Marc-Antoine Dupret

Stellar structure and evolution II

SPAT0045

Marc-Antoine Dupret

Associate Professor

email: ma.dupret@ulg.ac.be tel: 04 3669732

Bât. B5c, 1^er flour

Stellar formation

Stellar formation

1) Molecular clouds

• Mass : ~ 10⁴ – 10⁵ M_¯ . • Size : ~ 1 – 50 pc

•Temperature : ~ 10 – 15 K° • Number density : ~ 10⁸ - 10¹⁰ part./m³

3 types of interstellar clouds

H

mainly

2) Atomic clouds : HI regions

•Temperature : ~ 30 – 80 K°

Atomic H mainly

• Number density : ~ 10⁷ - 10⁹ m^-3

3) Ionized clouds : HII regions

•Temperature : ~ 8000 K°

H

mainly

• Number density :

~ 10² - 10⁶ m^-3 10^-25 _air

Most abundant but most difficult to detect (21 cm line)

• If the cloud density is high enough ( ) , the atoms can form molecules

³ / 10

m

m

Molecular clouds

We have a molecular cloud

• Most abundant molecule :

H

Problem : not easy to detect !!

Radio waves from CO

• Organic molecules mainly

•Mass : ~ 50 000 M

.

•Size : ~ 40 light years

Stellar formation

• The clouds do not collapse immediately because : - Force due to interstellar magnetic field - Internal pressure gradient

- Centrifugal force

But this equilibrium is not eternal

• Factors breaking the equilibrium : - Entering into a high density zone.

- Shock wave due to supernova explosion

Barnard 68, un globule de Bok composé de gaz et de poussière. Les forces de gravitéet de pression

sont encore en équilibre, mais le nuage risque à tout moment de s'effondrer

Gravitational contraction

Cloud collapse

Instability of a proto-stellar cloud (H

) Jeans criterion : the cloud is unstable if :

Isothermal sound speed

If _ff v_s <  , during the free fall time _ff, the pressure does not increase enough to counterbalance the increasing weight

collapse

Size of

perturbation v_s² = P/ ~ T/

Dispersion relation :

A slightly more rigorous approach :

How does the cloud reacts to a plane wave perturbation ?

w² < 0 : unstable w² > 0 : stable

Cloud collapse

Instability of the proto-stellar cloud (H

)

 ~ 10^-24 g/cm³, T ~ 10 K , =2

M_J ~ 400 M₀

 M_J

Fragmentation

Jeans criterion : the cloud is unstable if :

Isothermal sound speed

Size of

perturbation v_s² = P/ ~ T/

Cloud collapse

Step 1 : Isothermal collapse

Free fall time ~ (G)^-1/2 ~ 10⁶ years

>>

Thermal adjustment time scale

The isothermal collapse stops when _ff ~ _th

Isothermal collapse

0.018

Mass of smallest fragments ~ 0.03 M₀ >> planetary masses

Cloud collapse

Fragmentation Step 2 : Adiabatic collapse

When M ~ M_¯: free fall time scale ~ thermal adjustment time scale : Adiabatic collapse T_c, P_c

Supersonic collapse speed near the center

Appearance of a shock wave

“Hydrostatic equilibrium” below

Cloud collapse

Step 1 : Isothermal collapse

: no longer fragmentation

When T_c ~ 2000 K Dissociation : H₂ H + H

New collapse

New « hydrostatic equilibrium » below

10⁷ R₀

5UA

Appearance of a 2^nd shock wave Endothermic reaction ₁ =  ln P/ ln|_s

Dynamical instability

Step 2 : Adiabatic collapse

Cloud collapse

à 10⁷ R_¯ !

5UA

Cloud collapse

Step 2 : Adiabatic collapse

Dissociation : H₂ H + H

New collapse

₁ =  ln P/ ln|_s Dynamical instability

New hydrostatic equilibrium below

Appearance of a 2^nd shock wave

Lignes de naissance

L ~ 2 R



v

The kinetic energy of the accreted matter is converted into infrared radiation.

