XXIX IAC Winter School of Astrophysics

XXIX IAC Winter School of Astrophysics APPLICATIONS OF RADIATIVE TRANSFER TO STELLAR AND PLANETARY ATMOSPHERES Fundamental physical aspects of radiatie transfer: II.- Statistical Equilibrium Equations Artemio Herrero, Noember 3-4, 207

Outline II. Statistical Equilirium Equations Local Thermodynamic Equilibrium - Maxwell, Planck, Kirchhoff, Boltzmann and Saha Absorption and emission coefficients - Line Coefficients - Continuum Coefficients Statistical Equilibrium Equations Non-Local Thermodynamic Equilibrium LTE and NLTE

Local Thermodynamic Equilibrium We hae seen that to sole the Radiatie Transfer Equation we need to know the source function We hae seen that in TE η S = = B (T ) χ Can we use this?

Local Thermodynamic Equilibrium S (τ ) = B (T (τ )) S (τ 2 ) = B (T (τ 2 )) Local Thermodynamic Equlibrium (LTE): the laws of thermodynamic equilibrium are alid at the local temperature except that the radiation field itself is not the Planck function

Local Thermodynamic Equilibrium Maxwell-Botlzmann equation for the distribution of particle elocities f (υ )dυ = dn (υ ) m = N total 2π kt 3 2 4πυ 2 e mυ 2 2kT dυ (fast) ALWAYS The Saha and Botlzmann equations for the distribution of atomic energy leel populations * N j,k h N = 2π m k j+,k e B 2 3 2 Ne U jk U j+,k 3 χ jk 2 kt T e = 2.07 0 * nl g l χ lu = e n gu u 6 Ne U jk U j+,k 3 χ jk 2 kt T e k element j ionization state U j,k partition function kt χ j,k ionization energy of state j Kirchhoff's law η = B χ u,l excitation states in a gien ionization state But not: Planck's law I B sole the RTE (states with a lower energy are more populated) χ lu excitation energy from l to u. If l has a lower energy, χ lu is a positie quantity

Local Thermodynamic Equilibrium The distributions of ionization and excitation states are gien by the equations of Saha and Botzmann. These are the distributions of energy states that we obtain when we hae a closed and isolated system of interacting particles particles (collisions) eoling until equilibrium is reached Relatie population of ionization states according to Saha 0,j+,k Relatie population of excitation states according to Boltzmann χlu u l element k 0,j,k 0,j-,k

Local Thermodynamic Equilibrium Sometimes, Saha equation is expressed giing the population of an excitation state w.r.t. the population of the ground state of the next ionization state: 3 χ ijk g ijk * 6 ni, j,k = n0, j+,k N e 2.07 0 T 2 e kt = n0, j+,k N eφ (T) g0, j+,k and Boltzmann equation giing the population of an excitation state w.r.t. the total population of its ionization state: * ni, j,k g i, j,k χi, j,k N =U e j,k j,k kt where now i indicates an excitation state of ionization state j Note that the TE population of leel i depends on the real population of the ground state of the next ionization state.

Local Thermodynamic Equilibrium LTE is clearly inconsistent Howeer physical conditions may be faourable to LTE If radiation is trapped, the radiation field will be in equilibrium with matter If the amount of energy escaping by radiation is small compared with the radiatie energy content, LTE will be faoured. Intense emerging radiation fields will faour LTE departures The distributions of energy leels in ET are gien by particle collisions: if the number of collisions is sufficiently large, Maxwell-Botlzmann, Saha and Boltzmann equations will be alid When densities are high, the number of collisions will also be high. High densities will faour the hypothesis of LTE We can not say a priori whether LTE is alid, but we know that it will faoured by high densities and disfaoured by intense escaping radiation fields and we still hae to calculate the absorption coefficients

Absorption and emission coefficients We hae seen that to sole the Radiatie Transfer Equation we need to know the source function By definition, to know the source function we hae to know the absorption and emission coefficients Sν(τν) = ην / χν They were macroscopically defined. How do we calculate them? bound-free free-free bound-bound We hae three atomic processes o Absorption, spontaneous emission and estimulated emission Between two type of leels: bound and free Transitions can be radiatie and collisional o Radiatie transitions imply photon emission and absorption. They will modify the opacity and the atomic populations. plus scattering

