Diapositive 1 - astro.ulg.ac.bedupret/cours-structurevol1-ang2017.pdf · 1 . Ψ. 2 . Ψ. 3 . Else...

Stellar Structure SPAT0044

Marc-Antoine Dupret Associate Professor email: [email protected] tel: 04 3669732 Bât. B5c, 1er étage Slides: https://www.astro.ulg.ac.be/~dupret/

mailto:[email protected]

Goals of the course • Physical processes inside the stars

stars = laboratories of fundamental physics - Quantum physics (radiation - matter) and nuclear physics - Thermodynamics and quantum statistical physics

- Hydrodynamics

• Presentation of simple approximations enabling a

mathematical analysis of the stellar properties

• Link theory - observations

The Hertzsprung-Russell diagram

Lum

inos

ity

Color – spectral type Effective temperature ~ surface temperature

Stars of different masses,

different ages

Different positions in the HR diagram

L = Luminosity = Power radiated

by the star

Ejnar Hertzsprung Henry Norris Russell (1910)

Magnitude log (Luminosity L)

106 L¯

1 L¯

Luminosity of stars

103 L¯

10-3 L¯

Log (Effective Temperature)

Luminosity-flux-distance relation

• Luminosity L : Total power radiated by the star • In a star of radius R:

• b : energy flux crossing a surface perpendicular to the line of sight • [W/m2]

For a given luminosity, b decreases with the distance.

In order to measure the luminosity, the distance must be known

2

04L R F dλπ λ

∞= ⋅ ∫

24 r

Lbπ

=b

r

FrR2

2 =

Hertzsprung-Russell diagram

Stars whose distance was measured by the Hipparcos satellite

(parallax method)

d?

α α’

b MB – MV

MV

The monochromatic flux Fλ

Qtity of energy / u. time / u. surface / u. length (wavelength)

Fλ dλ : flux of energy between λ and λ+dλ

: Flux in the wavelength interval [λ1, λ2]

The bolometric flux F

Flux integrated over the whole spectrum :

Spectral nature of radiation

Luminosity and magnitude

- Luminosity L = 4πR² F (W)

- b = L / 4πr² (W/m²)

- Apparent magnitude m1 – m2 = -2,5 log (b1/b2)

- B = b (rpc/10)²

- Absolute magnitude M – m = 5 – 5 log rpc

The black body spectrum

Stellar spectrum ~ Black body

o Planck function

o Stefan-Boltzmann law

o Wien law

max0,29( )( )

cmT K

λ =

Black body laws

Definition 1 :

A: body emitting from its surface a bolometric flux F

The effective temperature of A is by definition :

Effective temperature: Definition

Presenter

Presentation Notes

ˈtɛmprɪtʃə

The effective temperature of an emitting body is the temperature of a hypothetical black body emitting the same bolometric flux.

For a spherical symetric body of radius R and luminosity L, we have by definition :

L = 4πR²σTeff4

Definition 2 :

Any body : F = σ Teff4

Black body : F = σ T4

Effective temperature: Definition

Presenter

Presentation Notes

haɪpəˈθɛtɪkəl language abuse


30000 K

Effective temperatures of stars

10000 K

Log (Effective Temperature) 3000 K

Size of stars

•Size of stars : 4 ²4 effT

RLF σ

π==

- Main Sequence stars have a restricted range of radii

- Supergiants are thousands times larger than the Sun

- Giants are ten-hundred times larger than the Sun

- White dwarfs are one hundred times smaller than the Sun

Cst T log 4 Rlog 2 Llog eff ++=


Stellar sizes

Log (Effective Temperature)


The main sequence

Stellar source of energy: Thermonuclear reactions

Longest Phase : Fusion of hydrogen into helium

Our Sun is a main sequence star

Most stars are in this phase

M ~ 100 M¯

They occupy a specific region of the HR diagram: The main sequence

M ~ 0.1 M¯

Red giants

Phase of core helium fusion into carbon and oxygen,

and/or shell hydrogen fusion

The core of a red giant is very dense,

its envelope is huge (van encompass the earth orbit)

M ~ 100 M¯

M ~ 0.1 M¯


White dwarfs

Slow cooling …

End of the life of our Sun: Small and dense (R ~ 0.01 R¯ , ρ ~ 1 ton / cm3)

M ~ 100 M¯

M ~ 0.1 M¯


Stellar masses Determination of the masses in binary systems: Third Kepler law

•Main sequence : 0.1 - 100 M

•Mass of giants : 1 - 10 M

•Mass of supergiants : ≥ 20 M

•Mass of white dwarfs : ≈ 0.5 M

•Mass-luminosity relation in main sequence stars

L ~ Mν(>1)

3000 K < Teff < 100000 K

0.08 M < M < 100 M

…0… < L < 106 L

…0… < R < 1000 R

Pop I – Pop II ⇒ X, Y, Z

Characteristics of stars

Our Sun

L = 4π R2σTeff4

Radius = 0.697 £ 106 km

Can be measured, knowing: - Its distance d - Its angular diameter θ

Luminosity = 3.826 £ 1026 W Can be measured, knowing :

- Its distance d - The flux f received on earth

R = d tg θ/2

L = 4π d2 f

Effective temperature = 5777 K ~ Surface temperature

Obtained from its definition :

Mass = 1.989 £ 1030 kg From the third Kepler law:

GM = 4π2 a3/P2

24 rdrdm ρπ=

drr dVdm 2 4 = = πρρ

Internal characteristics of stars

Spherical Symetry Only one space dimension: r

Masse

m

dm

Mass of each spherical shell of thickness dr

~ 1.4£103 kg/m3

Sun :

A gas sphere in hydrostatic equilibrium

The resultant force on each material volume

= 0

Sphere of gas in hydrostatic equilibrium

2 g r

GmdrdP

−=−= ρρ0 =∑F

dr

Equilibrium of the forces :

dmr

GmdrgrPM

rm

R

r

4

)()(

4∫∫ ==

πρ

Resultant force on an infinitesimal volume (1 m2 section) :

Pressure : -P(r+dr) + P(r) (surface force) Weight : -ρ g dr (volume force)

Integrated form :

-P(r+dr)

P(r)

g dm

( P(R) = 0 )

Sun, simplified model of constant density :

~ 5£1014 Pa

21 on accelerati Radial

rmG

drdP - −=

ρ

Dynamical time ~ free fall time

(pressure suppressed) GM

3

dynR t =22

dyntR

RGM

=

Sun : tdyn = 26 min << stellar evolution time

Out of equilibrium state:

Order of magnitude:

During the main part of the stellar life, hydrostatic equilibrium Exception : Explosion of a supernova (dynamical instability)

0 == ∑Fdm

γ

r£ (rP/ρ) = -(1/ρ2) rρ£r P = (1/2) r(Ω2) £ r(s2)

Rotational of the equilibrium equation s es = (1/2) r(s2)

For a rigid or cylindrical rotation (Ω(s)): rρ£rP = 0

Isobars = Iso-density = Isotherms : P(Ψ), ρ(Ψ), T(Ψ), …

Ψ = Total Potential =

Ψ1

Ψ2

Ψ3

Else (more complex differential rotation of real cases):

Isobars ≠ Iso-density ≠ Isotherms

Baroclinity

Fast rotation and hydrostatic equilibrium

s = r sinθ es = rs

Critical speed : if Veq = Ω R > Vcrit ¼ (GM/R)1/2 ,

Centrifugal force > gravity ) the envelope is expelled

Stars can rotate fast ) centrifugal force:

rP/ρ = -rφ + Ω2 s es

) Deformation of the star

Ω2 s es

Presenter

Presentation Notes

Break-up, disintegration

Stars can rotate fast ) centrifugal force:


rP/ρ = -rφ + Ω2 r sinθ es

) Deformation of the star

) All quantities (T, P, ρ, F, …) depend on 2 space dimensions (r, θ) ) much harder to solve

Critical speed : if Veq = Ω R > Vcrit ¼ (GM/R)1/2 ,

Centrifugal force > gravity ) the envelope is expelled

Some stars (Be stars, …) are not far away from that !

