Sec on 4.
Stellar evolu on (I)4.1 Virial theorem
4.2 Evolu on of density and temperature

titi

ti

4.3 Burning stages

Let’s understand these ques ons
with the words of physics• Why are stars so luminous?
• Why do stars show L ~ M ?
• Why do stars evolve?
• Why does the des ny of stars depend on the mass?
• Why do some stars explode?
• Why don’t normal star explode?
• Why does stellar core collapses?
• Why is the energy of supernova so huge?
•…
ti

ti4

Sec on 4.
Stellar evolu on (I)4.1 Virial theorem
4.2 Evolu on of temperature and densitytititi

4.3 Burning stages

Stellar life

(C: Essay Web)

M惑星状星雲> 10 Msun

1. Massive stars
軽い星

赤色巨星

~ 10 Myr

図の大きさは天体の大きさと一致していません(C: Essay Web)

白色矮星

H

HeFe

Si

C+O

Ne + Mg

Images are not to scale

Why do stars evolve??
“Evolu on” = Changes in the state with metiti

What happens when there is no more fuel for nuclear burningEtot: Total energy
Ω: Gravita onal energy
U: Internal energy

EU1⌦
2

U=01Etot = U + ⌦ = ⌦ =2Utititi

No nuclear burning- Total energy decreases- Contrac on (gravita onal energy decreases)- Temperature rises

EtotΩ

HHeC+O

Contrac on=> temperature rise

HeTriple alphaΤ ~ ρ1/3

ti

Images are not to scale=> C

Heated ironstars

http://iron.minatoseiki.co.jp/seizo.html

tt

Gets colder

Gets ho er

Sec on 4.
Stellar evolu on (I)4.1 Virial theorem
4.2 Evolu on of density and temperature

tititi4.3 Burning stages

Nuclear binding energy
Eb = [NmN + Zmp - mi] c2 > 0Larger binding energy= more stable

Fe has the largest
Eb/nucleon

Then, all the stars produce Fe? => NoStellar material does not always behave as ideal gas

PhaseMain reac ons

ProductsT

Nuclear sta s cal equilibrium
titi

ti

元素はいかにつくられたか（岩波書店）

ρ-T plane
密度 - 温度平面

Summary: Stellar evolu on (I)• Virial theorem (for ideal gas case)• Internal energy always relates with gravita onal energy• When stars lose energy, they contract• Temperature rises (“nega ve heat capacity”)• Evolu on of density and temperature• Rise in temperature due to contrac on T ~ ρ• Next burning stages => Onion-like structure• Do all the stars produce Fe?? => No.
1/3titititititi

Equa on of states plays an important roleAppendix

imes that of a typical reaction involving only strongTheinteractions.
Therate cannotbnet result
ofreactionhydrogenburning
ured in the laboratory and is only known from theory.
1 another4 proton to +from 3 He.
After some deuterium is produced, it rapidly reacts with4 H → He + 2 e + 2 ν .
quently three different branches are possible to complete the chain towards 4 He:

1a. H-burning (pp chain)

You may verify using Sec. 6.1.1 th
1 H + 1 H → 2 H + e+ + ν
nucleus two protons have to be con
2 H + 1 H → 3 He + γ
interactions (p → n + e+ + ν), whic!#!
##
not to include the neutrino energi!
##"!
!
$theenergy
thatisusedtoheat
the3 He + 3 He → 4 He + 2 1 H
3 He + 4 He → 7 Be + γ(includingthe γ-rays resulting fro#!
!
#!
#energies
oftheresultingnuclei.
T#!
pp1"!
$#smaller
than26.734
MeVanddep
7−771
8Be + e → Li + ν

7 Li + 1 H → 4 He + 4 He

pp2

Be + H → B + γ

8 B → 8 Be + e+ + ν

8 Be → 4 He + 4 He

pp3

The pp1 branch requires two 3 He nuclei, so the first two reactions in the chain have to takeTextbook
by Pols
wice. The alternative pp2 and pp3 branches
requireonly one 3 He nucleus and an already existi

Energy produc on rate(per gram)q ~ ρT4Since a simultaneous reaction between four protons is extremely unlikely, a chain of reactions isways necessary for hydrogen burning. This can take place in two distinct ways: either direct fusion
protons via the p-p chain, or by using already present CNO-nuclei as catalysts in the CNO cycle.
ydrogen burning in stars takes place at temperatures ranging between 8 × 106 K and 5.0 × 107 K,
pending on stellar mass and evolution stage.

he p-p chains

he first reaction is the so-called p-p reaction:1

H + 1 H → 2 H + e+ + νorp + p → D + e+ + ν .

