1. Lecture 3 Radiative and Convective Energy Transport I.Mechanisms of Energy Transport II. Radiative Transfer Equation III. Grey Atmospheres IV. Convective Energy Transport V. Mixing Length Theory

2. The Sun

3. I. Energy Transport In the absence of sinks and sources of energy in the stellar atmosphere, all the energy produced in the stellar interior is transported through the atmosphere into outer space. At any radius, r, in the atmosphere: 4  r 2 F(r) = constant = L Such an energy transport is sustained by the temperature gradient. The steepness of this gradient is in turn dependent on the effectiveness of the energy transport through the different layers

4. I. Mechanisms of Energy Transport 1. Radiation F rad (most important) 2. Convection: F conv (important in cool stars like the Sun) 3. Heat production: e.g. in the transition between the solar chromosphere and corona 4. Radial flow of matter: coronae and stellar winds 5. Sound waves: chromosphere and coronae Since we are dealing with the photosphere we are mostly concerned with 1) and 2)

5. I. Interaction between photons and matter Absorption of radiation: I I    dI s dI  = –   I dx   : mass absorption coefficient [  ] = cm 2 gm –1   = ∫ 0 s   dx Optical depth (dimensionless) Convention:  = 0 at the outer edge of the atmosphere, increasing inwards d   =   ds

6. Optical Depth I  I (s) dI  = –I d  → I (s) = I  e –  The intensity decreases exponentially with path length Optically thick:  > 1 Optically thin:  < 1 If  = 1 →  = I e ≈ 0.37 I 0 We can see through the atmosphere until  ~1

8. Optical depth, , of ring material small Optical path length small and you see central star and little nebular material Optical path length larger and you see nebular material as ring Optical depth, , of ring material larger Optical path length roughly the same one sees disk and no central star

9. Optical Depth The quantity  = 1 has a geometrical interpretation in terms of the mean free path of a photon:  = 1 = ∫  ds =  s s ≈ (  ) –1 s is the distance a photon will travel before it gets absorbed. In the stellar atmosphere the abosrbed will get re-emitted and thus will undergo a „random walk“. For a random walk the distance traveled in 1-D is s√N where N is the number of encounters. In 3-D the number of steps to go a distance R is 3R 2 /s 2. s At half the solar radius  ≈ 2.5 thus s ≈ 0.4cm so it takes ≈ 30.000 years for a photon do diffuse outward from the core of the sun.

10. Emission of Radiation I I  + dI dx j dI  = j  I dx j  is the emission coefficient/unit mass [ ] = erg/(s rad 2 Hz gm) j comes from real emission (photon created) or from scattering of photons into the direction considered.

11. II. The Radiative Transfer Equation Consider radiation traveling in a direction s. The change in the specific intensity, I, over an increment of the path length, ds, is just the sum of the losses (  ) and the gains (j ) of photons: dI  = –   I   j  I ds Dividing by   ds which is just d  = –I   j /  = –I + S dI d 

12. The Radiative Transfer Equation  appears alone in the previous equation therefore try solutions of the form I (  ) = f e b   Differentiating this function: = f  be b    e b  dI d  dfdf = bI   e b  dfdf d  Substituting into the radiative transfer equation: bI   e b  dfdf d  = – I   S

13. The Radiative Transfer Equation The first two terms on each side are equal if we set b = –1 and equating the second term: e –  dfdf d  or =  S f = ∫ S e t dt + c 0 0  t is a dummy variable I (  ) = fe b  =e –  ∫S (t )e t dt + c 0 e –  0  Set  = 0 → c 0 = I (0)

14. bringing e –  inside the integral: I (  ) =∫S (t )e – (  –t  + I  e –  0   =0 I (0) →  t  –t At point tn the original intensity, I (0) suffers an exponential extinction of e –  The intensity generated at t, S (t ) undergoes an extinction of e –(  –t ) before being summed at point  

15. I (  ) =∫S (t )e – (  –t  + I  e –  0  This equation is the basic intergral form of the radiative transfer equation. To perform the integration, S (  ), must be a known function. In some situations this is a complicated function, other times it is simple. In the case of thermodynamic equilibrium (LTE), S (T) = B (T), the Planck function. Knowing T as a function of x or  amounts to a solution of the transfer equation.