Influence of rotation

Affects the stellar structure and evolution : - 2D structure - Transport processes

Affects spectroscopic and photometric observables (Teff , …),

Because at the photosphere: F(θ), ge(θ)

State of the stellar matter

~ 5£1014 Pa

H H

H

H

H H

d0 ~ 10-10 m

Hydrogen atoms (neutral) in contact

~ 1.4£103 kg/m3

But is the atoms repulsion sufficient to support the

weight of the gas column ?

The matter is ionized

Presenter

Presentation Notes

ad absurdum argument

Mean distance between particles

The stellar matter is ionized: plasma

Size of nuclei ~ 10-10 m >> 10-15 m

Equation of state of an ideal gaz:

Estimation of the core temperature

Sufficient temperature for the thermonuclear fusion of hydrogen nuclei

State of the stellar matter

1) Radiation

Transport of energy from the center to the surface

Sources of opacity

Bound-bound transitions

In the stars: k T ~ h ν < kev ) electronic transitions

Bound-free Free-free Diffusion

Mean free path of photons ~ cm In stellar interiors !

Stellar interiors are completely opaque and in thermodynamic equilibrium

Transport of energy (from the center to the surface)

1) Radiation

In stellar interiors:

Mean free path of photons :

= (κρ)-1 ¼ 2cm <<

dr / dlnT, dr/dln P, …

Opaque interior

Matter-radiation thermodynamic equilibrium

Stefan law :

Planck law :

1) Radiation

dTTdTTL

3

4

4 )

σ

σ

=

+( =4T σ

T+dT

T

~ = (κρ)−1

Approximative proof

But the mean free paths in different directions where not

appropriately taken into account

Transport of energy from the center to the surface

1) Radiation Rigorous proof Solution of the radiative

transfer equation:

Radiative transfer equation

θ

* 2 π µ and integration

2/3

Rosseland mean

0

1) Radiation

κ % ) |dT/dr| %

dTTdTTL

3

4

4 )

σ

σ

=

+( =4T σ

T+dT

T

Gradient of T L

Tcenter >> Tsurface

Whatever the source of energy

For a fixed L

Example: partial ionization zones


2) Convection

m m

If (∆ρ)e < (∆ρ)m

Archimedes force upwards

Convectively unstable medium


2) Convection

ρe < ρm Convectively unstable medium

if and only if :

Temperature criterion ? Pressure equilibrium:

P=RρT/µ Ideal gas

Te /µe > Tm / µm m m


Pe = Pm

Pe = R ρe Te / µe = Pm = R ρm Tm / µm

ρe < ρm

Te > Tm If µe = µm

2) Convection

|dlnT/dr|e < |dlnT/dr|m (∆T/T)e > (∆T/T)m

∆ P < 0

(dlnT/dlnP)e < (dlnT/dlnP)m

Transport of energy !

m m

∆ T < 0


Convectively unstable medium if and only if : Te > Tm

2) Convection

m m

(dlnT/dlnP)e < (dlnT/dlnP)m

Efficient convection Negligible radiative loss

(dlnT/dlnP)e = ∂ lnT/∂ lnP|s ´ rad

(dlnT/dlnP)m ´ r > rad


Convectively unstable medium if and only if :

Convectively unstable medium if and only if :

Convection + radiation

rrad = radiative gradient = fictive gradient required to transport the whole energy by radiation

32163

TracL

drdT R

πρκ

−= ¥ T dlnP/dr = - T ρ g / P

4163

lnln

GmTacLP

PdTd R

πκ

= 4163

GmTacLP

rad πκ

≡∇

rrad > rad


Schwarzschild criterion: Convectively unstable medium

rrad > rad Schwarzschild criterion:

Convectively unstable medium

4163

GmTacLP

rad πκ

≡∇Convection typically appears when :

L/m Very high

Nuclear reactions only near the center

Convective core

κ (opacity) very high

Partial ionization zones near the surface

Convective envelope


Convective core

Convective envelope


A small (r - rad) is sufficient

In the deep layers where convective transport

is very efficient r ' rad ' 2/5

Fully ionized gas

Very high heat capacity

Fc = ρ cp Vc,r ∆T (Enthalpy flux) ¼ ρ cpT (P/ρ)1/2α2(r - rad)3/2

Mixing-length theory (very approximative !)


How efficient is this energy transport ?

Energy balance :

Equation of energy conservation

Volume source of energy inside stars : nuclear reactions

εn = εn(ρ, T, Xi)

is the power produced by nuclear reactions in the spherical shell of mass dm

εn dm

We define εn as the rate of energy generation by nuclear reactions per unit time and mass.

Energy lost by the spherical shell per unit time:

L(r+dr) – L(r) = dL = ∂ L/∂ r dr

dL = εn dm ) d L/d m = εn

At thermal equilibrium: produced energy = lost energy

) dL/dr = 4π r2 ρ εn


Energy balance :

dtdsT

dmdL

n −= ε

dmdtdsTdm

dtdvP

dtdu

dmdtdQdLdm

- n

=

+ =

=ε

Out of thermal equilibrium

εgrav « Energy supply from the gravitational contraction » (see the Virial theorem)

Expansion power

Internal energy variation


Baroclinity:

r£ (rP/ρ) = -(1/ρ2) rρ£r P = (1/2) r(Ω2) £ r(s2)

Rotational of the equilibrium equation rP/ρ = -rφ + Ω2 s es

s es = (1/2) r(s2)

For a rigid or cylindrical rotation (Ω(s)): rρ£rP = 0

Isobars = Iso-density = Isotherms : P(Ψ), ρ(Ψ), T(Ψ), …

Ψ = Total Potential =

Ψ1

Ψ2

Ψ3

Else (more complex differential rotation of real cases):

Isobars ≠ Iso-density ≠ Isotherms

Baroclinity


s = r sinθ es = rs

Fast rotation and energy conservation

Von Zeipel theorem: We consider the radiative zone of a rigidly rotating star P(Ψ), ρ(Ψ), T(Ψ), …

Ψ1

Ψ2

Ψ3

Non-constant on equipotentials

Whatever T(Ψ), it is impossible to have r¢ (Frad) = 0 everywhere !