(6.45)

T ~ 4 x 106 K

his involves the simultaneous β-decay of one of the protons during the strong nuclear interaction.
his is very unlikely and the p-p reaction therefore has an extremely small cross-section, about 10−20
mes that of a typical reaction involving only strong interactions. The reaction rate cannot be meared in the laboratory and is only known from theory.
After some deuterium is produced, it rapidly reacts with another proton to from 3 He. Subseuently three different branches are possible to complete the chain towards 4 He:
1 H + 1 H → 2 H + e+ + ν
2 H + 1 H → 3 He + γ!!
"!###!!

3 He + 3 He → 4 He + 2 1 H##$3 He + 4 He → 7 Be + γ!
"!
!

pp1! !####$#

7 Be + e− → 7 Li + ν

7 Be + 1 H → 8 B + γ

pp2

8 Be → 4 He + 4 He

7 Li + 1 H → 4 He + 4 He

8 B → 8 Be + e+ + ν

pp3(6.46)

he pp1 branch requires two 3 He nuclei, so the first two reactions in the chain have to take placewice. The alternative pp2 and pp3 branches require only one 3 He nucleus and an already existing 4 Heucleus (either present primordially, or produced previously by hydrogen burning). The resulting 7 Beti

Textbook by Pols

Textbook by Prialnik

The CNO cycle

16E
produconrateq~
ρT
1b.
Hburning(CNOcycle)
If some C, N, and O is already present in the gas out of which a star forms, and if the temperatu

7 Kcycle. This is a cyclic
is sufficiently high, hydrogen fusion can take place via theCNO
T ~so-called
1.5 x 10sequence of reactions that typically starts with a proton capture by a 12 C nucleus, as follows:!
12 C + 1 H → 13 N + γ

13 N → 13 C + e+ + ν

13 C + 1 H → 14 N + γ" 14N + 1 H → 15 O + γ

15 O → 15 N + e+ + ν

15 N + 1 H → 12 C + 4 He→ 16 O + γ

16 O + 1 H → 17 F + γ

17 F → 17 O + e+ + ν

17 O + 1 H → 14 N + 4 He

For example, if we denote by rpD the rate of 2 H + 1 H, one has rpD = rpp , etc. Note that describing the p-p reaction
Textbook by Prialnik
Textbookby Pols
21
‘slow’ and the H + H as ‘fast’ refers to the difference in cross-section factors #συ$ and not to the number of reactions pti

1(6.4aneously, the 3α reaction behaves as if it were a three-particle reaction and its rate is pr
4 He into a mixture of 12 C and 16 O, which ta
3eliumburning
consistsofthefusionofnα . The energy-generation rate can be written as
2. He-burning(triplealpha)8 K. Suchmperatures
T >10hightemperatures
are needed because (1) the Coulomb∼3 2%3α is= higherq3α X4 thanρ λ3αthat, of the H-burning reactions considered above, and (2) fusio
e fusionnderedthe
bytemperature
the fact that dependence
no stable nucleus
exists withmass
number= 8. Therefore
helit
whereis describedby thefactor
λ3α ,Awhich
depends onust
occur in twoross-sectionsof steps:
the two reactions (6.52). X4 ≈ Y is the mass fraction of 4 He. The
4 He +
8 Be is extremely high, with ν ≈ 40 at T 8 ≈ 1.0.
ensitivityof4theHe 3α↔ rate12 C has been created by the 3α reaction, it can captur
Whenasufficientamountof8 Be + 4 He → 12 C∗ → 12 C + γ
particle to form 16 O,
he 8 Be12nucleus
temporarily
formed in the first reaction has a ground state that is 92 ke
416C + He → O + γ, 4nergy than that of two separate He nuclei. It therefore decays back into two α particles
me 10−16 s. While extremely short, this time is long enough to build up a very small
91
8
oncentration of Be, which increases with temperature and reaches about 10−9 at T ≈ 1
e second reaction 8 Be(α, γ)12 C starts to occur at a significant rate, because of a reson∗Energyproduconrate
12e Gamow peak energy. The result is an excited compound nucleus C which subsequegram)the ground state(perof 12 Cwith emission of a γ photon. The corresponding energy leveq ~byρ2Fred
T40 Hoyle in 1954, because he could not otherwise explain thucleus was predicted
large amounts of carbon in the Universe. This excited state of 12 C was subsequentboratory experiments.
8KT~
1.5x10The net effect of the two reactions (6.52) is called the triple-α reaction,
412ti

