explaining puzzles of the h-r diagrampeople.physics.tamu.edu/belyanin/astr314/lecture10.pdf ·...

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Explaining puzzles of the H-R diagram:

The interiors of stars

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The mass-luminosity relation for 192 stars in double-lined spectroscopic binary systems.

L ~ M3.5 much stronger than inferred from L ~ R2 ~ M2/3

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Specific segments of the main sequence are occupiedby stars of a specific mass

Majority of stars are here

Abs

olut

e M

agni

tude

(M

)(L

umin

osity

)

Temperature (Color, B-V)Hotter

Brighter HR diagram

Spectral Type

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Age of the cluster from turnoff point

Turnoff point: stars of that mass are going to die and move away from the main sequence

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Stars spent most of their lives on the Main Sequence

At the end of its life the star moves away from the Main Sequence

More massive and more luminous stars die faster

Hypothesis: Stars on the Main Sequence live due to nuclear fusion of hydrogen!

• Stars stay on the main sequence until all hydrogen in the core is consumed • Then something should happen

A. Eddington (1920), G. Gamow (1928), H. Bethe (1939)

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Life of stars:Mass/gravity is everything

• Stars are born due to gravitational collapse of gas clouds

• Star’s life is a battle between thermal pressure generated by nuclear reactions and gravity

• Eventually, a star loses this battle, and gravity overwhelms

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Stars are gravitating spheres: they are held together by their own gravity. The gravity force acting on each volume element of a star is exactly balanced by gas pressure (Hydrostatic equilibrium)

This balance is steady

gravity

gas pressure

No gravity: the Sun will disperse in 1 day

No gas pressure: the Sun willcollapse in 20 minutes

Central pressure ~ 1010 atmospheres

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Hydrostatic equilibrium

Temperature in the center of a star

A =1 m2

=

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Internal structureCentral temperature for the sun Tc ≈ 1.5×107 K

Surface temperature of the sun Ts ≈ 5800 K

Heat transfer from the center to the surface!

Heat transfer determines both the internal compositionand the luminosity of the stars

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Internal source of energy

• Gravitational energy?• Chemical energy?• Nuclear reactions?

The Sun’s luminosity is L = 4x1026 Watt. Where does this energy come from?

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Chemical energy?This is the energy associated with breaking chemical bonds in molecules

1. Typical energy released per proton is ~ 1-10 eV

2. There are M/mp ~ 1057 protons in the Sun

Total available energy is Echem ~ 10x1057 = 1058 eV ~ 2x1039 J

Chemical energy will be radiated away during the time

But the Sun’s age is at least 4.6 billion years!Also, there is too hot for molecules in the sun

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Note:

If E is total energy stored in the sun (in J);

L is luminosity, or the rate with which this energy is spent(in J/sec);

Then the time it takes to spend all energy is T = E/L sec

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Gravitational energy?As the Sun radiates its thermal energy to outer space, it shrinks, and the central temperature is increased (!)

The energy source is the gravitational energy of a star

If the energy is radiated away with luminosity L = 4x1026 J/s,The Sun would radiate all its energy during the time

But the Sun’s age is at least 4.6 billion years!

Stellar Energy Sources

Recall that Gravitational Potential Energy is U = -G Mm/r. This says that as particles move toward each other, the potential energy is more negative.

For two particles may eventually end up in a bound orbit. The Virial Theorem says that the total energy in the bound system is E=(1/2)U. So, 50% of the energy is

available to be radiated away.

Gravitational Energy?

rMrdmi


Gravitational Energy

Now, instead of a point mass, dmi, consider a shell of thickness dr with mass dm = 4πr2 ρ dr

RMr

dr

ρ

Differential potential energy is

Integrating gives Potential Energy

ρ ≈ ρ = M/(4/3 πR3)

Mr ≈ 4/3 πr3 ρ

Approximate density as constant over

volume



Inserting these into the Potential Energy equation gives:

Solving gives:

Applying the Virial Theorem, E=U/2:

This is the amount of energy “Lost” in the gravitational collapse of system that ends up in a bound state.



Example, assume the Sun started as a spherical cloud of Hydrogen with a very, very large Radius, Ri >> R⊙. The energy radiated away is

ΔEg = - (Ef - Ei) ≈ -Ef ≈ (3/10) GM⊙2 / R⊙ = 1.1 x 1041 J.

