Lecture 6Sobolev Theory

Dilution Factor I

sinθc = R/r

ω∗ ≡ solid angle subtended by stellar disk

W ≡ ω∗/(4π) ≡ dilution factor

Dilution Factor II

ω∗ =

∫dΩ = 2π

∫ θc

0sinθ dθ

= 2π(−cosθ)∣∣∣θc0

= 2π [1− cosθc]

= 2π[1−

√1− (R/r)2

]W = ω∗/(4π)

= 1/2[1−

√1− (R/r)2

]W (R) = 1/2

W (r >> R) = 1/4(R/r)2

Flux from a Star I

sin(90− θ) = z/r = cosθ; sinθ = p/r

µ = cosθ =√

r2 − p2/r

Flux from a Star II

dµ = −1/2(r2 − p2)−1/22p/rdp

dµ = −(r2 − p2)−1/2p/rdp

F = 2π∫ π

0I cosθd(cosθ)

= −2π∫ 0

rI

√r2 − p2

rp

r√

r2 − p2dp

=2πr2

∫ r

0I p dp

Constant Direction Velocity SurfacesSurfaces of Constant Line-of-sight Velocity

Resonant Scattering Qualitative

Calculation of FluxResonant Scattering

Sobolev TheoryCastor, 1970

Mathematics IRadiation travels from positive to negative z, that is from left to rightSurfaces of constant velocity: vz = µv

I(r , µ) = I(p, z)

p2 + z2 = r2; z = −µ r

The transfer equation is:

dIν(p, z)

dz= κν(p, z)(Iν(p, z)− S(r))

Assume thatIν(rc , µ) = Ic for all ν, µ

Iν(p, z) =

∫ ∞z

S((p2 + z ′2)1/2)eτν(p,z)−τν(p,z′) dτν(p, z ′)

for p > rc or z > 0

Mathematics II

Iν(p, z) =

∫ (r2c−p2)1/2

zS((p2 + z ′2)1/2)eτν(p,z)−τν(p,z′) dτν(p, z ′)

+Iceτν(p,z)−τν(p,−(r2c−p2)1/2)) for p < rc , z < 0

τν(p, z) =

∫ z

−∞κν(p, z ′)dz ′

κν(p, z) = κo(r)φ(ν +ν0

cv(r)

rz − ν0

φ(x) is strongly peaked at x = 0 so

ν0

cv(r)

r(z0)z0 = −(ν − ν0)

ν0

cv((p2 + z2

o )1/2)

(p2 + z2o )1/2

z0 = −(ν − ν0)

Mathematics IIIThen let x be the argument of φ

x = ν − ν0 + ν0v0(r ′)

r ′z ′

∫ z

−∞φ(ν − ν0 + ν0

v0(r ′)r ′

z ′) dz ′ =

∫ x

−∞φ(x ′)

dz ′

dxdx

≈[∫ x

−∞φ(x ′) dx ′

]1/

dxdz ′|z0

dxdz ′

=∂

∂z(ν0

cv(r)

rz)

=ν0

c

[v(r)

∂z ′/(z ′2 + p2)1/2

∂z ′+

z ′

rdvdr

drdz ′

]

drdz

=zr

More Math

dxdz ′

=ν0

c

[v(r)

r+ z

d(v/r)

dz ′

]=

ν0

cv(r)

r

[1 + z(

v(r)

r)−1 d(v/r)

drdrdz ′

]=

ν0

cv(r)

r

[1 +

z2

r(v(r)

r)−1 d(v/r)

dr

]=

ν0

cv(r)

r

[1 +

z2

r(v(r)

r)−1 d(v/r)

dr

]=

ν0

cv(r)

r

[1 +

z2

r(v(r)

r)−1[1/r

d(v)

dr− v

r2 ]

]=

ν0

cv(r)

r

[1 +

z2

rrv

[1/rd(v)

dr− v

r2 ]

]

More Math I

dxdz ′

=ν0

cv(r)

r

[1 +

z2

r2 [r/vd(v)

dr− 1]

]=

ν0

cv(r)

r

[1 +

z2

r2 [d ln vd ln r

− 1]

]So Finally

τν(p, z) = κ0((p2 + z2)1/2)y(ν + ν0

cvr − ν0)

