2 . S te lla r s tr u c tu re e q u a ti o n s

(1)

2–1

2 . S te lla r s tr u c tu re e q u a ti o n s

2–2 Stellarstructureequations1

S te lla r s tr u c tu re e q u a ti o n s

dm dr=4πr2 ρ(r)(1)massconservation dP dr=−Gm(r)ρ(r) r2(2)hydrostaticequilibrium dL dr=4πr2 ǫ(r)(3)energyproduction dT dr=−3 4acκ(r)ρ(r) T3(r)L(r) 4πr2(a)radiative (4)energytransport dT dr=γad−1 γadT PdP dr(b)convective +Evolutionofchemicalabundances ∂Xi ∂t=mi ρ P jrji−P krik i=1,...,I

E q u a ti o n o f m a s s c o n s e rv a ti o n (o r m a s s c o n ti n u it y )

Generalassumptionsforthisandotherequations: 1.starsarestatic,i.e.wecanneglectvelocityandaccelerationterms 2.starsaresphericalsymmetric,i.e.sphericalcoordinates,variable

r m

=massinsidetheradius

r dm

=massintheshellofsize

dr dm =

4

π r

2

ρ dr

Indifferentialform dm dr

=

4

π r

2

ρ

S te lla r s tr u c tu re e q u a ti o n s

dm dr

=

4

π r

2

ρ ( r )

(1)massconservation dP dr=−Gm(r)ρ(r) r2(2)hydrostaticequilibrium dL dr=4πr2 ǫ(r)(3)energyproduction dT dr=−3 4acκ(r)ρ(r) T3(r)L(r) 4πr2(a)radiative (4)energytransport dT dr=γad−1 γadT PdP dr(b)convective +Evolutionofchemicalabundances ∂Xi ∂t=mi ρ P jrji−P krik i=1,...,I

(2)

E q u a ti o n o f h y d ro s ta ti c e q u ili b ri u m

Gravitationalforceonthevolumeelement:

F

G

=

−

G m r

2

A ρ dr

Buoyantforce(”Auftrieb”:Pressuredifference) ontheelement:

F

P

= A [ P ( r + dr )

−

P ( r )] = A dP dr dr

Forcesmustbebalancedinequilibrium:

A dP dr dr =

−

G m r

2

A ρ dr

⇒dP dr

= − G

mρ r2 2–6 Stellarstructureequations5

E q u a ti o n o f m o ti o n

Considerthecasethatgravitationalandpressureforcearenotinequilibrium. Thisresultsinanetforceandacceleration

F = ¨r dm = ¨r A ρ dr

Resultingequation:

F = F

G−

F

P

¨r A ρ dr =

−

G m r

2

A ρ dr

−

A dP dr dr

⇒

¨r ρ = − G

m r2

ρ −

dP dr

F re e fa ll (= d y n a m ic a l) ti m e s c a le

Letusassumethatthepressure(gradient)is switchedoff.

¨r ρ =

−

G m ρ r

2−dP////////////////// dr///////////// Notsurprisinglythistellsusthatthematterfalls intothecentre. Thetimescaleforthisisthefreefalltime

t

ff

= r ˙ r

F re e fa ll (= d y n a m ic a l) ti m e s c a le

Defineamassshellofmass

dm

initiallyatrest at

r

0enclosingamass

m

0.Thegravitational energyisconvertedintokineticenergy:

E

kin

= E

G 1 2

dm v

2

=

1 2

dm ˙ r

2

= G m

0

r dm

−

G m

0

r

0

dm

1 2dm///////////////////

˙ r

2

= G m

0

r

dm///////////////////−

G m

0

r

0dm/////////////////// ⇒

˙ r =

−s 2

G m

0

r

−

G m

0

r

0

(3)

F re e fa ll (= d y n a m ic a l) ti m e s c a le

Nowwelookatthevelocityoftheshellat

r

assumingthatitstarteditsfallat

r

0

=

∞.Wecanassume

m = m

0.

