Sec on 6.
星の進化 (1)
6.1 ビリアル定理

ti

6.2 温度と密度の進化

星の一生

惑星状星雲

軽い星

赤色巨星

赤色超巨星
重い星

白色矮星

超新星爆発
中性子星

星間空間
リゲル
図の大きさは天体の大きさと一致していません

ベテルギウス(C) Essay Web

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1. 重い星の場合
軽い星

惑星状星雲
* 太陽の10倍以上

赤色巨星

白色矮星

約1千万年
赤色超巨星
重い星

超新星爆発
中性子星

星間空間

図の大きさは天体の大きさと一致していません(C: Essay Web)

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ヘリウム

水素

炭素
酸素

約1千万年
鉄

図の大きさは天体の大きさと一致していません

ケイ素

ネオン
マグネシウムなぜ星は「進化」するのか？
「進化」= 時間とともにその姿を変化させる

核融合で輝いている星は、いつか燃料を使い尽くしてしまう=> そのとき星はどうなるか？

E

Etot: 全エネルギー
Ω: 重力エネルギー
U: 内部エネルギー

U
U=1
⌦
21
Etot = U + ⌦ = ⌦ =20
U

核融合をしないと、- 全エネルギーが下がり、- 収縮して、- 温度が上がる

Etot
Ω

水素

核融合のエネルギーを失う=> 収縮=> 温度が上がるΤ ~ ρ1/3
図の大きさは天体の大きさと一致していません

ヘリウム

ヘリウム

炭素
酸素

トリプルアルファ

=>炭素

熱した鉄

恒星

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

放置すると、、、

冷える

熱くなる

PhaseMain reac ons

Products

T

Nuclear sta s cal equilibrium
tititi

元素はいかにつくられたか（岩波書店）

原子核の束縛エネルギー
Eb = [NmN + Zmp - mi] c2 > 0

束縛エネルギーが大きい= より安定鉄が最も安定全ての星が鉄まで核融合を続けるの？ => No
理想気体ではなくなる効果が重要まとめ: 星の進化 (1)• ビリアル定理• 星の内部エネルギーは重力エネルギーと常に関係• 全エネルギーが下がる => 温度が上がる (負の比熱)• 温度と密度の進化• 星が収縮して温度が上がる T ~ ρ• より重い元素の核融合へ => たまねぎ構造• 全ての星が鉄を作る？ => No
1/3星の状態方程式が重要になる (次回のテーマ)

Appendix

imes that of a typical reaction involving only strongTheinteractions.
Therate cannot
bnet 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+ + ν
nucleus two protons have to be con
2 H + 1 H → 3 He + γinteractions (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"!
$#smaller
than26.734
MeVanddep
7−771
8Be + e → Li + ν

7 Li + 1 H → 4 He + 4 He

pp2Be + H → B + γ

8 B → 8 Be + e+ + ν

8 Be → 4 He + 4 He

pp3The 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 ~ 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~
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 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 pti1(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= 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 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
Whenasufficientamountof
8 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 ≈ 1e 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 c
Neon 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 by20 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 Hange
whichissufficienttobreakup
therelativelyfragileNenucleus
into8 beeon burning The next 23
nuclear burning cycle
might
expectedtobe
oxygenT~
7x10
K
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 +Atαreleased
Qfind= −4.73
MeVe protons and α↔
particles
themselves
at extremely
highoftemperaturescomp
20 Ne nucleus into 16 Oange
whichissufficient
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 + α16 O
Q=MeV
easeaisnetobtained
from theof −4.73
all thesereactions
and isismostly
roughlyQand≈ 2413Mg
Mwithenergy release
Q >combination
0. The composition
afterneon burning
9 K 16 20T~
1.5x
101224 Mg (tC+Creaction.
Themainproducts
after
exhaustionofallcarbonareO,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
thesidereactions
produceneutron-richisotopes
like16
he firstburning
reaction isAtendothermic,
the in,tworeactions
combine
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
mportant
ones fraction).
being:5% by mass
92
16 O + 16 O → 32 S∗ → 28 Si + α
Q = 9.59 MeV (∼ 960%)

5. O-burning

T~2-3x10K16≈ 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titititiofSiandS(together90%bymassfraction).
Thenetenergy
releasepee of secondary reactions that eventually lead to a composition
mostly cons+Q≈
16MeV.
Since
someofthesidereactionsinvolveβ-decaysandelectro1616
Silicon 90%
burningThefraction).
lightest andmostabundantnucleusin theashes
ofreacoxy
ogetherbymassThe
netenergyreleaseperO+
O
6. Si-burning（Nuclearstascalequilibrium）
excessof
thefinal
mixtureisfurtherincreased.
Coulombfor 28 Si involve+ 28 Si fusion
is prohibitively
high.
Insteadthe
siliccethe
someof thebarrierside reactions
β+ -decays
and electron
captures,ne9KT>
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
20
16
12The lightestandα)mostnucleus
intheashesof
oxygenburningis(γ, α) Ne(γ,α)O (γ, α) C(γ,α)
2αa seriesofphoto-disintegration(γ,α)
andα-capture(α,γ)
reactionswhen T 92828
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 s
4 He
28
2420
16
12 CHecapture
rMostnuclei,while
anotherpartcaptures
thereleasedto(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:+γ28 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
described
by final
nuclearequivalentsis ofthemostl
Saha=>ofequilibriummanyons
numberneutrons
andofprotons
present.
The
composition
then
28 Si + γ ↔ 24 Mg + α, an9ctions
areinequilibriumwitheachother,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
numberneutrons
andpresent.
The
final
compositionisthenmostl
energy tend to be 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/