The accretion disk

Initially, the star is entirely hidden by its accretion disk

Stationnary equatorial accretion disk:

Near the axis: Gravitational force >> Centrifugal force

- Matter is accreted near the poles where the angular momentum is low - The magnetic field acts against star-disk differential rotation

The accretion disk ensures the evacuation of angular momentum

j_cloud ~ 10²¹-10²² cm²s^-1 >> j_{T Tauri} ~ 10¹⁶-10¹⁷ cm²s^-1 >> j_Sun ~ 10¹⁵ cm²s^-1 Angular momentum per unit mass:

Gravitational force Centrifugal force=

(weak pressure gradient)

The accretion disk

The Hayashi track :

entirely convective models

) 2

, ( 4

1

r P T dm

dr



=  ₄

4 r m G dm

dP

−



= P

T r Gm dm

dT

ad 4

- 4

^ 

= 3 differential equations :

3 unknown functions : r(m), P(m), T(m) 2 parameters to be fixed : M et L

3 boundary conditions (1 at the center, 2 at the photosphere):

0 ) 0 ( = r

For fixed M and L One solution 1 point in the HR diag.

Fixed M, variable L one family of solutions curve in the HR diag.

This is the Hayashi track

General definition :

The Hayashi track of mass M is the curve in the HR diagram corresponding to the stellar models of mass M

entirely convective and in hydrostatic equilibrium (but not necessarily in

thermal equilibrium)

Hayashi tracks

of different masses

The Hayashi track:

entirely convective models

P ad

ln d

T ln

d  

Same mathematical question but with :

Forbidden region

Too strong Convection (L_c too large) Allowed region

Partially radiative model

Forbidden region Allowed

region

Right and left of the Hayashi track

P ad

ln d

T ln

d 

P ad

ln d

T ln

d 

Gravitational contraction

maintaining hydrostatic equilibrium Virial theorem

Total gravitational energy

4 r4

m G m

P = −





 ^{4 2}

1 r m

r =



Hydrostatic 

equilibrium Mass

Monoatomic perfect gas, degenerated or not,

non-relativistic

E_i = s₀^M u dm

E

= - 2 E

Integration by part

Total internal energy :

(Perfect gas)

E

= - 2 E

Helmholtz – Kelvin time : _HK = GM² / (R L)

E

= E

+ E

Total energy :

Conservation of total energy without internal production

No nuclear reactions !

L = - dE_tot /dt = – d E_i / dt – dE_g / dt

= dE_i /dt = -(1/2) dE_g / dt

L - s  dm

With nuclear reactions, replace L by

During the contraction, half of the released gravitational potential energy is converted into internal energy, the other half is radiated.

Gravitational contraction

maintaining hydrostatic equilibrium

Virial theorem

Simplified case : evolution at constant (low) mass

When a star becomes visible, it is entirely convective.

It goes down the Hayashi track

Gravitational contraction

maintaining hydrostatic equilibrium

Helmholtz – Kelvin time :



= GM

/ (R L)

Gravitational contraction

maintaining hydrostatic equilibrium

Minimal luminosity

(Kramers) and T_c



Appearance of a radiative core

As long as the star is entirely convective : descent along the

Hayashi track

16 4

3

GmT ac

L P

rad 

= 



 ~ ₀T^-3.5

Gravitational contraction

maintaining hydrostatic equilibrium

Minimal luminosity

Appearance of the radiative core when:

Next :

Crossing of the HR diagram towards the left

Henyey track

Gravitational contraction

maintaining hydrostatic equilibrium

Pre-main sequence

1.5 – 4 M_

Entirely convective

Entirely radiative

If M > 1.5 M

, the star becomes nearly entirely

radiative when T

> 6000-6500 K

Birthline

dM/dt ~ 10^-5 M₀/year

Gravitational contraction

maintaining hydrostatic equilibrium

But, in reality, the mass of the star is not constant during this phase !

Typical 2 steps modeling:

1. Evolution with constant accretion rate

(typically 10^-5M₀/ year)

The star moves on a

« birthline » of given accretion rate

2. Evolution with constant mass : 2.1 Hayashi track : low masses only 2.2 Henyey track

But, in reality, the mass of the star is not constant during this phase !

Typical 2 steps modeling:

1. Evolution with constant accretion rate

(typically 10^-5M₀/ year)

The star moves on a

« birthline » of given accretion rate

2. Evolution with constant mass : 2.1 Hayashi track : low masses only 2.2 Henyey track

Gravitational contraction

maintaining hydrostatic equilibrium

Starting or not of nuclear reactions ?

