Stars, Planets and Fluid Dynamics

Transcription

1 Laboratoire d Astrophysique de Toulouse-Tarbes, Observatoire Midi-Pyrénées, France 25 mai 2009

2 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

3 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

4 Stars... Introduction FIG.: La lumière des étoiles.

5 Why stellar physics? Stars are an essential component of the Universe They are the factories of metals They are a unique physics laboratory...

6 Outline Introduction Standard knowledge First difficulties 1 Introduction 2 Standard knowledge First difficulties

7 Outline Introduction Standard knowledge First difficulties 1 Introduction 2 Standard knowledge First difficulties

8 Modèles d étoiles Introduction Standard knowledge First difficulties Les modèles d étoiles existent depuis cinquante ans. On résout les équations suivantes : Φ = 4πGρ, potentiel gravitationnel P ρ Φ = 0 la statique (χ T) + ε = 0 Energie P P(ρ, T) χ χ(ρ, T) ε ε(ρ, T) Etat Opacité Energie nucléaire Remarque : il se peut que la situation soit instable convectivement, i.e. T < ( T) adiab. Dans ce cas les astrophysiciens font appel à leur théorie de la longueur de mélange.

9 Standard knowledge First difficulties La théorie de la longueur de mélange Flux total = Flux radiatif + Flux convectif F z = χ rad z T + C p δtv z 1 Bernoulli : 1 2 ρv2 δρgλ, Λ est la longueur de mélange. 2 δρ = aδt = aλ( z T z T ad ) 3 Flux = L 4πr 2 = χ rad z T + C p 2gaΛ 2 ( z T z T ad ) 3/2 We usually set Λ = αh p, α = O(1) to be determined. In the Sun α 1.7.

10 En Résumé Introduction Standard knowledge First difficulties L étoile est supposée sphérique et on résout des ED non-linéaires : avec la microphysique 1 d 2 rφ r dr 2 = 4πGρ dp dr = ρ dφ dr ( 1 d r 2 r 2 χ dt ) + ε = 0 radiatif dr dr dt dr = MLT convectif P P(ρ, T), χ χ(ρ, T), ε ε(ρ, T)

11 En image : the Sun s interior Standard knowledge First difficulties

12 Evolution with time Introduction Standard knowledge First difficulties L évolution est contrôlée par les réactions nucléaires : X t = f (X, Y, ρ, T...) Dans ces modèles la microphysique est très soignée et la structure du soleil peut être reproduite à moins de 1% près. Par exemple sur le profil de vitesse du son.

13 Standard knowledge First difficulties Diagramme de Hertzsprung-Russell

14 avec diamètre des étoiles Standard knowledge First difficulties

15 Standard knowledge First difficulties

16 Standard knowledge First difficulties

17 Outline Introduction Standard knowledge First difficulties 1 Introduction 2 Standard knowledge First difficulties

18 Convection Introduction Standard knowledge First difficulties MLT is very crude and two major inconveniencies for stellar evolution : 1 It is not predictive : what is α for other stars? 2 It cannot take into account penetrative convection

19 Numbers Introduction Standard knowledge First difficulties The Reynolds number for the Sun s convection. Typical kinematic viscosity in the Sun s conditions is 10 3 m 2 /s. Within a granule L=1000 km, V=1 km/s, Re= In the middle of the convection zone : L=10 5 km, V=50 m/s, Re=

20 Numbers Prandtl Introduction Standard knowledge First difficulties Heat diffusion is essentially due to photons : χ = 16σT 3 /3α Ross ; the thermal Prandtl number is small. The magnetic Prandtl is small too.

21 Standard knowledge First difficulties

22 Numbers Introduction Standard knowledge First difficulties So we are far from the situation reachable by DNS. Turbulence modelling is lacking... Let see a film... Observations of Sun s surface Simulation of Sun s surface

23 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

24 A new difficulty : Rotation Stars are all rotating. At some time, in their life, they were rapidly rotating, usually in their youth. First stars of the Universe were probably massive and fast rotators.