˙ r =

−v u u u t2 

G m

0

r

−Gm0//////////////////////////// r0/////////// 

=

−r 2

G m r

Note:thisistheescapevelocity.

t

ff

= r ˙ r = r

q 2Gm r

=

r

3 2

G m

Orifwelookatcompletestars

t

ff

=

r

R

3 2

G M =

s 1 8 3

π G ρ w it h ρ = M

4 3

π R

3 2–10 Stellarstructureequations9

F re e fa ll (= d y n a m ic a l) ti m e s c a le t

ff

=

r

R

3 2

G M

ExampleSun:

R

⊙

=

698

,

900

km

,

M

⊙

=

1

.

989·1030

kg t

FF

=

1130

s

Redgiant

R =

200

R

⊙:

t

FF

=

2003/2

t

⊙ FF

=

3

.

2·106

s =

37

d ay s

Whitedwarf

R =

10

,

000

km =

0

.

014

R

⊙:

t

FF

=

0

.

0143/2

t

⊙ FF

=

1

.

9

s

Starsreactondeviationsfromthehydrodynamicalequilibriumonthedynamical timescale.Since

t

FF≪

t

evolforalmostallevolutionaryphases,itissufficientto assumehydrodynamicalequilibrium.

C o lla p s e o f a n in te rs te lla r c lo u d

Considerthecollapseofaninterstellarcloud •gravitationalenergyisdissipatedintothermalen- ergy •butthecloudisopticallythintothermalradiation intheearlyphaseofthecollapse •⇒temperatureandpressureincreasenotvery much⇒approximatelyfreefall •inlaterphasesthecentralregionbecomesopti- callythick •⇒temperatureinthecentreincreasestoveryhigh values 2–12 Stellarstructureequations11

V ir ia l T h e o re m

HydrostaticEquilibrium:(Mr=m(r),massinsideradiusr)

dP dr =

−

G M

r

ρ r

2

dM

r

dr =

4

π r

2

ρ ∂ P ∂ M

r

=

−

G M

r 4

π r

4 Multiplicationwith4

π r

3 andintegration.Lefthandside: ZM 04

π r

3

∂ P ∂ M

r

dM

r

=

4

π r

3

P

M 0−

ZM 012

π r

2∂r ∂Mr

P dM

r

=

0−ZM 012

π r

21 4πr2ρ

P dM

r

=

−ZM 03

P ρ dM

r

(4)

V ir ia l T h e o re m

righthandside: −ZM 04

π r

3

G M

r 4

π r

4

dM

r

=

−ZM 0

G M

r

r dM

r Righthandside:Gravitationalpotentialenergy

dE

G=−GMr r.

E

Gincreases/decreases,asthestarexpands/contracts. 2–14 Stellarstructureequations13

V ir ia l T h e o re m

Lefthandside(assumingidealgas):

P ρ = R

g

µ T = ( c

p−

c

v

) T = ( γ

−1

) c

v

T

cp,cv:specificheat(permass)

γ =

cp cv

=

5 3(monoatomicgas)

P ρ =

2 3

c

v

T =

2 3

u

u=cvT:internalenergy(permass) −ZM 03

P ρ dM

r

=

−ZM 02

u dM

r

=

−2

E

I EI:totalinternalenergyofthestar ⇒

E

G

=

−2

E

I istheconsequenceofhydrostaticequilibrium(foramonoatomic,idealgas.)

V ir ia l T h e o re m

Totalenergy:

W = E

G

+ E

I

=

−

E

I

=

1 2

E

G

<

0 whenthestarslowlycontractsorexpands:

W

changesandthestarradiates energyaway. Energyconservation:dW dt

+ L =

0

L = dE

I

dt =

−1 2

dE

G

dt

idealgas:

L =

−1 2

˙ E

G

= ˙ E

I Contraction:Halfofthefreedenergywillberadiated,theotherhalfwill beusedtoheatthestar(increasetheinternalenergy). 2–16 Stellarstructureequations15

K e lv in -H e lm h o lt z ti m e s c a le

VirialTheorem:

L

isofthesamemagnitudeas dEG dt and dEI dt Definition

τ

KH:

τ

KH

= E

G

L

≈

E

I

L E

G≈

G M

r2

r

with

r =

1 2

R

and

M

r

=

1 2

M

⇒

E

G≈

G M

2 2

R

⇒

τ K H = G M

2 2

R L

(5)

K e lv in -H e lm h o lt z ti m e s c a le

Sun:

L =

3

.