Absorption and emission coefficients: Line coeffcients Assume we hae two bound leels, l and u in and atom (with l<u) What's the probability per unit time of absorbing (or emitting) a photon with frequency between and + d, traelling from (or to) a direction between ω and ω +dω? dω dp () = Blu I φ ()d 4π dω dpind () = Bul I ϕ ()d 4π dω spo dp () = Aulψ ()d 4π abs We will assume complete freq. redistribution (CFR): frequency and direction of a photon producing an absorption are uncorrelated with the same properties of the re-emitted photon This implies φ ψ ϕ For most astrophysical problems CFR works well. Sometimes, Partial freq. red. (PFR) shall be used where Blu (Bul ) is the probability that the radiation field I causes a transition l u (u l) per unit of intensity, Aul is the probability of an spontaneous u l transition dω and is the probability that the photon comes from direction dω 4π

Absorption and emission coefficients: Line coeffcients Assume we hae two bound leels, l and u in and atom (with l<u) The number of absorptions is: Number of atoms in the adequate initial state x transition probability x time Number of absorptions: nl dpabs dvdt Number of induced emissions: nu dpind dvdt (Absorbed energy) = energy of the absorption x number of absorptions ( ) de = nl dp dvdt nu dp dvdt = φ nl Blu nu Bul I ddω dvdt 4π From the definition of the absorption coefficient ( abs ) abs ind deabs = κ I ddω dadl And thus, for a bound-bound transition between l and u κ (l u) = φ nl Blu nu Blu 4π ( )

Absorption and emission coefficients: Line coeffcients Assume we hae two bound leels, l and u in and atom (with l<u) ( κ = φ nl Blu nu Bul 4π ε = φ nu Aul 4π ) and the Einstein coefficients are related by (easy to see in TE, where emissions=absorptions): 2 3 g l Aul = 2 Blu c gu ; g l Blu = gu Bul We define the cross-section as φ Blu σ lu () α lu () 4π

Absorption and emission coefficients: Line coeffcients How do we calculate the cross-sections or the Einstein coefficients? We hae to use quantum mechanics to calculate the transition probabilities Blu will be related to u H l where H hamiltonian describing the interaction u and l wae functions Largest contribution from first term of the hamiltonian, i.e., that corresponding to the dipole approximation 32π 4 Blu = u d l 2 3ch 2 Aul, Bul Further terms (electric quadrupole, magnetic dipole) much smaller forbidden transitions. Important if dipolar term is zero and collisions are ery low

Absorption and emission coefficients: Line coeffcients A traditional way to calculate the strength of transition is the classical oscillator. It results in π e2 σ= constat mc But we know that the spectral lines hae different intensities. Then π e2 σ= f = Blu mc 4π where f is known as the oscillator strength If there are subleels f lu = g l f l u g l l,u M=

Absorption and emission coefficients: Line coeffcients We find a lot of these atomic data in NIST database https://www.nist.go/pml/atomic-spectra-database Peter an Hoof Atomic line list http://www.pa.uky.edu/~peter/newpage/

Absorption and emission coefficients: Line coeffcients But bound-bound transitions are not infinitely sharp The intrinsic profile is a Lorentzian function: π e2 γ σ () = α () = mc Δω 2 + γ 2 ( ) 2 where Δω = (ω ω 0 ), ω =2π, and γ is the damping constant A lorentzian function is narrower and lower at the peak than a gaussian, but with more extended wings! this will hae an impact on final profiles and escape probabilities

Absorption and emission coefficients: Line coeffcients and therefore, for a bound-bound transition, or spectral line gl κ = σ lu () nl nu gu 3 gl 2 L ε = σ lu () 2 nu c gu L The structure is always of the form: cross section x atomic population

Absorption and emission coefficients: Line coeffcients Consider now the line source function in LTE ε nu Aul Aul 2 3 S = = = = 2 k nl Blu nu Bul Bul nl Blu c nl gu nu Bul nu g l nl* g l k BT n* = g e u u 3 S LTE 3 3 2 2 2 = 2 * = 2 = 2 = B c nl gu c g l gu k T c k BT B e e * nu g l gu g l

