Model Photospheres I.What is a photosphere? II. Hydrostatic Equilibruium III.Temperature Distribution in the Photosphere IV. The P g -P e -T relationship V. Properties of the Models VI. Models for cool stars

I. What is a Stellar Atmosphere? Transition between the „inside“ and „outside“ of the star Boundary between the stellar interior and the interstellar medium All energy generated in the core has to pass through the atmosphere Atmosphere does not produce any energy Two basic parameters: Effective temperature. Not a real temperature but the temperature needed to produce the observed flux via 4  R 2 T 4 Surface Gravity – log g (although g is not a dimensionless number). Log g in stars range from 8 for a white dwarf to 0.1 for a supergiant. The sun has log g = 4.44

What is a Photosphere? It is the surface you „see“ when you look at a star It is where most of the spectral lines are formed What is a Model Photosphere? It is a table of numbers giving the source function and the pressure as a function of optical depth. One might also list the density, electron pressure, magnetic field, velocity field etc. The model photosphere or stellar atmosphere is what is used by spectral synthesis codes to generate a synthetic spectrum of a star

Real Stars: 1. Spherical 2. Can pulsate 3. Granulation, starspots, velocity fields 4. Magnetic fields 5. Winds and mass loss

Our model: 1. Plane parallel geometry 2. Hydrostatic equilibrium and no mass loss 3. Granulation, spots, and velocity fields are represented by mean values 4. No magnetic fields

P + dP P dA r + dr M(r) The gravity in a thin shell should be balanced by the outward gas pressure in the cell r dm A P +dP dr P Gravity II. Hydrostatic equilibrium

F p = PdA –(P + dP)dA = –dP dA Pressure Force F G = – GM(r) dM r2r2 Gravitational Force r 0 M(r) = ∫  (r) 4  r 2 dr dM =  dA dr Both forces must balance: F P + F G = 0 – dP dA –G–G +  (r)M(r) r2r2 drdrdAdA = 0 dm A P +dP dr P Gravity

The pressure in this equation is the total pressure supporting the small volume element. In most stars the gas pressure accounts for most of this. There are cases where other sources of pressure can be significant when compared to P g. Other sources of pressure: 1.Radiation pressure: P R = 44 3c T4T4 = 2.52 ×10 –15 T4T4 dyne/cm 2 2. Magnetic pressure: P M = B2B2 88 3. Turbulent pressure: ~ ½  v 2 v is the root mean square velocity of turbulent elements

Footnote: Magnetic pressure is what is behind the emergence of magnetic „flux ropes“ in the Sun P phot P tube In the magnetic flux tube the magnetic field provides partial pressure support. Since the total pressure in the flux tube is the same as in the surrounding gas P tube < P phot. Thus  tube <  phot and the flux tube rises due to buoyancy force. + B2B2 88

T eff (K) SpP g (dynes/cm 2 ) P R (dynes/cm 2 ) B (Gauss) v 4000K5 V1 × A6 V1 × B8 V3 × B3 V3 × B0 V5 × Pressures B is for magnetic pressure = P g v is velocity that generates pressure equal to P g according to ½  v 2

We are ignoring magnetic fields in generating the photospheric models. But recall that peculiar A-type stars can have huge global magnetic fields of several kilogauss in strength. In these atmospheric models one has to treat the magnetic pressure as well.

 dA F gravity dx P P + dP F + dF x increases inward so no negative sign d  =   dx dP d  = g  dPdP –G–G  (r)M(r) r2r2 drdr = In our atmosphere dP = g  dx The weight in the narrow column is just the density × volume × gravity GM/r 2 = g (acceleration of gravity)

One way to integrate the hydrostatic equation P g ½ dP g = P g ½ g 00 d0d0  0 is a reference wavelength (5000 Å) P g (  0 ) = dt 0 g 3 2 00 0 ∫ oo Pg½Pg½ ( ( ⅔ P g (  0 ) = dlog t 0 g 3 2  0 log e –∞ ∫ log  o Pg½Pg½ ( ( ⅔ t0½t0½ Integrating on a logarithmic optical depth scale gives better precision P g (  0 ) = dt 0 00 0 ∫ oo gP g ½ 3/2 2 3

