Chapter 5 Stellar Energy Production - University of …eagle.phys.utk.edu/guidry/astro615/lectures/lecture_ch5.pdfChapter 5 Stellar Energy Production ... star birth and star death

Chapter 5

Stellar Energy Production

Stars have three primary sources of energy:

1. Heat left over from earlier processes,

2. Gravitational energy,

3. Thermonuclear energy.

We shall see that

• Gravitational energy plays an important role in various stages ofstar birth and star death.

• Objects like white dwarfs shine because of heat left over fromearlier thermonuclear and gravitational energy generation.

• However, the virial theorem indicates that the gravity andleft-over heat of formation can supply the energy of the Sun only ona 107 year timescale.

Thus, thermonuclear reactions are the onlyviable long-term stellar energy source.

95

96 CHAPTER 5. STELLAR ENERGY PRODUCTION

5.1 Nuclear Energy Sources

The luminosity of the Sun is

L⊙ ≃ 3.8×1033 erg s−1

and that of the most luminous stars is about106L⊙. Fromthe Einstein relationm = E/c2, the rate of mass conver-sion to energy that is required to sustain the Sun’s lumi-nosity is

∆m=1c2∆E→

∆m∆t

=1c2

∆E∆t

=L⊙c2 = 4.2×1012 g s−1,

and the most luminous stars require conversion rates a mil-lion times larger. Let us now discuss how nuclear reac-tions in stars can account for mass-to-energy conversionon this scale.

5.1. NUCLEAR ENERGY SOURCES 97

5.1.1 The Curve of Binding Energy

The binding energy for a nucleus of atomic numberZ andneutron numberN is

B(Z,N)≡ [ Zmp +Nmn︸︷︷︸

free nucleons

− m(Z,N)︸︷︷︸

bound system

]c2,

where

• m(Z,N) is the mass of the nucleus

• mp is the mass of a proton

• mn is the mass of a neutron.

The binding energy may be interpreted either as

• the energy released in assembling a nucleus from itsconstituent nucleons, or

• the energy required to break a nucleus apart into itsconstituents.

The more relevant quantity is often thebinding energy pernucleon,B(Z,N)/A, whereA = Z+N is the atomic massnumber.


0 50 100 150 200 2500

4

8

Nucleon Number A

Bin

din

g E

nerg

y / N

ucle

on

(M

eV

)

2

6

10

H1

He4

C12 Fe

56

Figure 5.1:The curve of binding energy. Only the average behavior is shown;local fluctuations have been suppressed, as has the isotopicdependence on(Z,N)for a givenA.

The average behavior of binding energy per nucleon asa function of the atomic mass numberA is shown inFig. 5.1. The general behavior of the curve of bindingenergy may be understood from simple nuclear physicsconsiderations, which are elaborated in the typeset chap-ter but we won’t go over here.


5.1.2 Masses and Mass Excesses

It is convenient to define themass excess,∆(A,Z), through

∆(A,Z) ≡ (m(A,Z)−A)mHc2,

• m(A,Z) is measured in atomic mass units (amu),

• A = Z+N is the atomic mass number, and

• the atomic mass unitmH (which is defined to be112

the mass of a12C atom) is given by

mH =1

NA

= 1.660420×10−24 g= 931.478 MeV/c2,

with NA = 6.02×1023 mole−1 Avogadro’s number.

The mass excess is useful because

• The number of nucleons is constant in low-energynuclear reactions (baryon number conservation).

• Therefore, the atomic mass numbers cancel on bothsides of any equation.

• Thus, sums and differences of masses(large num-bers) may equivalently be replaced by the corre-sponding sums and differences of mass excesses(small numbers).


Example: Using the definition of the mass excess, thebinding energy equation

B(Z,N)≡ [Zmp +Nmn−m(Z,N)]c2,

may be rewritten as

B(Z,N) = [Zmp +Nmn−m(Z,N)]c2

= [Zmp +(A−Z)mn−m(Z,N)]c2

= [Z∆p +(A−Z)∆n−∆(A,Z)]c2

= [Z∆p +(A−Z)∆n−∆(A,Z)]×931.478 MeV,

where we have abbreviated the mass excess of the neutronand proton by∆(1,0)≡ ∆n and∆(1,1)≡ ∆p, respectively.

In many mass tables, the mass excesses ratherthan the masses themselves are tabulated.


Example: Let us calculate the binding energy of4He. Therelevant mass excesses are

∆p = 7.289 MeV ∆n = 8.071 MeV ∆(4,2)= 2.425 MeV

and the binding energy of4He is then

B(Z,N) = [Z∆p +(A−Z)∆n−∆(A,Z)]c2

= 2×7.289+2×8.071−2.425= 28.3 MeV.

Thus more that 28 MeV of energy is requiredto separate4He into free neutrons and pro-tons.


5.1.3 QQQ-Values

TheQ-value for a reaction is thetotal mass of the reactantsminus the total mass of the products, which is equivalentto the corresponding difference in mass excesses,

Q= Mass of reactants− Mass of products

= Mass excess of reactants− Mass excess of products.

It is common to specify theQ-value in energyunits rather than mass units.


Example: For the nuclear reaction

2H + 12C→ 1H + 13C,

the tabulated mass excesses are

∆(2H) = 13.136 MeV ∆(12C) = 0 MeV

∆(1H) = 7.289 MeV ∆(13C) = 3.1246 MeV.

TheQ-value for this reaction is then

Q = ∆(2H)+∆(12C)−∆(1H)−∆(13C) = +2.72 MeV.

• The positiveQ indicates that this is anexothermic re-action: 2.72 MeV is liberated from binding energy inthe reaction, appearing as kinetic energy or internalexcitation of the products.

• Conversely, a negative value ofQ would indicateanendothermic reaction:additional energy must besupplied to make the reaction viable.


0 50 100 150 200 2500

4

8

Nucleon Number A

Bin

din

g E

nerg

y / N

ucle

on

(M

eV

)

2

6

10

H1

He4

C12 Fe

56

5.1.4 Efficiency of Hydrogen Fusion

Examination of the curve of binding energy suggests twopotential nuclear sources of energy:

• Fission of heavier elements into lighter elements.

• Fusion of lighter elements into heavier ones.

Since stars are composed mostly of hydrogen and helium,

• It is clear that their primary energy source must befusion of lighter elements to heavier ones.

• Coulomb repulsion between charged nuclei will in-hibit the possibility of fusion, so hydrogen (Z = 1)will be easier to fuse than helium (Z = 2).

• In particular, we shall see shortly that main sequencestars are powered by thermonuclear processes thatconvert four1H into 4He.


The total rest mass energy in one gram of material is

E = mc2 = 9×1020 erg,

Energy released in fusing 1 gm1H into 4He (Exercise 5.1):

∆E(fusion H→4He) = 6.3×1018 erg g−1.

Therefore, less than 1% of the initial rest mass is convertedinto energy in the stellar fusion of hydrogen into helium:

∆E(fusion H→4He)total rest-mass energy

=6.3×1018 erg g−1

9×1020 erg g−1 ≃ 0.007.

• Hydrogen fusion is actually a ratherinefficientsourceof energy.

• Furthermore, we shall see shortly that the fusion ratesin the cores of lower-mass main sequence stars suchas the Sun are actually quite small.

The solar luminosity is produced by a powersource equivalent toseveral 100 W lightbulbsper m3 of the solar core (Exercise 5.18).

• Fusion powers stars not because of its intrinsic effi-ciency, but rather because of the enormous mass ofstars, which implies that they havelarge reservoirsof hydrogen.


5.2 Thermonuclear Hydrogen Burning

The primary energy source of stars on the main sequenceis fusion of hydrogen into helium. Two general sets ofreactions can accomplish this:

1. Theproton–proton chain (PP chain)

2. TheCNO cycle

Generally we find that

• The proton–proton chain produces most of thepresent energy of the Sun and generally is dominantin stars of a solar mass or less.

• The CNO cycle quickly surpasses the proton–protonchain in energy production as soon as the mass ex-ceeds about a solar mass.

• The reason for this rapid switchover is that the PPchain and the CNO cycle have strong and very dif-ferent dependence on temperature.

5.2. THERMONUCLEAR HYDROGEN BURNING 107

5.2.1 The Proton–Proton Chain

The most important reactions of the PP chain are summarized in thefollowing figures:


106

years

4He

1H

1H

1010

years

2H

νβ+

1H

seconds

3He

γ − ray

1H

1H

1010

years

2H

νβ+

1H

seconds

3He

γ − ray

1H

1H

Figure 5.2: Main branch PP-I of the proton–proton chain. The inverses ofthetimes indicate the relative rates for each step under solar conditions and will bediscussed in more detail later.

H1

H1

+ H2

+ e++ νe

H2

H1

+ He3

+ γ

He3

He3

+ He4

+ Be7

+ γH1

He3

He4

+

Li7

+ νBe7

+ e-e

Li7

+ H1

He4

He4

+

B8

+ γBe7

+

B8

H1

Be8

He4

He4

+

e++ νeBe

8+

PP-II (15%)

Q = 25.7 MeV

PP-III (0.02%)

Q = 19.1 MeV

PP-I (85%)

Q = 26.2 MeV

+ H1

Figure 5.3:The three branches of the PP chain. The percentage contribution tosolar energy production and the effectiveQ-value are shown for each branch.


