Lecture 2 Solutions to the Transport Equation

Transcription

1 Lecture 2 Solutions to the Transport Equation

2 Equation along a ray I In general we can solve the static transfer equation along a ray in some particular direction. Since photons move in straight lines we want this to be a straight line (geodesic). Then the RTE is di ν ds = χ νi ν + η ν Now we just have to do the differential geometry to describe s in some coordinate system, which we saw last time. So let s make our life as simple as possible and assume that we can work in Cartesian coordinates and that our problem is one-dimensional. Then we can let ds dz and for the moment only look at rays headed directly toward us. Then the RTE becomes: di ν dz = χ νi + η ν

3 Equation along a ray II We can write the equation in a somewhat more illuminating form by dividing through by χ ν di ν dτ ν = I + S ν where dτ ν = χ ν dz is the optical depth and S ν = η ν /χ ν is the Source function Let us also suppress the frequency dependence. Then we have or di dτ = S I di dτ + I = S We can solve this equation with an integrating factor to obtain I = I e τ + S(1 e τ )

4 Equation along a ray III assuming that S is a constant. Consider the case I =. Then Then in the case τ ν << 1 or in LTE I ν = S ν (1 e τν ) I ν = τ ν S ν I ν = τ ν B ν = χ ν s B ν so the atmosphere is bright where χ ν is large and dim where χ ν is small. Hence you would expect to see an emission line spectrum. Now consider the thick case τ ν >> 1 independent of χ ν. I ν = S ν = B ν

5 Line Profile Natural Damping Power spectrum I(ω) = γ/2π (ω ω ) 2 + (γ/2) 2 This is a Lorentzian with width λ = 2πcγ ω 2 = 4πe2 = mc2 Pressure Broadening with lead to a bigger width and we also need to convolve this with the effects of Thermal Doppler Broadening. The Doppler Broadening will give a profile φ D = 1 π 1 λ D e [(λ λ ) 2 / λ D ] where λ D = λ c (2kT m )1/2

6 More on Solution Now consider I expand e τ 1 τ I = I e τ + (1 e τ )S I = I + τ(s I ) I = I + χ s (S I ) so if I > S we see absorption (and it is strongest where χ is strongest, i.e. in lines if I < S we see emission (and it is strongest where χ is strongest, i.e. in lines. Again in the thick case I = S

7 Line Profiles

8 Simple Stars Let s consider a star in LTE. Then S ν = B ν Let s cut the star into two regions bounded by τ = 1. In the upper layer I = B ν (τ > 1) and so S ν = B ν (τ < 1) I ν = I + χ ν s (B ν (τ < 1) B ν (τ > 1)) Now in stars we see absorption lines. Thus B ν (τ < 1) < B ν (τ > 1) which implies T (τ < 1) < T (τ > 1)

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10 z θ φ y x

11 General Rays So far we have considered just one ray directed along the z axis. Clearly for the formal solution each ray is independent of all the others. Let us consider a general ray in the direction defined by θ. But we want to measure optical depths: 1) from the outside in; 2) along the z axis. Thus the RTE becomes: cosθdτ(s) = dτ ν it is customary to define µ cosθ cosθ di ν dτ ν = I ν S ν

12 Solution for Plane-Parallel Atmosphere Let us assume that we have a semi-infinite slab and we wish to know the I(τ =, µ). Then we find: I(τ =, µ) = S ν (τ ν )e τν/µ dτ ν /µ Now τ ν /µ = τ(s) so this just says that the surface intensity is the sum over the path of the emission (source function).

13 Eddington Barbier Relation Let s assume I(τ =, µ) = I(τ =, µ) = S ν (τ ν ) = a ν + b ν τ ν S ν (τ ν )e τν/µ dτ ν /µ [a ν + b ν τ ν ]e τν/µ dτ ν /µ I(τ =, µ) = a ν + b ν µ = S ν (τ ν = µ) Thus, the measurement of the emergent intensity as a function of µ gives us information about S as a function of depth.

14 Eddington-Barbier and Limb Darkening

15 General Formal Solution I I(τ, µ) = 1 µ τ C µ di ν dτ ν = I ν S ν e (τ t)/µ S(t) dt + e (τ C)/µ I(C, µ) Clearly the Formal Solution is a boundary value problem. But the boundary depends on the sign of µ. For out-going rays, µ I is specified at the base of the atmosphere. For in-going rays, µ < I is specified at the surface of the atmosphere. We will generally assume that there is no external illumination and the I(τ =, µ < ) =. In that case, unless I(τ max, µ) increases exponentially with τ max the boundary term as τ max Then for large τ max we have: I(τ, µ) = 1 µ τ e (t τ)/µ S(t) dt ; µ

16 General Formal Solution II and the emergent intensity in the semi-infinite case is: I(, µ) = 1 µ e t/µ S(t) dt ; µ

17 More on the Source Function I In strict thermal equilibrium the RTE is: db ν ds = χ ν(b ν S ν ) = S ν = B ν (T ) While I ν = B ν in all directions implies S ν = B ν, the converse is NOT true. S ν = B ν I ν = B ν