Absorption and emission coefficients: Continuum Similarly, for bound-free and free-free transitions κ b f ε b f κ f f * k BT = σ ik () ni ni e 3 2 * = σ ik () 2 ni e c k BT k BT = σ kk ()ne nk e ε f f = σ kk ()ne nk 3 2 e 2 c k BT t0 ik ik t0+δt - - + + - + - + where i is the bound leel, k represents the next ionization state ik is the ionization energy from leel i and ni* indicates the population calculated using the formulae from TE Note : Bound-free transitions imply the encounter of an ion and a free electron. Their number is proportional to the product ne nk,where n is the actual population of k-ions IAC XXIXk WS

Absorption and emission coefficients: Continuum One of the most important contributors will be H. We obtain (Kramers, 923, also alid for hydrogenic ions): Bound-free: σ b f 4 Z = 2.85 029 5 3 g bf n cm2 nk ionization limit for leel n of Hydrogen: λ = n2 R where the Gaunt bound-free factor for H is gien by g bf = 0.3456 λ R 3 2 2 n (λ R) with R the Rydberg constant (R=.0968 05cm ), n the main quantum number and Z the ion charge Free-free: σ f f = 3.7 0 8 Z2 T 2 3 g ff cm 2 with g ff = + 0.3456 (λ R) 3 λk T hc B + 2

Absorption and emission coefficients: Continuum Other important contributors to the continuum opacity Opactities, cross-sections and oscillator strengths may be found in the pages of the Opacity Project (OP) and Iron Project (IP) : http://cdsweb.u-strasbg.fr/topbase/ home.html He Metals Electrons (Thompson scattering) σ e = 6.626 0 26 ne cm2 Molecules Hydrogen negatie ion The hydrogen negatie ion is of particular importance for solar type photospheres (first recognized by Wildt, 939)

Absorption and emission coefficients: Continuum χ total = χ j j

Statistical Equilibrium Equations We hae seen that the emission and absorption coefficients are of the form Coeff= cross-section x atomic population We hae seen how to calculate the cross sections We need to calculate the atomic populations: is LTE sufficient?

Statistical Equilibrium Equations We assume that the atomic (or molecular) leel populations are time independent. The system is in statistical equilibrium. dni = 0 i where i is any energy leel (of an atom, molecules or free particle) dt But leels are constantly being populated and depopulated. What happens is: ( processes depopulating i ) = ( processes populating i ) n P n P i ij j i j ji =0 j i where Pij is the probability that one atom in leel i suffers a transition to leel j ni (Rij + Cij ) n j (R ji + C ji ) = 0 j i (where j is any energy leel) j i ni (Rij + Cij ) + ni ( Rik + Cik ) n j (R ji + C ji ) nk ( Rki + Cki ) = 0 j i j i (where j is any bound energy leel)

Statistical Equilibrium Equations We hae one equation for each leel considered (and one set at each depth) N n P j n2 P2 ni Pi nn PN = 0 j=2 n P2 + n2 P2 j ni Pi2 nn PN 2 = 0 j 2! n Pi n2 P2i + ni Pij nn PNi = 0 A n = b j i! n PN n2 P2 N ni PiN + nn PNj = 0 j N One of the equations has to be replaced by an independent one, like the total number of particles or the charge conseration: n + n2 + + ni + + nn = Ak N tot

Statistical Equilibrium Equations n = b A H He Rij,Cij (H) Rij,Cij (He) NH C He 0 ni(he) He C NHe Rij,Cij (C) The solution is: ni(h) He H C H 0 H C 0 ni(c) NC n = A b

Statistical Equilibrium Equations Assume i < j. Rij will be the probability of haing an absorption from i to j : dω dω Rij =! P d =! Bij I φ d = 4π 0 Ω 4π 0 Ω abs = Bijφ d 4π 0 Rij = 4π 0! I Ω σ ij () d ω = Bijφ J d 0 J d (sometimes, particularly for analytical work, we find R = B J ) ij ij

Statistical Equilibrium Equations Radiatie rates: Rij = 4π 0 σ ij () J d n R ji = 4π i nj * σ ij () 2 3 k BT c2 + J e d 0 absorption σ ik () Rik = 4π J d 0 spontaneous + induced emis. n Rki = 4π i nj * σ ik () 2 0 c2 + J e 3 k BT d