(p,γ)

(p,γ)

(p,γ)

(p,α)β+

β+

9.97 m

12C

13C

15O

13N

15N

14N 6

6

7 8 9

7

8

9

Neutron Number

Pro

ton N

um

ber

C12

C13

C14

C15

N13

N14

N15

N16

O14

O15

O16

O17

F15

F16

F17

F18

Figure 5.4:The CNO cycle. The main part of the cycle is illustrated schematicallyon the left side. On the right side the main part of the cycle isillustrated with solidarrows and a side branch is illustrated with dashed arrows. The notation(p, i)means a proton capture followed by emission ofi; for example12C(p,γ)13N. Thenotationβ+ indicates beta decay by positron emission; for example,13N→ 13C +e+ + νe.

5.2.2 The CNO Cycle

The name of the carbon–nitrogen–oxygen or CNO cyclederives from the role played by carbon, nitrogen, and oxy-gen in the corresponding sequence of reactions. The CNOcycle is summarized in Fig. 5.4.


CNO CatalysisThe main part of the CNO cycle is termed theCN cycle.Summing net reactants and products around the CN cycle,

12C+4p→ 12C+ 4He+2β + +2ν.

(The γ-rays have been neglected since they do not corre-spond to a conserved quantity.)

• Therefore, the12C serves as acatalystfor the conver-sion of four protons to4He.

• It is required for the sequence to take place, but itis not consumed in the process because a12C is re-turned in the last step of the cycle.

We have written the CNO sequence as if the(p,γ) reactionon 12C were the first step, but it is a closed cycle and wemay consider any step to be the initial one.

• This implies that any of the carbon, nitrogen, or oxy-gen isotopes appearing in the cycle may be viewed asa catalyst that serves to convert protons into helium.

• The closed nature of the cycle also implies that

1. Any mixture of these isotopes will play the samecatalytic role.

2. If any one of the CNO isotopes is present ini-tially a mixture of the others will inevitably beproduced by the cycle of reactions.


CNO

PP

Sun

ε ~ T 17

ε ~ T 4

T6(K)

0 5 10 15 20 25 30 35

log

ε (

Me

V g

-1s

-1)

Figure 5.5: Rate of energy release in the PP chain and in the CNO cycle.T6

denotes the temperature in units of 106 K.

The rates of energy release from hydrogen burning for thePP chain and CNO cycle are illustrated in Fig. 5.5.

• Note that the CNO cycle has a much stronger tem-perature dependence than the PP chain.

• This temperature dependence implies that the star’smass on the main sequence is generally the most im-portant factor governing the competition of PP andCNO energy production, because that strongly influ-ences the core temperature.

We will see how to calculate these rates shortly.


The PP cycle can occur in any star containing H, but theCNO cycle requires the presence of C, N, or O as catalysts.

• Therefore, the CNO cycle should be relatively moreimportant in Pop I stars.

• However, this is generally of secondary importancefor the competition between the PP chain and CNOcycle in producing stellar energy.

• This issue is also of importance in understanding thevery first generation of stars that formed (Pop III).

– No CNO isotopes produced in the big bang.

– Thus the first generation of stars must have op-erated by the PP chain until some of those starscould produce carbon by triple-α (see below).

• These considerations are important for understand-ing the early Universe because relatively massivestars formed surprisingly early.

1. CNO is more efficient at producing energy inmassive stars than PP because of its very strongtemperature dependence.

2. Thus the pace of early structure evolution likelydepended on when the earliest stars producedand distributed sufficient CNO isotopes to allowa succeeding generation of stars to switch to themore efficient CNO cycle.

5.3. CROSS SECTIONS AND REACTION RATES 113

5.3 Cross Sections and Reaction Rates

To proceed with a more quantitative analysis of energyproduction in stars, we need to introduce the rudiments ofnuclear reaction theory as applied in stellar environments.Consider the representative nuclear reaction

α +X −→ Z∗ −→ Y +β ,

whereZ∗ denotes an excited compound nucleus as an in-termediate state [we will often write this equation in theshorthand notationX(α,β )Y].

• A compound nucleus is an excited composite formedin the initial collision that quickly decays into the fi-nal products of the reaction.

• In the above reaction the left side (α +X) is called theentrance channeland the right side (Y +β ) is calledtheexit channelfor the reaction.

• It is common to classify nuclear reactions accordingto the number of (nuclear) species in the entrancechannel; thus this is a2-body reaction.

• We shall often use 2-body reactions to illustrate but

– 1-body reactionslike A→ B+C and

– 3-body reactionslike A+B+C→ D

also play roles in stellar energy production.


Let us first consider a laboratory setting where the reactionis initiatedby a beam of projectilesα directed onto a target containing nucleiX.

• Thecross sectionσαβ (v) (a function of the velocityv) is

σαβ (v)≡ραβF(v)

=

(reactions per unit time per target nucleus

incident flux of projectiles

)

,

and has units of area (a commonly-used unit of cross section isthebarn (b), which is defined to be across section of10−24 cm2).

• The incident particle fluxF(v) is given by

F(v) = nαv,

wherenα is the number density of projectilesα in the beam andv is their relative velocity.

• The number of reactions per unit time (reaction rate) per targetnucleusραβ is

ραβ = F(v)σαβ = nαvσαβ ,

and thetotal reaction rate per unit volumerαβ (v) results frommultiplying ραβ by the number densitynX of target nucleiX:

rαβ (v) = ραβ nX = nαnXvσαβ (v)(1+δαX)−1,

and has units ofcm−3 s−1

in the CGS system.

5.3. CROSS SECTIONS AND REACTION RATES 115

• The factor involving the Kroeneckerδ (δab is one if a = b andzero isa 6= b) is introduced to prevent overcounting when thecolliding particles are identical.

1. The productnαnXvσαβ (v) is the rate per unit volume for the2-body reaction

2. nαnX is the number of unique particle pairs(α,X) containedin the unit volume.

3. But for the collision of identical particles (α = X), the num-ber of independent particle pairs(α,α) is notN2

α but 12N2

α .

4. Therefore, for identical particles the rate expression must bemultiplied by a factor of1/(1+ δαX) = 1

2 to avoid doublecounting.

5. More generally, one finds that forN identical particles afactor of1/N! is required to prevent double counting.

• Normally we will work in thecenter of mass coordinate system,so velocities, energies, momenta, and cross sections will be cen-ter of mass quantities, unless otherwise noted:

E = ECM =

(mX

mα +mX

)

Elab v =√

2E/µ µ ≡mαmX

mα +mX

,


0 4 × 10 7 8 × 107 .1 2 × 108

Velocity (cm/sec)

0

3 × 108

2 × 108

1 × 108

1 × 107 K

2 × 107

K

Velo

city D

istr

ibution f

(v)

Figure 5.6: Maxwellian velocity distribution for two temperatures; the dashedarrows indicate the mean velocities for each distribution.

5.4 Thermally-Averaged Reaction Rates

The preceding equations assume a beam of monoenergeticparticles.

• In a stellar environment we instead have a gas in ap-proximate thermal equilibrium.

• If we assume that the gas can be described classi-cally, at equilibrium it has aMaxwell–Boltzmann dis-tribution ψ(E) of energies

ψ(E) =2

π1/2

E1/2

(kT)3/2exp(−E/kT).

This distribution is illustrated for two different tempera-tures in Fig. 5.6 as a function ofv =

√

2E/µ.

5.4. THERMALLY-AVERAGED REACTION RATES 117

• A thermally-averaged cross section〈σv〉αβ may be defined byaveraging the velocity-dependent cross section over the veloci-ties in the gas,

〈σv〉αβ ≡∫ ∞

0ψ(E)σαβ (E)v dE,

=

√

8πµ

(kT)−3/2∫ ∞

0σαβ (E)e−E/kTE dE,

where the second form follows from substitutingv =√

2E/µ.

• The units of〈σv〉αβ arecm3 s−1 = cm2 · (cm/s) or cross sec-tion times velocity. We may then introduce a thermally-averagedreaction rate:

rαβ = nαnX

∫ ∞

0ψ(E)σαβ (E)v dE= nαnX〈σv〉αβ

= ρ2N2A

XαXX

AαAX

〈σv〉αβ = ρ2N2A YαYX〈σv〉αβ ,

where in the interest of compact notation

1. We drop explicit display of theδ -function factor necessarywhen the colliding particles are identical, and

2. In the second line we introduce the mass fractionsXi andthe abundances (molar fractions)Yi defined in Ch. 3.

• The units ofrαβ arecm−3 s−1 (rate per unit volume),

• The clear physical interpretation ofrαβ = nαnX〈σv〉αβ is thatthe total rate for this 2-body reaction is the (thermally averaged)rate for a singleα to react with a singleX to produceY + β ,multiplied by the number ofαs per unit volume and the numberof Xs per unit volume.


5.5 Parameterization of Cross Sections

To proceed, we require the cross section evaluated in thecenter of mass system in order to calculate the thermally-averaged rates. This cross section may be parameterizedin the general form

σαβ (E) = πgλ-2ΓαΓβΓ2 f (E),

• The energy widthsΓi ≡ h̄/τi are expressed in termsof the corresponding mean lifeτi for decay of thecompound system through channeli,

• the entrance channel is denoted byα,

• the exit channel byβ ,

• the total width isΓ = ∑i Γi, where the sum is over allopen channelsi,

• the probability to decay to channeli is Pi = Γi/Γ,

• the reduced deBroglie wavelength is defined throughλ-2 = h̄2/2µE,

• The statistical factorg contains information on thespins of projectile, target, and compound nucleus(typically of order 1)

• The detailed reaction information resides inf (E).