Thermal desequilibrium :

Stationary case : V ¢ rS ≠ 0 Meridional circulation

Large scale transport process: 1) Transport of angular momentum

2) Transport of chemical elements

Characteristic time for the establishment of the meridional circulation: Eddington-Sweet time: Next, the circulation adapts to ensure the transport of angular momentum required by the boundary conditions (winds, accretion, tidal effects in close binaries, …)

r¢F/(ρT) = - V ¢ rS < 0

r¢F/(ρT) = - V ¢ rS > 0

Fast rotation and energy conservation

Meridional circulation

Equations of structure

24 rdrdm ρπ=Mass

2rmG

drdP ρ

−=Hydrostatic Equilibrium

32163

TracL

drdT

πρκ

−=Radiation

PT

rGm

drdT

2ad- ρ∇=

Transport of energy

)(4 2gravnr

drdL εερπ +=Conservation

of energy

4 equations 7 unknowns (m, ρ , P, T, L, κ , ε )

ρ = ρ(T, P, Xi) , κ = κ(T, P, Xi), ε = ε(T, P, Xi)

Xi : chemical composition

Convection (efficient)

241

rdmdr

ρπ=

4 4 rmG

dmdP

π−=

342643

TracL

dmdT

πκ

−=

PT

rGm

dmdT

4ad 4 -

π∇=

gravndmdL εε +=

Equations of structure

Mass

Hydrostatic Equilibrium

Radiation

Convection (efficient)

Transport of energy

Conservation of energy

4 equations 7 unknowns (m, ρ , P, T, L, κ , ε )

ρ = ρ(T, P, Xi) , κ = κ(T, P, Xi), ε = ε(T, P, Xi)

Xi : chemical composition

Boundary conditions

4 differential equations 4 boundary conditions

Masse : m(0) = 0 ou r(0) = 0 Luminosity : L(0) = 0

Center

Photosphere Continuous match with the atmosphere model

Temperature : T = Teff = (L/4π R2σ)1/4 Pressure : P = P(Teff, log g, Xi) = f(L, R, M, Xi)

¼ 0

State equation and chemical composition

Stars are mainly composed of: - Hydrogen (initial mass fraction X ~ 0.7) - Helium (initial mass fraction Y ~ 0.28) - Heavy elements (« metals »: Fe, C, N, O, …) total mass fraction Z ~ 0.015 for the Sun)

~ Perfect gas

The gas is nearly fully ionized in the stellar interior but Coulombian interaction between ions and electrons does not affect significantly their trajectories: (3/2) kT = Ekin > e2/d (electrostatic potential energy)

ugaz m

Tkk TnPµ

ρ ==Perfect gas Mean molecular weight (dimensionless)

n = ne + ∑i ni = ∑i (1+Zi)ni

Mixing of free electrons and nuclei

(fully ionized case)

Pgaz = Pe + ∑i Pi

ni = ρi/(Αi mu) = ρ Xi /(Αi mu)

µ = ( ∑i Xi (1+Zi) / Αi )-1

Equation of State

Pressure = normal momentum flux

Pressure integral

θ

All particles in the same direction :

Isotropic gas: Fraction of particles with direction between θ and θ+dθ :

Pressure of particles with angles in [θ, θ+dθ ] and momentum in [p, p+dp] :

Total pressure:

p(ν) = hν/c Momentum of a photon:

Radiation pressure

General non-isotropic case Tensor

Density of radiative energy / u. volume and frequency:

Planck function

Pressure = momentum flux

Equation of State

a = 4 σ/c

Total pressure 4

31 aTm

TkPu

+= µρ

In « hot » stars, PR = (1/3) aT4 can become significant

compared to Pgaz.

Equation of State

Total pressure 4

31 aTm

TkPu

+= µρ

But, gas of fermions

Pauli exclusion principle

Not more than 2 fermions in the phase volume: dpxdpydpz

dx dy dz = h3

Equation of State

Volume occupied in the phase space:

Equation of State

Quantum degeneracy of the electron gas

Maxwell-Boltzmann distribution

of momentum :

Number of electrons

Volume occupé dans l’espace des phases:

4π p2 dp dV

Equation of State


Volume occupied in the phase space:


of momentum :

Number of electrons

Degeneracy appears when ρ or T

Fermi-Dirac statistics

Equation of State



of momentum :

Degeneracy criterion

Equation of State


No degeneracy if

Mean molecular weight / electron

The degeneracy appears first for the electrons

Equation of State


A simple example: completely degenerated gas

(T ! 0 limit)

Fermi momentum

n

Quantum degeneracy

Non-relativistic case

Mean molecular weight / electrons

Equation of State

v = p / me (non-relativistic case)

Quantum degeneracy Mean molecular weight /

electrons

Equation of State

v ~ c (relativistic case)

Extremely relativistic case

Partly relativistic case

Equation of State

Quantum degeneracy

Calculus of the electronic density and the mean molecular weight per electron

ni = ρi/(Αi mu) = ρ Xi /(Αi mu)

µe = ( ∑i Xi Zi / Αi )-1

Hydrogen (X), Helium (Y), metals (for the metals, Zi/Αi ' ½)

Equation of State

Pressure and density are quasi insensitive to the temperature

) ∂ln P/∂ln T|ρ ' 0

Fermi-Dirac distribution

x

(mekT)1/2

Equation of State


n

Radiation pressure:

ugR m

TkPaTPµ

ρ=≥= 4

31

Degeneracy:

(MKSA)

Relativistic degeneracy: Crystalisation

of ions

(g/cm3)

Equation of State

4

31 aTm

TkPu

+= µρ

Degenerated :

3/5)(e

eKP µ

ρ≈ 3/4)(e

eKP µ

ρ∗≈

Non-relativistic degeneracy Relativistic degeneracy

Non-degenerated :

Equation of State

Opacity

Mean Rosseland opacity

32163

TracL

drdT

πρκ

−=

Parts of the spectrum which have the highest weight: - Maximum of the Planck distribution : ν ¼ kT/h - Less opaque regions (harmonic mean)

κ = κ(T, ρ, Xi) κν = f(ν, T, ρ, Xi)

Depends on the frequency

Planck function (Black body)

Opacity

Bound-bound transitions

Sources of line broadening:

- Natural :

- Thermal : Doppler effect due to the thermal agitation

- Collisions (Stark effect) : Perturbation of the potential due to

nearby ions and free electrons

Life-time of the state

Absorption if ν ¼ (E2-E1)/h

ν = ∆E/h

E

E + ∆E

e-

Opacity

Bound-bound transitions ν = ∆E/h

E

E + ∆E

e-

Calculation of the opacity : Nbr / m3

Cross-section of one transition

- Cross section determination for all transitions ! …

- Determination of the occupation of all electronic states

Cross section: σν n = κν ρ = Σi,j ni σν i,j

T < 106 K up to 50% T ~ 107 K less than 10%

Contribution of the bound-bound transitions to the total opacity :

Bound-free transitions

Absorption 8 ν ¸ χ/h, Ecin (e-) = hν - χ

σν / ν-3

For a given transition : σν,i / ν-3 For each new transition (ν ¸ χi/h) discontinuity

Cross section

Opacity

ν > χ/h

E

E + χ

e-

Continu


κλ / λ3

For a given transition : σλ,i / λ3 For each new transition (λ · hc/χi) discontinuity

Ionisation of the Hydrogen atom

Opacity κλ

Opacity


ν > χ/h

E

E + χ

e-

Continu

Grosso-modo … At a given temperature : hν ~ kT

Opacity calculation : - Cross section determination for all possible ionizations …

- Determination of the occupation of all electronic states :

free electrons, states of ionisation (Saha eq.) and excitation (Boltzmann)

Transitions with χi >> kT never occur because too few photons have enough energy Transitions with χi << kT never occur because too few ions are at the required initial ionization state (Saha)

transitions with χi ' kT occur


Opacity


Absorption 8 ν ¸ χ/h

Total opacity : complicated !

