m.r. burleigh 2601/unit 4 department of physics and astronomy lifecycles of stars option 2601

M.R. Burleigh 2601/Unit 4

DEPARTMENT OF PHYSICS AND ASTRONOMY

LIFECYCLES OF STARSLIFECYCLES OF STARS

Option 2601Option 2601


Stellar PhysicsStellar Physics

Unit 1 - Observational properties of Unit 1 - Observational properties of starsstars

Unit 2 - Stellar SpectraUnit 2 - Stellar Spectra Unit 3 - The SunUnit 3 - The Sun Unit 4 - Stellar StructureUnit 4 - Stellar Structure Unit 5 - Stellar EvolutionUnit 5 - Stellar Evolution Unit 6 - Stars of particular interestUnit 6 - Stars of particular interest


DEPARTMENT OF PHYSICS AND ASTRONOMY

Unit 4Unit 4

Stellar StructureStellar Structure


StarbirthStarbirth


Young StarsYoung Stars


Globular ClustersGlobular Clusters


Star DeathStar Death



Hydrostatic equilibriumHydrostatic equilibrium Equations of stateEquations of state Energy transport (not derived)Energy transport (not derived) Energy sourcesEnergy sources Stellar modelsStellar models Mass-Luminosity relationMass-Luminosity relation Eddington LimitEddington Limit


r = R

r

Centre (r = 0)

(r)

P + dP P dr

Hydrostatic EquilibriumHydrostatic Equilibrium


Equation of hydrostatic equilibrium: 2r

rrGmdrdP

R R drrrrdmM 0 024

Only need to know (r) to determine mass of star radius R

Hydrostatic EquilibriumHydrostatic Equilibrium

(1)(1)

rrdr

dm 24However:

(2)(2)


E.g. Sun’s central pressureE.g. Sun’s central pressure

G=6.67x10G=6.67x10-11-11 Nm Nm22KgKg-2-2, M, M=1.989x10=1.989x103030kg, kg, RR=6.96x10=6.96x1088mm

(ave) (ave) =3M =3M /4 /4RR33=1410kgm=1410kgm-3-3

Surface pressure = 0Surface pressure = 0

Let r = dr = RLet r = dr = R and M(r)= Mand M(r)= M

PPcc~G M~G M (ave) (ave) / R / R = 2.7x10= 2.7x101414NmNm-2-2


Assume material is a perfect gas

Obeys perfect gas law:

rkTrnrP

Number density of particles

Boltzmann’s constant (1.381 10-23JK-1)

Equations of StateEquations of State

(3)(3)


n(r) is dependant on density and composition: Hmr

rrn

mH = 1.67 10-27 kg = mass of a hydrogen atom = mean molecular weight

2

1

21

43

2

1

ZYX

Mass fractions of: H He Metals (all other heavier elements)

Hmr

rkTrrP

Equations of StateEquations of State


In massive stars, radiation pressure also contributes to the total pressure:

3

4rTarPrad

a = 7.564 10-14Jm-3K-4 = radiation constant ( = ¼ ac)

Radiation PressureRadiation Pressure


E.g. Sun’s central temperatureE.g. Sun’s central temperature

Use PUse Pcc and and estimates, assume estimates, assume ~ ½ ~ ½

Then TThen Tc c ~ P~ PccmmHH//(ave) (ave) k ~ 1.2x 10 k ~ 1.2x 1077KK

Gas dissociated into ions & electrons but Gas dissociated into ions & electrons but overall electrically neutral… a overall electrically neutral… a plasmaplasma


Energy TransportEnergy Transport

T(r) depends on how energy is T(r) depends on how energy is transported from interior transported from interior surface surface

Three processes…Three processes…1.1.Conduction – collision of hot energetic Conduction – collision of hot energetic

atoms with cooler… poor in gasesatoms with cooler… poor in gases2.2.Convection – mass motions of fluids, Convection – mass motions of fluids,

need steep temp. gradient… happens in need steep temp. gradient… happens in some regions of most starssome regions of most stars


Energy TransportEnergy Transport

Three processes…Three processes…1.1.ConductionConduction2.2.ConvectionConvection3.3.Radiation – high energy photons flow Radiation – high energy photons flow

outward losing energy by scattering and outward losing energy by scattering and absorption… opacity sources at high T absorption… opacity sources at high T are i) electron scattering and ii) are i) electron scattering and ii) photoionizationphotoionization


(r) (opacity) depends only upon N(r), T(r) and (r)

L(r) at the surface is the star’s bolometric luminosity

Radiative Transport EquationRadiative Transport Equation

drdT

rrrTr

rL

3

64 32

(4)


For the SunFor the Sun

LL ~ 9.5x10~ 9.5x102929// Joules s Joules s-1-1

However, we do not know However, we do not know very well very well Ranges from 10Ranges from 10-3 -3 << << << 10 << 1077

Therefore… Therefore… – 101022 22 << L<< L << 10<< 1032 32 Joules sJoules s-1-1

Measured value is 3.9x10Measured value is 3.9x102626 implies implies ~ 2.4x10~ 2.4x1033


The Virial TheoremThe Virial Theorem

Considers total energy in a starConsiders total energy in a star Gravitational contractionGravitational contraction Gravitational potential energy Gravitational potential energy kinetic kinetic

energy energy Kinetic energy in bulk Kinetic energy in bulk Heat Heat


Take the equation of hydrostatic equilibrium:

2r

rrGmdrdP rr

drdm 24and

rmG

dmdP

rr

mGdmdP 3

44

4

But: dmdr

PrdmdP

rPrdmd

3.444 233

r

mG

dm

drPrPr

dm

d 3.44 23

The Virial TheoremThe Virial Theorem


Integrate over the whole star:

MMM

dmr

Gmdm

PPr 000

3 34

P, and r are

functions of m

Zero at both limits (P(m) = 0 marks the boundary of the star)

Twice the thermal (kinetic) energy

-2U

Gravitational binding energy

022 UU

0 UE

UETotal energy of a star:


Gravitational contraction

½ excess must be lost by radiation

21

UBut, using Virial theorem:

1) Star gets hotter

2) Energy is radiated to space

3) Total energy of the star decreases (becomes more –ve more tightly bound)

Gravitational ContractionGravitational Contraction


Stellar Thermonuclear ReactionsStellar Thermonuclear Reactions

Light elements “burn” to form heavier Light elements “burn” to form heavier elementselements

Stellar cores have high enough T and Stellar cores have high enough T and for nuclear fusionfor nuclear fusion

Work (after 1938) by Hans Bethe and Work (after 1938) by Hans Bethe and Fred HoyleFred Hoyle



Energy release can be calculated from Energy release can be calculated from E=mcE=mc22 – e.g. 4 x e.g. 4 x 11

11H atoms H atoms 1 x 1 x 4422He atomHe atom

4 x 1.6729x104 x 1.6729x10-27-27kg = 6.6916x10kg = 6.6916x10-27-27kgkg 1 x 6.6443x101 x 6.6443x10-27-27kgkg E = 4.26x10E = 4.26x10-12-12JJ



In the Sun ~10% of its volume is at the In the Sun ~10% of its volume is at the T and T and required for fusion required for fusion

Total energy available is…Total energy available is…– Energy per reaction x mass/mass in each Energy per reaction x mass/mass in each

reactionreaction

EEtot tot = 4.26x10= 4.26x10-12 -12 x 2x10x 2x102929/6.6916x10/6.6916x10-27 -27 = = 1.27x101.27x104444JJ

LL = 3.9x10= 3.9x102626 t ~ 3.3x10 t ~ 3.3x101717s ~ 10s ~ 101010yrsyrs


CNO cycle

HeCHN

eNO

OHN

NNC

eCN

NHC

42

126

11

157

157

158

158

11

147

147

11

136

136

137

137

11

126


Proton – proton chain (PPI, T < 2 107K)

1.44MeV

5.49MeV

12.9MeVHHHeHeHe

HeHH

eHHH

11

11

42

32

32

32

11

21

21

11

11


The PPI ChainThe PPI Chain


a

b

c

PPI Chain PPI Chain

HHHeHeHe

HeHH

eHHH

11

11

42

32

32

32

11

21

21

11

11


The CNO CycleThe CNO Cycle


CNO CycleCNO Cycle

HeCHN

eNO

OHN

NHC

eCN

NHC

42

126

11

157

157

158

158

11

147

147

11

136

136

137

137

11

126


Triple AlphaTriple Alpha

CHeBe

BeHeHe126

42

84

84

42

42

High level reactions ~108K

FeuptoMgNeO 2412

2210

168 ,,

Addition of further alphas


n = number density of particles

Hmr

rrn

= mean molecular weight

Hydrostatic equilibrium:

2r

rrGmdrdP (1)

Mass equation: rrdrdm 24 (2)

Equation of state: rkTrnrP (3)

Stellar Models: EquationsStellar Models: Equations


Radiation pressure: rTa

rPrad4

3

a = radiation constant = 7.564 10-14Jm-3K-

4

= ¼ ac

= opacity

ε = rate of energy production (Js-1kg-1)

dr

dTrr

rTrrL

364 32

Radiative transport: (4)

Energy generation: rrrdrdL 24 (5)


e.g.e.g. r = 0r = 0 M(r) = 0M(r) = 0 L(r) = 0L(r) = 0

r = Rr = R M(r) = MM(r) = M L(r) = LL(r) = L T(r) = TT(r) = Teffeff

And (r), P(r) 0

Need to apply boundary conditions to the equations to use them, i.e. fix/know values at certain values of r (centre or surface)

Boundary ConditionsBoundary Conditions


From (1) write dP P and dr r

Then: P = PS – PC = 0 - PC

Surface Centre

and r = R

For a perfect gas P T

RM

PC

RM

T

RM

T

C

C


From (4)

Also:

Substitute

432 CCC RT

R

TTRL

3R

MM

TRL C

44

RM

TC

3

44

MMRM

RL

Observed relationship is L M3.3

( is dependant on T and )


Hydrostatic equilibrium assumes no net outward motion of material from the star, but the outward flow of radiation imparts a force on the material

Momentum of radiation = cr

L24

T = cross-section of electron-photon scattering = 6.7 10-29m2

2r

MGmHThis is opposed by gravitational force =

Force =cr

LT24

Eddington LimitEddington Limit


The forces are equal at the Eddington limit

T

HTH GMcmL

cr

L

r

MGm

4

4 22

SunE M

ML 38103.1 erg s-1

So if L > LE material is expelled

2max 20

SunMM



Hydrostatic equilibriumHydrostatic equilibrium Equations of stateEquations of state Energy transport (not derived)Energy transport (not derived) Energy sourcesEnergy sources Stellar modelsStellar models Mass-Luminosity relationMass-Luminosity relation Eddington LimitEddington Limit

m.r. burleigh 2601/unit 4 department of physics and astronomy lifecycles of stars option 2601

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