5.5. PARAMETERIZATION OF CROSS SECTIONS 119

The parametersΓ appearing in

σαβ (E) = πgλ-2ΓαΓβΓ2 f (E),

have the units of̄h divided by time, which is energy.

• They are calledenergy widthsbecause states withshort lifetimes for decay (large decay rates) corre-spond to spectral peaks (resonances) broad in energy,by a∆E·∆t ≃ h̄ uncertainty principle argument.

• Conversely, states with long decay lifetimes (smalldecay rates) correspond to narrow resonances. Thelimiting case is a state that is completely stable,which then corresponds to a vanishing decay rate andhas a sharply defined energy.

• The factorf (E) is generally either

1. Resonant, in which case it is strongly peaked inenergy because of a narrow (quasibound) statein the compound nucleus, or

2. Nonresonant, because the reaction energy is farfrom a resonance, or there are no resonances(quasibound states) in the channel of interest.

• The total rates will typically be a sum of contribu-tions for resonant and nonresonant pieces.


5.6 Resonant Cross Sections

In the simplest case of an isolated resonance,f (E) can beexpressed in theBreit–Wigner form

f (E)res =Γ2

(E−Er)2+(Γ/2)2,

where the resonance energyEr is related to a correspond-ing excitation energyE∗ for a quasibound state in the com-pound nucleus through

Er = (mZ−mα −mX)c2 +E∗.

The corresponding Breit–Wigner cross section is

σαβ = πgλ-2ΓαΓβΓ2 f (E)

= πgλ-2 ΓαΓβ(E−Er)2 +(Γ/2)2,

and will generally exhibit a strong peak nearEr.

5.6. RESONANT CROSS SECTIONS 121

ELab

(MeV)

S(E

)

10-4

10-2

100

102

104

106

10-12

10-10

10-8

10-6

10-4

10-2

σ(E

) and E

xp/E

S(E)

σ(E)

Exp/E

0.1 0.3 0.5 0.7

Figure 5.7:Cross sectionσ(E) in barns for the reaction12C(p,γ)13N. TheS-factorS(E) is defined later. The curve labeled Exp/E is the factor exp(−2πη)/E.

As an example, we consider the reaction12C(p,γ)13N il-lustrated in Fig. 5.7. This reaction has a resonance corre-sponding to a state in13N at an excitation energy of 2.37MeV that is strongly excited at a laboratory proton energyof 0.46 MeV.


If the Maxwell–Boltzmann distributionψ(E) and the widthsΓi varyslowly over a resonance, we may assume

ψ(E)→ ψ(Er) Γα → Γα(Er) Γβ → Γβ (Er),

and we obtain he resonant velocity-averaged cross section

〈σv〉αβ =π h̄2g2µ

√

8πµ

(kT)−3/2e−Er/kT

× Γα(Er)Γβ (Er)∫ ∞

0

1(E−Er)2 +(Γ/2)2 dE.

If the resonance is broad orEr is small, thepreceding assumptions maybe invalid and itmay be necessary to integrate over the reso-nance energy numerically using the data.

The integrand peaks nearEr; extending the lower limit of the integralto minus infinity and assuming the widths to be constant gives

〈σv〉αβ = 2.56×10−13 (ωγ)r

(µT9)3/2exp(−11.605Er/T9) cm3 s−1,

whereEr is in MeV, T9 indicates temperature in units of109 K, and

(ωγ)r ≡ gΓαΓβ

Γ

has units of MeV and is tabulated for reactions of interest.

5.7. NONRESONANT CROSS SECTIONS 123

5.7 Nonresonant Cross Sections

Most (but not all) reactions relevant for stellar energy pro-duction are, not surprisingly, exothermic

Q≡ (mass of reactants)− (mass of products)

= mαc2+mXc2−mβ c2−mYc2 > 0.

• Typical Q-values for the reactions of interest are∼ 1 MeV, with this energy going into kinetic energyof particles in the exit channel and any internal exci-tation of the products.

• This additional energy leads to a marked asymmetryin the entrance and exit channels for charged particlereactions of interest in stellar energy production.

• In the entrance channel, the thermal energies avail-able are set by the temperatures through

kT = 8.6174×10−8T keV,

with the temperature expressed in kelvin.

• Hydrogen burning in main sequence stars typicallyoccurs in a temperature range107 K <

∼ T <∼ 109 K,

implying kinetic energies in the plasma of1 keV <

∼ kT <∼ 100 keV.

• Thus, for averageQ∼ 1 MeV noted above, we oftenhave

E(entrance) << E(exit)

for reactions of interest.


V(r)

r0

Nuclear

Potential

Coulomb

Barrier

Attra

ctive

Repuls

ive

Incident Charged Particle

Figure 5.8:The Coulomb barrier for charged-particle reactions.

5.7.1 Coulomb Barriers

Because of the low average energies in the entrance chan-nel, charged-particle reactions are strongly influenced bythe Coulomb barrier

ECB = 1.44ZαZX

R(fm)MeV,

whereZi is the atomic number of particlei, the barrierseparation distanceR is

R≃ 1.3(A1/3α +A1/3

X ) fm,

andAi is the atomic mass number (in amu) of particlei.As is common in nuclear physics, energies have been ex-pressed in MeV and distances in fermis (fm).


Example: Consider a proton scattering from a28Si nu-cleus. From

ECB = 1.44ZαZX

R(fm)MeV R≃ 1.3(A1/3

α +A1/3X ) fm

we obtain

ECB = 1.44(1)(14)

1.3(11/3+281/3)MeV = 3.8 MeV.


Table 5.1: Coulomb barriers for p + X

X ZαZβ R (fm) ECB (MeV)11H 1 2.6 0.55126 C 6 4.3 2.02814Si 14 5.2 3.85626Fe 26 6.3 6.0

Some typical Coulomb barriers for proton reactions p + Xare shown in Table 5.1, where we note that

• Entrance channel energies for hydrogen fusion instars (which lie in the range 10−3 to 10−1 MeV)are typically orders of magnitude lower than theCoulomb barrier.

• As we shall see, this implies adramatic temperaturedependencefor hydrogen fusion reactions.

• On the other hand, exit channel energies (approxi-mately 1 MeV in typical cases) are comparable tothe barrier energies for fusion of protons with lighterions.


5.7.2 Barrier Penetration Factors

Classically, characteristic energies in a stellar plasma are too small tosurmount the Coulomb barrierfor typical charged-particle reactions.

• However, it is possible forquantum-mechanical tunnelingtotake place for energies that lie below the height of the barrier,albeit with small probability.

• AssumingECB >> E, the barrier penetration probability for acollision with zero relative orbital angular momentum (termeds-waves in scattering theory) is

P(E) ∝ e−2πη ,

where the dimensionless Sommerfeld parameterη is defined by

η =ZαZXe2

h̄v=

√µ2

ZαZXe2

h̄E1/2.

• For representative values of the parameters,P(E) <∼ exp(−12).

Thus, charged particle reactions crucial to power generation instars arehighly improbable events.

• Since the reaction probability for barrier penetration will be dom-inated by the probability to penetrate the barrier, we take as anentrance channel width for nonresonant reactions

Γα ≃ e−2πη ,

which clearly has a strong energy dependence.


Lab Proton Energy (keV)

S(E

) (

keV

-barn

)

0 100 200 300 4001

10

100

12C p, γ( )

Figure 5.9:S-factor versus laboratory proton energy for a typical capture reaction.This is the portion of theS-factor for the region below the resonance in Fig. 5.7.

5.7.3 Astrophysical S-Factors

Exit channel energies for charged-particle reactions arecomparable to the barrier energies; thus we assume thatΓβ is a weakly varying function ofE, as isΓ = Γα + Γβ ,and we express the nonresonant cross section as

σαβ (E) = πgλ-2ΓαΓβΓ2 f (E)≡

S(E)

Ee−2πη ,

whereλ-2 ∝ 1/E andΓα ∝ e−2πη have been used.

• The factorS(E) ≡ σαβ (E)Ee2πη , is termed theas-trophysical S-factor.

• It is assumed to vary slowly withE and contains allenergy dependence not introduced explicitly throughthe factorsλ-2∼ 1/E or exp(−2πη).

The S-factor for a(p,γ) reaction is illustrated in Fig. 5.9.


• Because of the low energies corresponding tokT thatcharacterize stellar plasmas, reaction cross sectionsare often needed atlower energies than can feasiblybe measured in a laboratory.

• Then, experimental measurements at higher energymust beextrapolated to the lower energy of interest.

• This is usually done by assuming no resonance at thelower energy and plotting

S(E)≡ σαβ (E)Ee2πη ,

which has smooth behavior and therefore is moreeasily extrapolated than the full cross section.

Then from

〈σv〉αβ =

√

8πµ

(kT)−3/2∫ ∞

0σαβ (E)e−E/kTE dE,

σαβ (E) = πgλ-2ΓαΓβΓ2 f (E)≡

S(E)

Ee−2πη ,

the thermally-averaged nonresonant cross section is

〈σv〉αβ =

√

8πµ

(kT)−3/2∫ ∞

0S(E)e−E/kT−2πη dE

=

√

8πµ

(kT)−3/2∫ ∞

0S(E)e−E/kTe−bE−1/2

dE

whereb≡ 2π(µ/2)1/2ZαZXe2/h̄ andS(E) is in erg cm2.


exp (- E /kT ) Barrier

Penetration

Gamow

Window

Energy

Figure 5.10:The Gamow window.