Role of electronic degeneracy

Question : During the gravitational contraction, is T_c always increasing ? Virial theorem : P/ increases

Ideal gas : P/ / T Yes for a non-degenerated ideal gas But … No for a degenerated gas

r r’ = x r = r + dr = (1+dr/r) r

 ’ = /x³

P=Weight P’ = Weight’

= P/x⁴

d/ = ’(m)/(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 Homologous contraction

dr/r = x-1 = cst

Equation of state : d/ =  dP/P -  dT/T

dT/T = (/) dP/P - (1/) d/

= (4-3)/(3) d/ dP/P = (4/3) d/ r r’ = x r = r + dr = (1+dr/r) r

 ’ = /x³

P P’ = P/x⁴

d/ = ’(m)/(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 Homologous contraction

dr/r = x-1 = cst

Starting or not of nuclear reactions ?

Role of electronic degeneracy

dT/T = (4  -3)/(3  ) d  / 

For a non-degenerated  = 1 ,  = 1 dT/T = (1/3) d/ ideal gas

For a degenerated  = 3/5 ,  = 0 dT/T = - ∞ d/

non-relativistic gas

d/ =  dP/P -  dT/T

T

T

The gravitational energy released during the contraction is not

sufficient to accelerate the degenerated electrons (Fermi energy ).

The missing energy is drawn from the kinetic energy of the non- degenerated ions, so that the temperature decreases.

Starting or not of nuclear reactions ?

Role of electronic degeneracy

Degenerated : dT/T = -  d/ non-relativistic gas

 : Degeneracy parameter

Non-degenerated gas dT/T = (1/3) d/

Cas général : dT/T = (4-3)/(3) d/

Slope 1/3

Slope 2/3

T^-3/2 = cte

General case : dT/T = (4-3)/(3) d/

Starting or not of nuclear reactions ?

r r’ = x r = r + dr = (1+dr/r) r

Homologous expansion with change of mass :

Fixed density: d/ = 0 dr/r = 1/3 dm/m dP/P = 2/3 dm/m Ideal gas: dT/T = dP/P - d/ = dP/P = 2/3 dm/m

The larger the mass, the larger the central temperature (at fixed density)

 = dm/(4 r²dr)

dP/P = (1+dm/m)²/(1+dr/r)⁴ - 1 = 2 dm/m – 4 dr/r P = ^smM Gm/(4 r⁴)dm

m(r) m’(r’) = q m(r) = (1+dm/m) m(r)

‘ = dm’/dr’ (1/(4r’²)) = (q/x)dm/dr (1/(4x²r²)) = (q/x³) 

d/ = (1+dm/m)/(1+dr/r)³ – 1 = dm/m – 3 dr/r P’ = ^smM Gm’/(4 r’⁴)dm’ = (q²/x⁴) P

Cas général : dT/T = (4-3)/(3) d/

Evolution of T

as a function of 

dT/T= 2/3 dm/m

Non-degenerated gas

dT/T = (1/3) d/ Degenerated : dT/T = -  d/ non-relativistic gas

Effect of the mass on the ( 

, T

) trajectory

Required temperature for hydrogen burning :

T

~ 10

K

Critical mass : M = 0.08 M

If M < 0.08 M

, the star never reaches

the temperature required to initiate hydrogen burning (because of electron degeneracy)

Brown dwarf

Starting or not of

nuclear reactions ?

Total gravitational energy E_{i, tot} = ∫₀^M u dm = 3/2 ∫₀^M (P/) dm

E

= - 2 E

-1/2 dE

/dt = L = dE

/dt

Cooling of brown and white dwarfs

Total internal energy

Virial theorem

The internal energy increases during the contraction !!

-1/2  E

=  E

Virial theorem :

E_e : Electrons internal energy

E_ions << E_e The whole released gravitational energy

increases the electrons kinetic energy

/

= cst.

= - 3 dr/r

E_g = - 2 E_{i, tot}

Cooling of brown and white dwarfs

L = - dE_tot /dt = - d E_ions /dt – d E_e /dt – d E_g /dt

= - d E_ions /dt

The whole radiated energy is drawn from the ions kinetic energy.

Decoupling Hydrostatic pressure, gravity :

electron gas Thermal aspects : ion gas

Temperature decreases

Cooling of brown and white dwarfs

The whole released gravitational energy increases the electrons kinetic energy.

L = - d E

/dt = - c

M dT

/dt E

= ∫

u

dm = c

T

M

Isothermal core

(very efficient electronic conduction)

The temperature decreases slowly

Cooling of brown and white dwarfs

The whole radiated energy is drawn from the ions kinetic energy.

Homologous contraction or expansion, maintaining hydrostatic equilibrium

dT/T = (4-3)/(3) d/ Non-deg. gas : dT/T = (1/3) d/

State eq. : d/ =  dP/P -  dT/T

Gaz parfait

Non-degenerated ideal fully ionized gas : dQ = - c_v dT !!!