We know the Sun has a Luminosity of 3.826 x 1026 W (J/s). Dividing ΔEg by the luminosity gives an estimate for how long it would take the Sun to radiate away

its gravitational potential energy.

t = ΔEg / L⊙ ≈ 107 yr.

Called the “Kelvin-Helmholtz” timescale (people who first worked it out). Does this make sense ?

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Nuclear reactions?

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Nuclear reactions?

• Fission: decay of heavy nuclei into lighter fragments

•Fusion: synthesis of light nuclei into a heavier nucleus

Energy released per proton is ~10-20 MeV!!

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Energy is released in fusion reaction if the sum of masses of initial nuclei is larger that the mass of the final nucleus

mp + mp

MD + me < 2 mpDeuterium

Positron (antielectron)

neutrino

Deuterium has larger binding energy than protons (more tightly bound)

ΔM = 2 mp- MD - me

Energy released E = ΔM c2

Famous Einstein’s relation: E = mc2

hydrogen

hydrogen

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What is binding energy?

It exists due to attractive forces between parts of a compound system: protons and neutrons in a nucleus, electrons and ion in an atom, Earth and moon, etc.

Binding energy is negative!: Ub = -|Ub|

Total energy of a system is the sum of energies of its parts plus binding energy:

E = E1 + E2 + Ub = E1 + E2 - |Ub|

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Energy is released in fission reaction if the mass of an initial nucleus is larger that the sum of masses of all final fragments

MU > MRb + MCs + 3 mn

Rubidium and Cesium are more tightly bound, or have larger binding energy than Uranium.

It is energetically favorable for Uranium to split.

When is the energy released in fission reactions?

ΔM = MU – (MRb + MCs + 3 mn)

Energy released E = ΔM c2 Famous Einstein’s relation: E = mc2

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There are no heavy elements on the stars

|Ub|

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Energy ProductionEnergy generation in the sun

(and all other stars):

Nuclear Fusion= fusing together 2 or more lighter nuclei to produce heavier ones.

Nuclear fusion can produce energy up to the production of iron;

For elements heavier than iron, energy is gained by nuclear fission.

Binding energy due to strong force = on short range, strongest of the 4 known forces: electromagnetic, weak, strong, gravitational

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Proton-proton cycle: four hydrogen nuclei fuse to form one helium nucleus

Hydrogen Fusion

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Einstein’s relation: E = mc2

Energy released in one cycle:(Binding energy)

Hans Bethe 1939

0.007, or 0.7% of the rest energy of protons (4mpc2) is released

This is 107 times more efficient than chemical reactions!

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There is more than enough nuclear fuel for 1010 years!

Does nuclear fusion provide enough energy to power the Sun?

Assume 1056 protons in the core:

million tons of hydrogen are fused every second on the Sun!

How much hydrogen should be fused per second to provide the Sun’s luminosity?

Nuclear fusion efficiency:0.7% of the hydrogen mass is converted into radiation in the p-p cycle

Matter-antimatter annihilation has even higher efficiency: 100% !!

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Proton-proton cycle

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Proton-proton cycleStep 1

Step 2

Step 3

All positrons annihilate with electrons creating gamma-quanta

Nuclear Fusion is governed by strong nuclear force. But, to fuse two hydrogen atoms, they must overcome the Coulomb barrier and come at ~ 1 fm from each other.

Recall that the electric (repulsive!) force between two protons is (1/4πε0) q2/r2.

1H + 1H --> 2H + positron + neutrinoStep 1:

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Protons should be hot!

How hot??

Make assumption that the energy to overcome the Coulomb barrier is the thermal energy of the gas. Refer to relative velocity of two nuclei (protons) using their reduced

mass, μm.

The point where the total energy is zero is where the two nuclei get to their closest approach and then are repelled by the Coulomb force:

(1/2)μmv2 = (3/2)kT = (1/4πε0) (Z1Z2e2)/r

T = (Z1Z2e2) / 6πε0 kr

For two hydrogen atoms, Z1=Z2=1 and r≈1 fm = 10-15 m. This then gives T≈1010 K.

But, the center of the Sun is only ~1.6 x 107 K.

How does nuclear fusion take place ?

George Gamow 1930s

Answer is Quantum Mechanics. Recall that we learned that the Heisenberg Uncertainty principle states that the uncertainties in the position and momentum of a

particle are related by

ΔxΔp > h/2

Quantum Mechanics allows that the location can not be

known precisely. Allows particle to “exist” past

Coulomb barrier, and eventually fuse !