5.7.4 The Gamow Window

The energy dependence of

〈σv〉αβ =√

(8/(πµ)(kT)−3/2∫ ∞

0S(E)e−E/kTe−bE−1/2

dE

resides primarily in the factor

FG≡ e−E/kTe−bE−1/2,

which is termed theGamow window.

• The first factorexp(−E/kT), arising from theMaxwell–Boltzmannvelocity distribution,decreases rapidly with energy.

• The second factorexp(−bE−1/2), arising from thebarrier pen-etration factor,increases rapidly with energy.

• The product is stronglylocalized in energy(Fig. 5.10).

Only if the energies fall within the relevant Gamow win-dow will charged particle reactions in stellar environmentshave significant probability to occur.


The maximum of the Gamow peak is found at (see Exer-cise 5.4)

E0 = 1.22(Z2αZ2

X µT26 )1/3 keV.

• For many reactions of interest this corresponds toonly tens of keV.

• This implies that laboratory cross sections often mustbeextrapolated to these low energiesto calculate as-trophysical processes because it is very difficult to doreliable experiments at such low energies.

Useful approximate expressions for the width of theGamow peak and for the nonresonant cross section canbe obtained by

• assuming the Gamow peak to be a Gaussian havingthe same peak position and curvature at the peak asthe realistic Gamow peak (see Exercise 5.5).

In this approximation the width of the Gamow peak is

∆ =4

31/2(E0kT)1/2 = 0.75(Z2

αZ2XµT5

6 )1/6 keV

and the cross section is

〈σv〉αβ ≃0.72×10−18S(E0)a2

µZαZXT2/36

exp(−aT−1/36 ),

in units of cm3 s−1, where a = 42.49(Z2αZ2

X µ)1/3 andS(E0) is evaluated at the energy of the Gamow peak inunits of keV barns.


From the gaussian approximation we find that for the in-teraction of two protons at a temperature ofT6 = 20 (thatis, T = 20×106 K),

kT = 1.7 keV E0 = 7.2 keV ∆ = 8.2 keV,

and from the earlier table the corresponding Coulomb bar-rier is about 500 keV.

5.8. LIBRARIES OF CROSS SECTIONS 133

5.8 Libraries of Cross Sections

The Gamow window is not Gaussian.

• By using expansions to characterize the deviation from Gaus-sian behavior of the realistic curve, correction terms may be de-rived that give a more accurate representation of the thermally-averaged cross section.

• One parameterization that incorporates such correction termsand is often used in reaction rate compilations is

〈σv〉= a( f0+ f1T1/3+ f2T2/3+ f3T + f4T4/3+ f5T5/3)e−bT−1/3

T2/3

wherea, b, and fn parameterize the cross section.

The Caughlan and Fowler compilation used in Exercises for this chap-ter is parameterized in this manner.


5.9 Total Rate of Energy Production

The total reaction rate per unit volumerαβ is given by

rαβ = ρ2N2A

XαXX

AαAX

〈σv〉αβ = ρ2N2A YαYX〈σv〉αβ ,

The corresponding totalrate of energy production per unitmassis then given by the product of the rate and theQ-value, divided by the density:

εαβ =rαβ Q

ρ,

which has CGS units oferg g−1s−1. TheQ-value enteringthis expression is defined by

Q≡ (mass of reactants)− (mass of products)

but with the proviso thatif a reaction produces a neutrinothat removes energy from the star without appreciable in-teraction in the core, this neutrino energy should be sub-tracted from the totalQ-value.

5.10. TEMPERATURE AND DENSITY EXPONENTS 135

5.10 Temperature and Density Exponents

It is often useful to parameterize the energy productionrate of a star in the power-law form

ε = ε0ρλ Tν ,

and characterize the behavior of the energy production interms of thetemperature exponentν and thedensity expo-nentλ .

• Although the energy production is not universallyof this form, this approximation with constant expo-nents is usually valid for a limited range of tempera-tures and densities.

• Since the energy production mechanisms for stars areoften operative only in a very narrow range of tem-peratures and densities, this can provide a useful pa-rameterization for the regions of physical interest.

From the preceding expression, we may define tempera-ture and density exponents for an arbitrary energy produc-tion functionε(ρ ,T) through

λ =

(∂ lnε∂ lnρ

)

Tν =

(∂ lnε∂ lnT

)

ρ.


Table 5.2: Density and temperature exponents

Stellar process Density (λ ) Temperature (ν)

PP chain 1 ∼ 4

CNO cycle 1 ∼ 16

Triple-α 2 ∼ 40

Temperature and density exponents are displayed in Ta-ble 5.2 for the PP chain and CNO cycle, and for the triple-α process that burns helium to carbon in red giant stars.

• Notice in Table 5.2 the exquisite temperature depen-dence exhibited by these reactions.

• This enormous sensitivity of energy production totemperature is a central feature of stellar structureand stellar evolution.

5.11. REACTION SELECTION RULES 137

5.11 Reaction Selection Rules

Sometimes it is possible to make statements about the astrophysicalsignificance of various nuclear reactions based on selection rules andconservation laws, without having to calculate any detailed rates.

• For example, angular momentum is conserved in all reactionsand the angular momentumJJJ of a compound nucleus state pop-ulated in a two-body reaction must satisfy

jjj1+ jjj2+ lll = JJJ.

where jjj1 and jjj2 are the angular momenta associated with thecolliding particles andlll is the angular momentum of relativeorbital motion in the entrance channel.

• Likewise, isotopic spin (an abstract approximate symmetry) isconserved to a high degree in strong interactions and the isotopicspins in a two-body reaction must approximately satisfy

ttt1+ ttt2 = TTT,

wherettt1 andttt2 are the isotopic spins associated with the collid-ing particles andTTT is the isotopic spin of the state that is popu-lated in the reaction.

• Parity (symmetry of the wavefunction under space reflection)is maximally broken in the weak interactions, but is essentiallyconserved in the strong and electromagnetic reactions. More-over, in a nuclear reaction that does not involve the weak force,the parities must satisfy

(−1)l π( j1)π( j2) = π(J).

whereπ =± denotes the parity andj i the angular momentum ofthe states.


Compound nucleus states with angular momentum,isospin, and parity quantum numbers that do not satisfythese selection rules will generally not be populated sig-nificantly in reactions.

For nuclei with even numbers of protons and neutrons(even–even nuclei) the ground states always have angularmomentum and parityJπ = 0+.

• Therefore, if the colliding particles in the entrancechannel are even–even nuclei in their ground states,the angular momentum and the parity of the state ex-cited in the compound nucleus areboth determinedcompletely by the orbital angular momentum of theentrance channel:

J = l π(J) = (−1)l .

• Resonance states satisfying this condition are said tohavenatural parity.

5.11. REACTION SELECTION RULES 139

Consider the reactionα + 16O→ 20Ne∗

(where the * onthe Ne indicates that it is in an excited state).

• Under normal astrophysical conditions theα-particleand16O will be in their ground states and thus willeach haveJπ = 0+.

• Therefore, parity conservation requires that any stateexcited in20Ne by this reaction have parity

π(20Ne) = (−1)l = (−1)J.

• We conclude that states in20Ne having Jπ =

0+,1−,2+,3−, . . . may be populated (because theyare natural parity) but that population of states having(say)Jπ = 2− or 3+ is forbidden by parity conserva-tion.

• In the20Ne spectrum there is a state at 4.97 MeV ofexcitation relative to the ground state havingJπ =2−.

• This state cannot be excited in the capture reactionα + 16O→ 20Ne+ γ because to do so would violateparity conservation (it is not a natural parity state).

As we shall see in the later discussion of helium burning,if this seemingly obscure state had the opposite parity, wewould not exist (!).


5.12 Reactions of the Proton–Proton Chain

The slowest reaction in the PP chain,

H1

H1

+ H2

+ e++ νe

H2

H1

+ He3

+ γ

He3

He3

+ He4

+ Be7

+ γH1

He3

He4

+

Li7

+ νBe7

+ e-e

Li7

+ H1

He4

He4

+

B8

+ γBe7

+

B8

H1

Be8

He4

He4

+

e++ νeBe

8+

PP-II (15%)

Q = 25.7 MeV

PP-III (0.02%)

Q = 19.1 MeV

PP-I (85%)

Q = 26.2 MeV

+ H1

and therefore the one that governs the overall rate at which the chainproduces power, is the initial step

1H+ 1H−→ 2H+e+ +νe.

This reaction is very slow because it proceeds by the weakinteraction rather than the strong interaction (the appear-ance of the neutrino on the right side is the sure sign thatthis is a weak interaction). There is a reason why weakinteractions are called weak:their reaction rates are typi-cally many orders of magnitude smaller than correspond-ing strong interaction rates.

5.12. REACTIONS OF THE PROTON–PROTON CHAIN 141

The reaction1H+ 1H−→ 2H+e+ +νe is nonresonant.

• The reaction rate is found to be

rpp = 12n2

p〈σv〉pp =1.15×109

T2/39

X2ρ2exp(−3.38/T1/39 ) cm−3 s

−1,

whereX is the hydrogen mass fraction.

• As shown in Exercise 5.8, the temperature exponent is

νpp = 11.3/T1/36 −2/3,

implying thatνpp≃ 4 for T6 = 15.