Degenerated gas : contraction is blocked : dQ = c_v dT

Gravothermal specific heat

Provided heat

Heat is supplied to the star, how does it react ? …

How does the star react to the onset

of nuclear reactions ?

Gaz parfait

When heat is supplied, the temperature decreases ! Why ?

The provided heat leads to a double expansion work (to maintain the hydrostatic equilibrium).

The remainder of required energy to produce this work is drawn from the internal energy reservoir

The temperature decreases Local Virial theorem

Gravothermal specific heat

Homologous contraction or expansion in hydrostatic equilibrium Non-degenerated ideal fully ionized gas : dQ = - c_v dT !!!

How does the star react to the onset

of nuclear reactions ?

Why ?

The ion and electron gas are decoupled.

The degenerated electron gas freezes the star

(because the mass-radius relation of a degenerated sphere is stable).

The provided heat increases the kinetic energy of the ions The temperature increases

Degenerated gas, the star is “frozen” : dQ = c_v dT

Homologous contraction or expansion in hydrostatic equilibrium

Gravothermal specific heat

How does the star react to the onset

of nuclear reactions ?

Loop

1) Contraction u T

2) dQ>0 (nuclear reactions) T P expansion 3) Pdv > dQ du < 0 T

4)  dQ<0 contraction 5) Thermal equilibrium

How does the star react to the onset

of nuclear reactions ?

Expansion of the core

T

|dT/dm|

< 0

L

The starting of nuclear reactions leads to an expansion of the core, a decrease of the central

temperature and of the luminosity and a decrease of the total stellar radius.

How does the star react to the onset

of nuclear reactions ?

C₁₂, O₁₆ -> N₁₄ Towards

CNO equilibrium

Towards P-P equilibrium

If M > 1.2 M₀, 2 steps : 1)

2) Cycle reaches equilibrium C₁₂, O₁₆ N₁₄

2 hooks in the HR diag.

Henyey track

The onset of nuclear reactions leads to an expansion of the core,

a decrease of the central temperature and of the luminosity and a decrease of the

total stellar radius.

Concrete example : Pre-Main- Sequence evolution of a 2 M ₀ star

ZAMS

When the core becomes radiative, T ds/dt = du/dt + P dv/dt = - dL/dm < 0 its contraction is accelerated and L

But the contraction of the envelope is stopped

Radiative zone Convective zone

ZAMS

Nuclear reactions (CNO) Appearance of a convective core

and drop of L, 2 hooks in the HR diag.

Radiative zone Convective zone

Concrete example : Pre-Main-

Sequence evolution of a 2 M ₀ star

ZAMS



_g < 0 expansion work, _g > 0 contraction (released gravitational energy)

Concrete example : Pre-Main-

Sequence evolution of a 2 M ₀ star

2 observational cases :

T Tauri stars

- Low masses ( < 2-3 M

) - On the Hayashi track

- Highly variable and active (X, radio), winds - Accretion disk (one half)

Herbig Ae/Be stars

- Larger masses ( 1.5 - 8 M

) - On the Henyey track

- Circumstellar gas emission lines

Secular stability

At thermal equilibrium: dQ/dt = 0, dT/dt = 0, s  dm = L

The thermal equilibrium is broken, maintaining hydrostatic equilibrium, does the star come back to its initial state ?

At disequilibrium (small perturbation % equilibrium) :

Homologous perturbation : d(X/X)/dr = 0 (X=T, , L, , , , … )

T T’ = T + T

 ’ = + 

L L’ = L +  L

 ’ =  + 

Is the onset of nuclear reactions a

thermally stable process ?

Non-degenerated ideal gas

_T ~ 5 (p-p)

~ 15 (CNO)

Degenerated gas

_ ~ 1

_ ~ 1

_T ~ -3.5 (Kramers)

c

= - c

Stable c

= c

Unstable

The thermal equilibrium is broken, maintaining hydrostatic equilibrium, does the star come back to its initial state ?

Is the onset of nuclear reactions a thermally stable process ?

Secular stability

The Zero Age Main Sequence (ZAMS)

Location of models in thermal equilibrium for core H burning, with a fixed chemical composition

Duration of the main sequence phase:

 ~ (Q

/4) N

(M/10) / L

~ 10 £ 10

years (sun)

Q_pp ~ 25 Mev

>> 

Thermal equilibrium From now on, stellar evolution is due

to changes of the internal chemical composition ( essentially).