Numerical Procedure Problem: we must now   as a function of   since   appears in the integrand.  is dependent on temperature and electron pressure. Thus we need to know how T and P e depend on  0. Guess the function P g (  0 ) and perform the numerical integration New value of P g (  0 ) is used in the next iteration until convergence is obtained A good guess takes 2-3 iterations

III. Temperature Distribution in the Solar Photosphere Two probes of depth: Limb Darkening Wavelength dependence of the absorption coefficient Limb darkening is due to the decrease of the continuum source function outwards I (  ) = ∫S e –  sec  sec  d  0 ∞ The exponential extinction varies as  sec , so the position of the unit optical depth along the line of sight moves upwards, i.e. to smaller .  ds The increment of path length along the line of sight is ds = dx sec 

22 11 dz  =1 surface Top of photosphere Bottom of photosphere Temperature profile of photosphere z=0 Temperature z The path length dz is approximately the same at all viewing angles, but at larger the optical depth of  =1 is reached higher in the atmosphere Limb Darkening

Solar limb darkening as a function of wavelength in Angstroms Solar limb darkening as a function of position on disk

At 1.3 mm the solar atmosphere exhibits limb brightening Horne et al Temperature profile of photosphere and chromosphere z=0 Temperature z chromosophere In radio waves one is looking so high up in the atmosphere that one is in the chromosphere where the temperature is increasing with heigth

Limb darkening in other stars Use transiting planets At the limb the star has less flux than is expected, thus the planet blocks less light No limb darkening transit shape

The depth of the light curve gives you the R planet /R star, but the „radius“ of the star depends on the limb darkening, which depends on the wavelength you are looking at To get an accurate measurement of the planet radius you need to model the limb darkening appropriately If you define the radius at which the intensity is 0.9 the full intensity: At =10000 Å, cos  =0.6,  =67 o, projected disk radius = sin  = 0.91 At =4000 Å, cos  =0.85,  =32 o, projected disk radius = sin  = 0.52 → disk is only 57% of the „apparent“ size at the longer wavelength 

The transit duration depends on the radius of the star but the „radius“ depends on the limb darkening. The duration also depends on the orbital inclination When using different data sets to look for changes in the transit duration due to changes in the orbital inclination one has to be very careful how you treat the limb darkening.

Possible inclination changes in TrEs-2? Evidence that transit duration has decreased by 3.2 minutes. This might be caused by inclination changes induced by a third body But the Kepler Spacecraft does not show this effect.

One possible explanation is that this study had to combine different data sets taken at different wavelength band passes (filters). But the limb darkening depends on wavelength. At shorter wavelengths the star „looks“ smaller. The only star for which the limb darkening is well known is the Sun

In the grey case had a linear source function: S = a + b  I (  ) = ∫S e –  sec  sec  d  0 ∞ Using: I (  ) = a + b cos  This is the Eddington-Barbier relation which says that at cos  =  the specific intensity on the surface at position equals the source function at a depth  I c = I c (0) (1 –  +  cos  ) Limb darkening laws usually of the form:  ≈ 0.6 for the solar case, 0 for A-type stars I c (0) continuum intensity at disk center

I (  ) = ∫S e –  sec   sec  0 ∞ d log  log e Rewriting on a log scale: Contribution function No light comes from the highest and lowest layers, and on average the surface intensity originates higher in the atmosphere for positions close to the limb. Sample solar contribution functions

Wavelength Variation of the Absorption Coefficient Since the absorption coefficient depends on the wavelength you look into different depths of the atmosphere. For the Sun: See into the deepest layers at 1.6  m Towards shorter wavelengths  increases until at = 2000 Å it reaches a maximum. This corresponds to a depth of formation at the temperature minimum (before the increase in the chromosphere)