Estimate quantum mechanical effect on temperature. Use that the “wavelength” of a particle is λ=h/p, which means we can rewrite the Kinetic energy as:

Nuclear Energy

(1/2)μmv2 = p2/2μm = (h/λ)2 / 2μm

Then we can set the distance of closest approach equal to one wavelength (where the height of the potential barrier is equal to the Kinetic Energy).

(1/4πε0) Z1Z2e2 / λ = (h/λ)2 / 2μm

Solving for λ and substituting r=λ into

T = (Z1Z2e2) / 6πε0 kr

we have

Tquantum = Z1Z2e4 μm / (12π2ε02h2k)

For two protons, μm = mp/2 and Z1=Z2=1 which gives Tquantum ≈ 107 K. That’s better, and quantum mechanics matters !

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Step 2

Takes 6 seconds to occur

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Step 3

Takes 1 million years to occur

More on stellar nucleosynthesis

An Element is specified by a # of protons, Z.

How do protons stay together in nuclei ?

Turns out, then need neutrons to glue the nuclei together or the +e charge of the protons would break the nuclei apart.

An Isotope of an element is identified by the # of neutrons, N, in a nucleus.

The number of nucleons (protons + neutrons) is A = Z + N.

A is refereed to as the mass number. Masses of atomic particles are:

mp = 1.67262158 x 10-27 kg = 1.00727646688 u

mn = 1.67492716 x 10-27 kg = 1.00866491578 u

me = 9.10938188 x 10-31 kg = 0.0005485799110 u

1 u is the mass of a Carbon-12 nucleus ÷ 12 (definition).

But, a carbon 12 nucleus might have a mass of 6mp + 6mn = 12.096 u > 12 u.

What is “missing” is the binding energy of C

Common notations


Nuclear Fusion releases energy. It converts mass into energy. Recall Relativity, E=mc2. 1 u = 931.494013 MeV/c2.

Note that the mass of hydrogen in the ground state, mH = 1.00782503214 u. This says that mH < mp + me = 1.00783. The difference is actually -13.6 eV.

The Sun is fusing He from H. A He-4 nucleus has a mass of 4.0026 u.

4 Hydrogen atoms have a mass of 4.0313 u.

Δm = 0.028697 u, or 0.7% of the total energy.

This is an energy of E=Δmc2 = 26.731 MeV. This is the binding energy of a He-4 nucleus. To break apart a He-4 nucleus takes this much energy.

Nuclear Energy

Stellar Nucleosynthesis

There certain conservation laws in nature, that must be obeyed. Some are

Conservation of electric charge : all products have same net charge as reactants.

Conservation of Lepton number : Leptons are “light things”, electrons, positrons, muons, neutrinos. Must have same lepton # before and after a reaction.

Example: e- + e+ → 2γ

On LHS an electron and positron annihilate.

Initial total charge is qi = -1 + 1 = 0.

Total lepton # is Li = 1 (for e-) + -1 (for e+) = 0.

Final charge qf = 0 and final lepton # Lf = 0.

Two photons are required to conserve momentum (another conservation law).

Conservation laws


There are several paths that Stars can use to fuse He from H.

The Proton-Proton (PP) chain

4 1H → 4He + 2e+ + 2νe + 2γ

Fusion Reactions in Stars

21

1H + 1H→ 2H + e+ + νe

1H + 2H→ 3He + γ3He + 3He → 4He + 2 1H

The intermediate steps involve the intermediate products Deuterium (2H) and helium-3 (3He)

1 1 1

1 1 2

2 2 2 1

Each reaction has its own rate because each has its own Coulomb Barrier to overcome.

2

1

Stellar NucleosynthesisFusion Reactions in Stars

A variant is the PP II chain (starting with step 3 in the PP)3He + 4He → 7Be + γ2 2 4

7Be + e- → 7Li + νe4 3

7Li + 1H → 2 4He3 1 2

Yet another variant is the PP III chain (starting with Step 2 in PPII)

7Be + 1H → 8B + γ4 51

8B → 8Be + e+ + νe5 4

8Be → 2 4He4 2

There are other possibilities. You can imagine others, then you work out what the timescales are (how long does the reaction take). This determines if it matters.

Stellar NucleosynthesisFusion Reactions in Stars


The Carbon-Nitrogen-Oxygen (CNO) Cycle

This cycle uses C, N, and O as a catalyst for the fusion of He. Proposed by Hans Bethe in 1938. There are several variants.