Mass-Luminosity relation

Low masses

Intermediate masses

Very large masses

The Zero Age Main Sequence (ZAMS)

Mass-Luminosity relation, consequences :

1) M L

2) M Lifetime

The Zero Age Main Sequence (ZAMS)

Typical internal structures

The size increases with the mass

< 0

The density decreases with the mass

> 0

The temperature slightly increases with the mass

Slope of the main sequence

The Zero Age Main Sequence (ZAMS)

Evolution: Main sequence phase

Evolution of the global parameters

Assuming  increases everywhere homologous transformation

 increases

The luminosity increases during the evolution

Nuclear reactions control: R

So that T and thus L do not increase too much

> 0

But in reality,  increases only in the core

Solution : contraction of the core and expansion of the envelope The core must contract, but without too large increase of

the pressure and thus the temperature.

Contraction : Weight

Expansion : Weight

Evolution: Main sequence phase

Evolution of the global parameters

Low masses, p-p (M < 1.2 M₀)

High masses, CNO (M > 1.2 M₀)

 ~ 6 ,  ~ X²T⁶:

L , R , Teff ?

 ~ 15 ,  ~ X  T¹⁵:

L , R , Teff

 increases in the core contraction of the core

and expansion of the envelope

Evolution: Main sequence phase

Evolution of the global parameters

Effect of the metallicity

Z  in the envelope, the core does not change much.

|dT/dm| T_eff

M=1.4 M₀ M=1.45 M₀

M=1.4 M₀ M=1.45 M₀

Z=0.01 Z=0.02

M=1.4 M₀ M=1.45 M₀

M=1.4 M₀ M=1.45 M₀

Z=0.01 Z=0.02

M=1.6 M₀

Evolution: Main sequence phase

Convective envelopes (low masses)

Difficulty : Modeling of the super-adiabatic zone (r >> r_ad)

r - r

3D MLT

Mixing-length theory

l =  H_p

 : mixing-length parameter

 |dT/dm| T_eff

 M_ZC

 r_ZC Convective zone base :

T ~ cst. T_ZC ~ cst.

F_c =  c_p V_c,rT (enthalpy flux)

~  c_pT (P/)^1/2²(r - r_ad)^3/2

H_p = |dr/dln P|

Mixing-length:

Evolution: Main sequence phase

Energy transfer 3D simulation

of convection

Convective cores (high masses)

Very efficient convection

F

/ l

c

T (r - r

)

r ~ r

Evolution of the convective core’s mass :

The larger the mass, the larger the convective core’s mass

Most of the time,

the mass of the convective core decreases as the star evolves, leaving above a -gradient zone

T ^

r

X+1

16

3

GmT ac

L P

rad



 



Evolution: Main sequence phase

Convective cores and envelopes

Convective core Convective

envelope

Convective cores (high masses)

If the mass of the convective core increases as the star evolves because : - Starting of the CNO cycle (around 1.3 M₀)

- r_ad because P_R (very massive stars) X discontinuity 1+X_e > 1+X_i



> 

i

Evolution: Main sequence phase

Discontinuity of X 1+X_e > 1+X_i



> 

i

r

> r

Semi-convection

Partial Mixing

Evolution: Main sequence phase

Convective cores (high masses)

If the convective core mass increases (starting of CNO cycle or massive star)

Where is the convective zone boundary ??

Convective penetration – overshooting

When a convective bubble

enters in a radiative zone:

r < r

zone where:

 T < 0 F_c < 0

Cooling r ~ r_ad Growth of the adiabatic

mixed (chemically homogeneous) region

Convective cores (high masses)

Evolution: Main sequence phase

Very well constrained by asteroseismology !

Overshooting at the top of convective cores

Simple parameterization :

 r

= 

H

_ov = 0.1 – 0.2

Increases : - Duration of the main sequence - Luminosity

Very well constrained by asteroseismology !

Evolution: Main sequence phase

Differential rotation in stars

Conservation of angular momentum:

If  does not depend on 

Global :

Local :

is constant over time

Evolution: Main sequence phase

Differential rotation

Rotational mixing

Evolution: Main sequence phase

Turbulence due to the shear

Meridional circulation due to Thermal imbalance (Von Zeipel paradox) Exchange of angular momentum due to

winds, tidal interactions, accretion, … Transport of angular momentum

Transport of chemicals Diffusive process

Gratton-öpik circulation cell

Rotational mixing

Turbulence (shear) + meridional circulation

Partial mixing

~ Diffusive process

Convective overshooting versus rotational mixing Convective overshooting

Inertia of

convective bubbles

Growth of the mixed adiabatic zone

Depends on rotation rate Independent of rotation

Evolution: Main sequence phase

Les étoiles peuvent tourner vite ) force centrifuge:

La Rotation

rP/ = -r + ² r sin e_s ) Déformation de l’étoile ) Toutes les grandeurs (T, P, , F, …) dépendent

de 2 dimensions spatiales (r, ) ) bcp. plus lourd à résoudre Vitesse critique : si V_eq =  r > V_crit ¼ (GM/R)^1/2 ,

la force centrifuge dépasse la gravité ) enveloppe éjectée Pour certaines étoiles (étoiles Be, …), on n’en est pas loin !