Opacity


ν > χ/h

E

E + χ

e-

Continu

For a given transition : σν,i / ν-3

Dependence with respect to the temperature : Grosso modo … κν / ν-3 , hν ~ kT κ / T-3

Kramers law (approximative !) :

5.324 )1(1034.4 −+×≅ TXZbf ρκ

Heavy elements are the main contributors because of their high ionisation potential

The initial ionization states are occupied in the high temperature layers

Z = mass fraction of heavy elements

m2/kg

Opacity


Free-free absorptions

E cin, fin(e-) = E cin, init(e-) + hν

hν e-

Z

e- An ion is needed to ensure the momentum conservation

5.321 )1()(1068.3 −++≅ TXYXff ρκ

Kramers law(approx.)

σλ / λ3/v

Initial velocity of the electron

λ ~ hc/(kT)

v ~ (3kT/me)1/2

κ / T-3.5

Inverse Bremsstrahlung

Limitation: Degenerated gas

High number of impossible transitions (many occupied states)

Smaller opacity

m2/kg

Opacity

)1(02.0 / 0.04 e

Xes

+≅= µκ

Electron scattering

hν e-

e-

hν

Elastic collision electron – photon

Non-relativistic case :

Thomson diffusion

Relativistic case :

Compton diffusion

Electromag. wave oscillation of the electron

reemission in another direction

Classical interpretation:

Independent of T and ν !

2-28 m 10 0.665 ×=Tσ

m2/kg

Opacity

Bound-Free transitions

Free-Free transitions

Approximation Kramers laws

5.324 )1(1034.4 −+≅ TXZbf ρκ

5.321 )1()(1068.3 −++≅ TXYXff ρκ

Electron scattering )1(02.0 Xes +≅κ

Bound-Bound transitions

m2/kg

m2/kg

m2/kg

In the 3 last cases: κ / nbr. e-/m3 / 1+X

Opacity

Electronic conduction

If the electron gas is highly degenerated

Very high kinetic energy of the electrons

They efficiently transport energy by conduction

« As if the opacity was lower »

Opacity

Opacity

M

HeII Fe H

log R = log ρ – 3 log T + 18 ~ constant from the center to the surface

Polytropic spheres

Non-relativistic degenerated gas : P = K1 ρ5/3

Relativistic degenerated gas : P = K1 ρ4/3

Polytrope (definition) : P = K ργ = K ρ1+1/n

Hydrostatic equilibrium

Poisson

equation

Presenter

Presentation Notes

short-circuit

Poisson

Change of variables dimensionless variables

Equation of

Lane-Emden

Polytropic spheres

Cases with analytical solutions :

Equation de

Lane-Emden

Other important cases: n = 3/2 (γ = 5/3, non-relativistic degenerated)

n = 3 (γ = 4/3, relativistic degenerated) n = 1 (γ = 1, isothermal sphere, infinite radius)

n < 5 ( γ > 6/5 ) Finite radius, n ¸ 5 Infinite radius

Polytropic spheres

A constant density model (n=0)

Huge convective instability

Such a model is not realistic, the density must increase towards the centre

so that dlnT/dlnP ~ 1 – dlnρ/dln P is small enough.

Polytropic spheres

Hierarchy of polytropes : n = 0 : constant density, dln T/dln P = 1 (convectively unstable)

n = 3/2 (γ = 5/3) : non-relativistic degenerated or convective zone

n = 3 (γ = 4/3) : relativistic degenerated or radiation pressure

n = 3.25 : ~ radiative envelope

n = 5 (γ = 6/5) : Limite finite radius infinite n = 1 (γ = 1) : isothermal sphere, infinite radius

|rρ| |rT|

deg. relativistic deg. non- relativistic

Isothermal sphere

Polytropic spheres

Un modèle simple d’enveloppe radiative

Kramers : a = 1 , b = -4.5

Hypothèses:

- Enveloppe radiative :

- Opacités : - L, m constants (ε = 0, ρ petit)

Polytrope n = 3.25

Mass

Mass-radius relation of polytropes

Polytropic spheres

Mass – radius relation

of polytropes (fixed K !)

Polytropes : degenerated gas, white dwarfs

0 · n < 1 : M R

1 < n < 3 : M R

m(r) m’(r) = q m(r)

ρ(r) ρ’(r) = q ρ(r)

Weight(r) = Weight’(r) = q2 Weight(r) sr

R Gmρ/r2 d r

P(r) P’(r) = K (ρ’)1+1/n = q1+1/n P(r)

n < 1 P’(r) > Poid’(r)

Increasing the mass by a factor q , keeping the same radius

The pressure increases more than the weight dilatation

n > 1 P’(r) < Poid’(r)

The pressure increases less than the weight contraction

Mass - Radius relation of polytropes (fixed K)

Non-relativistic degenerated gas, n = 3/2 :


Relativistic degenerated gas, n = 3 : M = Constant !

Chandrasekhar’s limiting mass

When the state of matter tends towards complete non-rel. degeneracy:

R ~ constant Cooling at constant radius of white dwarfs

Presenter

Presentation Notes

Subrahmanyan Chandrasekhar est un astrophysicien et un mathématicien d'origine indienne. Il est lauréat du prix Nobel de physique de 1983. (19 octobre 1910 à Lahore, Punjab, Indes britanniques - 21 août 1995 à Chicago).

Lum

inos

ity

Effective temperature

Cooling at constant radius of white dwarfs


M is fixed to Mch, R is free

r(m) r’(m) = x r(m)

ρ(m) ρ’(m) = ρ(m)/x3

Weight(m) Weight’(m) = Weight(m)/x4

P(m) P’(m) = K (ρ’)4/3 = K ρ4/3/x4 = P(m)/x4

P(m) = Weight(m) P’(m) = Weight’(m)

Polytropes : relativistic degenerated gas

Let’s multiply by a constant factor x the radius of each sphere

The star remains in hydrostatic equilibrium!!