Solar Temperature distributions Best agreements are deeper in the atmosphere where log  0 = –1 to 0.5 Poor aggreement is higher up in the atmosphere

Temperature Distribution in other Stars The simplest method of obtaining the temperature distribution in other stars is to scale to a standard temperature distribution, for example the solar one. T(  0 ) = S 0 T סּ (  0 ) In the grey case: T(  ) = [  + q(  )] ¾ ¼ T eff T(  ) = T eff T סּ eff Tסּ()Tסּ() In the grey case the scaling factor is the ratio of effective temperatures

Scaled solar models agree well (within a few percent) to calculations using radiative equilibrium. They also agree well when applying to giant stars. Numerically it was easier to use scaled solar models in the past. Now, one just uses a grid of models calculated using radiative equilibrium

When solving the hydrostatic pressure equation we start with an initial guess for P g (  0 ). We then require that the electron pressure P e (  0 ) = P e (P g ) in order to find  0 (  0 ) =  0 (T,P e ) for the integrand. The electron pressure depends on the temperature and chemical composition. IV. The P g –P e –T Relation N 1j N 0j =  j (T) PePe N 1j = number of ions per unit volume of the jth element N 0j = number of neutrals  (T) = 0.65 u 1 u 0 T 5/2 10 –5040I/kT See Saha equation from 2nd lecture

Neglect double ionization. N 1j = N ej, the number of electrons per unit volume that are contributed by the jth element.  j (T) PePe = N ej N 0j N ej N j – N ej = The total number of jth element particles is N j = N 1j + N 0j. Solving for N ej  j (T)/P e N ej NjNj =  j (T)/P e The pressures are: P e =  N ej kT j P g =  (N ej + N j )kT j

Taking ratios:  N ej kT  (N ej + N j )kT PePe PgPg =  Aj Aj PePe PgPg =  j (T)/P e  j (T)/P e  Aj Aj  j (T)/P e  j (T)/P e 1 +  Nj Nj PePe PgPg =  j (T)/P e  j (T)/P e  Nj Nj  j (T)/P e  j (T)/P e 1 + Using the number abundance A j = N j /N H N H = number of hydrogen

This is a transcendental equation in P e that has to be solved iteratively.  ( T ) are constants for such an iteration. P e and P g are functions of  0. This equation is solved at each depth using the first guess of P g (  0 ). log  –4 –3 –2 0 1 –1 For the cooler models the temperature sensitivity of the electron pressure is very large with d log P e /d log T ≈ 12 since the absorption coefficient is largely due to the negative hydrogen ion The absorption coefficient  is largely due to the negative hydrogen ion which is proportional to P e so the opacity increases very rapidly with depth.

Hydrogen dominates at high temperatures and when it is fully ionized P g ≈ 2P e At cooler temperatures P e ~ P g ½ Where does the later come from? Assume the photosphere is made of single element this simplifies things: P e = P g  (T)/P e  (T)/P e P e =  (T)P g – 2  (T)P e =  (T)(P g – 2P e ) 2 P g >> P e in cool stars P e ≈  (T)P g 2

Completing the model P g (  0 ) = dlog t 0 g 3 2  0 log e –∞ ∫ log  o Pg½Pg½ ( ( ⅔ t0½t0½ We can now can compute this Take T(  0 ) and our guess for P g (  0 ) Compute P e (  0 ) and  0 (  0 ) Above equation gives new P g (  0 ) Iterate until you get convergence (≈ 1%) Can now calculate geometrical depth and surface flux