Textbook by Prialnikcleus is formed which can then decay via many different channels. The most important cNeon burning The next nuclear burning cycle might be expected to be oxygen fusion, b
the following:t somewhat lower temperature (T 9 ≈ 1.5) a process called ‘neon burning’ is initiated by
20 Ne. At this temperature a sufficient number of photons have energies inisintegrationof
12 C + 12 C → 24 Mg∗ → 20 Ne + α
Q = 4.616 MeV 20(∼ 50%)
16 O and 4 H
angewhichissufficient
tobreakup
therelatively
fragileNenucleus
into
8 be
eon burning The next 23
nuclear burning cycle
might
expected
tobeoxygen
T~
7x10K
20 Ne nucleus, thus:→ capture
Na + pof the Q=
2.238 MeV(∼ 50%)
mmediately
followedbytheαparticlebyanothersomewhat lower temperature (T ≈ 1.5) a process called ‘neon burning’ is init

3. C-burning

92016
20 Ne + γofisintegration
Ne.
this temperaturea sufficientnumberphotons haveeO +At
αreleased
Qfind= −4.73
MeVe protons and α↔
particlesthemselves
at extremely
highoftemperaturescomp
20 Ne nucleus into 16 O
angewhichissufficient
tobreakup
therelatively
fragilese neededheliumburning,and will almost immediately react with othe
4.20Ne-burning
Nefor+ αhydrogen→1224 Mgand+γ
Q=
9.31
MeV
24 Mg. Examples are 23 Na(p, α)20 Ne, 20 Ne(α, γ)he mixture, fromC toby
Mgnucleus,
and chainsmmediately
followed
the capture of the α particle by another 2024Neth

13 N(e+ ν)13 C(α, n)16 O, where the neutron will immediately react further. TheC(p,γ)
he first reaction is endothermic, but effectively the two reactions combine to 2 20 Neoverall→ 1620 Ne + γ ↔ 16 O + α
16 O
Q=MeV
easeaisnetobtained
from theof −4.73
all thesereactions
and isismostly
roughly
Qand≈ 2413Mg
Mwith
energy release
Q >combination
0. The composition
afterneon burning
9 K 16 20T~
1.5x
101224 Mg (tC+Creaction.
Themainproducts
after
exhaustion
ofall
carbonareO,Ne
and20
245% by massfraction).
Ne +α → Mg + γ
Q = 9.31 MeV% by mass fraction). These most abundant nuclei have equal numbers of protons and neut
21,22 Ne, 23 Na and 25,26 Mg, so th
meofthesidereactionsproduceneutron-richisotopes
like16
he firstburning
reaction isAtendothermic,
the in,tworeactionscombine
2 20
OxygenT 9 ≈ 2.0 fusionbutof effectively
O nuclei sets
which
is in many
ways toanalogburning the overall composition has a ‘neutron excess’ (n/p > 1, or µe > 2).
16 O
arbon
fusionreaction
described
above.
in this casethere
are severalchannels
ith a net
energyreleaseQ > 0.
The Also
composition
afterneonburningreaction
is mostly
mportantones fraction).
being:
5% by mass
92
16 O + 16 O → 32 S∗ → 28 Si + α
Q = 9.59 MeV (∼ 960%)