18 More on the Source Function II At the surface I ν (µ ) = B ν (T ) and I ν (µ < ) =. Thus B = 1 J = 1/2 I ν dµ = 1/2B ν (T ) 1 1 H = 1/2 I ν µ dµ = 1/4B ν (T ) 1 F ν = 4πH ν = πb ν B ν dν = σ π T 4 σ = 2π5 k 4 Effective Temperature of a star defined by 15h 3 c 2 = Stephan s constant F = σt 4 eff or L = 4πR 2 σt 4 eff In LTE S ν = B ν (T (r)), then we get I ν from the formal solution. Another standard case is S ν = J ν : pure, monochromatic (coherent), isotropic scattering. This follows from integrating the RTE over all

19 More on the Source Function III directions and equating power absorbed to power emitted in a given volume. If we combine thermal emission and scattering we get that the RTE is given by di ν ds = κ ν (I ν B ν ) σ ν (I ν J ν ) = (κ ν + σ ν )(I ν S ν ) S ν = (1 ɛ ν )J ν + ɛ ν B ν (1) ɛ ν κ ν κ ν + σ ν χ ν κ ν + σ ν

20 Moment Equations I For a plane-parallel atmosphere the RTE is: µ di ν dz = χ ν(i ν S ν ) Let S ν and χ ν be independent of µ Recall 1 J ν = 1/2 I ν dµ H ν = 1/2 K ν = 1/2 Integrating the RTE over µ gives I ν µ dµ I ν µ 2 dµ dh ν dz = χ ν(j ν S ν )

21 Moment Equations II or multiplying the RTE by µ and integrating dk ν dz = χ νh ν The net flux integrated over all frequencies is H = H ν dν In Radiative equilibrium (only radiation carries energy) dh dz = χ ν (J ν S ν ) dν = For combined thermal emission and scattering S ν = ɛ ν B ν + (1 ɛ ν )J ν J ν S ν = ɛ ν (J ν B ν )

22 Moment Equations III Thus, Radiative Equilibrium scattering drops out. κ ν (J ν B ν ) dν =

23 Two-Stream Eddington Approximation I Assume Where I ± = constant Then I(µ) = J ν = 1/2 H ν = 1/2 { I + µ I µ < I ν dµ = 1/2(I + + I ) I ν µ dµ = 1/4(I + I ) 1 K ν = 1/2 I ν µ 2 dµ = 1/6(I + + I ) = 1/3J ν 1 From the moment equations we have dk dτ = H 1/3 dj dτ = H

24 Two-Stream Eddington Approximation II Then at τ = J = 3Hτ + constant J = 1/2I + H = 1/4I + J = 2H J(τ) = 3H(τ + 2/3) In RE J = B σt 4 π B = 3H(τ + 2/3) 4 = 3σT eff (τ + 2/3) 4π T = T eff at τ = 2/3

25 Exponential Integral Solution I The solution to the RTE is really not I ν, but rather J ν, since once we have J ν we know S ν and we just have to perform a formal solution. From the general expression for I ν (τ, µ) we can write J ν = 1/2 1 1 [ 1 µ [ 1 µ τ τmax + 1/2 e (t τ)/µ S(t) dt τ ] + e (τmax τ)/µ I(τ max, µ) dµ ] e (τ t)/µ S(t) dt + e τ/µ I(, µ) dµ For simplicity consider zero incident intensity at both boundaries. Define: E 1 (x) = 1 e x/µ µ dµ

26 Exponential Integral Solution II where E 1 is the first exponential integral. Let y = 1/µ, then E 1 (x) = where z = xy. For large x e xy y dy = e z x z dz E 1 (x) e x /x Then we can write J(τ) = 1/2 τmax E 1 ( t τ )S(t) dt + incident terms J(τ) = Λ τ {S}

27 Exponential Integral Solution III Similarly 1 H ν (τ) = 1/2 + 1/2 + 1/2 1 I(τ, µ)µ dµ [ τ 1 1 [ τmax τ ] e (τ t)/µ S(t) dt + e τ/µ I(, µ) dµ ] e (t τ)/µ S(t) dt + e (τmax τ)/µ I(τ max, µ) dµ For simplicity again consider zero incident intensity at both boundaries. Define: Then E n (x) = x n 1 x e z z n E n(x) = E n 1 (x) dz

28 Exponential Integral Solution IV and E n (x) = [ e x xe n 1 (x) ] /(n 1) Then we can write τ τmax H(τ) = 1/2 E 2 (τ t)s(t) dt + 1/2 E 2 (t τ)s(t) dt τ + incident terms Similarly, H(τ) = 1/2 τmax E 2 ( t τ )sgn(t τ)s(t) dt H(τ) = Φ τ {S} τmax K (τ) = 1/2 E 3 ( t τ )S(t) dt K (τ) = X τ {S}

29 Exponential Integral Solution V Can show Λ τ (a + bt) = a + bτ + 1/2 [be 3 (τ) ae 2 (τ)] Φ τ (a + bt) = 4/3b + 2 [ae 3 (τ) be 4 (τ)] X τ (a + bt) = 4/3(a + bτ) + 2 [be 5 (τ) ae 4 (τ)]