16. Radiative Transfer Equation for Spherical Geometry After all, stars are spheres! z x y To observer dzdz r  dr rd  dI   dz = – I + S dI dz = ∂I ∂z dr dz ∂I ∂∂ dd dz +

17. Radiative Transfer Equation for Spherical Geometry Assume I has no  dependence and dr = cos  dz r d  = –sin  ∂ I ∂r∂r cos    dr – ∂ I ∂∂ sin    r = –I + S This form of the equation is used in stellar interiors and the calculation of very thick stellar atmospheres such as supergiants. In many stars (sun) the photosphere is thin thus we can use the plane parallel approximation

18. Plane Parallel Approximation To observer  ds To center of star  does not depend on z so there is no second term d I cos    dr = –I + S Custom to adopt a new depth variable x defined by dx = –dr. Writing d  for   dx: d I cos   d  = I – S The increment of path length along the line of sight is ds = dx sec 

19. The optical depth is measured along x and not along the line of sight which is at some angle . Need to replace  by –  sec  The negative sign arises from choosing dx = –dr c  Therefore the radiative transfer equation becomes: I (  ) =–∫S (t )e – ( t –   sec  sec  d  The integration limit c replaces I (0) integration constant because the boundary conditions are different for radiation going in (  > 90 o ) and coming out (  < 90 o ). In the first case we start at the boundary where  =0 and work inwards. So when I = I in, c=0. In the second case we consider radiation at the depth  and deeper until no more radiation can be seen coming out. When I = I out, c=∞

20. Therefore the full intensity at the position  on the line of sight through the photosphere is: I (  ) = ∫S e – ( t –   sec  sec  dt ∞  = I out (  ) + I in (  ) = – ∫S e – ( t –   sec  sec  dt 0  Note that one must require that S e –   goes to zero as  goes to infinity. Stars obviously can do this!

21. An important case of this equation occurs at the stellar surface: I in (  ) = 0 I out (  ) = ∫S (t )e –t sec  sec  dt  ∞ Assumption: Ignore radiation from the rest of the universe (other stars, galaxies, etc.) This is what you need to compute a spectrum. For the sun which is resolved, intensity measurements can be made as a function of . For stars we must integrate I over the disk since we observe the flux.

22. The Flux Integral ∫ I  cos  d  F=F= ∫ I  cos  sin  d  F  = 2    = ∫ I out  cos  sin  d  2    + ∫ I in  cos  sin  d  2    Assuming no azimuthal (  ) dependence

23. The Flux Integral Using previous definitions of I in and I out ∫ F  = 2    ∫  ∞ S e – ( t –   sec  sin  dt  d  ∫ – 2   ∫  0 S e – ( t –   sec  sin  dt  d  

24. The Flux Integral If S is isotropic – 2  ∫   ∫  e – ( t –   sec  sin  dt  d  S  Let w = sec  and x = t –  F  = 2  ∫   ∫  ∞ e – ( t –   sec  sin  dt  d  S ∫   e – ( t –   sec  sin  dt  d  ∫ ∞  e –xw w2w2 dw =

25. The Flux Integral ∫ ∞  e –xw wnwn dw Exponential Integrals E n (x) = F    = 2  ∫  ∞ S  E 2  t  –  )dt  – 2  ∫  0 S  E 2   – t )dt In the second integral w = –sec  and x =  – t . The limit as  goes from  /2 to  is approached with negative values of cos  so w goes to ∞ not –∞

26. The Flux Integral The theoretical spectrum is F at  = 0: F   = 2  ∫  ∞ S  t   E 2  t )dt F is defined per unit area In deriving this it was assumed that S is isotropic. It most instances this is a reasonable assumption. However, in stars there are Doppler shifts due to photospheric velocities, stellar rotation, etc. Isotropy no longer holds so you need to do an explicit disk integration over the stellar surface, i.e. treat F locally and add up all contributions.