Statistical Equilibrium Equations For the collisional rates, we hae Assume again i < j σ ij will be the probability that an atom in state i will suffer a transition to leel j due to an impact with an electron of elocity υ per unit length. Then 0 0 Cij = σ ij (υ ) υ ne f (υ )dυ = π a02 Qij υ ne f (υ )dυ = ne qij (T ) The problem is to know Qij (or qij ). If not, we can use an Regermorter formula: qij (T ) = C0T with u0 = 2 Eij ( ) 4.5 f E E 2 u e u0 Γ (u ) ij H ij e 0 0 and Γ e (u0 ) max g,0.276e 0 E (u0 ) u kt E is the exponential integral The expression is alid for ions (for neutral atoms Γ e (u0 ) is different)

Statistical Equilibrium Equations An important property of collisions is the ratio between upwards and downwards transitions We know that in TE we must hae C ij ni C ij = n j C ji * * C ji n = n * j i But C ij, C ji depend only on atomic properties and the electrons elocity distribution Thus C ij C ji n = n j * holds also in non-equilibrium conditions i (as far as the particle elocity distribution is maxwellian)

Non-Local Thermodynamic Equlibrium Now we see the difficulty of departing from LTE: Rij (τ ) = 4π σ ij () σ ij () J (τ )d = 4π 4π 0 0 We want to calculate the Sole the RTE di µ = I S dτ emergent flux (or intensity) F+ (0), I + (0) Need to know J (τ ), I (τ ) Need Rij (τ ) = 4π 0 σ ij ()!Ω I (τ ) dω d J (τ )d η Need S = f (n,σ,τ, J ) χ Need to know σ Need to know n (theory or lab)

Non-Local Thermodynamic Equlibrium There are basically four ways to sole the coupling between the SEE and the RTE.- Assume LTE: S = B 2.- Use Λ-iteration (or J-n steps): J 0 = Λ S0 = Λ B ni0 S J = Λ S ni S2 - It's too slow (particularly for optically thick transitions) 3.- Use an approximation for the lambda operator: the Accelerated Lambda Iteration ( ) ( ) J (τ ) = Λτ S = Λ S + Λ Λ S = Λ S + J J = Λ S + ΔJ Where Λ S is an approximation for Λ S and ΔJ is the error in the approximation (known from preious iterations). In its simplest form (Scharmer, 98) S (τ ) τ > γ Λ S = / 2S (τ = γ ) τ γ - and we start soling from the bottom. It inoles matrix inersion with dimension N leels N leels * τ 4.- Linearization (Newton-Raphson iteration) dn f f (x ) = n n=0 dx (x x0 ) n = f (x0 ) + x=x0 df dx (x x0 ) + f (x0 ) + f (x0 )δ (x0 ) = 0 δ (x0 ) = x=x0!! n A b!!!! for the SEE, A 0 n0 = b0 n0 A 0 b0 + x j j ( ) ( ) δ xj = 0 j!! n A b ( x j ) δ xj = 0 0 - and the RTE has also to be linearized. It is expensie, but ery stable f (x0 ) f (x0 )

LTE and NLTE Consider the line source function in NLTE η 2 3 2 3 2 3 S = = 2 = 2 = 2 * χ c gu nl c gu bl nl c b kt l B g n g b n* e l u l u u bu In the Wien region, with >> k BT bu bu LTE 2 3 S = 2 = B = S bl c bl k T bl B e bu Using now the Eddington-Barbier approximation bu LTE bu LTE NLTE NLTE H (τ = 0) = S (τ = 2 3) = S (τ = 2 3) = H (τ bl bl ibu > bl H NLTE (0) > H LTE (0) more flux, less absorption ibu < bl H NLTE (0) < H LTE (0) less flux, more absorption ibu = bl H NLTE (0) = H LTE (0) same flux, same absorption = 0)

LTE and NLTE What s the typical situation in NLTE? A solar example 5890 A Left: results for the VALIII solar model (it includes chromosphere) Right: simplified NaI Gotrian diagram from Schlieder et al., 202