Lecture 11

Energy transport

Review: Nuclear energy

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

ix

  L    

r ix

  0

X i X x

 

T

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

dL r dr

 4 

r

2 

Proton-proton chain (PPI)

1 1

H

 1 1

H

 1 2

H



e

  

e

1 2

H

 1 1

H

 2 3

He

  2 3

He

 2 3

He

 2 4

He

 2 1 1

H

The net reaction is: 4 1 1

H

 2 4

He

 2

e

  2 

e

 2  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: Energy 

p

 

n



e

  

e

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

Proton-proton chain (PPII and PPIII)

Alternatively, helium-3 can react with helium-4 directly: 3 4 7 2

He

 2

He

 4

Be

  4 7

Be



e

  3 7

Li

 

e

3 7

Li

 1 1

H

 2 2 4

He

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

Yet another route is via the collision between a proton and the beryllium-7 nucleus 7 4

Be

 1 1

H

 5 8

B

  5 8

B

 4 8

Be



e

  

e

4 8

Be

 2 2 4

He

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

The PP chain

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

pp

 2 .

38  10 4  5

X

2

T

6  2 / 3

e

 33 .

80

T

6  1 / 3

W

/

kg

 5   10 5

kg

/

m

3

T

6 

T

/ 10 6

K

Near T~1.5x10

7 K (i.e. the central temperature of the Sun): 

pp

 1 .

07  10  7  5

X

2

T

6 4 W/kg



pp

Example

 1 .

07  10  7  5

X

2

T

6 4 W/kg 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 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: 12 6

C

 1 1

H

 13 7

N

  13 7

N

 13 6

C



e

 13 6

C

 1 1

H

 14 7

N

  14 7

N

 1 1

H

 15 8

O

  15 8

O

 15 7

N



e

  

e

 

e

15 7

N

 1 1

H

 12 6

C

 2 4

He



CNO

 8 .

67  10 25  5

XX CNO T

6  2 / 3

e

 152 .

28

T

6  1 / 3

W

at T~1.5x10

7 K 

CNO

/

kg

 8 .

24  10  27  5

XX CNO T

6 19 .

9 W/kg

Helium collisions

• 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

P

 

kT



m H

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

T

 1  4  0  2 4 3 

Z

1 2

Z

2 2

e

4

kh

2  1 .

9  10 7   

m H

 

Z

1 2

Z

2 2 K As H burning progresses, the temperature increases and eventually He burning becomes possible

The triple-alpha process

The burning of helium occurs via the triple alpha process: 2 4

He

 2 4

He

 4 8

Be

4 8

Be

 2 4

He

 12 6

C

  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  3   5 .

09  10 11  5 2

Y

3

T

8  3

e

 44 .

027

T

8  1

W

/

kg

 3   3 .

85  10  8  5 2

Y

3

T

8 41 .

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

Nucleosynthesis

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

12

C

6  2 4

He

 16

O

8   16 8

O

 2 4

He

 20 10

Ne

 

Nucleosynthesis

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

Break

Summary

We have now established four important equations: Hydrostatic equilibrium: Mass conservation: Equation of state:

dP

 

GM r



dr dM r

 4 

r

2 

dr P

 

kT



m H r

2 Energy production

dL r dr

 4 

r

2  There are 5 variables (P, energy transportation.

 ,M r , T and L r ) and 4 equations. To solve the stellar structure we will need to know something about the

Energy 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 transport

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

dP rad dr

   

c F rad

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

dT dr

  3   64 

L r r

2

T

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