• The rate of change for proton number because of this reaction isgiven by the usual radioactive decay law,

dndt

=−1τ

n,

whereτ is the mean life.

• Thus, themean life for a protonwith respect to being convertedin the PP chain is

τp =−np

dnp/dt=

np

2rpp

,

where a factor of two appears in the denominator because twoprotons are destroyed in each reaction.


For typical solar conditions we may take a temperatureof T6 = 15, a central density ofρ = 100 g cm−3, and ahydrogen mass fraction in the center ofX = 0.5.

• This yields an estimate ofτp≃ 6×109 years.

• This is remarkably long and sets the scale for themain sequence life of the Sun.

The slowness of the initial step in the PP-chain is ultimately relatedto the diproton (2He) not being a bound system.

• If the diproton were bound, the first step of the PP-chain couldbe a strong interaction and the lifetime would be much shorter.

• Instead, the first step must wait for a highly improbable event:a weak decay of a proton from a broadp–p resonance having avery short lifetime.

• In contrast, the mean life for the deuterium produced in thefirststep of the PP chain and consumed in the next step (p+ d→3He+ γ) is about one minute under the conditions prevailing inthe solar core.

• The final fusion of two helium-3 isotopes to form helium-4 ismuch slower (τ ∼ 106 years), but is orders of magnitude fasterthan the first step.

• Thus, the initial step of the PP chain governs the rate of thereac-tion and in turn sets the main sequence lifetime for stars runningon the PP chain.

5.12. REACTIONS OF THE PROTON–PROTON CHAIN 143

H1

H1

+ H2

+ e++ νe

H2

H1

+ He3

+ γ

He3

He3

+ He4

+ Be7

+ γH1

He3

He4

+

Li7

+ νBe7

+ e-e

Li7

+ H1

He4

He4

+

B8

+ γBe7

+

B8

H1

Be8

He4

He4

+

e++ νeBe

8+

PP-II (15%)

Q = 25.7 MeV

PP-III (0.02%)

Q = 19.1 MeV

PP-I (85%)

Q = 26.2 MeV

+ H1

The relative importance of PP-I versus PP-II and PP-III depends onthe competition between the reactions

3He(3He,2p)4He and 3He(4He,γ)7Be.

• Over the range of temperatures where PP is expected to be im-portant, the first reaction generally is faster than the second byabout four orders of magnitude, ensuring the dominance of PP-Iover PP-II and PP-III.

• The branching between PP-II and PP-III depends on the compe-tition between electron capture and proton capture on7Be.

• At the temperature of the Sun, electron capture dominates andPP-II is much stronger than PP-III.

• At somewhat higher temperatures, PP-III will begin to makemuch larger relative contributions (however, at higher tempera-tures the CNO process will quickly become more important thanthe PP chains in stellar energy production).


Table 5.3: EffectiveQ-values

Process Qeff (MeV) % Solar energy

PP-I 26.2 83.7

PP-II 25.7 14.7

PP-III 19.1 0.02

CNO 23.8 1.6

The effectiveQ-values for the PP chain depends on which subchain isfollowed, since the energy carried off by neutrinos is different in the3 cases. The effectiveQ-values are listed in Table 5.3.

• The average energy released per PP chain fusion in the Sun is

∆Epp = 0.85

(26.2

2

)

+(0.15)(25.7) = 15 MeV,

where PP-III has been ignored and the factor of 2 in the denom-inator of the first term results from the first two steps of PP-Ineeding to run twice to provide the two3He isotopes requiredfor the last step.

• Although PP-III has negligible influence on solar energy produc-tion, the preceding discussion indicates that it produces muchhigher energy neutrinos than PP-I or PP-II.

• The PP-III chain is highly temperature dependent because it isinitiated by proton capture on aZ = 4nucleus (Coulomb barrier).

• Therefore, detection of the high-energy neutrinos emitted fromthe PP-III chain can provide a very sensitive probe of the centraltemperature of the Sun. We shall return to this issue when wediscuss the solar neutrino anomaly.

5.13. REACTIONS OF THE CNO CYCLE 145

5.13 Reactions of the CNO Cycle

Because of Coulomb barriers andS-factors, the slowest reaction inthe CNO cycle

(p,γ)

(p,γ)

(p,γ)

(p,α)β+

β+

9.97 m

12C

13C

15O

13N

15N

14N 6

6

7 8 9

7

8

9

Neutron Number

Pro

ton N

um

ber

C12

C13

C14

C15

N13

N14

N15

N16

O14

O15

O16

O17

F15

F16

F17

F18

is typicallyp+14N→15O+ γ,

which hasS= 3.3 keV barnsin the energy range of interest.


• The corresponding mean life for14N against this reaction in thecore of the Sun is approximatelyτ14-N = 5×108 years.

• The concentration of14N at the core of the Sun in the Stan-dard Solar Model is2.6×1022 cm−3 (corresponding to an molarabundanceY of 0.6%).

• The hydrogen concentration is3×1025 cm−3 and previously themean life for consumption of a proton by PP chain fusion wasestimated to be6×109 years.

• These numbers imply that the ratio of PP chain to CNO cyclereactions in the core of the Sun is approximately

(rate for PP

rate for14N+p

)

=

(τ14-N

τpp

)(3×1025

2.6×1022

)

≃ 100,

• We conclude that for conditions prevailing in the Sun the PPchain dominates the CNO cycle.

Detailed calculations within the Standard Solar Model in-dicate that the Sun is producing 98.4% of its energy fromthe PP chain and only 1.6% from the CNO cycle.


1H 4He

14N

15N

15O and 13N offscale

12C

13C

4 8 12 16 20

0

-2

-4

-6

-8

-10

Log Time (seconds)

Log M

ass F

raction X

Energy

Equilibrium

Figure 5.11:The CNO cycle run to completion starting with only hydrogen anda small amount of12C. Isotopic mass fractions are shown as solid lines (the massfractions for13N and15O are of order 10−14 or smaller and are offscale on this plot).The dashed line is the integrated energy release on an arbitrary log scale. Thetemperature and density were assumed constant atT6 = 20 andρ = 100 g cm−3,respectively.

In Fig. 5.11 we illustrate several remarks made earlier by implement-ing a calculation of the CNO abundances carried to hydrogen deple-tion (conversion of all hydrogen to helium) for a star with a constanttemperature ofT6 = 20and constant density of100 g cm−3.

• We have assumed an initial mixture having only two isotopes:1H (0.995 mass fraction) and12C (0.005 mass fraction).

• Even though we start with only a trace amount of one CNO iso-tope (12C), the cycle eventually generates an equilibrium abun-dance of all CNO isotopes and steadily releases energy by con-verting all the hydrogen to helium.


1H 4He

14N

15N

15O and 13N offscale

12C

13C

4 8 12 16 20

0

-2

-4

-6

-8

-10

Log Time (seconds)

Log M

ass F

raction X

Energy

Equilibrium

• Once the cycle is in equilibrium, the mass fractions of the CNOisotopes remain essentially constant, so they may be viewedascatalyzing the conversion of hydrogen to helium, as discussedearlier.

• Notice also the result (which is a general one) that the CNO cy-cle run long enough to equilibrate tends to produce14N as thedominant CNO isotope, even though there was no initial abun-dance of this isotope at all in this simulation (to understand why,see Exercise 5.14).

• It is thought that most of the14N found in the Universe has beenproduced by the CNO cycle.


The effectiveQ-value for the CNO cycle is 23.8 MeV (Ta-ble 5.3). The rate of energy production is

εCNO =4.4×1025ρXZ

T2/39

exp(−15.228/T1/39 ) erg g−1 s

−1,

and the corresponding temperature exponent is

νCNO = 50.8/T1/36 −2/3,

which givesνCNO≃ 18 for T6 = 20.

This remarkably strong temperature depen-denceimplies that, were the Sun only slightlyhotter, the CNO cycle instead of the PP chainwould be the dominant energy productionmechanism.


5.14 Timescale for Main Sequence Lifetimes

Since the main sequence is defined by stable burningof hydrogen, the rate of hydrogen fusion determines atimescale for life on the main sequence.

• Comparison of stellar evolution simulations with ob-servations suggest thatstars leave the main sequencewhen∼10% of their original hydrogen has beenburnedto helium.

• Let us define atimescaleτnuc for main sequence life-times by forming the ratio of the energy releasedfrom burning 10% of the hydrogen and the luminos-ity.

• The energy available from the burning of one gramof hydrogen to helium is∼ 6×1018 ergs. Therefore,

τnuc =EH/10

L= 6×1017 XM

Ls,

– EH is the energy available from fusing all the hy-drogen in the star,

– L is the present luminosity in erg s−1,

– X is the original hydrogen mass fraction,

– M is the mass of the star in grams.

5.14. TIMESCALE FOR MAIN SEQUENCE LIFETIMES 151

Example: Inserting values characteristic of the Sun, wefind for the solar main sequence timescale

τ⊙nuc =EH/10

L∼ 2.2×1017 s∼ 1010 yr.

Expressing this timescale in solar units, we may write forany star

τnuc = 1010(

MM⊙

)(L⊙L

)

yr,

and utilizing the mass–luminosity relation

LL⊙≃

(MM⊙

)3.5

,

the main sequence timescale may be expressed forM ≥M⊙ as

τnuc≃ 1010(

MM⊙

)−2.5

yr.


Mass / M

Luminosity

(Solar Units)

Surface Temperature (K)

106

104

102

100

10-2

Time on the Main

Sequence (106 years)

5 10 15 20 25

Figure 5.12:Main sequence lifetimes, temperatures, and luminosities.