827·1033erg s ⇒τKH=1.6·107 years OnlyforafewmillionyearstheSuncansustainitsradiationenergylossby gravitationalenergy(contraction). ageoftheEarth(geology):4.5billionyears ⇒TheremustbeanothersourceofenergyfortheSun. 2–18 Stellarstructureequations17

S te lla r s tr u c tu re e q u a ti o n s

dm dr

=

4

π r

2

ρ ( r )

(1)massconservation dP dr

=

−Gm(r)ρ(r) r2(2)hydrostaticequilibrium dL dr=4πr2 ǫ(r)(3)energyproduction dT dr=−3 4acκ(r)ρ(r) T3(r)L(r) 4πr2(a)radiative (4)energytransport dT dr=γad−1 γadT PdP dr(b)convective +Evolutionofchemicalabundances ∂Xi ∂t=mi ρ P jrji−P krik i=1,...,I

E q u a ti o n o f e n e rg y p ro d u c ti o n ℓ

=luminosityinsidetheradius

r dℓ

=luminosityproducedinshell

dℓ = dV ǫ =

4

π r

2

dr ǫ ǫ :

theenergygeneratedpervolumeand persecond

q :

theenergygeneratedpermassandper second

q dm = ǫ dV q

4

π r

2

ρ dr = ǫ

4

π r

2

dr q = ǫ/ ρ

centreofstar:

ℓ (

0

) =

0 surface:

ℓ ( R ) = L dℓ =

4

π r

2

dr ǫ

⇒dℓ dr

=

4

π r

2

ǫ =

4

π r

2

ρ q

2–20 Energytransport1

T ra n s p o rt o f E n e rg y : R a d ia ti o n

Afewestimates: Meanfreepathofaphoton:

l =

1

n σ σ :

atomicabsorptioncrosssection,

n :

particledensity Introducingtheabsorptioncrosssectionperunitmass

κ

.Units:

cm

2

g

−1 .

l =

1

ρ κ

withthemassdensity

ρ

.

(6)

T ra n s p o rt o f E n e rg y : R a d ia ti o n

Typicalvaluefor

κ

instellarmatter:

κ =

1

cm

2

g

−1 . Lightscatteringatelectrons(Thomsonscattering)setsalowerlimitforionised matter.Valueforfullyionisedhydrogen:

κ =

0

.

4

cm

2

g

−1 . MeandensityoftheSun:

ρ

⊙

= M

⊙ 3 4

π R

3 ⊙

=

1

.

4

g cm

3 TypicalmeanfreepathintheSun:

l <

1 0

.

4×1

.

4

cm

≈2

cm

⇒Propagationofphotonsintheinteriorofstarscanbetreatedas“randomwalk” withfrequentabsorptionsandre-emissions–analoguestodiffusionprocesses. 2–22 Energytransport3

T ra n s p o rt o f E n e rg y : R a d ia ti o n

TemperaturegradientwithintheSun:

∆ T ∆ R = T

core−

T

surface

R

⊙≈1

.

6·107

K

7·1010

cm

≈2·10−4

K cm

Temperaturechangeoverthemeanfreepathofaphoton:

∆ T = l

×

∆ T ∆ R =

2

cm

×2·10−4

K cm =

4·10−4

K

Veryminimaltemperaturedifference.(Local)radiationfield“sees”almost isothermallayer(exception→stellaratmosphere).

