly (80%) composed of nitrogen molecules, each of which consists of 28 nucleons (protons and neutrons), where the mass of one nucleon is about one atomic mass unit (1.6605402 × 10–24 g). The radius of a typical nitrogen molecule is about 1 Å, and the density of air at sea level is roughly 1.2 × 10–3 g cm–3. Calculate the mean free path for collisions between air molecules under such conditions. The air is at room temperature, i.e. about 300K, which allows you to estimate the root-mean-squared speed, vrms, for air molecules. From that information, calculate the average time between collisions of atoms.

Solution:For nitrogen, σ = π(2r)2 = π(2 × 10–8)2 cm2 ≈ π × 4 × 10–16 cm2

The number density is n = ρ/m(N2) = 1.2 × 10–3 g cm–3/(28 × 1.6605402 × 10–24 g) = 2.581 × 1019 cm–3 So the mean free path is l = 1/nσλ = 1/(2.581 × 1019 × π × 4 × 10–16 ) ≈ 3 × 10–5 cmvRMS = (3kT/m)½ = (3 × 1.38 × 10–16 × 300/4.65 × 10–23)½

≈ 5.17 × 104 cm/sThe average time between collisions is t = l/v = (1.25 × 10–4 cm)/(5.17 × 104 cm/s) ≈ 2 × 10–9 s

Question. According to a “standard model” for the Sun, the central density is 162 g cm–3 and the Rosseland mean opacity κ is 1.16 cm2 g–1.a. Calculate the mean free path for a photon at the centre of the Sun.b. If the mean free path remains constant for the photon’s journey to the surface of the Sun, how long, on average, would it take for photons to escape from the Sun?

Solution:The mean free path for a photon is l = 1/κλρ = 1/(1.16 × 162) ≈ 5 × 10–3 cmThe distance traveled by a photon between collisions is d2 = Nl2

In order to escape from the Sun a photon has to travel a distance of 6.9598 × 1010 cmNumber of collisions is N = d2/l2 = (6.9598 × 1010)2/(5 × 10–3)2 = 1.9376 × 1026, each taking t = l/c = 5 × 10–3 cm/3 × 1010 cm/s = 1.7 × 10–13 s Time to escape the Sun is Nt = 1.9376 × 1026 × 1.7 × 10–13 s) = 3.23 × 1013 s ≈ 106 years