25 The VLTI at the European Southern Observatory in Chile

26 Fig. 2. FIG.: Achernar seen with VLTI (Domiciano de Souza et al. AA, 2003) V

27 FIG.: Altair seen by NPOI (Peterson et al. ApJ 2006), Ω 0.9Ω B

28 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

29 Hot poles and cool equator Gravity darkening ρ ρ(φ), T T(Φ), Φ = φ g + φ c F = χ T = χ(φ)t (Φ) Φ = Cte g eff This is von Zeipel law : Diffusive flux is approximately proportional to local effectif gravity, = T eff g 1/4 eff.

30 A schematic view... Ω FIG.:... of a massive star in rapid rotation.

31 To summarize Introduction Rotating stars are not spherically symmetric : they need 2D-models They contain a vectorial quantity : angular momentum Their stably stratified radiative zones are not in hydrostatic equilibrium Poles are hotter than the equator They lose angular momentum with mass...

32 Evolutionary questions Rotation implies Mixing of stably stratified zones (baroclinic flows) evolution of angular momentum (spin down flows) less dense core slower evolution Contamination of surface layers with nuclear synthesis product Enrichement in metals of the ISM Crucial to first stars and supernovae explosions, GRB,...

33 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

34 Steady Homogeneous Fully Radiative Baroclinic Star Star in a box Take a self-gravitating fluid inside a spherical container of radius R, rotating with mean angular velocity Ω = Ωe z. Use of spherical coordinates Outer boundary r = R is not an isobar

35 Equations Introduction Governing equations : φ = 4πGρ ρtv s = F + ε ρ (2Ω v + v v) = p ρ ( φ 1 2 Ω2 r 2 sin 2 θ ) + F v (ρv) = 0 where Viscous force : F v = µ ( v + 13 ) ( v) Energy flux (Radiative) : F = χ T Equation of state : p = R M ρt + a 3 T4

36 Stellar microphysics Nuclear reactions (pp-chain) ε = ε 0 X 2 ρ 2 T 2/3 e bt 1/3 with ε 0 = (cgs) and b = 3600 (values adopted for the CESAM code, Morel, 1997). Opacity (Kramer s type law) then χ = 16σT3 3κρ κ = κ 0 T β ρ η = χ 0 T β+3 ρ η 1 with κ 0 = (cgs), β = 1.98 and η = 0.14 (Christensen-Dalsgaard & Reiter, 1995).

37 Boundary conditions At the surface Matching of the gravitational potential with the vacuum solution. Pressure at the pole. Stress-free boundary conditions for the velocity field. u r = σ rθ = σ rϕ = 0 Temperature T r + σ TT = 0 equivalent to black body radiation with σ T = 3ρκ/16. We use a constant value of σ T for the entire surface.

38 Dimensionless equations φ = ρ P ρtu s = χ T + Λε E ρ (2Ω u + u u) = p ρ φ eff + EF u (ρu) ( = 0 p = π c ρt + 1 β c β c T 4)

39 Dimensionless equations φ = ρ P ρtu s = χ T + Λε E ρ (2Ω u + u u) = p ρ φ eff + EF u (ρu) ( = 0 p = π c ρt + 1 β c β c T 4)

40 Dimensionless numbers E = µλ ρ c R 2 P = µr M χ c π c = R MT c λ 2 R 2 Λ = ε cr 2 χ c T c β c = π c π c + at4 c λ 2 3ρ c R 2 where λ = (4πGρ c ) 1/2 is the free-fall time scale.

41 Dimensionless numbers E = µλ ρ c R 2 P = µr M χ c π c = R MT c λ 2 R 2 Λ = ε cr 2 χ c T c β c = π c π c + at4 c λ 2 3ρ c R 2 where λ = (4πGρ c ) 1/2 is the free-fall time scale.

42 Spectral methods (1) Efficient discretization : require much less points than finite differences methods for similar accuracy. Special care of discontinuities. Multi-domain approach. Models produced are well suited for the calculation of oscillations.

43 Spectral methods (2) Gauss-Lobatto collocation points for the radial part. Associated with Chebyshev polynomials. Pseudo-spectral method. Spectral precision in the real space. Spherical harmonics for the angular part. Decomposition of scalar quantities. f (r, θ) = f l (r)y 0 l (θ) Vector fields. u(r, θ) = u l (r)r 0 l + v l (r)s 0 l + w l (r)t 0 l where R 0 l = Y 0 l e r, l l S 0 l = Y0 l θ e θ and T 0 l = Y0 l θ e ϕ

44 Algorithm Introduction Because of nonlinearities, equations are solved iteratively. We can write the equations in the form where L n is a linear operator. L n (y n+1 ) = RHS(y n ) Both the operator and the right hand side term depend on the previous solution y n. Solve for the next iteration y n+1 by inverting the linear operator.