A n a lo g y : T ra n s fe r o f H e a t b y R a n d o m M o ti o n s

Considerthetransferofenergyacrossan (imaginary)surface. •Ifthereisatemperaturegradient,thenpar- ticlescrossingfromabovewillhaveadiffer- entthermalenergy

u ( x )

,tothosebelow. •Assumethattheparticlesaremovingwith velocity

v

•Alsoassumethattheparticlestraveladis- tance

l

beforetheyinteract–

l

isthemean freepath •Andassumethatonaverage1

/

6ofthepar- ticlesinthegasaretravellingintheposi- tive/negative

x

direction Note:stricttreatmentgivesthesameresults 2–24 Energytransport5

A n a lo g y : T ra n s fe r o f H e a t b y R a n d o m M o ti o n s

Withtheseassumptionswecanwritetherateofenergytransfer(heatflux

j

) acrossthesurfaceas

j ( x ) =

1 6

v u ( x

−

l )

−1 6

v u ( x + l ) =

−1 3

v l du dx

Theenergygradientcanbere-written

du dx = du dT dT dx = C

V

dT dx

withtheheatcapacityforconstantvolume

C

V

=

du dT Therateofenergytransport(=fluxdensity)acrossthesurfaceisproportionalto thetemperaturegradient.

j ( x ) =

−

K dT dx w it h K =

1 3

v lC

V

K

isthecoefficientofthermalconductivity(ofthegas)

(7)

E n e rg y T ra n s fe r o f b y R a n d o m M o ti o n s o f P h o to n s j ( x ) =

−

K dT dx w it h K =

1 3

v lC

V Photonsmovewiththespeedoflight(

v = c

).Assumingablackbodyradiation fieldtheenergydensityfollowfromtheStefan–Boltzmannlaw

u

rad

= a T

4 ⇒

C

V

= du dT =

4

a T

3

a =

4

σ /c

istheradiationconstant.

j ( x )

isthe(radiative)energyflux

F

. Thustheequationofenergytransportbyphotonstakestheform

F ( x ) =

−

K

rad

dT dx w it h K

rad

=

4 3

c la T

3 Movetosphericalsymmetry

F ( r ) =

−

K

rad

dT dr

E n e rg y T ra n s fe r o f b y R a n d o m M o ti o n s o f P h o to n s

Themeanfreepathofphotonsis

l =

1 ρκ.

K

rad

=

4 3

c la T

3

=

4

a c

3

ρ κ T

3 Inequilibriumtheflux4

π r

2

F

streamingthroughashellistheluminosity

ℓ

producedinsidetheshell.

F ( r ) = ℓ

4

π r

2

=

−

K

rad

dT dr =

−4

a c

3

ρ κ T

3

dT dr

⇒dT dr

= −

3κρ 16πacℓ r2T3 Notethatthisequationcanalsobeappliedtoenergytransportbyheat conduction.(Unimportantinmosttypesofstars,butimportantinwhitedwarfs.)

S te lla r s tr u c tu re e q u a ti o n s

dm dr

=

4

π r

2

ρ ( r )

=

−Gm(r)ρ(r) r2(2)hydrostaticequilibrium dL dr

=

4

π r

2

ǫ ( r )

(3)energyproduction dT dr

=

−3 4acκ(r)ρ(r) T3(r)ℓ(r) 4πr2(a)radiative (4)energytransport dT dr=γad−1 γadT PdP dr(b)convective +Evolutionofchemicalabundances ∂Xi ∂t=mi ρ P jrji−P krik i=1,...,I 2–28 Energytransport9

E n e rg y T ra n s fe r b y C o n v e c ti o n

Granulation

Schematicviewoftheconvectioncells attheSun’ssurface Convectionisthetransportofheatbybulkmotionofgas.Pocketsofhotgasrise intoacoolerenvironmentandreleasetheirexcessenergy.Fallingpocketsmove intoawarmerenvironmentandreleasetheircoolergasthere.