27. The Mean Intensity and K Integrals J    = 1/2 ∫  ∞ S  E 1  t  –  )dt  + 1/2 ∫  0 S  E 1   – t )dt K    = 1/2 ∫  ∞ S  E 3  t  –  )dt  + 1/2 ∫  0 S  E 3   – t )dt

28. The Exponential Integrals ∫ ∞  e –xw wnwn dw Exponential Integrals E n (x) = ∫ ∞  wnwn dw E n (0) = = 1 –n 1 ∫ ∞  e –xw wnwn dw dE n dx = 1d ∫ ∞  e –xw w n –1 dw =– dE n dx = – E n –1

29. The Exponential Integrals Recurrence formula E n +1 ( x ) = e –x –xE n (x)

30. The Exponential Integrals For computer calculations: E 1 ( x ) = e –x –xE n (x) E 1 (x) = – ln x – 0.57721566 + 0.99999193x – 0.24991055 x 2 + 0.05519968 x 3 – 0.00976004 x 4 + 0.00107857 x 5 for x ≤ 1 E 1 (x) = x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 x 4 + b 3 x 3 + b 2 x 2 + b 1 x + b 0 1 xe x a 3 = 8.5733287401 a 2 = 18.0590169730 a 1 = 8.6347608925 a 0 = 0.2677737343 b 3 = 9.5733223454 b 2 = 25.6329561486 b 1 = 21.0996530827 b 0 = 3.9584969228 E n (x) = 1 xe x [ 1 – n x n(n+1) x2x2 + – … [ xe x 1 ≈ Assymptotic Limit: x >1 From Abramowitz and Stegun (1964) Polynomials fit E 1 to an error less than 2 x 10 –7

31. Radiative Equilibrium Radiative equilibrium is an expression of conservation of energy In computing theoretical models it must be enforced Conservation of energy applies to the flow of energy through the atmosphere. If there are no sources or sinks of energy in the atmosphere the energy generated in the core flows to the outer boundary No sources or sinks in the atmosphere implies that the divergence of the flux is zero everywhere in the photosphere. In plane parallel geometry: d dx F(x)= 0or F(x) = F 0 A constant F(x) = ∫F 0 (  )d  =F 0 for flux carried by radiation 0 ∞

32. d ∫  ∞ S  E 2  t  –  )dt –– ∫  0 S  E 2   – t )dt Radiative Equilibrium ∫  ∞ [ [ == F0F0 22 This is Milne‘s second equation It says that in the case of radiative equilibrium the solution of the radiative transfer equation is found when S is known that satisfies this equation.

33. Radiative Equilibrium Other two radiative equilibrium conditions come from the transfer equation : d I cos  dx =   I –   S Integrate over solid angle d dx ∫ I  cos  d  =   I  d  –   S  d  ∫∫ Substitue the definitions of flux and mean intensity in the first and second integrals dF dx = 4   J – 4   S

34. Radiative Equilibrium Integrating over frequency But in radiative equilibrium the left side is zero! d dx ∫  ∞ F  d = 4  J  d – 4  S  d ∫  ∞ ∫  ∞ ∫  ∞  J  d = ∫  ∞  S  d Note: the value of the flux constant does not appear

35. Radiative Equilibrium Using expression for J ∫  ∞  ∫  ∞ J    = ½ ∫  ∞ S  E 1  t  –  )dt + ½+ ½ ∫  0 S  E 1   – t )dt [ ½= 0 S  E 1  t  –  )dt ∫  0 S  E 1   – t )dt + ½ [ d First Milne equation

36. If you multiply radiative equation by cos  you get the K-integral d I cos 2  dx   I cos  –   S  cos  d d K = d  F0F0 ∫  ∞ 1st moment 2nd moment ∫ ∫ = 44 dd dd Radiative Equilibrium ∫  ∞ ∫  ∞ [ ½ S  E 3  t  –  )dt ∫  0 S E 3   – t )dt + ½ [ d d  d = F0F0 44 Third Milne Equation And integrate over frequency

37. Radiative Equilibrium The Milne equations are not independent. S that is a solution for one is a solution for all three The flux constant F 0 is often expressed in terms of an effective temperature, F 0 =  T 4. The effective temperature is a fundamental parameter characterizing the model. In the theory of stellar atmospheres much of the technical effort goes into iterative schemes using Milne‘s equations of radiative equilibrium to find the source function, S (  )