45 Results Introduction The following figures have been calculated using the values : Polar pressure : p s = 10 5 (scaled by central pressure) Angular velocity : Ω = 0.07 ( 82% of the critical velocity) Central temperature : T c = K (approx. solar value) Ekman number : E = 10 8 σ T = 1.5 (Boundary condition for temperature) β c = 1 (Neglect radiation pressure) Prandtl number : P = 0 (Neglect advection of temperature) The resulting star has the physical parameters : Mass : 1.105M Radius : R Luminosity : 0.829L Central density : 85.6 g/cm 3

46 Spectral convergence max Ck over l φ p T w u v k max Cl over k φ p T w u v l

47 Isobars Introduction

48 Square of the Brunt-Väisälä frequency (N 2 )

49 Differential rotation (1)

50 Differential rotation (2) δω/ω Ω=0.01 Ω=0.02 Ω=0.03 Ω=0.04 Ω=0.05 Ω=0.06 Ω= π/4 π/2 3π/4 π θ

51 Meridional circulation

52 Spheroidal geometry The star is flattened by the centrifugal force Spheroidal coordinates ζ, θ defined as { r = r(ζ, θ) θ = θ The coordinates are chosen so that ζ = 1 defines the real surface of the star. We use the following general expression for r(ζ, θ) r = aζ + b + A(ζ)F(θ) + B(ζ)G(θ)

53 Boundary conditions Boundary condition for gravitational potential cannot be expressed easily at the surface of the star. Use of an external domain with spherical boundary.

54 Numerical method Determining the shape of the surface We don t know a priori the shape of the surface, so it should be calculated. The surface of the star is an isobar (constant pressure). At each iteration, we use the pressure profile to update the shape of the surface. R(θ) = α [ log p(ζ = 1, θ) log p(ζ = 1, 0) ]

55 Numerical method Operators Introduction We decompose the operators in a spherical-like part and a non-spherical residual which will take part in the right hand side of the equation. Example (Poisson s equation) : We decompose φ = ρ φ = g ζζ φ + ns φ where g ζζ is a component of the metric tensor, then we solve φ n+1 = 1 ) g (ρ n ns φ n ) + (1 gζζ φ n g where g = max(g ζζ ).

56 Results Differential rotation Introduction

57 Results Meridional circulation Introduction

58 Rotation and Convection

59 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

60 Introduction Why should we bother about binary stars? 70% of the stars belong to a multiple system... Binary systems offer new windows and new constraints on stellar parameters (masses) and stellar evolution (formation)

61 Novelties with binary stars They are born together = same age We may know the masses We can measure the periastron precession in some cases

62 How to measure the masses law : G(M 1 + M 2 ) = a 3 Ω 2. and Kepler s

63 Eclipsing binaries Introduction These are the stars where we get the most precise parameters.

64 Difficulties Introduction exchange angular momentum during evolution = some additional mixing Broken axisymmetry Mass exchange...

65 An illustration Introduction

66 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

67 Some additional difficulties The magnetic field is certainly everywhere present. On the Sun : it is the clear manifestation of the solar dynamo

68 Les taches solaires

69 L activité magnétique du soleil FIG.: le cycle de 11 ans du soleil

70 Other stars Introduction

71 and a new puzzle Introduction

72 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

73 Stellar formation Introduction Another puzzle attached to stars is their formation and their eventual planets. Standard scenario : The gravitational collapse of a gas cloud with some angular momentum generates many proto-stars with a disc... Each disc may transform into a set of planets... Planets are not all the same : terrestrial type and giant type ; separated by the snow line But exoplanets show many many hot-jupiters = planetary migration What is the structure of a giant planet? Answer needed to constrain planet formation

74 Some illustrations Introduction

75

76 Simulation du core-collapse Supernova Supernova 2D (Simulation du groupe d E. Müller)

77 Outline Introduction 1 Introduction 2 Standard knowledge First difficulties

78 Introduction Fluid dynamics is at the heart of nowadays stellar physics

79 THE END