Influence de la rotation

Affecte la structure et l’évolution de l’étoile:

- Structure 2D

- Processus de transports

Affecte les observables spectro- et photométriques (T_eff , …),

car à la photosphère: F(), g_e()

La baroclinité:

r£ (rP/) = -(1/²) r£r P = (1/2) r(²) £ r(r² sin²) Rotationel de l’équation d’équilibre ^{rP/ = -r + 2 r sin e}s

r sin e_s = (1/2) r(r² sin²)

Pour une rotation solide ou cylindrique ((r sin)): r£rP = 0 Isobares = Iso-densité = Isothermes : P(), (), T(), …

 = Potentiel total = (Rot. solide)

₁

2

3

Sinon (rotation différentielle plus complexe):

Isobares  Iso-densité  Isothermes

Baroclinité

La Rotation

Le paradoxe de Von Zeipel:

Considérons la zone radiative d’une étoile en rotation solide P(), (), T(), …

1

2

3

Non-constant sur les équipotentielles

Quel que soit T(), il est impossible d’avoir r¢ (F_rad) = 0 partout ! Déséquilibre thermique :

Cas stationaire : V ¢ rS  0

Circulation méridienne

La Rotation

Circulation méridienne

Mécanisme de transport à grande échelle:

1) Transport du moment cinétique

2) Transport des

éléments chimiques

Temps caractéristique associé à son établissement:

temps d’Eddington-Sweet:

Ensuite, la circulation s’adapte pour assurer le transport du moment angulaire requis par les conditions aux limites (vents, accrétion,

effet de marée dans les binaires proches, …)

r¢F/(T) = - V ¢ rS < 0 r¢F/(T) =

- V ¢ rS > 0

La Rotation

La rotation différentielle dans les étoiles

- Rompt la symétrie sphérique

- Varie en fonction de la profondeur - Crée un déséquilibre thermique

Si conservation du moment angulaire :

Si  ne dépend pas de 

Globale : Locale :

Circulation méridienne

est constant avec le temps

Evolution: Phase de séquence principale

Evolution: Phase de séquence principale

La rotation interne dans les étoiles

Conservation locale du moment angulaire

Mais en réalité, il existe de nombreux mécanismes de transports et pertes de moment angulaire (champs magnétique, circulation méridienne, turbulence, ondes

internes …)

L’astérosismologie nous permettra d’en savoir plus !

The heaviest elements go down The lightest elements go up

Microscopic diffusion

Evolution: Main sequence phase

Gravitational settling

Very quick near the surface, very slow deep in the star

High cross-section pushed up Low cross section go down

Radiative forces

Example: Iron, Nickel, … T = E/k ~ 2£10⁵ K :

e^- transitions from layer M

Opacity peak Accumulation of Iron

Diffusion

No diffusion Helium

Hydrogen

Processus non-standards: Diffusion microscopique

1) Les forces entrant en jeu

Pression partielle sur les particules i :

1.1 Les collisions

Force moyenne exercée sur la particule i par collisions :

En l’absence de r P , évolution vers l’homogénisation

1.2 La gravité

Force exercée sur la particule i par l’attraction gravifique :

Gravité + collisions

Les particules les plus lourdes sombrent, les plus légères flottent.

Processus non-standards: Diffusion microscopique

1) Les forces entrant en jeu

1. 3 Forces radiatives

Impulsion fournie / u. de vol. , de fréq. et de temps par les photons sur les ions i :

« Accélération » (force / u. de masse) de la particule i :

Processus non-standards: Diffusion microscopique

1) Les forces entrant en jeu

Contribution de toutes les forces sur la particule i :

2) Vitesses de diffusion

3) Evolution des abondances

Conservation des particules i :

Libre parcours moyen:  = ( n)^-1 Vitesse thermique

Si élément minoritaire (c_i << 1)

Processus non-standards: Diffusion microscopique

4) Diffusion et gradients de température

Le libre parcours moyen dépend de la température car  / T^-2 (ions), v_T / T^1/2 augmente avec T déplacement des régions chaudes vers les régions froides

Très approximatif, le signe peut être opposé dans certains cas !