5. O-burning

T~2-3x10K
16≈ 2.0 fusion of O nuclei sets in, which is in many wa

Oxygen burning At T→
9 31 P + p
Q = 7.68 MeV (∼ 40%)
arbon fusion reaction described above. Also in this case there are several reaction
imilar to ones
carbonbeing:burning, the p and α particles are immediately captured by other nucmportanttitititiofSiandS(together90%
bymassfraction).
The
netenergyreleasepee of secondary reactions that eventually lead to a composition
mostly cons+
Q≈
16
MeV.
Since
someofthesidereactionsinvolveβ-decaysandelectro1616
Silicon 90%
burning
Thefraction).
lightest andmostabundantnucleusin theashes
ofreacoxy
ogetherbymassThenet
energyreleaseperO+
O
6. Si-burning（Nuclearstascal
equilibrium）
excessofthe
final
mixtureisfurtherincreased.
Coulombfor 28 Si involve+ 28 Si fusion
is prohibitively
high.
Insteadthe
siliccethe
someof thebarrierside reactions
β+ -decays
and electroncaptures,
ne9K
T>
4x10
a series ofphoto-disintegration
mixtureis furtherincreased. (γ, α) and α-capture (α, γ) reactions when T 9
4 He ofHightemperature=>
photo-dissociaon nucleus
Siliconburning
lightestandmost part
abundantin the ashes‘melts’intolighter The
nuclei,while
anothercaptures
the released
to oxyma
the Coulomb barrier for 28 Si + 28 Si fusion is prohibitively high. Instead silic
28 S
28 Si (γ,24 Mgabundant
201612The lightestandα)mostnucleus
intheashesofoxygenburningis(γ, α) Ne(γ,α)
O (γ, α) C(γ,α)
2αa seriesofphoto-disintegration(γ,α)
andα-capture(α,γ)
reactionswhen T 928
28
er for 28
Si + Si fusion
is prohibitively
high. Instead
silicon burning
procee
432
36
40
4456
‘melts’ intolighter
anotherparttheγ)released
Si (α,γ) nuclei,S (α, γ)while
Ar (α,γ) Ca(α,capturesγ) Ti (α,.>. . Ni He to ma
isintegration (γ, α) and α-capture (α, γ) reactions when T 9 ∼ 3. Part of the s4 He28
2420
1612 C
He
capturerMostnuclei,while
anotherpart
capturesthereleasedto(γ,makeheavierSi(γ,α)Mg(γ,α)Ne(γ,α)O(γ,α)α) 2αof these reactions are in equilibrium with eachother,e.g. 28 Sinuclei:+γ
28 Si20(α,3236Ar
40 Caby44 Ti (α,56 Niof thenuclei
canbe (α,describednuclear
equivalentsof the Sahaγ)
S(α,γ)
γ)
(α,γ)
γ)
.
.
.
4abundances
1612Mg (γ, α) Ne (γ, α) O (γ,9 α) C (γ, α) 2αequilibrium. For T > 4 × 10 K a state close to nuclear statistical equilibrium
2 S (α, γ) 36Ar (α, γ) 40 Ca (α, γ) 44 Ti (α, γ) . . . 56 Ni
28 Si + γ
Mostofthesereactionsareinequilibrium
with
each
other,
e.g.
where the most abundant nuclei are those with thelowest binding energy, c(Ex.)
abundancesof the nuclei
canbe reac
described
by final
nuclearequivalentsis ofthemostl
Saha=>ofequilibrium
many
ons
numberneutrons
andofprotons
present.
Thecomposition
then
28 Si + γ ↔ 24 Mg + α, an9ctions
areinequilibrium
with
each
other,e.g.
equilibrium.
For andT >e4− -captures× 10 K a during
state close
to nuclear
statistical(dueto β-decays
previousburningcycles). equilibrium
nucleidescribed
bynuclei
nuclearthelowest
Saha equation
for ioni
where canthe bemostabundant
areequivalents
those withofthebinding energy,
9>Nuclei
4 × 10ofwith
Ka highstate binding
closeprotons
to nuclear
statisticalequilibrium
(NSE)
canbe
rea
numberneutronsandpresent.
The
finalcomposition
isthenmostl
energy tend to be produced (Fe, Co, Ni)
6.5emission
bundant
nuclei areandthosewith the during
lowestprevious
binding burningenergy, cycles).
constrained by the(due to Neutrinoβ-decayse− -capturess and protons present. The final composition is then mostly 56 Fe because n/