12C + 1H→ 13N + γ13N → 13C + e+ + νe

13C + 1H → 14N + γ14N + 1H → 15O + γ15O → 15N + e+ + νe

15N + 1H → 12C + 4He267 1

7

8

7

76

7

6

7 1

1

1

Primary CNO Cycle

Variant CNO Cycle (occurs 0.04% of the time)

15N + 1H → 16O + γ16O + 1H → 17F + γ17F → 17O + e+ + νe

17O + 1H → 14N + 4He

Starting with the last line of the Primary CNO cycle

2718

9 8

9

817

18


Triple Alpha Process

Fusion of Three 4He nuclei to form 12C. Called “alpha” because an alpha particle is a 4He nucleus.


4He + 4He⇌ 8Be

8Be + 4He → 12C + γ

2

Note that as H → He the mean molecular weight increases. The ideal gas law predicts that the central pressure, PC, will then decrease. The star will no longer be in equilibrium and will

collapse, raising the temperature. At some point He begins to “burn” (fuse to C).

2 4

4 2 6


Carbon and Oxygen Burning

After sufficient Carbon has been produced, further fusion occurs. Generally this is done by adding 4He nuclei. These elements are called “α” elements

(created by α-particle capture).


12C + 12C→6 6

16O + 2 4He ***20Ne + 4He23Na + 1H23Mg + n ***24Mg + γ{8 2

10 2

111

12

12

12C + 4He → 16O + γ826

2

16O + 4He → 20Ne + γ1028

At higher temperatures still:

16O + 16O→

24Mg + 2 4He ***28Si + 4He31P + 1H31S + n32S + γ

8 {812 2

2

1

16

15

16

16

*** reactions are endothermic - they absorb energy rather than release it. These are rare.

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There are no heavy elements on the stars

|Ub|

Most Abundant Elements in the Solar Photosphere.

Element Atomic # Log Relative Abundance

H 1 12.00

He 2 10.93 ± 0.004

O 8 8.83 ± 0.06

C 6 8.52 ± 0.06

Ne 10 8.08 ± 0.06

N 7 7.92 ± 0.06

Mg 12 7.58 ± 0.05

Si 14 7.55 ± 0.05

Fe 26 7.50 ± 0.05

S 16 7.33 ± 0.11

Al 13 6.47 ± 0.07

Ar 18 6.40 ± 0.06

Ca 20 6.36 ± 0.02

Ng 11 6.33 ± 0.03

Ni 28 6.25 ± 0.04

Note: These are all α elements. Stars are very efficient at making α elements !

Most abundant cosmic elements are H, He, O, C, Ne, N, Mg, Si, Fe. True for cosmos and

the Sun.


Where do elements with Z > 26 come from ?

s-Process Nucleosynthesis

One way is by s-process, s- for “slow”. Free neutrons do not feel Coulomb Barrier and collide with nuclei. Occasionally they stick, making larger nuclei. If neutron flux is not too

great, these heavier nuclei decay before more neutron captures.

Technetium (Tc) has no stable isotopes (all decay). But it is found in the atmospheres of giant stars. Most abundant isotope, 99Tc has a half-life of 200,000 yrs, much less than

lifetime of star. Must be forged in the star and dredged up.

However, there are occasions when the neutron flux is much, much higher... especially when nucleosynthesis stops in stars, causing the cores to collapse, which increases the neutron

density.


43

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The solar neutrino problem

,000 years

Neutrino have zero or very small mass and almost do not interact with matter

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Neutrino image of the Sun

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The Davis experiment

400,000 liters of perchlorethyleneburied 1 mile deep in a gold mine

About 1 Chlorine atom per day is converted into Argon as a result ofinteraction with solar neutrino

Much more difficult than finding a needle in a haystack!!

There are 1032 Cl atoms in a tank!

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Sudbury neutrino observatory: 1000 tons of heavy water D2O

,000 ton of ultra-pure water13,000 detectors

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Observed neutrino flux is 2 times lower than the theoretical prediction!

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The problem has been finally solved just recently:

Neutrinos “oscillate”! They are converted into other flavors: mu and tau neutrinos

Neutrinos should have massParticle physics models should include this effect

explaining puzzles of the h-r diagrampeople.physics.tamu.edu/belyanin/astr314/lecture10.pdf ·...

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