We are often interested in the geometric depth scale (i.e. where the continuum is formed). This can be computed from dx = d  0 /  0  The Geometric Depth x (  0 ) = ∫ 1  0 (t 0 )  (t 0 ) dt 0 0 00 The density can be calculated from the pressure ( P = (  /  )KT )  = N H (hydrogen particles per cm 3 ) x  A j  j grams/H particle) where  j is the atomic weight of the j th element N H =  A j N – N e kT  A j P g – P e =

x(  0 ) = ∫  A j kT(t 0 )t 0  0 (t 0 )  A j  j [P g (t 0 )t 0 – P e (t 0 )] d log t 0 d log e –∞–∞ log  0 x(  0 ) = ∫  A j kT(p)  A j  j 1 g –∞–∞ PgPg dp p A more interesting form is to integrate on a P g scale with dP g =  gdx The thickness of the atmosphere is inversely proportional to the surface gravity since T(P g ) depends weakly on gravity This makes physical sense if you recall the scale height of the atmosphere: Scale height H = kT/  g

Computation of the Spectrum The spectrum F   = 2  ∫  ∞ S  t   E 2  t )dt F   = 2  ∫ ∞ S    E 2  ) –∞–∞  (  0 )  0 d log t 0  0 (  0 ) d log e It is customary to integrate on a log  scale Flux contribution function

Flux Contribution Functions as a Function of Wavelength Flux at 8000 Å originates higher up in the atmosphere than flux at 5000 or 3646 Å But cross the Balmer jump and the flux dramatically increases. This is because there is a sharp decrease in the opacity across the Balmer jump.

Flux Contribution Functions as a Function of Effective Temperature A hotter star produces more flux, but this originates higher up in the atmosphere T= K T= 8090 K T= 4620 K

Computation of the Spectrum There are other techniques for computing the flux → Different integrals. Integrating flux equation by parts: F   = 2  ∫  ∞ S    E 2  )d  F   =  S (0) + ∫  ∞ E 3  )d  d S   dt The flux arises from the gradient of the source function. Depths where dS/d  is larger contribute more to the flux

V. Properties of Models: Pressure Relationship between pressure and temperature for models of effective temperatures 3500 to K. The dashed line marks where the slope exceeds 1 –1/  ≈ 0.4 and implies instability to convection d log T/d log P g = 0.4 Convection gradient Temperature Cannot scale T(P g ), unlike T(  0 )!!!

Teff = 8750 K Increasing the gravity increases all pressures. For a given T the pressure increases with gravity dlogP g dlog g = 0.85 dlogP g dlog g = 0.62 Effects of gravity

P g (  0 ) = dlog t 0 g 3 2  0 log e –∞ ∫ log  o Pg½Pg½ ( ( ⅔ t0½t0½ P g ≈ C(T) g ⅔ since pressure dependence in the integral is weak. So dlog P/dlog g ~ 0.67 In general P g ~ g p In cool models p ranges from 0.64 to 0.54 in going from deep to shallow layers In hotter models p ranges from 0.85 to 0.53 in going from deep to shallow layers Recall P e ≈ P g ½ in cool stars → P e ≈ constant g ⅓ P e ≈ 0.5 P g in hot stars → P e ≈ constant g ⅔

Properties of Models: Chemical Composition In hot models hydrogen takes over as electron donor and the pressures are indepedent of chemical composition In cool models increasing metals → increasing number of electrons → larger continous absorption → shorter geometrical penetration in the line of sight → gas pressure at a given depth decreases with increasing metal content Gas pressureElectron pressure

Qualitatively: Using  A j for the sum of the metal abundances PePe PgPg   A j = PePe PgPg   N j kT = PePe PgPg PP  N j kT ≈ PePe Since P H, the partial pressure of Hydrogen dominates the gas pressure  N j =  ( N 1 + N 0 ) j, the number of element particles is the sum of ions and neutrals and P e =N e kT =  N 1j kT for single ionizations

  A j PePe PgPg ≈  N 1j  (N 1 + N 0 ) j In the solar case metals are ionized  N 1j >> N 0j   A j PePe PgPg ≈  dP g = g 00 d0d0 g P e  0 /P e d0d0 == g P g  A j  0 /P e  0 is dominated by the negative hydrogen ion, so  0 /P e is independent of P e