(8)

E n e rg y T ra n s fe r b y C o n v e c ti o n

Whenisthethermalstratificationstable/unstableagainstconvection? GedankenexperimentbyKarlSchwarzschild: •Considerapocketofgasat

r

whichrisesa smalldistance

δr

•Weassumethatthisisanadiabaticprocess, i.e. –rapidadjustmentofpressure.Thegasbub- bleisinpressureequilibriumwithitsenvi- ronment –Noexchangeofthermalenergybetween bubbleandenvironment. 2–30 Energytransport11

E n e rg y T ra n s fe r b y C o n v e c ti o n

Whenisthethermalstratificationstable/unstableagainstconvection? GedankenexperimentbyKarlSchwarzschild: •Iftherisingpocketafteradiabaticexpansion hashigherdensitythenthesurroundingen- vironmentitwillsinkback–thestratification isstableagainstconvection •Iftherisingpocketafteradiabaticexpansionis lessdensethenthesurroundingenvironment itwillcontinuerising–thestratificationisun- stable

E n e rg y T ra n s fe r b y C o n v e c ti o n

Whenisthethermalstratificationstable/unstableagainstconvection? GedankenexperimentbyKarlSchwarzschild: •Initialvaluesinthepocketofgas:

P

1

,T

1

,ρ

1(= environmentvaluesat

r

) •Valuesafterriseby

δr

:

P

∗ 2

= P

1

+ δP

,

T

∗ 2

= T

1

+ δT

,

ρ

∗ 2

= ρ

1

+ δρ

•Environmentalvaluesat

r + dr

:

P

2

= P

1

+ ∆ P

,

T

2

= T

1

+ ∆ T

,

ρ

2

= ρ

1

+ ∆ ρ

•Stable:

ρ

∗ 2

< ρ

2i.e.

∆ ρ < δρ

Unstable:

ρ

∗ 2

> ρ

2i.e.

∆ ρ > δρ

E n e rg y T ra n s fe r b y C o n v e c ti o n

Firstlawofthermodynamics

dU = dQ

−

P dV dU

=Changeofinternalenergyofamasselement

dQ

=heataddedtothatelement

dW

=workbythatelementonitssurrounding

dW = pdV

;

V =

1

/ρ

=thespecificvolume Idealgas:

P V = k µ m

H

ρ T = n R T dU =

3 2

k µ m

H

T =

3 2

n R T C

P

=

dQ dT Pspecificheatatconstantpressure

C

V

=

dQ dT Vspecificheatatconstantvolume

C

P

= C

V

+ n R

(9)

E n e rg y T ra n s fe r b y C o n v e c ti o n

Adiabaticprocess:Noexchangeofheatwithsurrounding:

dQ =

0

dU =

−

P dV P dV + V dP = n R dT

idealgasequation

dU = C

V

dT

thatis

dT = dU C

V

pdV + V dP = n R C

V

P dV γ = C

P

C

V

γ dV V =

−

dP P

→

P V

γ

= K

→

P = K ρ

γ

γ :

adiabaticexponent,thefactor

K

dependsonthethermodynamicstate (entropy)ofthesystem–butisconstantunderadiabaticconditions. 2–34 Energytransport15

E n e rg y T ra n s fe r b y C o n v e c ti o n

Adiabaticexpansion

P = K ρ

γ

dP dρ = K γ ρ

γ−1

= K γ ρ

γ

ρ = γ P ρ

⇒

dρ ρ =

1

γ dP P

Thusfortheadiabaticallyexpandinggasbubble

δρ ρ =

1

γ δP P

Ontheotherhandwehavetheidealgaslaw:

P = ρ µ m

amu

k T

⇒

dP dr = k µ m

amu

T dρ dr + ρ dT dr

E n e rg y T ra n s fe r b y C o n v e c ti o n dP dr = k µ m

amu

T dρ dr + ρ dT dr

Multiplyby

∆ r

→

∆ P , ∆ ρ , ∆ T

anddivideby

P =

ρ µmamu

k T

gives

∆ P P = ∆ ρ ρ + ∆ T T

⇒

∆ ρ ρ = ∆ P P

−

∆ T T

Fortheadiabaticallyexpandinggasbubblewederived

δρ ρ =

1

γ δP P

Theconditionforthelayerbeingunstableagainstconvectionwas

∆ ρ > δρ

,i.e

∆ P P

−

∆ T T >

1

γ δP P

E n e rg y T ra n s fe r b y C o n v e c ti o n ∆ P P

−

∆ T T >

1

γ δP P

Weareassumingpressureequilibriumbetweenbubbleandenvironment,i.e.