What happens when M > Mch ?

r(m) r’(m) = x r(m)

ρ(m) ρ’(m) = ρ(m)/x3


P(m) P’(m) = P(m)/x4

P’(m) < Poid’(m) The collapse continues

P(m) < Poids(m) contraction

No hydrostatic equilibrium


Polytropes : relativistic degenerated gas

Let’s consider a star in hydrostatic equilibrium

r(m) r’(m) = x r(m)

ρ(m) ρ’(m) = ρ(m)/x3


P(m) P’(m) = P(m) (ρ’/ρ)Γ1 = P(m)/x3Γ1

If Γ1 > 4/3 (x<1) , P’(m) > Poid’(m),

The star comes back to equilibrium: Dynamically STABLE

P(m) = Poids(m)

Dynamical stability criterion

Not necessarily a polytrope

Initial state :

Homologous adiabatic contraction of the gas P’/P = (ρ’/ρ)Γ1

If Γ1 < 4/3 (x<1) , P’(m) < Poid’(m), The star collapses:

Dynamically UNSTABLE

The Mass – Radius relation



slowly accretes matter from its red giant companion

1st scenario (the most frequent): White dwarf of C-O in a binary system

What happens when the Chandrasekhar’s limiting mass is reached ?

The C-O core’s mass reaches the Chandrasekhar’s limit !

the C-O core’s mass increases

2nd scenario : During shell burning of H and He,

The core contracts and reaches huge densities

Nuclear reactions in a degenerated medium thermal runaway The entire star burns and explodes like a huge thermonuclear bomb.

Type Ia supernova L ~ 109 - 1010 L¯

Very massive stars

Main scenario:

Core fusion of silicon, contraction,

Shell fusion of silicon, contraction, …

) The iron core contracts,

becomes degenerate and grows in mass.

When MFe ' Chandrasekhar’s mass

) collapse until the nuclei are in contact

) rebound, chock wave

) Explosion, type II Supernova

Onion Structure

What happens when the Chandrasekhar’s limiting mass is reached ?

Nuclear reactions

Main source of energy during the life of a star

Role

Calculus of the rate of energy generation by nuclear reactions ε

/ u. mass and time

Stellar nucleosynthesis: Creation of the

nuclei of the univers :

Calculus of the time evolution of

abundances during the life of stars

• Momentum conservation

Movement eq. of the “relative particle” in a fixed potential :

Classical (Newton):

Quantum (Schroedinger) :

Conservation laws

Reduced mass

(2 bodies problem, non-relativistic)

• Charge conservation

• Nuclei conservation

Nuclear reactions

• Total angular momentum

J = ∑ L + ∑ S = Cst.

“Mass” excess = - binding energy

Sum of spins

• Energy

Sum of orbital angular momenta

Provided energy

• Lepton number

Binding energy per nucleon

Nuclear reactions

Conservation laws

• Energy

Provided energy Binding energy / nucleon

Reaction rate = nbr. of reactions

/ u. time and volume

Nbr. of particle couples

/ u. volume2 Relative speed

Cross section

Speed distribution

Rate of energy generation ε = power / u. mass

Sum over all reactions

Nuclear reactions

Total kinetic energy in the center of mass frame:

Probability density of E : (Boltzmann, Fermi-Dirac, …)

The whole difficulty is the determination of the cross section σ(E)

Nuclear reactions

Determination of the cross section σ(E) Different factors:

1) Geometrical factor Pλ

Collisional cross section of a nucleus : Pλ = λ2 /(4π)= (h/(mµ v))2/(4π) ~ 1/E

De Broglie

2) Gamov factor PG

Crossing of the Coulomb barrier

Cross section of a nucleus in the quantum diffusion process with no angular momentum

(l=0)

Nuclear reactions

The Coulomb barrier

for T ~ 107 K

r0 : nuclear radius

More than 1000 times too small

Probability e-1000 (Max-Boltz.) !

Impossible

ECb

rmin

Nuclear reactions

Determination of the cross section σ(E)

2) Gamov factor PG

Fortunately …

quantum tunneling Nuclei can cross the potential barrier thanks to

ECb

rmin

Nuclear reactions

The Coulomb barrier

2) Gamov factor PG


Wave function :

Schroedinger eq. :

Crossing probability of the Coulomb barrier

Square potential barrier

2) Gamov factor PG


Nuclear reactions

Quantum tunneling

Coulomb potential Crossing probability of the Coulomb

barrier (WKB approximation) :

ECb

rmin

Nuclear reactions

2) Gamov factor PG


Crossing probability of the Coulomb barrier (WKB approximation) :

ECb

rmin

Nuclear reactions

2) Gamov factor PG


Quantum tunneling

3) Nuclear factor S(E)

Nuclear properties, strong (and sometimes weak) interaction

S(E) slowly varies with E

Except near a resonance

Experimental determination of S(E)

Nuclear reactions


f

m

m

Nuclear reactions


b

Gamow peak

m

Nuclear reactions


m

Nuclear reactions


Gamow peak

Gaussian approximation

m

Nuclear reactions

Gamow peak

Presenter

Presentation Notes

standard deviation Integral d’une gaussienne normalisée : racine(pi)

Light nuclei: ν ∼ 6 ( < σ v > / T6 )

Heavy nuclei:

ν ∼ 20-30 !! ( < σ v > / T30 )

Extremely sensitive to the temperature !!

T7 = T (K) / 107

Nuclear reactions

Resonant reactions

Discrete set of states-energies of the nuclei ~ electronic levels of atoms

Shell-model: Each nucleon in the average potential generated by the

others

Nuclear reactions

Presenter

Presentation Notes

This observation, that there are certain magic numbers of nucleons: 2, 8, 20, 28, 50, 82, 126 which are more tightly bound than the next higher number, is the origin of the shell model 82 : Plomb (lead=liiid)

Discrete set of states-energies of the nuclei ~ atoms

Quasi- stationary levels

Stationary levels

If E is near En

Resonant reaction

Nuclear reactions

Resonant reactions

Around the resonance energy:

For E ~ Eres , σ ∼ σmax = (2l+1) λ2 /(4π) = (2l+1) π (~/(mµ v))2

σ very high for the nuclei colliding with the required angular momentum and a near-resonance energy

S(E) is sharp around Eres

Γ

Even more sensitive to the temperature

ex : reaction 3α : ε ~ T40 !!!

Must remain inside the integral

Γ = ~/τ Nuclear reactions

Resonant reactions

Hydrogen burning

1. Proton-proton chain (T > 107 K)

eeHHH ν++→+ 0

1

2

1

1

1

1

1

γ+→+ HeHH 32

21

11

HHeHeHe 11

42

32

32 2+→+

PPI

ppI chain

Hydrogen burning


eeHHH ν++→+ 0

1

2

1

1

1

1

1

γ+→+ HeHH 32

21

11

HHeHeHe 11

42

32

32 2+→+

PPI

But : - Energy loss by neutrinos ! - Out of equilibrium !