Some main sequence lifetimes are illustrated in Fig. 5.12

• The Sun has a main sequence lifetime of about10billion years,but

• A 20 M⊙ stays on the main sequence for about5.5million yearsand

• A 100 M⊙ star lives on the main sequence for onlyabout100,000 years.

• Conversely, for main sequence stars withM << M⊙,we may estimate that the main sequence lifetimegreatlyexceeds the present age of the Universe.

5.15. THE TRIPLE-ALPHA PROCESS 153

1H 2H

3He 4He

6Li 7Li

9Be

10B 11B

12C

0 1 2 3 4 5 6

0

1

2

3

4

5

6

H

He

Li

Be

B

C

Neutrons

Pro

tons

A=5 is

oto

pes

A=8 is

oto

pes

Figure 5.13:The green boxes indicate stable isotopes. There are no stable mass-5or mass-8 isotopes.

5.15 The Triple-Alpha Process

Main sequence stars produce their power by hydrogen fusion.Thisbuilds up a thermonuclear ash of helium in the core of the star.

• The star continues to fuse hydrogen to helium in a shell sur-rounding the central core of helium that is built up.

• This hydrogen shell burningadds gradually to the accumulatingcore of helium and the star remains on the main sequence untilabout 10% of its initial hydrogen has been consumed.

• Fusion of the helium to heavier elements is difficult because

– There is alarge Coulomb barrier,

– There areno stable mass-5 and mass-8 isotopesto serve asintermediaries for the production of heavier elements (Fig. 5.13).


1H 2H

3He 4He

6Li 7Li

9Be

10B 11B

12C

0 1 2 3 4 5 6

0

1

2

3

4

5

6

H

He

Li

Be

B

C

Neutrons

Pro

tons

A=5 is

oto

pes

A=8 is

oto

pes

• Thus, helium fusion can occur only at very high temperaturesand densities: temperatures in excess of about108 K and densi-ties of102−105 g cm−3.

• Such conditions can result when stars exhaust their hydrogenfuel and their cores begin to contract.

• Because there areno stable mass-8 isotopes,the resulting fusionof helium must involve atwo-step processin which

1. Two helium ions (alpha-particles) combine to form highlyunstable8Be,

α +α→ 8Be

2. This in turn combines with another helium ion to form car-bon.

α + 8Be→ 12C

• The resulting sequence, which is crucial to the power generatedby red giant stars and to the production of most of the carbonand oxygen in the Universe, is called thetriple-α process.


Our bodies are composed of about 65% oxy-gen and 18% carbon.We owe our very exis-tence to the (highly improbable) triple-α pro-cess,as we discuss further below!


The burning of helium to carbon by the triple-α process may beviewed as taking place inthree basic steps:

1. A small transient population of8Be is built up by He + He fu-sion,

4He+ 4He↔ 8Be.

which hasQ≃ 92 keV.

2. A small transient population of12C in an excited state is built upby the reaction

4He+ 8Be↔ 12C∗.

On general grounds, to produce a finite popu-lation of 12C∗ this reaction must be resonant.Otherwise it would be too slow to competewith the decay of8Be to twoα-particles.

3. A small fraction of the12C∗ excited states decay electromagnet-ically by

12C∗→ 12C+2γ

to the ground state of carbon-12.

This (highly improbable) sequence of reac-tions has the net effect of converting threehelium ions to12C, with an energy releaseQ = +7.275 MeV. Let us consider each ofthese steps in more detail.


5.15.1 Equilibrium Population of Beryllium-8

• The mean life for decay of8Be back into two alpha-particlesis τ ≃ 10−16 seconds, which corresponds to a width ofΓ8-Be =

h̄τ−1 = 6.8 eV.

• The capture will be too slow to compete with this decay backinto alpha-particles unless the corresponding resonance peak over-laps substantially with the Gamow peak.

• Thus, we expect this rate controlling step of the triple-α processto be significantonly when the energy of the Gamow peak iscomparable to theQ-value of 92 keVfor the reaction, and thisin turn sets the required conditions for the triple-α process toproceed.

• The maximum of the Gamow peak is given approximately by

E0 = 1.22(Z2αZ2

X µT26 )1/3 keV.

which implies a temperature of1.2×108 K for E0 = 92 keV.

• Therefore,only for temperatures of order108 K can the initialstep of the triple-α reaction produce a sufficient equilibrium con-centration of8Be to allow the subsequent steps to proceed.


• The preceding simple estimate ignores details suchas electron screening that must be considered for amore precise estimate of the temperature for heliumburning, but it sets the correct order of magnitude.

• Also note that this temperature estimate raises thequestion ofwhy helium was not consumed in bigbang nucleosynthesisby the triple-α mechanism.The answer:

– Thetemperaturewas high enough but

– Thedensitywas too low.

– Both high temperatures and high densities,which awaited the formation of stars,were re-quired to produce significant amounts of carbonby the triple-α mechanism.


We may estimate the equilibrium concentration of8Be un-der various conditions by application to nuclei (rather thanatoms) of asuitable modification of the Saha equations.The required changes are

1. Replace the number densities of ions and electronswith the number densities ofα-particles, and thenumber density of neutral atoms with the numberdensity of8Be.

2. Replace the statistical factorsg for atoms with cor-responding statistical factors associated with nuclei.This is trivial for the present case: the ground statesof both 8Be and4He have angular momentum zeroandg = 1 in both cases.

3. The ionization potentials entering the atomic Sahaequations are replaced byQ-values in the nuclearcase. In the present example,Q = 91.78 keV for8Be→ αα (“ionization” of 8Be to twoα-particles).

4. The electron mass in the atomic Saha equations isreplaced by the reduced massmαmα/2mα = 1

2mα .

The resultingnuclear Saha equationis

n2α

n(8Be)=

(πkTmα

h2

)3/2

exp(−Q/kT).

The situation where such equations are applicable istermednuclear statistical equilibrium (NSE).


Example: Helium flashesare explosive helium burningevents that can occur in red giant stars (see Ch. 9).

Typical helium flash conditions in lower-mass red giantstars correspond to a temperature ofT9≃ 0.1 and a densityof ρ ≃ 106 g cm−3.

For a triple-α powered helium flash in a pure helium core,we obtain from

n2α

n(8Be)=

(πkTmα

h2

)3/2

exp(−Q/kT).

thatn(8Be)

nα= 7×10−9,

corresponding to an equilibrium8Be concentration ofn(8Be) = 1021 cm−3 during the flash (see Exercise 5.10).


γ2

+

0+

12C

α + 8Be

0+

γ

0.29 MeV

Q = 7.367 MeV

0.00 MeV

4.44 MeV

7.65 MeV

Figure 5.14:Nuclear energy levels in12C for the final steps of the triple-α reac-tion. Levels are labeled byJπ and energy relative to the ground state. The 0+ stateat 7.65 MeV is the Hoyle resonance.

5.15.2 Formation of the Excited State in Carbon-12

The next step of the triple-α process,

α + 8Be→ 12C+ γ

has Q = 7.367 MeV and proceeds through an angular momentumJ = 0 resonance in12C at an excitation energy relative to the12Cgroundstate of 7.654 MeV, as illustrated in Fig. 5.14.

Hoyle resonance: The existence of this reso-nance waspredicted by Hoyleto explain theenergy production in red giant stars.


Ground

StateCompound Nucleus

Resonance

StateE r

Q

Γ

Ep

JπE r = Q + Ep

0

Figure 5.15:Relationship ofQ-value, resonance energyEr, and center of massenergyEp when an isolated resonance is maximally excited in a reaction.

• As illustrated in Fig. 5.15, the population of the Hoyle state isoptimized if the center of mass energy plus theQ-value is equalto the resonance energy relative to the ground state of12C.

• Once this excited state is formed, the dominant reaction isarapid decay back toα + 8Be, but a small fraction of the time theground state of12C may instead be formed by twoγ-ray decays,as also illustrated in Fig. 5.14.

• If nuclear statistical equilibrium is assumed, the concentrationof 12C∗ excited states is given by

n(

12C∗)

n3α

= 33/2(

h2

2πmαkT

)3

exp[(3mα −m∗12)c2/kT],

wherem∗12 is the mass of12C in the excited state (see Exercise5.9).


5.15.3 Formation of the Ground State in Carbon-12

• The preceding considerations determine a dynamical equilib-rium

4He+ 4He+ 4He ←→ 4He+ 8Be ←→ 12C∗.

• This produces an equilibrium population of12C∗, almost all ofwhich decays back to4He+ 8Be.

• However, the excited state of12C can decay electromagneticallyto its ground state with a mean life of

τ(

12C∗→ 12C(gs))

= 1.8×10−16 s,

• This implies thatone in about every 2500 excited carbon nucleithat are produced decay to the stable ground state.

• Because this decay probability is so small, it does not influencethe equilibrium appreciably and we may represent the entiretriple-α process schematically as

4He+ 4He+ 4He ↔︸︷︷︸

equilibrium

4He+ 8Be ↔︸︷︷︸

equilibrium

12C∗ −→︸︷︷︸leakage

12C(gs).

where left-right arrows indicate approximate nuclear statisticalequilibrium and the one-way arrow indicates a leakage from theequilibrium that is a small perturbation and therefore doesnotdisturb it significantly.