∆ P = δP ∆ T T <

1−1

γ δP P = γ

−1

γ

δP P

uselogarithmicderivatives: ⇒

P T dT dP < γ

−1

γ

⇒

dl n T dl n P <

γ

−1

γ

Notethatboth,temperatureandpressuregradient,arenegativeinastar. ⇒

dT dr

> γ

−1

γ

T P

dP dr

(10)

E n e rg y T ra n s fe r b y C o n v e c ti o n

Schwarzschildcriterionforconvectivestability: ⇒∇

=

dlnT dlnP

<

γ−1 γ ∇willbeusedasabbreviationforthelogarithmicgradientdlnT dlnPthroughoutthe restofthelecture. •Iftheactualtemperaturegradientissteeperthanthecriticalvalueonthe righthandside,thenthestratificationisunstableandconvectionwill dominatetheenergytransfer •Otherwisethestratificationisstableandenergytransferisradiative 2–38 Energytransport19

E n e rg y T ra n s fe r b y C o n v e c ti o n

Adiabaticindex

γ

isrelatedtothenumberofdegreesoffreedom(d.o.f.

s

)

γ =

1

+

s 2 s 2

=

1

+

2

s

Classicalone-atomicgashashas3d.o.f.⇒

γ =

5 3 Ifgascanalsoabsorbenergythroughinternald.o.f.(rotationandvibrationin molecules)orbyionisation,dissociation ⇒

γ

decreasesandcanapproachvaluesclosetoone ⇒criticaltemperaturegradientbecomessmall Energytransportbyconvectionisveryefficient.Thetemperaturegradientin convectiveregionsisclosetotheadiabaticone–verycloseininterior convectionzones.Lessgoodinoutersurfaceconvectionzone→treatmentwith mixinglengththeory(chapter10.4inCarroll&Ostlie).

E n e rg y T ra n s fe r in th e S u n

•Thesolarcoreisstable,energy transportbyradiation •Surfaceconvectionzone(→gran- ulationatsurface) •Thesurfaceconvectionzoneof theSunandothercoolmainse- quencestarsisdrivenbyionisa- tion/recombination •Matteratthebottomofthezoneisfullyionised,neutralatthetop.Infusionof thermalenergyispartlyusedforionisation→

γ

decreasesbelow5/3→ criticalT-gradientbecomessmall •Inadditionopacity

κ

intheionisationzoneislarge→radiativegradientis steep 2–40 Energytransport21

S te lla r s tr u c tu re e q u a ti o n s

dm dr

=

4

π r

2

ρ ( r )

=

4

π r

2

ǫ ( r )

=

−3 4acκ(r)ρ(r) T3(r)ℓ(r) 4πr2(a)radiative (4)energytransport dT dr

=

γad−1 γadT PdP dr(b)convective +Evolutionofchemicalabundances ∂Xi ∂t=mi ρ P jrji−P krik i=1,...,I

(11)

C o n v e c ti v e te m p e ra tu re g ra d ie n t

Howtofindthetruetemperaturegradient∇? Inrealitythereisbothconvectiveandradiativetransport. ∇ad

=

adiabaticgradient ViolationoftheSchwarzschildcriterion:∇

>

∇e;∇e

=

gradientinsidetheeddie onlysomefractionofenergyistransportedbyradiation:∇

<

∇rad;∇rad

=

gradientifradiativetransferonly ascendingeddy:becauseof∇

>

∇ethetemperatureexceedsthatofthe environment⇒additionalradiativecooling:∇e

>

∇ad ⇒∇rad

>

∇

>

∇e

>

∇ad limitingcases:

F

conv≪

F

rad∇→∇rad

F

conv≫

F

rad∇→∇ad 2–42 Energytransport23

C o n v e c ti v e e n e rg y

Ascendingeddieshaveheatexcessw.r.t.thesurroundingwhichwillbereleased atsometime. Energyexcesspervolume:

ρ c

P

∆ T

.