Energy of the neutrino

Somme sur les réactions

4

Hydrogen burning


Presenter

Presentation Notes

“Mass” excess

Crossing of the Coulomb barrier immediately followed by: p n + e+ + ν (β-decay)

Spin change of proton transformed into neutron Gamow-Teller β-decay

Very low probability because bad covering of the wave functions

( s ψf* ψi dV )

Very long reaction time: τH ~ 109 years

e+ + e- γ Eneutrino ' 0.262 Mev lost

eeHHH ν++→+ 0

1

2

1

1

1

1

1p-p reaction


Hydrogen burning

Presenter

Presentation Notes

Discovered by Hans Bethe (1939)

Deuterium life-time:

τD ~ 1 sec. γ+→+ HeHH 32

21

11

Deuterium burning

In earth’s oceans: D/H = 1.5 £ 10-4

(Big-bang + solar heating of comets’ ice ?)

First: Burning of initial deuterium, next equilibrium.


Hydrogen burning

Presenter

Presentation Notes

This is thought to be as a result of natural isotope separation processes that occur from solar heating of ices in comets. Like the water-cycle in Earth's weather, such heating processes may enrich deuterium with respect to protium. Kinetics -> kineeetiks

He3 + He3 reaction

Deuterium at equilibrium

HHeHeHe 11

42

32

32 2+→+

First, very slow reaction !!

Decreases as T

For too low T, He3 does not have time to reach equilibrium He3 peak


Hydrogen burning

HHeHeHe 11

42

32

32 2+→+

Decreases as T

τeq (years)

Sun

m/M

1. Proton-proton chain

He3 + He3 reaction

Hydrogen burning

PPII and PPIII chains

γ+→+ BeHeHe 74

42

32

PPIII γ+→+ BHBe 8

511

74

eeBeB ν++→ 01

84

85

HeBe 42

84 2→

PPII ν0

0

7

3

0

1

7

4+→+

−LieBe

γ+→+ BeHLi 8

4

1

1

7

3

HeHeBe 4

2

4

2

8

4+→

T6

X = Y

Hydrogen burning


Lithium problem

Non-standard hydrodynamical mixing processes below the convective envelope:

diffusion, meridional circulation, g-waves, …

Lithium abundance at the surface of the Sun ~ 140 times

smaller than the initial value PPII γ+→+ BeHLi 8

4

1

1

7

3

The study of young clusters (Hyades, …) shows that the surface abundance of Li

decreases with time.

TLi ~ 2 106 K

Hydrogen burning


Different energies of the neutrinos :

PPII ν00

7

3

0

1

7

4+→+

−LieBe

PPIII eeBeB ν++→ 01

84

85

PPI eeHHH ν++→+ 0

1

2

1

1

1

1

1

Eν = 0.263 Mev

Eν = 0.8 Mev

Eν = 7.2 Mev

Loss : 2 £ 0.263 = 2 %

Loss : 0.263 + 0.8 = 4 %

Loss : 0.263 + 7.2 = 27.9 %

26.73

26.73

26.73

Hydrogen burning


2. CNO cycle T > 17 x 106 K

Hydrogen burning

CNO cycle

Presenter

Presentation Notes

Catalyseur = catalyst


γ+→+ OHN 15

8

1

1

14

7

Main loop :

Slowest reaction: At equilibrium, most of the C and O has been transformed

into 14N

Secondary loop

104 times less probable Main role: conversion O16 N14

Hydrogen burning

104 times less probable

1. C12 N14 O16 N14 (much slower)

2. Complex because : - Presence of a convective core which

homogenizes and of which mass is varying

- Some elements never reach the equilibrium value

Initially : C12 : N14 : O16 = 5.5 : 1 : 9.5

At equilibrium (typically) : C12 : N14 : O16 = 0.15 : 15 : 0.3


Hydrogen burning

T15

T5 Log ε

T6

The CNO dominates for T > 17 £ 106 K

Massive stars

εpp / T5 εCNO / T15


Hydrogen burning

Helium burning:

3α reaction (T ~ 108 K)

He4 + He4 À Be8*

He4 + Be8* ! C12 + γ C12 + He4 ! O16 + γ O16 + He4 ! Ne20

Resonant reaction ! Very sensitive to T

Life time ~ 10-16 s (Be8*/He4) ~ 10-9

(Negligible)

C12 + He4 ! O16 negligible

2) C12 + He4 ! O16

1) He4 + Be8* ! C12

Finally : C12 ¼ O16

C12

E3α = 3 ∆ MHe4 – ∆ MC12

= 7.375 Mev

<< 26.7 Mev (pp)

α Captures

Helium burning

3 α process and α captures (T > 100 106 K)

γO HeC +→+ 16

8

4

2

12

6

γCHeBe +→+ 12

6

4

2

8

4

BeHeHe 8

4

4

2

4

2⇔+

ν00

0

1

18

8

4

2

14

7++→+ eO HeN

γNe HeO +→+ 20

10

4

2

16

8

γMg HeNe +→+ 24

12

4

2

20

10

γNe HeO +→+ 22

10

4

2

18

8

-----

Evolution: high masses Advanced phases of nuclear reactions

in high mass stars

HNaCC 1

1

23

12

12

6

12

6+→+

n Mg CC 1

0

23

12

12

6

12

6+→+

HeNeCC 4

2

20

10

12

6

12

6+→+

Q (Mev)

Carbon burning (T ~ 6-8 x 108 K)

12C + 1H → 13N + γ 13N → 13C + e+ + ν 13C + 4He → 16O + 1n

Source of neutrons

<Q> ~ 13 Mev εcc / exp(-84/T9

1/3)

Photo-disintegration of neon (T ~ 1.2-1.5 x 109 K) HeOγNe 4

2168

*2010 Ne +→→+ kT ~ 0.12 MeV Q = 4.73 Mev

« Thermodynamic » equilibrium

As much reactions in each sense ! reaction rate :

T % ) equilibrium towards more de photo-disintegration

~ Saha

The liberated α nuclei are immediately captured:

MgNe 2412

42

2010 He →+ Mev4.58 O 2 24

1216

82010 ++→ MgNe)

Advanced phases of nuclear reactions

in high mass stars

Oxygen burning (T ~ 2 x 109 K)

Q (Mev)

For example …

<Q> ~ 16 Mev

Final product: 2/3 of 28Si for 1/3 of 32S


in high mass stars

Q (Mev)


in high mass stars

Oxygen burning (T ~ 2 x 109 K)

<Q> ~ 16 Mev

Final product: 2/3 of 28Si for 1/3 of 32S

Global result …

Silicon « burning » (T ~ 3 x 109 K)


in high mass stars

γSHeSi +⇔+ 32

16

4

2

28

14

γ+⇔+ CaHeAr 40

20

4

2

36

18

γ+⇔+ ArHeS 36

18

4

2

32

16

• •

Complicated network !

kT <~ Q for the weakly linked nuclei ) photodisintegrations and a captures towards more stable nuclei ) rearrangement: Fe, Ni %, …

α captures:

Fe


in high mass stars

Silicon « burning » (T ~ 3.3 x 109 K)

Nuclear thermodynamic equilibrium (T ~ 7 x 109 K) kT <~ Q for most reactions ) nuclear thermodynamic equilibrium ) Equal rates of direct and inverse reactions ) Abundance ratios given by « Saha-like » equations:

All abundances are obtained if np and nn are given ) 2 additional constraints are required : Fixed because equilibrium well before the β disintegrations


in high mass stars

1) kT « not too large » ) most abundant elements : maximum of fA,Z = QA,Z / A ) mainly Ni, next Fe 2) If enough time, β decays ) Z/N & (dNi = - Ni dt/τi ) ) mainly 56Fe 3) kT > Q ) photodisintegrations ) 4He, … %


in high mass stars

Nuclear thermodynamic equilibrium (T ~ 7 x 109 K)

Core of a pre-supernova

Hydrogen Helium Carbon Oxygen Silicon Iron

Onion-like structure

Duration of ultimate phase (15 M¯) : Burning of Carbon : ~ 6000 ans Neon : ~ 7 ans Oxygen : ~ 2 ans Silicon : ~ 6 jours


in high mass stars

Nuclear thermodynamic equilibrium (T ~ 7 x 109 K)

The stability valley Nucleosynthesis : s-process

Slow neutron capture

β- decay

Nucleosynthesis : s-process

Life-time β- decay << ∆t neutron capture

Nucleosynthesis : s-process

Life-time β- decay >> ∆t neutron capture

1) Neutrons source: Electrons capture by the nuclei: endothermal reaction

2) Rapid neutrons capture process: Just after the nuclear reactions due to the chock wave passage, the

heavy nuclei capture very quickly the produced free neutrons:

r-process

Nucleosynthesis : r-process

Nucleosynthesis : r-process

Equations de structure

241

rmr

ρπ=

∂∂Conservation

de la masse

4 4 rGm

mP

−=

∂∂

πEquilibre

hydrostatique

342643

TracL

mT

πκ

−=∂∂

Radiation

PT

rGm

mT

4ad 4 -

π∇=

∂∂

Convection

Transfert d’énergie

ε=∂∂mLConservation

de l’énergie

+ 4 conditions aux limites

ρ = ρ ( P, T, Xi ) κ = κ ( P, T, Xi ) ε = ε ( P, T, Xi)

Méthode de résolution numérique


Posons:

Système à résoudre:

Non-linéaire !

Méthode des différences finies ) N couches: i=1, … N

(+ 4 conditions aux limites)

Système de 4N-2 équations linéaires

Algorithme de Newton-Raphson ) Linéarisation du système autour de l’estimation à l’étape k:

Calcul d’un modèle statique: composition chimique fixée en chaque couche

Jacobienne



Posons:

Système à résoudre:

Non-linéaire !

Méthode des différences finies ) N couches: i=1, … N

Algorithme de Newton-Raphson ) Linéarisation du système autour de l’estimation à l’étape k:

Calcul d’un modèle statique: composition chimique fixée en chaque couche

Jacobienne

(+ 4 conditions aux limites) CL 1 CL 2 1-2 1-2 1-2 1-2 2-3 2-3 2-3 2-3 3-4 3-4 3-4 3-4

VARIABLES Système de 4N-2 équations linéaires

E Q U A T I O N S

Calcul de l’évolution temporelle d’un modèle L’évolution temporelle découle essentiellement de la variation de composition chimique due aux réactions nucléaires, phénomènes de transport (diffusion, …)

Modèle au temps tk ! Modèle au temps tk+1 = tk + δt

1) Calcul de la nouvelle composition: Xi (m, tk) ! Xi (m, tk+1) = Xi(m, tk) + δXi

) µ change ) équation d’état change ) structure change

2) Avec la nouvelle composition chimique, le modèle en tk n’est plus solution des équations ) on itère (méthode de Newton-Raphson) jusqu’à la convergence On obtient ainsi le modèle en tk+1

Cinétique des réactions nucléaires: Attention aux échelles de temps ! Certaines réactions sont très rapides (combustion du deuterium, …) ) forcer la mise à l’équilibre, ou utiliser une méthode implicite


Calcul de l’évolution temporelle d’un modèle L’évolution temporelle découle essentiellement de la variation de composition chimique due aux réactions nucléaires, phénomènes de transport (diffusion, …)

Modèle au temps tk ! Modèle au temps tk+1 = tk + δt

1) Calcul de la nouvelle composition: Xi (m, tk) ! Xi (m, tk+1) = Xi(m, tk) + δXi

) µ change ) équation d’état change ) structure change

2) Avec la nouvelle composition chimique, le modèle en tk n’est plus solution des équations ) on itère (méthode de Newton-Raphson) jusqu’à la convergence On obtient ainsi le modèle en tk+1

Cinétique des réactions nucléaires: γ+→+ HeHHpd 3

221

11 :Attention aux échelles de temps !

Certaines réactions sont très rapides (combustion du deuterium, …) ) forcer la mise à l’équilibre, ou utiliser une méthode implicite

Exemple du deuterium

eeHHHpp ν++→+ 01

21

11

11 :


Systèmes planétaires Contraction gravifique

Formation des étoiles

Formation des étoiles

Contraction gravifique Instabilité du nuage proto-stellaire (H2)

Critère de Jeans : le nuage est instable si

ρ ~ 10-24 g/cm3, T ~ 10 K , µ=2 MJ ~ 1000 M¯

Vitesse du son isotherme

Si τff vs < λ , le milieu n’a pas le temps de contrebalancer la perturbation sur l’échelle de temps de chute libre τff effondrement

Longueur d’onde de la perturbation vs

2 = P/ρ = < T/µ

Contraction gravifique

Contraction isotherme

Etape 1 : Effondrement gravifique ~ isotherme Temps de chute libre ~ (Gρ)-1/2 ~ 108 ans >> temps d’ajustement thermique ~ λ/c (milieu optiquement mince)

Fragmentations

MJ ~ ρ-1/2

Contraction gravifique Etape 1 : Effondrement gravifique ~ isotherme

Fragmentations car MJ

Etape 2 : Effondrement gravifique ~ adiabatique Quand M ~ 1 M¯, temps de chute libre ~ temps d’ajustement thermique :

Contraction adiabatique Tc, Pc • Apparition d’un 1er front d’onde de choc

(Tc ~ 2000 K) H2 H + H

• Apparition d’un 2er front d’onde de choc

• Equilibre hydrostatique

Etape 3 : Contraction gravifique tout en maintenant l’équilibre hydrostatique

Théorème du Viriel

Energie potentielle gravifique totale

44 rmG

mP

−=

∂∂

π ρπ 241rm

r

=∂∂Equilibre

hydrostatique Masse

Gaz parfait ionisé dégénéré ou non, non-relativiste

Ei = s0M u dm Eg = - 2 Ei

Intégration par partie

Etape 3 : Contraction gravifique tout en maintenant l’équilibre hydrostatique


(Gaz parfait) Eg = - 2 Ei

Temps d’Helmoltz – Kelvin : τHK = GM2 / (2 R L)

Etot = Ei + Eg Energie totale :

Conservation de l’énergie globale sans production interne

Pas de réactions nucléaires !

L = - dEtot /dt = – d Ei / dt – dEg / dt = dEi /dt = -(1/2) dEg / dt

L - s ε dm

Si réactions nucléaires, remplacer L par

La moitié de l’énergie potentielle gravifique libérée lors de la contraction de l’étoile conduit à une augmentation de son énergie interne, l’autre moitié est rayonnée.

