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

Goals of this lecture• Standard proper$es of stars• Stellar structure and proper$es• Stellar evolu$on• Origin of the elements in the Universe• Nucleosynthesis in stars and supernovae• Explosion mechanism of supernovae• Topics in $me-domain astronomy• Radia$on from explosive phenomena• Mul$-messenger astronomy

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

Stellar life

(C: Essay Web)

1. Massive stars
軽い星

赤色巨星

惑星状星雲
M > 10 Msun

~ 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

HHeFe

SiC+O

Ne + Mg

Images are not to scale

Why do stars evolve??
“Evolu$on” = Changes in the state with $me
What happens when there is no more fuel for nuclear burningEtot: Total energy
Ω: Gravita$onal energy
U: Internal energy

EU
U=1
⌦21
Etot = U + ⌦ = ⌦ =20
U

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

EtotΩ

HHeC+O

Contrac$on=> temperature rise

HeTriple alphaΤ ~ ρ1/3Images are not to scale=> C

Heated ironstars

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

Gets colder

Gets ho_er

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

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

Larger 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
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.
1/3

Equa$on of states plays an important roleLet’s understand these ques$ons
Knowing = Understanding
with the word of physics• Why do some stars explode?
• Why don’t normal star explode?
• Why do stars show L ~ M ?
• Why do stars evolve?
• Why does the des$ny of stars depend on the mass?
• Why does stellar core collapses?
• Why is the energy of supernova so huge?
•…
4Thermodynamics

Electromagne$sm

Classical
mechanics

Sta$s$cal
mechanics

Astrophysics
Hydrodynamics

Rela$vity

Quantummechanics
Nuclear physics

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 with
4 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+ + νnucleus two protons have to be con2 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❍
✟
✟
❍
✟
❍
energiesoftheresultingnuclei.
T❍
✟
pp1
✙
✟
❥
❍
smallerthan26.734
MeV
anddep7−771
8Be + e → Li + ν

7 Li + 1 H → 4 He + 4 He

pp2Be + H → B + γ

8 B → 8 Be + e+ + ν8 Be → 4 He + 4 He

pp3

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

Energy produc$on rate
(per gram)q ~ ρT4

Since 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:
1H + 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 10−20mes 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 + ν7 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 Be

Textbook by Pols

Textbook by Prialnik

The CNO cycle

16E
produc$onrateq~
ρT1b.
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 x1210
sequence of reactions that typically starts with a proton capture by a 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 He1(6.4For 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
Textbook by Pols
‘slow’ and the 2 H + 1 H as ‘fast’ refers to the diﬀerence in cross-section factors ⟨συ⟩ and not to the number of reactions paneously, the 3α reaction behaves as if it were a three-particle reaction and its rate is pr4 He into a mixture of 12 C and 16 O, which ta
3eliumburning
consistsofthefusion
ofnα . The energy-generation rate can be written as
8 K. Such
2. He-burning(triplealpha)mperatures
T >10hightemperatures
are needed because (1) the Coulomb∼ 3 2
ϵ3α is= higher
q3α X4 thanρ λ3αthat, of the H-burning reactions considered above, and (2) fusio
e fusionnderedthe
bytemperature
the fact that dependence
no stable nucleus
exists withmassnumber= 8. Therefore
helit
whereis describedby thefactor
λ3α ,Awhichdepends 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
8oncentration of Be, which increases with temperature and reaches about 10− 9 at T ≈ 1e second reaction 8 Be(α, γ)12 C starts to occur at a significant rate, because of a reson∗
Energyproduc$onrate
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 leveucleus was predictedq ~byρ2Fred
T40 Hoyle in 1954, because he could not otherwise explain thlarge amounts of carbon in the Universe. This excited state of 12 C was subsequentboratory experiments.
8K
T~
1.5x
10
The net eﬀect of the two reactions (6.52) is called the triple-α reaction,
Textbook by Prialnik412

cleus is formed which can then decay via many diﬀerent channels. The most important c
Neon burning The next nuclear burning cycle might be expected to be oxygen fusion, bthe 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
ange
which
issuﬃcienttobreakuptherelatively
fragileNenucleus
into
8 be
eon burning The next 23
nuclear burning cycle
might
expectedtobe
oxygenT~
7x10K
20 Ne nucleus, thus:→ capture
Na + pof the Q=2.238 MeV(∼ 50%)
mmediately
followedbytheα
particlebyanothersomewhat lower temperature (T ≈ 1.5) a process called ‘neon burning’ is init

