m.r. burleigh 2601/unit 4 department of physics and astronomy lifecycles of stars option 2601
Post on 29-Jan-2016
234 Views
Preview:
TRANSCRIPT
M.R. Burleigh 2601/Unit 4
DEPARTMENT OF PHYSICS AND ASTRONOMY
LIFECYCLES OF STARSLIFECYCLES OF STARS
Option 2601Option 2601
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
DEPARTMENT OF PHYSICS AND ASTRONOMY
Unit 4Unit 4
Stellar StructureStellar Structure
M.R. Burleigh 2601/Unit 4
StarbirthStarbirth
M.R. Burleigh 2601/Unit 4
Young StarsYoung Stars
M.R. Burleigh 2601/Unit 4
Globular ClustersGlobular Clusters
M.R. Burleigh 2601/Unit 4
Star DeathStar Death
M.R. Burleigh 2601/Unit 4
Star DeathStar Death
M.R. Burleigh 2601/Unit 4
Star DeathStar Death
M.R. Burleigh 2601/Unit 4
Star DeathStar Death
M.R. Burleigh 2601/Unit 4
Stellar StructureStellar Structure
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
r = R
r
Centre (r = 0)
(r)
P + dP P dr
Hydrostatic EquilibriumHydrostatic Equilibrium
M.R. Burleigh 2601/Unit 4
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)
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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)
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
(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)
M.R. Burleigh 2601/Unit 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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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:
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
Stellar Thermonuclear ReactionsStellar Thermonuclear Reactions
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
M.R. Burleigh 2601/Unit 4
Stellar Thermonuclear ReactionsStellar Thermonuclear Reactions
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
M.R. Burleigh 2601/Unit 4
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
Stellar Thermonuclear ReactionsStellar Thermonuclear Reactions
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
M.R. Burleigh 2601/Unit 4
The PPI ChainThe PPI Chain
M.R. Burleigh 2601/Unit 4
a
b
c
PPI Chain PPI Chain
HHHeHeHe
HeHH
eHHH
11
11
42
32
32
32
11
21
21
11
11
M.R. Burleigh 2601/Unit 4
The CNO CycleThe CNO Cycle
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
Triple AlphaTriple Alpha
CHeBe
BeHeHe126
42
84
84
42
42
High level reactions ~108K
FeuptoMgNeO 2412
2210
168 ,,
Addition of further alphas
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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)
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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 )
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
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
M.R. Burleigh 2601/Unit 4
Stellar StructureStellar Structure
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
top related