Formal solvers of the RT equation

Transcription

1 Formal solvers of the RT equaton

2 Formal RT solvers Runge- Kutta (reference solver) Pskunov N.: 979, Master Thess Long characterstcs (Feautrer scheme) Cannon C.J.: 970, ApJ 6, 55 Short characterstcs (Hermtan scheme) Bellot Rubo et al.: 998, ApJ 506, 805 Short characterstcs (Bezer attenuaton operator) de la Cruz Rodríguez & Pskunov: o3, ApJ 764, 33

3 Why we need a formal solver? What s a formal solver? We assume that we know the source functon and the opactes along our ray. Ths s suffcent to compute the ntenstes (as functon of wavelength). In practce, local opactes and source functon may also depend on the ntenstes comng from dfferent drectons. Ths wll requre teratons. In the next lecture we wll talk about how to get to self- consstency.

4 Solvng RT wth RK Smple mnded approach: dy = f( x) y+ g( x); y( x ) = y k = f( x ) y + g( x ) 0 0 h hk h k = f x + y g x h hk h k3 = f x + y + + g x + ( ) ( ) ( ) k = f x + h y + hk + g x + h 4 3 h y = y + k + k + k + k 6 ( ) + 3 4

5 More clever RK. Prevous example suffers from all problems nherent to RK, specally when dealng wth complex medum where f and g have sharp varatons. Instead one can solve RT analytcally: τ ʹ ʹ Δτ ( t) I ( ) I ( ) e τʹ ʹ B ( t) e τʹ ʹ = τʹ + dt τ ʹ In partcular, ths s useful for a half- nfnte medum where we can easly use Gauss quadratures for the ntegral: I (0) B ( t) e t dt ω B ( τ ), 0 = N = =

6 The nodes and weghts for Laguerre polynomals: Nodes E E E E E E E E-07 Weghts The only problem s that values of T are not known n τ,ι We can fnd them solvng ODE for optcal depth: =, x = 0 dτ k ( x) ( x) τ 0 ρ = Advantages: smple boundary condton, RHS does not depend on unknown functon and RHS s always non- negatve.

7 4th order Runge-Kutta for the geometrcal depth = ; x0 = 0 dτ α ( x) k k = ; k = ( x ) α x hk α = ; k = ( + ) 3 4 α 3 ( x + hk ) α ( x + hk ) h x = x + k + k + k + k 6 ( ) We ntegrate the equaton for x from 0 to each of the τ, consecutvely. For each x we fnd the temperature and then ntensty usng Gauss quadratures.

8 Requrements for a formal solver RK s good to study the propertes of your envronment, selectng the grd etc. For practcal applcatons the solver must be quck and stable. It should be able to acheve good accuracy on the prescrbed grd. The formal solver should not propagate/amplfy errors whch may be deadly of we need teratons. These requrements force us to use fnte dfferences schemes.

9 Method classfca6on RT s solved along rays or characterstcs that do not necessarly concde wth the selected grd. Indvdual ray can be followed through the whole medum boundary- to- boundary or over a short part extendng the length of one grd cell. RT solvers based on complete rays are known as long characterstcs methods. RT solvers that follow radaton through a sngle grd cell at a tme are called short characterstcs methods. In D there s obvously no dfference between short and long characterstc methods

10 Method classfcaton Long characterstcs Short characterstcs

11 Comparson of long versus short Long: Use both boundary condtons Get ntenstes n both drectons Mean of the two ntenstes s actually a component of J Expansve n D or 3D f the geometrcal grd does match the ray drectons Short: Fast Follow the geometrcal grd no matter what Need two- drectonal ntegraton to evaluate J Suffers from numercal lght defocussng

12 Lght defocussng

13 Feautrer RT solver Equaton of radatve transfer (agan) di = ( I S ) α where x s a geometrcal dstance along the ray Let s splt the ntensty n two flows: I + n the drecton of ncreasng x and I - n the opposte drecton. The RT equaton can be wrtten for each drecton: + di + = α ( I S) di = α ( I S)