The Sun

Central temperature

~ 15 10⁶ K

Proton-proton chain

15 millions °

5800°

T°K

pp_I dominates

pp_III negligible ^T6

 c²/c²

Very well constrained by helioseismology

Sound speed : c

= P 

/  ~ T/

Rotation

Convective envelope :

Differential rotation in latitude Radiative core : rigid rotation !!

Because of angular momentum

transport by gravitational waves ?

The sun

Different neutrino energies

PPII



0 7

3 0

1 7

4

Be +

e → Li + PPIII

B →

Be +

e + 

PPI ^H + ^H → ^H +

^e + 

1 2

1 1

1 1

1 E_ = 0.263 Mev

E_ = 0.8 Mev E_ = 7.2 Mev

The solar neutrino problem

T₆

The flux of neutrinos at different energies is an excellent indicator of the temperature !

The sun

Spectre des neutrinos solaires

Exercice : Calculer la quantité d’hydrogène consommé en une seconde

L

= 3.9 10

erg/s =  M c

 M = 3.9 10

/(3 10

)

= 4.33 10

g

Chaque seconde, 4.33 10

tonnes de matière disparaît

 m = m(4p) – m(He) = (4.031280-4.002603) uma

= 0.028677 uma = 7.1136‰ de m(4p)

La masse de 4.33 10

tonnes représente donc 7.1136‰

de la masse d’hydrogène transformée

Masse totale d’hydrogène transformée par seconde

609 millions de tonnes

Exercice : Calculer le nombre de neutrinos reçus sur Terre (par cm

et par seconde)

Chaque seconde, 609 millions de tonnes d’hydrogène sont transformées en hélium

Nombre de protons = 6.09 10

/1.66 10

N

= 3.67 10

Chaque fois que 4 protons sont transformés, 2 neutrinos sont émis

Nombre de neutrinos = N

/2 = 1.83 10

/s

Surface de la sphère à la distance d de la Terre = 4  d

(d = 150 10

km) = 2.83 10

cm

Nombre de neutrinos par seconde et par cm

65 milliards

360 mille milliards par seconde

Homestake

(R. Davis, 1968-…) 380 000 l C

Cl

37

Cl + 



Ar + e



= 35 jours E

> 0.814 Mev 1 

/ 2.5 jours !

37

Ar + e 

Cl

+ 

37

Cl



Cl + photon

2 km sous terre

Détection sur terre des neutrinos

Homestake

(R. Davis, 1968-…)

37

Cl + 



Ar + e



= 35 jours E

> 0.814 Mev 1 

/ 2.5 jours !

37

Ar + e 

Cl

+ 

37

Cl



Cl + photon

1 SNU = 10

captures / sec. / atome de Cl

Modèles théoriques prédisent:

8-10 SNU

Résultats de l’expérience:

4 SNU

Détection sur terre des neutrinos

Gallex ^(1992-…)

E

> 0.233 Mev

Tous les neutrinos pp,

Be,

B

71

Ga + 



Ge + e



= 11.5 jours

71

Ge + e 

Ga

+ 

71

Ga



Ga + photon

Détection sur terre des neutrinos

SuperKamiokande (M. Koshiba 1996-…) 50 000 m

eau très pure

Diamètre 40 m – Hauteur 40 m 11 000 photodétecteurs (50 cm)

Effet Cerenkov Directionnel

Temps réel E

> 6 Mev

25 

/ jour !

1000 tonnes D

O

12 m diameter Acrylic Vessel

18 m diameter support structure; 9500 PMTs (~60% photocathode coverage) 1700 tonnes inner shielding H₂O

5300 tonnes outer shielding H₂O

Sudbury Neutrino Observatory ^(1998-…)

Tous types de neutrinos



, 

, 

Photo, Courtesy of Sudbury Neutrino Observatory

1 Kton of D²O 9500 PM tubes

The flux of detected electronic neutrinos is 2-3 times smaller than what it should be in a standard solar model !!

