Lecture 11Lecture 11

Energy transport

Review: Nuclear energyReview: Nuclear energy

• If each reaction releases an energy the amount of energy released per unit mass is just

TXXr xiixix 0

24 rdr

dLr

• The sum over all reactions gives the nuclear reaction contribution to in our fifth fundamental equation:

Proton-proton chain (PPI)Proton-proton chain (PPI)

HHeHeHe

HeHH

eHHH e

11

42

32

32

32

11

21

21

11

11

2

The net reaction is: 2224 42

11

eeHeH

But each of the above reactions occurs at its own rate. The first step is the slowest because it requires a proton to change into a neutron:

eenp Energy

This occurs via the weak force. The rate of this reaction determines the rate of Helium production

Proton-proton chain (PPII and PPIII)Proton-proton chain (PPII and PPIII)

HeHLi

LieBe

BeHeHe

e

42

11

73

73

74

74

42

32

2

Alternatively, helium-3 can react with helium-4 directly:

HeBe

eBeB

BHBe

e

42

84

84

85

85

11

74

2

Yet another route is via the collision between a proton and the beryllium-7 nucleus

This reaction only occurs 0.3% of the time in the Sun.

In the Sun, this reaction occurs 31% of the time; PPI occurs 69% of the time.

The PP chainThe PP chain

The nuclear energy generation rate for the PP chain, including all three branches:

kgWeTX Tpp /1038.2

3/1680.333/2

62

54

KTT 66 10/

Near T~1.5x107 K (i.e. the central temperature of the Sun):

W/kg1007.1 46

25

7 TXpp

355 /10 mkg

ExampleExample

W/kg1007.1 46

25

7 TXpp

If we imagine a core containing 10% of the Sun’s mass, composed entirely of hydrogen (X=1), calculate the total energy produced by the PP reaction.

The CNO cycleThe CNO cycle

There is a second, independent cycle in which carbon, nitrogen and oxygen act as catalysts. The main branch (accounting for 99.6% of CNO reactions) is:

HeCHN

eNO

OHN

NHC

eCN

NHC

e

e

42

126

11

157

157

158

158

11

147

147

11

136

136

137

137

11

126

kgWeTXX TCNOCNO /1067.8

3/1628.1523/2

6525

W/kg1024.8 9.1965

27 TXXCNOCNO at T~1.5x107 K

Helium collisionsHelium collisions

Recall that the temperature at which quantum tunneling becomes possible is:

2

422

21

20 3

4

4

1

kh

eZZT

• As hydrogen is converted into helium, the mean molecular weight increases.

• To keep the star in approximate pressure equilibrium, the density and temperature of the core must rise

Hm

kTP

As H burning progresses, the temperature increases and eventually He burning becomes possible

K 109.1 22

21

7 ZZmH

The triple-alpha processThe triple-alpha process

The burning of helium occurs via the triple alpha process:

CHeBe

BeHeHe126

42

84

84

42

42

The intermediate product 8-beryllium is very unstable, and will decay if not immediately struck by another Helium. Thus, this is almost a 3-body interaction

kgWeTY T /1009.51

8027.4438

325

113

W/kg1085.3 0.418

325

83 TY

Note the very strong temperature dependence. A 10% increase in T increases the energy generation by a factor 50.

NucleosynthesisNucleosynthesis

At the temperatures conducive to helium burning, other reactions can take place by the capturing of -particles (He atoms).

NeHeO

OHeC2010

42

168

168

42

126

NucleosynthesisNucleosynthesis

The binding energy per nucleon describes the stability of a nucleus. It is easier to break up a nucleus with a low binding energy.

BreakBreak

SummarySummary

We have now established four important equations:

2r

GM

dr

dP r

24 rdr

dM r

Hm

kTP

Hydrostatic equilibrium:

Mass conservation:

Equation of state:

There are 5 variables (P,,Mr, T and Lr) and 4 equations. To solve the stellar structure we will need to know something about the energy transportation.

24 rdr

dLr Energy production

Energy transportEnergy transport

Radiation: the photons carry the energy as they move through the star, and are absorbed at a rate that depends on the opacity.

Convection: buoyant, hot mass will rise

Conduction: collisions between particles transfer kinetic energy of particles. This is usually not important because gas densities are too low.

Radiation transportRadiation transport

When we considered the properties of radiation, we found an equation relating the pressure gradient to the radiative flux:

radrad F

cdr

dP

From this we can derive an expression for the temperature gradient, assuming a blackbody.

In regions of high opacity, or high radiative flux, the temperature gradient must be steep to transport the energy outward.

3264

3

Tr

L

dr

dT r