lecture 10 energy production. summary we have now established three important equations: hydrostatic...

Lecture 10Lecture 10

Energy production

SummarySummary

We have now established three important equations:

2r

GM

dr

dP r

24 rdr

dM r

Hm

kTP

Hydrostatic equilibrium:

Mass conservation:

Equation of state:

There are 4 variables (P,,Mr and T) and 3 equations. To solve the stellar structure we will need to know something about the energy production and transportation

• Can be modified to include radiation pressure

• Breaks down at high temperatures

and at high densities.

Stellar luminosityStellar luminosity

Where does the energy come from? Possibilities:

• Gravitational potential energy (energy is released as star contracts)

• Chemical energy (energy released when atoms combine)

• Nuclear energy (energy released when atoms form)

Gravitational potentialGravitational potential

The gravitational potential energy for a system of two particles:

r

MmGU

As M and m are brought closer together, the potential energy becomes more negative.

Assume the Sun was originally much larger than it is today, and contracted. How much gravitational potential energy is released?.

Chemical energyChemical energy

Chemical reactions are based on the interactions of orbital electrons in atoms. Typical energy differences between atomic orbitals are ~10 eV.

e.g. assume the Sun is pure hydrogen. The total number of atoms is therefore

5727

30

1019.11067.1

1099.1

H

Sun

m

Mn

If each atom releases 10 eV of energy due to chemical reactions, this means the total amount of chemical energy available is

JeV 3958 1010

This is ~100 times less than the gravitational potential energy available, and would be radiated in only 100,000 years at the present solar luminosity

Atomic nucleiAtomic nuclei

Usually we express the mass of nuclei in atomic mass units u, defined to be 1/12 the mass of 12C (the 12 is the mass number A. 12C has 6 protons, 6 neutrons and 6 electrons)

1u=1.66054x10-27kg

um

um

um

e

n

p

0005486.0

00866.1

00728.1

• The mass of 1 proton + 1 electron is 1.0078285u• Note 6 p + 6 n + 6 e- = 12.099. Does this make

sense?

The EquationThe Equation

Einstein showed that mass and energy are equivalent, and related by:

2mcE

Thus atomic masses can be expressed as energies

2MeV/c 49432.9311 u

When equating rest masses to energies, it is customary to omit the factor c2 and leave it implicitly assumed.

Binding energyBinding energy

The mass of an atom (protons+neutrons+electrons) is not equal to the mass of the individual particles.

e.g. the hydrogen atom is less massive than the sum of the proton and electron mass by 1.46x10-8 u

This mass difference corresponds to an energy:

eVE 6.13

This energy is the binding energy. It is the energy released when an electron and proton are brought together.

Binding energyBinding energy

There is also a binding energy associated with the nucleons themselves.

• The mass difference is 0.099u, equivalent to 92.22 MeV!• This is the binding energy of the C-12 atom

• Recall Carbon-12: 6 p + 6 n + 6 e- = 12.099.

FusionFusion

Making a larger nucleus out of smaller ones is a process known as fusion.

For example:remnants mass low HeHHHH

The mass of 4 H atoms is 4.031280u. The mass of He is 4.002603u.

• The mass difference is 0.028677u, equivalent to 26.71 MeV. As we will see, the energy of the typical low-mass remnants

amounts to only ~1 MeV. ~0.7% of the H mass is converted into energy

Nuclear energy: fusionNuclear energy: fusion

In contrast with chemical reactions, nuclear reactions (which change one type of nucleus into another) typically release energies in the MeV range, 1 million times larger.

E.g. Assume the Sun was originally 100% hydrogen, and converted the central 10% of H into helium.

This would release an energy:

J

cME Sun

44

2

103.1

007.01.0

Assuming the Sun’s luminosity has been constant at 3.8x1026 W, it would take ~10 billion years to radiate all this energy.

Nuclear energy can sustain the solar luminosity over its likely lifetime

Nuclear energy: fusionNuclear energy: fusion

Nuclear energy is sufficient to sustain the Sun’s luminosity. But fusion is difficult to achieve! Can it actually occur naturally in the Sun?

BreakBreak

Coulomb repulsionCoulomb repulsion

r

eZZUC

221

04

1

The repulsive force between like-charged particles results in a potential barrier that gets stronger as the particles get closer:

• The strong nuclear force becomes dominant on very small scales, 10-15 m.

• What temperature is required to overcome the Coulomb barrier?

Statistical mechanicsStatistical mechanics

• If the gas is in thermal equilibrium with temperature T, the atoms have a range of velocities described by the Maxwell-Boltzmann distribution function.

• The number density of gas particles with speed between v and v+dv is:

dvvekT

mndvn kT

mv

v22

2/3

42

2

The most probable velocity:m

kTv 22

The average kinetic energy: kTvm2

3

2

1 2

Quantum tunnelingQuantum tunneling

The answer lies in quantum physics. The uncertainty principle states that momentum and position are not precisely defined:

2

xpx

•The uncertainty in the position means that if two protons can get close enough to each other, there is some probability that they will be found within the Coulomb barrier.

This is known as tunneling. The effectiveness of this

process depends on the momentum of the particle

Quantum tunnelingQuantum tunneling

• So the protons don’t need to get anywhere near 10-15m before they can begin to tunnel past the barrier

• Without this quantum effect, fusion would not be possible in the Sun and such high luminosities could never be achieved.

Approximately: tunneling is possible if the protons come within 1 deBroglie wavelength of each other:

kT

h

v

h

p

h

3

For two protons, at T~107 K m

T7

3

1057.31013.1

Quantum tunnelingQuantum tunneling

What temperature is required for two protons to come within one de Broglie wavelength of each other?

kT

hdeBroglie

3

Review: Nuclear fusionReview: Nuclear fusion

•Fusion is possible if the particle energy (3/2 kT) is equal to or greater than the Coulomb potential energy:

r

eZZUC

221

04

1

where the distance r can be taken to be the de Broglie wavelength:

kT

h

3

Fusion is therefore possible at temperatures

K 109.1 22

21

7 ZZm

TH

Nuclear energyNuclear energy

• The energy production rate must be related to the rate of collisions.

• Since the full rate equation can be very complex, it is common to approximate it as a power-law over a particular temperature range

TXXrr xiix 1

0

• Here, Xi and Xx are the mass fractions of the incident and target particles, respectively

• For two-body interactions (i.e. p+p collisions), ~1• The parameter can have a wide range of values

Nuclear energyNuclear 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:

Nuclear reactionsNuclear reactions

So – what are the specific reactions we’re talking about?? The probability that four H atoms will collide at once to form a single

He atom is exceedingly small. Even this simple fusion reaction must occur via a number of steps.