38. III. The Grey Atmosphere The simplest solution to the radiative transfer equation is to assume that  is independent of frequency, hence the name „grey“. It occupies an „historic“ place and is the starting point in many iterative calculations. Electron scattering is the only opacity source relevant to stellar atmospheres that is independent of frequency. Integrate the basic transfer equation over frequency and denote: I = ∫I d 0 ∞ J = ∫J d 0 ∞ S = ∫S d 0 ∞ K = ∫K d 0 ∞ F = ∫F d 0 ∞ dIdI cos  dd = –I  + S Where d  =  dx, the „grey“ absorption coefficient

39. The Grey Atmosphere This grey case simplifies the radiative equilibrium and Milne‘s equations: F(x) = F 0 J = S dKdK dd = F0F0 44 The Eddington approximation, hemispherically isotropic outward and inward specific intensity: I(  ) = I out (  ) for 0 ≤  ≤  /2 I in (  ) for  ≤  ≤ 

40. The Grey Atmosphere J =J = 1 44 ∫ IdId = 1 2 ∫ I out 0  /2 sin  d  + 1 2 ∫ I in  /2  sin  d  = 1 2 I out (  ) + I in (  ) F(  ) =  I out (  ) – I in (  ) K(  ) = 1 6 I out (  ) + I in (  ) S(  ) = J(  ) = 3K(  ) But since the mean intensity equals the source function in this case

41. The Grey Atmosphere We can now integrate the equation for K: K(  ) =  F0F0 44  F0F0 66 Where the constant is evaluated at  = 0. Since S = 3K we get Eddington‘s solution for the grey case: S(  ) = 3F03F0 44 (  + ⅔) The source function varies linearly with optical depth. One gets a similar result from more rigorous solutions dKdK dd = F0F0 44

42. III. The Grey Atmosphere Using the frequency integrated form of Planck‘s law: S(  ) = (  )T 4 (  ) and F 0 =  T 4 the previous equation becomes T(  ) = (  + ⅔) ¾ ¼ At  = ⅔ the temperature is equal to the effective temperature and T (  ) scales in proportion to the effective temperature. T eff Note that F(0) =  S(⅔), i.e. the surface flux is  times the source function at an optical depth of ⅔.

43. III. The Grey Atmosphere Chandrasekhar (1957) gave a complete and rigorous solution of the grey case which is slightly different: S(  ) = 3F03F0 44 [  + q(  )] T(  ) = [  + q(  )] ¾ ¼ T eff q (  ) is a slowly varying function ranging from 0.577 at  = 0 and 0.710 at  = ∞ Eddington: „This, however is a lazy way of handling the problem and it is not surprising that the result fails to accord with observation. The proper course is to find the spectral distribution of the emergent radiation by treating each wavelength separately using its own proper values of j and .“

44. IV. Convection In a star the heat flux must be sufficiently great to transport all the energy that is liberated. This requires a temperature gradient. The higher the energy flux, the larger the temperature gradient. But the temperature gradient cannot increase without limits. At some point instability sets in and you get convetion In hot stars (O,B,A) radiative transport is more efficient in the atmosphere, but core is convective. In cool stars (F and later) convective transport dominates in atmosphere. Stars have an outer convection zone.

46. IV. Convection  T  rr 11 ** P*P* PP ** P*P* 22 PP Consider a parcel of gas that is perturbed upwards. Before the perturbation  1 * =  1 and P 1 * = P 1 For adiabatic expansion: P =    = C P /C V = 5/3 C p, C v = specific heats at constant V, P After the perturbation: P 2 = P 2 *  2 * =  1 ( P2*P2* P1*P1* ) 1/ 

47. IV. Convection Stability Criterion: Stable:  2 * >  2 The parcel is denser than its surroundings and gravity will move it back down. Unstable:  2 * <  2 the parcel is less dense than its surroundings and the buoyancy force will cause it to rise higher.