Integrating:   A j d0d0 g  0 /P e ½Pg½Pg 2 = ∫ 0 00 g and T are constants  A j ) P g =c 0 –½  A j ) P e =c 0 ½ For metals being neutral:  N 1J <<  (N 1 + N 0 ) j can show  A j ) P g =c 0 –⅓  A j ) P e =c 0 ⅓

Properties of Models: Effective Temperature Note scale change of ordinate In hotter models opacity increases dramatically More opacity → less geometrical penetration to reach the same optical depth We see less deep into the stars → pressure is less But electron pressure increases because of more ionization

This is seen in the models If you can see down to an optical depth of  ≈ 1, the higher the effective temperature the smaller the pressure

Properties of Models: Effective Temperature For cool stars on can write: P e ≈ C e  T At high temperatures the hydrogen (ionized) has taken over as the electron donor and the curves level off

A grid of solar models

Log  Temperature Depth (km)

Log  The mapping between optical depth and a real depth

Wavelength (Ang) Amplitude (mmag) Why do you need to know the geometric depth? In the case the pulsating roAp stars, you want to know where the high amplitude originates

Log  Log (Pressure) P gas P electron P e ≈P g ½

Note bend in Main Sequence at the low temperature end. This is where the star becomes fully convective

VI. Models for Cool Objects Models for very cool objects (M dwarfs and brown dwarfs) are more complicated for a variety of reasons, all related to the low effective temperature: 1.Opacities at low temperatures (molecules, incomplete line lists etc.) not well known 2.Convection much stronger (fully convective) 3.Condensation starts to occur (energy of condensation, opacity changes) 4.Formation of dust 5.Chemical reactions (in hotter stars the only „reactions“ are ionization which is give by ionization equilibrium)

Much progress in getting more complete line lists for water as well as molecules. Models have gotten better over time, but all models produce a lack of flux (over opacity) in the K-band. M8V Allard et al

Dust Clouds The cloud composition according to equilibrium chemistry changes from: Zirconium oxide (ZrO 2 )Perovskite and corundum (CaTiO 3, Al 2 O 3 ) Silicates: forsterite (Mg 2 SiO 4 )Salts: (CsCl, RbCl, NaCl) Ices: (H 2 O, NH 3, NH 4 SH) M → L → T dwarfs At Teff < 2200 K the cloud layers become optically thick enough to initiate cloud convection.

T eff = 2600 K : No dust formation T eff = 2200 K : dust has maximum optical thick density T eff = 1500 K: Dust starts to settle and gravity waves causing regions of condenstation Allard et al Intensity variation due cloud formation and granulation.

Of course these models do not include rotation and Brown Dwarfs can have high rotation rates. Jupiter is as a good approximation as to what a brown dwarf atmosphere really looks like

Most cool star models have a use a more complete line lists for molecules, including water, and also include dust in the atmosphere T eff = 2900 K, log g=5-0 model compared to GJ Å Discrepancies are due to missing opacities Å 4000 Å 3m3m Optical Infrared

Comparison of Models NextGen: overestimating T eff Ames-Cond/Dusty: underestimating T eff BT-Settl: Using Asplund Solar abundances

Stars with „normal“ opacities Condensation Dust clouds Allard et al. 2010

Now days researchers just download models from webpages. Kurucz model atmospheres have become the „industry standard“, and are continually being improved. These are used mostly for stars down to M dwarfs. The Phoenix code is probably more reliable for cool objects. Kurucz (1979) models - ApJ Supp. 40, 1 R.L. Kurucz homepage: The PHOENIX homepage P.H. Hauschildt: Holweger & Müller 1974, Solar Physics, 39, 19 – Standard Model Allens Astrophysical Quantities (Latest Edition by Cox)

TEFF GRAVITY LTE ITLE SDSC GRID [+0.0] VTURB 0.0 KM/S L/H 1.25 OPACITY IFOP CONVECTION ON 1.25 TURBULENCE OFF BUNDANCE SCALE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE ABUNDANCE CHANGE EAD DECK6 72 RHOX,T,P,XNE,ABROSS,ACCRAD,VTURB E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E Sample grid from Kurucz