3. C-burning

92016
20 Ne + γofisintegration
Ne.
this temperaturea suﬃcientnumberphotons haveeO +Atαreleased
Qfind= −4.73
MeV
e protons and α↔
particles
themselves
at extremely
highoftemperaturescomp
20 Ne nucleus into 16 O
ange
which
issuﬃcient
tobreakup
therelatively
fragilese neededheliumburning,
and will almost immediately react with othe
20 Nefor+ αhydrogen→1224 Mgand+γQ=
9.31
MeV
4.
Ne-burning
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 2024Ne
th13 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 + α16 O
Q=
4.73
MeV
easeaisnetobtained
from theof −allthesereactions
and isismostly
roughlyQand≈ 2413Mg
Mwith
energy release
Q >combination
0. The composition
afterneon burning
9 K 16 201224 Mg (tT~
1.5x10C+C
reaction.
Themainproducts
afterexhaustion
ofallcarbon
areO,Neand20
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
meof
thesidereactionsproduceneutron-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 toanalog
burning the overall composition has a ‘neutron excess’ (n/p > 1, or µe > 2).
16 O
arbon
fusionreaction
described
above.
in this casethere
are several
channels
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-burningT~
2-3x10K16≈ 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 nuc
mportant

ofSiandS(together90%
bymassfraction).
Thenetenergyreleasepee of secondary reactions that eventually lead to a composition
mostly cons+Q≈
16MeV.
Since
someofthesidereactions
involveβ-decaysandelectro
1616
Silicon 90%
burningThefraction).
lightest andnucleus
in theoxy
ogetherby mass
Themostnet abundant
energy release
perO ashes+ Oofreac
6.
Si-burning（Nuclear
sta$s$calequilibrium）
excessof
thefinal
mixtureisfurtherincreased.
2828+Coulomb
for Si involve+ Si fusion
is prohibitively
high.
Insteadthe
silic
cethe
someof thebarrierside reactions
β -decays
and electron
captures,
ne9K
T>
4x10
a series ofphoto-disintegration
mixtureis furtherincreased. (γ, α) and α-capture (α, γ) reactions when T 9
4 He of
Siliconburning
lightest
andmost part
abundant
nucleusin the ashes
Hightemperature=>
photo-dissocia$on‘melts’intolighter The
nuclei,while
anothercaptures
the released
to oxyma
the Coulomb barrier for 28 Si + 28 Si fusion is prohibitively high. Instead silic
28 S
28 Si (γ,24 Mgabundant
20
16
12The lightestandα)most
nucleusintheashesof
oxygenburningis(γ, α) Ne(γ,α)
O (γ, α) C(γ,α)
2αa seriesofphoto-disintegration(γ,α)
andα-capture(α,γ)
reactionswhen T 92828er for 28
Si + Si fusion
is prohibitivelyhigh. Insteadsilicon burningprocee
43236
404456‘melts’ intolighter
anotherparttheγ)released
Si (α,γ) nuclei,
S (α, γ)while
Ar (α,γ) Ca(α,capturesγ) Ti (α,.>. . Ni He to ma
isintegration (γ, α) and α-capture (α, γ) reactions when T 9 ∼ 3. Part of the s
4 He
28
24
2016
12 C
He
capturerMostnuclei,
while
anotherpart
capturesthereleasedto(γ,makeheavierSi(γ,α)Mg(γ,α)Ne(γ,α)O(γ,α)α) 2αof these reactions are in equilibrium with eachother,e.g. 28 Sinuclei:+γ
28 Si20(α,3236Ar
40 Caby44 Ti (α,56 Niof thenuclei
canbe (α,describednuclear
equivalentsof the Sahaγ)
S(α,γ)γ)(α,γ)γ).
.
.
4abundances
1612Mg (γ, α) Ne (γ, α) O (γ,9 α) C (γ, α) 2αequilibrium. For T > 4 × 10 K a state close to nuclear statistical equilibrium
2 S (α, γ) 36Ar (α, γ) 40 Ca (α, γ) 44 Ti (α, γ) . . . 56 Ni
28 Si + γ
Mostofthesereactionsareinequilibriumwitheachother,e.g.
where the most abundant nuclei are those with thelowest binding energy, c(Ex.)
abundancesof the nuclei
canbe reac$ons
describedTheby final
nuclearequivalentsis ofthemostlSaha
numberneutrons
andofprotons
present.
composition
then=>ofequilibrium
many
28 Si + γ ↔ 24 Mg + α, an9ctionsareinequilibriumwitheachother,e.g.
equilibrium.
For andT >e4−-captures× 10 K a during
state close
to nuclear
statistical(dueto β-decays
previousburningcycles). equilibrium
nucleidescribedbynuclei
nuclearthelowest
Saha equation
for ioni
where canthe bemostabundant
areequivalents
those withofthebinding energy,9> 4 × 10of K
a state andcloseprotons
to nuclear
statisticalequilibrium(NSE)
can berea
numberneutronspresent.
Thefinalcomposition
isthenmostl
Nuclei with high binding
energytend
tobe
produced(Fe,Co,Ni)
6.5
emissionbundant
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/