14 We defne two new varables U=½(I + + I - ) and V=½(I + - I - ). Now we can add/subtract the two equatons of RT and dvde the results by : di + di + du + = α ( I S ) = α ( I S ) = α V di di + dv + = α ( I S ) = α ( I S ) = α ( U S )

15 nd order form of RT We substtute the dervatve of V n the nd equaton usng the expresson for V from the st equaton: V = ( ) α du dv = α ( U S ) The equatons for U and V can be combned nto a sngle nd order ODE: d du α α = ( ) ( U S )

16 Boundary Condtons Boundary condtons are set n the two ends of the medum. For the smallest x we can wrte: du ( ) α + A = V = ( I I ) = For the opposte end we have: du = ( I + I ) I = U I ( ) B + + A+ α + = V = ( I I ) = = ( I + I ) + I = U + I + B

17 Fnte dfferences equaton have famlar form (note the sgn n the defntons of a and c ): au + bu cu = d for =, K, N a c + ( α ) + ( α ) = x x x x,, + ( α ) + ( α ) = x x x x, +, + + b = a + c + α d, = α S,,

18 For = we can wrte a lnear boundary condton: b U U + U ( τ τ ) du = U + = dτ ( τ τ) ( A = U + U I ) ( τ τ) ( α + α) x x c ( α + α) U + ( x x) U = ( α + α) ( ) = x x I A d

19 or we can wrte quadratc boundary condton: U+ U du δτ d U U δτ U = dτ τ dτ τ b ( α + α) ( x x) δτ U U + δτ ( U I ) + δτ ( U S ) A c + δτ + δτ U = A = δτ I + δτ S The case of =N s smlar d

20 For sem- nfnte medum boundary condton at looks a bt dfferent: du + = U I dτ du d S = + ds = + St () = S( τ) + ( t τ) dτ dt t= τ cn = δτ Ths s known as dffuson b boundary condton N = + δτ ( S, N S, N) dn = ( S, N + S, N) + δτ + ( t ) I τ S () t e dt U S τ τ + ds dτ dτ I S deep

21 How does ths work? Select drecton. Setup a D grd and compute opactes, sources functon and optcal steps. Compute the Feautrer coeffcents ncludng those gven by the boundary condtons. Solve 3- dagonal SLE As free lunch you get the contrbuton of ths ray to the angle- averaged ntensty J whch s U and to the flux dvergence whch s V.

22 Hermtan method Taylor expanson for the ntensty n pont τ : 4 n n δ d I I+ = I + n n! dτ I I n= di d I d I d I δ δ δ dτ dτ dτ 6 dτ ( δ ) d I d I d I dτ dτ dτ 3 4 ' 3 = " = + δ + δ 3 4 δ ( ) ' ' ( " " ) ' I = α I S ( ) ( ) ( ) " ' ' I = α α I S S + α I S + δ δ I = I + I + I + I I I = q I + p

23 Attenuaton operator solver Soluton of RT over one grd cell can be wrtten: ( τ+ τ) I ( τ ) = e I ( τ ) + + τ+ + τ ( τ+ t) S () t e dt where τ s the optcal path along the ray Suppose S slowly changes wth τ whch can be approxmated by a lnear functon. Then we can take the ntegral analytcally! ( ) ( τ τ ) ( ) + S() τ = S ( ), + S ( ), + τ+ τ τ+ τ + ( τ+ ) = ( τ) + η, I I e τ τ τ τ

24 Source Func6on Approxma6on Quadratc approxmaton for the source functon s better than lnear, but Bezer splnes are a much more robust alternatve February, 05 3D Radatve Transport

25 How does ths work? Short characterstcs result n recurrence relaton between I and I + Select drecton For startng grd ponts ncomng ntensty s gven by boundary condtons Compute the opacty and the source functon n the grd ponts and nterpolate for the up- stream and down- stream ponts (for quadratc schemes). Compute ntenstes for all ponts n the next layer.

26 Comparson of the solvers

27 Home work 4 Compute spectral synthess usng a method of your choce for a statc D model atmosphere of the Sun. For a fxed geometrcal depth grd and wavelength grd you are gven a D array of opactes and D array of source functon. The boundary condtons: no radaton enters through the surface and the flux spectrum at the deepest atmosphere pont s gven.