ν0c

vr

[1 +

z2

r2 [d ln vd ln r

− 1]

]−1

y(x) =

∫ x

−∞φ(x ′) dx ′

y(+∞) = 1y(−∞) = 0

τν(p, z) = τν(p,∞)y

More Math II

In the limit:

φ→ δ

y → Θ(z − z0)

τν(p,∞) =

πe2

mc (gf )lu(Nlgl− Nu

gu)

νoc

vr

[1 +

z2

r2 [d ln vd ln r

− 1]

]−1

Iν(p, z) =

∫ 1

y(z)S((p2 + z2)1/2)eτν(p,∞)(y(z)−y(z′))

τν(p,∞)dy(z ′)

p > rcorz > 0

(1)

More Math III

Iν(p, z) =

∫ y(−(r2c−p2)1/2)

y(z)S((p2 + z2)1/2)eτν(p,∞)(y(z)−y(z′))

τν(p,∞)dy(z ′)

+Iceτν(p,∞)(y(z)−y(−(r2c−p2)1/2)) p < rc , z < 0

Recall,y(z) = y(ν +

ν0

cvr

z − ν0)

and that if z > z0 the integrals nearly vanish. If z < z0 then the lowerlimit→ 0. In this case most of the contribution to Iν comes at z = z0.Thus we can pull S out of the integral and evaluate it at z = z0

Iν(p, z) = S((p2 + z20 )1/2)

(1− eτν(p,∞)(y(z)−1))

)p > rcorz > 0

More Math IV

Iν(p, z) = S((p2 + z20 )1/2)

(1− eτν(p,∞)(y(z)−y(z′))

)+Iceτν(p,∞)(y(z)−y(−(r2

c−p2)1/2)) p < rc , z < 0

Flux I

F = 4π∫ rc

0

[S((p2 + z2

0 )1/2)(1− e−τ(p,∞)y(−(r2

c−p2)1/2))

+Ice−τ(p,∞)y(−(r2c−p2)1/2)

]2πp dp

+4π∫ ∞

rc

[S((p2 + z2

0 )1/2(1− e−τ(p,∞))]

2πp dp

Define

Fc = 4π∫ rc

02πpIc dp = 4π2r2

c Ic

Flux II

Fν − Fc

Fc=

1r2c

∫ ∞0

S((p2 + z20 )1/2)

Ic

(1− e−τ(p,∞)

)2p dp

− 1r2c

∫ rc

0

[1− e−τ(p,∞)y(−(r2

c−p2)1/2)]

2p dp

− 1r2c

∫ rc

0

S((p2 + z20 )1/2

Ic

[e−τ(p,∞)y(−(r2

c−p2)1/2)

−e−τ(p,∞)]

2p dp

I Term 1: Emission from entire core as if it were transparentI Term 2: Radiation removed from continuum by material in front of

the coreI Radiation that would have come from the part of the envelope that

is occulted by the core

Flux III

y(−((r2c − p2)1/2) =

0 if ∆ν < 01 if ∆ν > 0

So absorption and occultation terms are images of each other aroundthe line center modulo the factor S/Ic in the occultation term.

Shell

Homologous Lines

SYNOW Assumptions I

S = WIc

W =12

[1−

√1− rc

r

2]

v ∝ r

so photons come into resonance with lines i.e., y = 0,1 and weassume τ ≡ τ(r) which is specified to be for example a power-law orexponential in SYNOW. Then the flux for z ≥ 0 is

F (z) = 2π∫ rc

0Icp dp +

∫ ∞rc

S(r)[1− e−τ(r)

]p dp

= πr2c Ic + 2π

∫ ∞rc

S(r)[1− e−τ(r)

]p dp

SYNOW Assumptions II

Then the flux for z < 0 is

F (z) = 2π∫ p0

0Icp dp + 2π

∫ ∞p0

S(r)[1− e−τ(r)

]p dp

+2π∫ rc

p0

Ice−τ(r)p dp

= πp20Ic +

∫ ∞rc

S(r)[1− e−τ(r)

]p dp

p0 =

√r2c − z2 for −rc < z < 0

0 for z ≤ −rc