48. IV. Convection Stability Criterion: 2*2* 11 = ( P2P2 P1P1 ) 1/  = { P 1 + dP dr rr } P1P1 1/  = {} 1 + rr 1 P dP dr  2 * =  1 +  ( 11 P ) 1 dP dr rr  2 =  1 + dd dr rr  ( 11 P ) 1 dP dr dd > This is the Schwarzschild criterion for stability

49. IV. Convection By differentiating the logarithm of P = k  TT we can relate the density gradient to the pressure and temperature gradient:  ( T P ) 1dT dr ( 1 – ) () dP dr star > The left hand side is the absolute amount of the temperature gradient of the star. Both gradients are negative, so this is the algebraic condition for stability. The right hand side is the „adiabatic temperature gradient“. If the actual temperature gradient exceeds the adiabatic temperature gradient the layer is unstable and convection sets in: () dT dr () dT dr star < ad The difference in these two gradients is often referred to as the „Super Adiabatic Temperature Gradient“

50. Brunt-Väisälä Frequency The frequency at which a bubble of gas may oscillate vertically with gravity the restoring force: The Convection Criterion is related to gravity mode oscillations:  is the ratio of specific heats = C v /C p g is the gravity  ( ) 1 dlnPdlnP dr dlndln –N 2 = g Where does this come from?

51.  T  rr 11 ** P*P* PP ** P*P* 22 PP  2 * =  1 +  ( 11 P ) 1 dP dr rr  2 =  1 + dd dr rr Difference in density between inside the parcel and outside the parcel:  =  (  P ) 1 dP dr rr – dd rr =  (  ) 1 dlnPdlnP rr dlndln rr  – =  ()  dlnPdlnP rr rr  – 1 dlndln The Brunt-Väisälä Frequency

52.  =  A  r =  () dlnPdlnP dr – 1 dlndln A Buoyancy force: F B = –  Vg V = volume F B = –g  AV  r In our case x =  r, k = g  AV, m ~  V () dlnPdlnP dlndln N 2 = g  AV/V  = gA = g  dr – 1 The Brunt-Väisälä Frequency is the just the harmonic oscillator frequency of a parcel of gas due to buoyancy F B = –kx  2 = k/m For a harmonic oscillator:

53. V. Mixing Length Theory Convection is a difficult problem for which we still have no good theory. So how do most atmospheric models handle this complicated problem? By reducing it to a single free parameter whose value is left to guess work. This is mixing length theory. Physically it is a piece of junk, but at this point there is no alternative although progress has been made in hydrodynamic modeling. Simple approach: 1.Suppose the atmosphere becomes unstable at r = r 0 → mass element rises for a characteristic distance L (mixing length) to r + L 2. Cell releases excess energy to the ambient medium 3.The cell cools, sinks back, absorbs energy and rises again In this process the temperature gradient becomes shallower than in the purely radiative case.

54. V. Mixing Length Theory Recall the pressure scale height: H = kT/  g The mixing length parameter  is given by  = L H  = 0.5 – 1.5

55. V. Mixing Length Theory ½  v 2 = g  L = g  H The convective flux is given by:  =  C p v  T v is the convective velocity   =   → v ≈ g½(  H) ½ (  ½  =  C p  T 3/2 gHgH  ½  =  C p g  ½  H)  – dd dx cell dd dx average 3/2 Or in terms of the temperature gradients

56. Scale height H = kT/  g The Sun: H~ 200 km A red giant: H ~ 10 8 km How large are the convection cells? Use the scale height to estimate this:

57. Schwarzschild (1976) Schwarzschild argued that the size of the convection cell was roughly the width of the superadiabatic temperature gradient like seen in the sun.

58. In red giants this has a width of 10 7 km

59. Convective Cells on the Sun and Betelgeuse Betelgeuse =  Ori From computer hydrodynamic simulations The Sun 1000 km ~10 7 km

61. Numerical Simulations of Convection in Stars: http://www.astro.uu.se/~bf/movie/movie.html

65. http://www.aip.de/groups/sternphysik/stp/box_simulation.html With no rotation:With rotation:

66. cells small large total number on surface motions and temperature differences relatively small Time scales short (minutes) Relatively small effect on integrated light and velocity measurements cells large small total number on surface motions and temperature differences relatively large Time scales long (years) large effcect on integrated light and velocity measurements

67. RV Measurements from McDonald AVVSO Light Curve