Naissance ou non des réactions nucléaires ?

Rôle de la dégénérescence

Question : lors de la contraction gravifique, Tc augmente-t-elle toujours ?

Theorème du Viriel ) P/ρ augmente Pour un gaz parfait : P/ρ / T oui pour un gaz parfait non-dégénéré

Mais … ce n’est plus le cas pour un gaz dégénéré.

r r’ = x r = r + dr = (1+dr/r) r

ρ ρ’ = ρ/x3

P=Poids P’ = Poids’ = P/x4

dρ/ρ = ρ’(m)/ρ(m) – 1 = 1/(1+dr/r)3 - 1 = – 3 dr/r

dP/P = P’(m)/P(m) – 1 = 1/(1+dr/r)4 - 1 = – 4 dr/r

Contraction homologue

dr/r = x-1 = cst


Equation d’état générale : dρ/ρ = α dP/P - δ dT/T

dT/T = (α/δ) dP/P - (1/δ) dρ/ρ = (4α-3)/(3δ) dρ/ρ

dP/P = (4/3) dρ/ρ


r r’ = x r = r + dr = (1+dr/r) r

ρ ρ’ = ρ/x3

P P’ = P/x4

dρ/ρ = ρ’(m)/ρ(m) – 1 = 1/(1+dr/r)3 - 1 = – 3 dr/r

dP/P = P’(m)/P(m) – 1 = 1/(1+dr/r)4 - 1 = – 4 dr/r

Contraction homologue

dr/r = x-1 = cst


dT/T = (4α-3)/(3δ) dρ/ρ

Pour un gaz parfait : α = 1 , δ = 1 dT/T = (1/3) dρ/ρ non-dégénéré


Pour un gaz dégénéré : α = 3/5 , δ = 0 dT/T = - 1 dρ/ρ non-relativiste

dρ/ρ = α dP/P - δ dT/T

T

T

L’énergie gravifique libérée lors de la contraction n’est pas suffisante pour accélérer les électrons dégénérés (énergie de Fermi ). L’énergie manquante est puisée dans l’énergie cinétique des ions non- dégénérés de sorte que la température diminue.

Pour un gaz dégénéré : dT/T = - ∞ dρ/ρ non-relativiste

ψ : paramètre de dégénérescence

Pour un gaz parfait : dT/T = (1/3) dρ/ρ

Cas général : dT/T = (4α-3)/(3δ) dρ/ρ

Pente 1/3

Pente 2/3 ρT-3/2 = cte



r r’ = x r = r + dr = (1+dr/r) r

Dilatation homologue avec changement de masse

A densité fixée : dρ/ρ = 0 dr/r = 1/3 dm/m dP/P = 2/3 dm/m

Pour un gaz parfait: dT/T = dP/P - dρ/ρ = dP/P = 2/3 dm/m

Plus la masse est élevée, plus la température centrale est élevée (à densité fixée)

ρ = dm/(4π r2dr)

dP/P = (1+dm/m)2/(1+dr/r)4 - 1 = 2 dm/m – 4 dr/r

P = smM Gm/(4π r4)dm

m(r) m’(r’) = q m(r) = (1+dm/m) m(r)

ρ‘ = dm’/dr’/(4πr’2) = (q/x)dm/dr /(4πx2r2) = (q/x3) ρ

dρ/ρ = (1+dm/m)/(1+dr/r)3 – 1 = dm/m – 3 dr/r

P’ = smM Gm’/(4π r’4)dm’ = (q2/x4) P

Pour un gaz parfait : dT/T = (1/3) dρ/ρ

Pour un gaz dégénéré : dT/T = - ∞ dρ/ρ non-relativiste


Evolution de Tc en fonction de ρc

dT/T =

2/3 dm/m

Effet de la masse sur la trajectoire (ρc, Tc)


Température requise pour la combustion de l’hydrogène :

Tc ~ 107 K

Masse critique : M = 0.08 M¯

Si M < 0.08 M¯, l’étoile n’atteint jamais la température requise pour initier la combustion de l’hydrogène (à cause de la dégénérescence)

Naine brune

Energie potentielle gravifique totale

Ei, tot = s0M u dm = 3/2 s0

Μ (P/ρ) dm

Eg = - 2 Ei, tot

-1/2 dEg/dt = L = dEi /dt

Refroidissement des naines brunes et blanches

Energie interne totale


L’énergie interne totale augmente lors de la contraction d’un gaz dégénéré (naine blanche) aussi !!

-1/2 δ Eg = δ Ei,tot

Les naines blanche : refroidissement

Ei, tot : Energie interne totale (ions + électrons)

Eg : Energie potentielle gravifique


Ee : Energie interne des électrons

Eions << Ee

Toute l’énergie gravifique libérée augmente l’énergie cinétique des électrons

δρ/ρ = cst.

= - 3 δr/r

Eg = - 2 Ei, tot

Les naines blanches : refroidissement

Toute l’énergie gravifique libérée augmente l’énergie cinétique des électrons

L = - dEtot /dt = - d Eions /dt – d Ee /dt – d Eg /dt = - d Eions /dt

Toute l’énergie rayonnée est puisée dans l’énergie cinétique des ions

Découplage

Hydrostatique, gravité : Gaz d’électrons

Thermique : Gaz d’ions

La température diminue

Les naines blanches : refroidissement

L = - d Eions /dt = - cv M dT0 /dt

Toute l’énergie rayonnée est puisée dans l’énergie cinétique des ions

Eions = s0M uions dm = cv T0 M

Cœur quasi isotherme (conduction électronique très efficace)

La température diminue lentement

Panorama des scénarios possibles d’évolution des étoiles

Naine brune

Combustion de l’Hydrogène

M < 0.08 M¯

T ' 107 K

Tem

péra

ture

cen

trale

temps


Naine brune

Naine blanche Combustion

de l’Hydrogène

Combustion de l’Helium

0.08 M¯ < M < 0.8 M¯

M < 0.08 M¯

T ' 107 K

T ' 108 K

Tem

péra

ture

cen

trale

temps


Naine brune


de l’Hydrogène


Combustion du Carbone

Naine blanche 0.8 M¯ < M < 9 M¯

0.08 M¯ < M < 0.8 M¯

M < 0.08 M¯

T ' 108 K

T ' 107 K

T ' 6 £ 108 K

Tem

péra

ture

cen

trale

temps


Tem

péra

ture

cen

trale

Naine brune


de l’Hydrogène


Combustion du Carbone

Combustion du Silicium

Naine blanche

Explosion en Supernova

M > 9 M¯

0.5 M¯ < M < 9 M¯

0.08 M¯ < M < 0.5 M¯

M < 0.08 M¯

T ' 108 K

T ' 107 K

T ' 6 £ 108 K

T ' 3 £ 109 K

temps

Diapositive 1 - astro.ulg.ac.bedupret/cours-structurevol1-ang2017.pdf · 1 . Ψ. 2 . Ψ. 3 . Else...

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Transcript of Diapositive 1 - astro.ulg.ac.bedupret/cours-structurevol1-ang2017.pdf · 1 . Ψ. 2 . Ψ. 3 . Else...