• Thus in the approximate equilibrium

4He+ 4He+ 4He ↔ 4He+ 8Be ↔ 12C∗ −→ 12C(gs).

The production rate for12C in its ground state is then the prod-uct of the equilibrium12C∗ population and the decay rate to theground state,

dn(

12C∗)

dt= n

(12C∗

)

︸︷︷︸

Number density12C*

×1

τ (12C∗→ 12C(gs))︸︷︷︸

Electromagnetic decay rate

=n3

ατ (12C∗→ 12C(gs))

33/2(

h2

2πmαkT

)3

× exp[−(m∗12−3mα)c2/kT],

where we have used

n(

12C∗)

n3α

= 33/2(

h2

2πmαkT

)3

exp[(3mα −m∗12)c2/kT],

We see that the rate of carbon production depends primarily ontwo things:

1. An activation energy given by(m∗12−3mα

)c2 = 0.3795 MeV

that must be borrowed to create the12C∗ intermediate state.

2. The mean life for the decay12C∗→ 12C(gs), which is

τ(

12C∗→ 12C(gs))

= 1.8×10−16 s.


Table 5.4: Parameters for the triple-α reaction

T (K) kT (keV) Q/kT exp(Q/kT)

5×107 4.309 −88.08 5.6×10−39

1×108 8.617 −44.04 7.5×10−20

2×108 17.234 −22.02 2.7×10−10

The strong temperature dependence for the triple-α reac-tion results primarily from the exponential Boltzmann fac-tor in

dn(

12C∗)

dt=

n(

12C∗)

τ (12C∗→ 12C(gs))

=n3

ατ (12C∗→ 12C(gs))

33/2(

h2

2πmαkT

)3

× exp[−(m∗12−3mα)c2/kT],

because at helium burning temperatures the average ther-mal energykT is typically much less than the activationenergy ofQ = −379.5 keV. This is illustrated in Ta-ble 5.4, where we see that doubling the temperature in thevicinity of 108 K changes the Boltzmann factor by tens oforders of magnitude.


5.15.4 Energy Production in the Triple-ααα Reaction

The total energy released in the triple-α reaction isQ =7.275 MeVand the energy production rate is given by

ε3α =5.1×108ρ2Y3

T39

exp(−4.4027/T9) erg g−1s−1,

whereY is the helium abundance. From

λ =

(∂ lnε∂ lnρ

)

Tν =

(∂ lnε∂ lnT

)

ρ.

• This implies density and temperature exponents

λ3α = 2 ν3α = 4.4/T9−3,

• The quadratic dependence on the density occurs be-cause the reaction iseffectively 3-body,since threealpha-particles must fuse to form the carbon; the de-pendence on temperature is very strong, as we havealready seen in connection with Table 5.4.

• For T8 = 1, we obtain a temperature exponentν3α ≃40, implying thata helium core is a very explosivefuel. a

aAn explosion is runaway burning (whether ordinary chemicalburning or ther-monuclear burning). If a thermonuclear fuel has a large temperature exponent, inthe corresponding temperature range the rate of burning canincrease enormouslyif the temperature increases only a little. This greatly increases the probability thatburning becomes explosive once it is initiated.

5.16. BURNING OF CARBON TO OXYGEN AND NEON 167

Temperature (K)

107

108

109

10-22

10-24

10-20

10-18

10-16

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100

102

Rate

(cm

3 m

ole

-1 s

-1)

12C + α 16O + γ

16O + α 20Ne + γ

Helium

Burning

Figure 5.16:Radiative alpha capture rates for12C and16O.

5.16 Burning of Carbon to Oxygen and Neon

• Once carbon has been formed by the triple-α sequence, oxygencan be produced by the radiative capture reaction

4He+ 12C→ 16O+ γ.

• There areno resonances near the Gamow windowfor this re-action, so the rate isslow (and experimentally uncertain). Thecurrently accepted rate is shown in the figure above.

• The uncertainty in this rate hassignificant astrophysical conse-quencesbecause it determines the ratio of carbon to oxygen toneon production in stars.


• In addition to the impact on abundance of these elements in theUniverse, the carbon–oxygen ratio in stellar cores can havelargeinfluence on various aspects of late stellar evolution.

For example, the composition of whitedwarfs and of the cores of massive stars latein their lives depend critically on this rate, soit can have a large impact on how stars dieand what is left behind when they do.

5.16. BURNING OF CARBON TO OXYGEN AND NEON 169

Temperature (K)

107

108

109

10-22

10-24

10-20

10-18

10-16

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100

102

Ra

te (

cm

3 m

ole

-1 s

-1)

12C + α 16O + γ

16O + α 20Ne + γ

Helium

Burning

Figure 5.17:Radiative alpha capture rates for12C and16O.

• Once oxygen is produced by4He+ 12C→ 16O+ γ, neon can beformed by4He+16O→ 20Ne+γ. The rate is plotted in Fig. 5.17.

• It is slowunder helium burning conditions becauseit is nonres-onantand has alarger Coulomb barrierthanα capture on12C.

• This implies thatlittle neon is producedduring helium burning.

The primary residue of helium burning is acarbon–oxygen core.

• The carbon is produced by the triple-α sequence and the oxygenby radiative capture on the carbon, with the ratio of carbon tooxygen depending strongly on the uncertain rate for the radiativecapture reaction.


Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars

12C survives because no

thermal resonance. Some 16O produced by tail of

9.58 MeV resonance and

7.12, and 6.92 MeV sub-

threshold resonances.

Little 20Ne because

4.97 MeV state in

Gamow window is

unnatural parity state

Resonance with allowed

quantum numbers in

Gamow window

1− can't decay to

ground state because

of isospin symmetry

2− Can't be populated

because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

Figure 5.18:Summary of helium burning.

5.17 The Outcome of Helium Burning

The outcome of helium burning is summarized inFig. 5.18. This outcome is a remarkable example ofhow fundamentally different our Universe would be if justa few seemingly boring details of nuclear physics wereslightly different.

5.17. THE OUTCOME OF HELIUM BURNING 171

Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars




7.12, and 6.92 MeV sub-


Little 20Ne because

4.97 MeV state in

Gamow window is



quantum numbers in

Gamow window

1− can't decay to


of isospin symmetry


because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

• The proportion of carbon to oxygen in the Universe is deter-mined by the competition between the carbon-producing triple-α reaction and the carbon-depleting, oxygen-producing radia-tive capture reaction4He+ 12C→ 16O+ γ.

• Further, that much carbon or oxygen exists at all is cruciallydependent on the slowness of the neon-producing reaction4He+16O→ 20Ne+ γ.


Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars




7.12, and 6.92 MeV sub-


Little 20Ne because

4.97 MeV state in

Gamow window is



quantum numbers in

Gamow window

1− can't decay to


of isospin symmetry


because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

• If, contrary to fact, a resonance existed near the fusion windowfor the reaction4He+ 12C→ 16O+ γ, the corresponding ratewould be large and almost all carbon produced by triple-α wouldbe converted rapidly to oxygen, leaving little carbon in theUni-verse.


Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars




7.12, and 6.92 MeV sub-


Little 20Ne because

4.97 MeV state in

Gamow window is



quantum numbers in

Gamow window

1− can't decay to


of isospin symmetry


because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

• A similar fate would follow if the 0+ excited state in12C at 7.65MeV were a little higher in energy, since this would greatly slowthe triple-α rate by virtue of the Boltzmann factor in

dn(

12C∗)

dt=

n3α

τ (12C∗→ 12C(gs))33/2

(h2

2πmαkT

)3

× exp[−(m∗12−3mα)c2/kT],

and any carbon that was produced would be converted rapidly tooxygen through4He+ 12C→ 16O+ γ.


Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars




7.12, and 6.92 MeV sub-


Little 20Ne because

4.97 MeV state in

Gamow window is



quantum numbers in

Gamow window

1− can't decay to


of isospin symmetry


because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

• Conversely, if the resonance at 7.65 MeV in12C did not exist,the triple-α reaction would not work at all in red giant stars andthere would be little of either carbon or oxygen in the Universe.


Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars




7.12, and 6.92 MeV sub-


Little 20Ne because

4.97 MeV state in

Gamow window is



quantum numbers in

Gamow window

1− can't decay to


of isospin symmetry


because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

• Finally, if the neon-producing reaction4He+ 16O→ 20Ne+ γ.

were (contrary to fact) resonant—which it would be if the parityof a single excited state in neon were positive instead of negative,

– Most of the carbon and oxygen produced by helium burningwould be transformed by this reaction to neon.

– Neon is a noble gas and therefore essentially chemically in-ert, in contrast to the rich chemistry of carbon that makesbiology as we know it possible.


Gamow windows

Narrow resonance

Broad resonance

8Be thermal

equilibrium

12C is created

through thermal

resonance

Helium Burning

Reactions in

Red Giant Stars




7.12, and 6.92 MeV sub-


Little 20Ne because

4.97 MeV state in

Gamow window is



quantum numbers in

Gamow window

1− can't decay to


of isospin symmetry


because of parity

conservation

2.94 2+

0 0+7.65 0+

9.61 3−

4.44 2+

0 0+

0 0+

0 0+

9.58 1−

8.87 2−

7.12 1−

6.13 3−6.92 2+

6.05 0+

5.78 1−

7.00 4−

5.62 3−

1.63 2+

6.72 0+

4.97 2−

4.25 4+

20Ne

16O

12C

8Be4He α α

α

α

Q = 4.73

Q = 7.16

γ

γ

Q = 7.37Q = -0.09

• Thus, our very existence appears to depends on theparity ofobscure nuclear states in atoms that have nothing whatsoever todo with the chemistry of life!