Neutrinos are oscillating ! If we consider all kinds of neutrinos, the total detected flux is in agreement with the solar model

Evolution : phase de séquence principale

Evolution de la composition chimique

Faibles masse : Chaîne P-P

La majorité des éléments « minoritaires » : D, L

, B

, B est brulée au cœur lors de la phase pré-séquence principale

T_Li ~ 2 10⁶ K , T_Be ~ 3 10⁶ K, T_B ~ 5 10⁶ K

Valeurs d’équilibre

(suffisemment profond)



0 0 7

3 0

1 7

4

Be +

e → Li +



+

→

+ H Be

Li

4 1

1 7

3

 +

→ + H B Be

7 4

1. Chaîne proton-proton (T > 10

K)

Temps de réaction très court :

_pd ¼ 1 sec.

+ 

→

+ H He H

1 1

Combustion du deuterium

Sur terre : D/H = 1.5 £ 10^-4 Première phase de combustion du

Deuterium initial, puis équilibre.

Evolution de la composition chimique

Phénomène de spaliation

1. Chaîne proton-proton (T > 10

K)

Combustion He₃ + He₃

Deuterium à l’équilibre

H He

He

He ₂³ ₂⁴ ₁¹

3

2 + → +2

Au début, réaction très lente !!

Décroit quand T

Pour T trop petit, He³ n’a jamais le temps d’atteindre sa valeur d’équilibre,

donc bosse

Evolution de la composition chimique

1. Chaîne proton-proton

Combustion He₃ + He₃ H He

He

He ₂³ ₂⁴ ₁¹

3

2 + → +2

Evolution de la composition chimique

Décroit quand T

_eq (années) Soleil

Problème du Lithium

Mécanismes hydrodynamiques non-standards de transport des régions profondes vers la surface : diffusion, circulation méridienne, ondes internes, …

L’abondance de Lithium à la surface du Soleil est 140 * plus

petite que sa valeur initiale PPII

^{Li +}

^{H →}

^{Be + }

L’étude d’amas jeunes (Hyades, …) montre que l’abondance en surface

diminue au cours du temps T_Li ~ 2 10⁶ K

1. Chaîne proton-proton

Evolution de la composition chimique

2. Cycle CNO T > 15 10

K

1. C₁₂ N₁₄

O¹⁶ N¹⁴ (beaucoup plus lent)

2. Evolution complexe car :

- Présence d’un coeur convectif qui homogénise et dont la taille change - Certains constituants n’atteignent

jamais la valeur d’équilibre

Evolution de la composition chimique

Au départ :

C¹² : N¹⁴ : O¹⁶ = 5.5 : 1 : 9.5 A l’équilibre (typiquement) : C¹² : N¹⁴ : O¹⁶ = 0.15 : 15 : 0.3

12

C

14

N

Post main sequence evolution

Step 1:The whole star contracts due to the thermal imbalance (Virial theorem)

After all hydrogen has been consumed in the core:

Step 2: Hydrogen burning starts in a shell above the helium core

H He He

I fix the mass M_c of the isothermal core and let the radius R_c vary.

For each R_c , I have one solution and thus one value of the pressure P₀ at the top of the core.

The curve P₀ (R_c) behaves as follows, with a maximum :

Post main sequence evolution:

Evolution of the helium core

Assumption (maybe wrong!): the helium core is in thermal equilibrium No energy production below the H-burning shell.

dL/dm =  = 0 et L(0) = 0 L = 0 dT/dr = 0

The helium core is isothermal Question : is the helium core able

to support the envelope above it ?

The helium core: Schönberg – Chandrasekhar limit of the relative core mass q=M

/M

Post main sequence evolution

Virial theorem for the helium core (P > 0 at the top !):

Isothermal helium core

M_c : Helium core mass M : Total mass

Envelope

The isothermal core is not always able to support the envelope!!

Support condition:

Schönberg – Chandrasekhar limit of the relative core mass

The helium core: Schönberg – Chandrasekhar limit of the relative core mass q=M

/M

Post main sequence evolution

M_c : Helium core mass M : Total mass

Isothermal helium core

If P_{0, max} > Weight of the envelope (P_e), a solution R_c exists such that P₀(R_c) = P_e This is so when M_c / M_tot < q_sc ~ 0.1

q_sc ~ 0.1 is thus a mass ratio limit beyond which an isothermal core cannot support the envelope above it. What happens in this case ?

The core cannot be in thermal equilibrium : dT/dr < 0 fast contraction of the core

Evolution of the helium core :

Schönberg – Chandrasekhar limit of the relative core mass

Support condition: The curve P₀ (R_c) has a maximum :

If P_{0, max} < Weight of the envelope (P_e), there is no solution whatever R_c This is so when M_c / M_tot > q_sc ~ 0.1