∆ T

resultsfromthethedifferenceofthe temperaturegradients:

∆ T = dT dr

e−

dT dr ∆ r

(A)

∆ r

distancetravelled

v

meanvelocityofeddyenergyflux:

F

conv

= ρ c

P

v ∆ T

(B) Tofind

v ∆ T

isaseriousproblem.Noselfconsistenttheoryisavailable! Simpleapproach:mixinglengththeory (Prandtl,Biermann,Vitense):Alleddiestravelacertaindistance

l

m,dissolveand releasetheirenergy. Howlargeis

l

m?

M ix in g le n g th th e o ry

Thepressurescaleheight:

H

P

:=

−

P

dr dP Inhydrostaticequilibrium:dP dr

=

−

ρ g T = co n st

and

µ = co n st

gives:

p ( r ) = p ( r

0

) e

−r−r0 HP Examples: Earthatmosphere9km solaratmosphere180km whitedwarfatmosphere250m Mixinglengthapproach:

l

m

= α

m

H

P 2–44 Energytransport25

C h e m ic a l c o m p o s it io n

Thechemicalcompositiondirectlyinfluences

κ ,ǫ ,µ .. .

.Materialishighlyionised inthestellarinterior.

X

i:massfractionofelementi P i

X

i

=

1 numberfraction

n

i

X

i

=

mini ρ oftenonlythreecomponentsareconsidered:hydrogen,heliumand“metals”:

X := X

H

Y := X

He

Z :=

1−

X

−

Y

typicalvaluesfrominterstellarclouds:

X =

0

.

70

Y =

0

.

28

Z =

0

.

02 elementalcompositionischangedbynuclearreactions

X

i

= X

i

( M

r

,t )

(12)

C h e m ic a l c o m p o s it io n C h a n g e s w it h ti m e

a)radiativeregions: Noexchangeofmaterialbetweenneighbouringshells.reactionrate

r

lm: Frequencyofanuclearreactiontoconvertanucleus

l

intoanucleus

m

. Manydifferentreactionsarepossibleforasinglechemicalelement

i

,both destructiveones(

r

ikandconstructiveones(

r

ji) ∂ni ∂t

=

P j

r

ji−P k

r

ik ∂Xi ∂t

=

mi ρhP j

r

ji−P k

r

iki energyproduction,

e

pq:Energypernucleusconverted:

ǫ =

P p,q

ǫ

pq

=

1 ρP p,q

r

pq

e

pq energyproductionpermassofanelement:

q

pq

=

epq mp

r

pq

= ρ

ǫpq epq

= ρ

ǫpq qpqmp 2–46 Energytransport27

C h e m ic a l c o m p o s it io n

∂Xi ∂t

=

P jǫji qji−P kǫik qik e.g.hydrogenburning:H→He ∂X ∂t

=

−ǫH qH∂Y ∂t

=

−∂X ∂t b)convectiveregions: convectiveregionsarealwayshomogeneous:∂Xi ∂Mr

=

0 forastationaryconvectionzone: ∂Xi ∂t

=

1 m2−m1Rm2 m1∂Xi ∂t

dm

C h e m ic a l C o m p o s it io n

Boundariesofaconvectionzonemaychange:

∂ X

i

∂ t =

1

m

2−

m

1Zm2 m1

∂ X

i

∂ t dm + ∂ m

2

∂ t X

i2−

X

i −

∂ m

1

∂ t X

i1−

X

i chemicalcompositionofaconvectionzonemaychangewithoutnuclearburning! 2–48 Stellarstructureequations1

S te lla r s tr u c tu re e q u a ti o n s

dm dr

=

4

π r

2

ρ ( r )

=

4

π r

2

ǫ ( r )

=

−3 4acκ(r)ρ(r) T3(r)ℓ(r) 4πr2(a)radiative (4)energytransport dT dr

=

γad−1 γadT PdP dr(b)convective ∂Xi ∂t

=

mi ρP j

r

ji−P k

r

ik

i =

1

,. .. ,I

(5)evolutionofabundances