The Anthropic Principle and Helium Burning: Somewould argue, based on the observed diversity of life onEarth and how quickly it arose after formation of theplanet, that life in the Universe is inevitable.

• But this point of view assumes the existence of thechemicals on which life (as we know it) is built.

• The preceding discussion suggests that the existenceof the building blocks of life depends on arcane factson the MeV scale (nuclear physics) that have nothingto do with the physics of eV scales (chemistry) thatgoverns life.

• The very possibility of biochemistry may be an acci-dent of physical parameter values in our Universe.

• Such considerations lie at the basis of the (simplest)anthropic principle:

The Universe has just the right value of con-stants and just the detailed physics like theenergy and parity of obscure nuclear statesrequired for life because, if it didn’t, therewould be no life in the Universe and there-fore no one to ask the question.

• It is not clear whether this line of thinking, as intrigu-ing as it is, enhances our scientific understanding ofthe Universe (or other possible universes).


If a star is massive enough, more advanced burnings arepossible by virtue of the high temperatures and pressuresthat result as the core contracts after exhausting its fuel.Typical burning stages in massive stars and their charac-teristics are illustrated in the following table and figure.

5.18. ADVANCED BURNING STAGES 179

Table 5.5: Burning stages in massive stars (Woosley)

Nuclear Nuclear Ignition Minimum main Period infuel products temperature sequence mass 25M⊙ star

H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years

C Ne, Na, Mg, O 6×108 K 4M⊙ 600 years

Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

5.18 Advanced Burning Stages

Stellar Burning Shells

Hydrogen

Oxygen

Carbon

Silicon

Iron

Helium Center of 25 Solar

Mass Star

25 M

T = 4 ×109 K

ρ = 107 gcm

-3

T = 2 ×107 K

ρ = 102 gcm

-3



H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years


Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

Carbon burning: Carbon burns at a temperature ofT ∼5×108 K and a density ofρ ∼ 3×106 g cm−3, primarilythrough

12C+ 12C−→ 20Ne+α12C+ 12C−→ 23Na+ p12C+ 12C−→ 23Mg+ p

As indicated in the table above, such reactions are possiblefor stars having masses larger than about 4M⊙.

• Burning stages beyond that of carbon require condi-tions that are probably realized only for stars havingM >

∼ 8M⊙ or so.

• At the required temperatures, a new feature comesinto play becausethe most energetic photons can dis-rupt the nuclei produced in preceding burning stages.



H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years


Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

Neon burning: At T ∼ 109 K, neon can burn by a two-step sequence. First, a neon nucleus isphotodisintegratedby a high-energy photon(which become more plentifulas the temperature is increased since the average photonenergy is∼ kT)

γ + 20Ne−→ 16O+α.

Then the alpha-particle produced in this step can initiate aradiative capture reaction

α + 20Ne→ 24Mg+ γ.

This burning sequence produces a core of16O and24Mg.



H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years


Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

Oxygen burning: At a temperature of2×109 K, oxygencan fuse through the reaction

16O+ 16O−→ 28Si+α

The silicon thus produced can react only at temperatureswhere the photon spectrum is sufficiently “hard” (has sig-nificant high-energy components) that photodissociationreactions begin to play a dominating role.



H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years


Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

Silicon burning: At T ∼ 3×109 K, silicon may be burnedto heavier elements.

• At these temperatures the photons are quite energeticand those in the high-energy tail of the Maxwell–Boltzmann distribution can readily photodissociatenuclei.

• A network of photodisintegration and capture reac-tions in approximate nuclear statistical equilibriumwith each other develops and the population in thisnetwork evolves preferentially to those isotopes thathave the largest binding energies.

• Since from the plot of the binding energy curve themost stable nuclei are in the iron group, silicon burn-ing carried to completion under equilibrium condi-tions tends to produce iron-group nuclei.


Temperature (K)

108

109

1010

10-22

10-24

10-20

10-18

10-16

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100

102

Ra

te (

s-1

)28Si + γ 24Mg + α

Helium

Burning

Figure 5.19:The temperature dependence of the rate-controlling step insiliconburning. For reference, the typical range of temperatures corresponding to heliumburning is indicated.

• The initial step in silicon burning is a photodisintegration like

γ + 28Si−→ 24Mg+α Q =−9.98 MeV,

which requires a photon energy of9.98 MeVor greater.

• From Fig. 5.19 we infer that silicon burning has an extremelystrong temperature dependence (nearT9 = 3 the T exponent isν ∼ 50), and it requires temperatures more than an order of mag-nitude larger than for helium burning.


• Theα particles thus liberated can now initiate radiative capturereactions on seed isotopes in the gas; a representative sequenceis

α + 28Si←→ 32S+ γ

α + 32S←→ 36Ar + γ...

α + 52Fe←→ 56Ni + γ

• The reactions in this series are typically inequilibrium or quasiequi-librium, and are much faster than the initial photodisintegration.

• Thus the photodisintegration of silicon is therate-controllingstepin silicon burning.


-8

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Log R

ate

20Ne(α,γ)24Mg

24Mg(α,γ)28Si

28 Si(α,γ)

32 S

32 S(α,γ)36 Ar

36 Ar(α,γ)

40 Ca

40 Ca(α,γ)

44 Ti

20Ne(γ,α)16O

36Ar(γ,α)32S

24Mg(γ,α)20Ne

40Ca(γ,α)36Ar

52Fe(γ,α)48Cr

48Cr(γ,α)44Ti

32S(γ,α)28Si

28Si(γ,α)24Mg

44Ti(γ,α)40Ca

44Ti(α,γ)48Cr

20Ne(12C,α)28Si

48Cr(α,γ)52Fe

52Fe(α,γ)56Ni

1 2 3 4 5 6 7 8 9 10

Temperature (109 K)

Figure 5.20:Some rates for competing capture reactionsA(α ,γ)B and photodis-integration reactionsA(γ ,α)B that are important for silicon burning. The photodis-integration rates are given in units of s−1. Theα-capture reaction rates are given inunits of cm3 mole−1 s−1. To convert theα-capture rate for the reaction ofα +A tounits of s−1, it must be multiplied by a molar density factorρYαYA, whereρ is themass density,Yα is theα-particle abundance, andYA is the abundance of isotopeA.

• The rates for some of the competing capture and photodisinte-gration reactions in silicon burning are illustrated in Fig. 5.20.

• Note the steep temperature dependence of the photodisintegra-tion reactions, and that for high enough density andα-particleabundance many of the photodisintegration rates will becomecomparable to the rates for their inverse capture reactionssome-where in the temperature range 109–1010K.


• Since the iron group nuclei are the most stable in the Universe,silicon burning represents the last stage by which fusion and ra-diative capture reactions can build heavier elements underequi-librium conditions.

• One might think that we could make still heavier elements byincreasing the temperature, so that the required extra energy forfusion presented by higher Coulomb barriers could be providedby the kinetic energy of the gas.

• But this becomes self-defeating in statistical equilibrium becausethe higher temperatures will also lead to increased photodissoci-ation and iron-group nuclides are still the equilibrium product.

• In subsequent chapters we will address the issue of other mecha-nisms by which stars can produce the elements heavier than ironthat our Universe obviously contains.



H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years


Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

5.19 Timescales for Advanced Burning

As is apparent from the table above, the timescales for ad-vanced burning are greatly compressed relative to earlierburning stages. These differences are particularly strikingfor very massive stars, which rush through all stages atbreakneck speed. For example, the25M⊙ example usedfor the above table

• Takes only about 10 million years to advance throughits hydrogen and helium burning phases,

• Completes its burning of oxygen in only six months,and

• Transforms its newly-minted silicon into iron groupnuclei in a single day.

5.19. TIMESCALES FOR ADVANCED BURNING 189


H He 4×106 K 0.1M⊙ 7×106 years

He C, O 1.2×108 K 0.4M⊙ 5×105 years


Ne O, Mg 1.2×109 K ∼ 8M⊙ 1 years

O Si, S, P 1.5×109 K ∼ 8M⊙ ∼ 0.5 years

Si Ni–Fe 2.7×109 K ∼ 8M⊙ ∼ 1 day

These timescales are set by

• The amount of fuel available,

• The energy per reaction derived from burning thefuel, and

• The rate of energy loss from the star, which ulti-mately governs the burning rate.

• The last factor is particularly important because en-ergy losses are large when the reaction must run athigh temperature.

Each factor separately shortens the timescale for advancedburning; taken together they make the timescales for themost advanced burning almost instantaneous on the scaleset by the hydrogen burning.


An Analogy: To get a perspective on how short the ad-vanced burning timescape is, imagine the lifetime of the25M⊙ star to be compressed into a single year.

• Then the hydrogen fuel would be gone by about De-cember 7 of that year,

• The helium would burn over the next 24 days,

• The carbon would burn in the 42 minutes before mid-night,

• The neon and oxygen would burn in the last severalseconds before midnight, and

• The silicon would be converted to iron in the last1/100 second of the year (with a quite impressiveNew Year’s Eve fireworks display in the offing—seethe later discussion of core-collapse supernovae).

Documents

Chapter 5 Stellar Energy Production - University of …eagle.phys.utk.edu/guidry/astro615/lectures/lecture_ch5.pdfChapter 5 Stellar Energy Production ... star birth and star death