Stellar atmospheres
1 The Eddington approximation and the two-stream model
A stellar atmosphere is a thin region with total depth much less than the stellar radius.
Thus we usually model it as a -at slab. The opical depth is measured vertically downward
from the surface.
Letzs start by looking at a |grey} atmosphere in which ® has no frequency dependence.
Then we can drop the subscript º :
For a ray travelling at angle µ to the normal,
dI
= ¡®I + j
ds
where ds = ¡dz= cos µ and d¿ = ®dz: Thus
dI
cos µ
= I ¡S
d¿
We now de®ne the moments:
Z
1
mean intensity: J =
Id4¼
Z
1
Flux: H =
I¹d4¼
Z
1
radiation pressure K =
I¹2 d4¼
Now we take moments of the radiative transfer equation. First, integrate the equation over
solid angle to obtain:
dH
= J ¡S
d¿
Next multiply by ¹ before integrating, to get
dK
=H
d¿
Now in an atmosphere we do not expect any energy generation, and thus H =constant ( and
J = S): . Thus from the last equation,
K = H¿ + constant
1
(1)
To make further progress, we need to assume something about the angular distribution of
radiation. The simplest model is the two-stream approximation, in which
½
I1 if 0 < µ < ¼=2 (outgoing radiation)
I =
I2 if ¼=2 < µ < ¼ (incoming radiation)
Then the moments are
J
H
K
=
1
2
=
1
2
=
1
2
½Z
½Z
½Z
1
I1 d¹ +
0
1
Z
0
I2 d¹
¡1
I1 ¹d¹ +
0
1
2
Z
I1 ¹ d¹ +
0
0
=
I2 ¹d¹
¡1
0
Z
¾
2
¾
I2 ¹ d¹
¡1
1
(I1 + I2 )
2
=
¾
1
(I1 ¡ I2 )
4
=
1
J
3
The third relation is not unexpected! This is the Eddington approximation.
At the surface, we have
I2 = 0 at ¿ = 0
Thus, at ¿ = 0; J = 2H = 3K: Thus the constant in equation (1) equals 2H=3: Thus
K = H (¿ + 2=3)
and
J = H (3¿ + 2)
With J = ¾T 4 , this becomes a relation for temperature versus depth:
µ
¶
T e4
3
4
T =
1+ ¿
2
2
(2)
where the effective temperature T e is de®ned to be the temperature of a surface emitting an
amount of radiation equal to that from the star. The effective temperature equals the actual
temperature at an optical depth ¿ = 23 : Thus we may interpret "the surface" of the star to be
at ¿ = 2=3:
See R&L p 42-45 for an alternative treatment of this topic.
2 Limb darkening
Using the formal solution of the transfer equation, with ¿ ! 1 at the bottom of the
atmosphere, and assuming LTE (S º = Bº (T ) ), we have:
Z 1
Iv (µ; 0) =
Bº (T (¿ )) e ¡¿ º sec µ d¿ º sec µ
0
Now we expand B º in a Taylor series about the optical depth ¿ ¤ (unspeci®ed for the
moment):
¯
dB º ¯¯
Bº (T (¿ )) = Bº (T (¿ ¤ )) +
(¿ º ¡ ¿ ¤ ) + ¢ ¢ ¢
d¿ ¯¿ ¤
Putting this into the integral, the ®rst term is
£
¤1
Bº (T (¿ ¤ )) ¡e¡ ¿ º sec µ 0 = Bº (T (¿ ¤ ))
2
In the second term, we integrate by parts:
Z 1
Z 1
(¿ º ¡ ¿ ¤ ) e¡ ¿ º s ec µ d¿ º sec µ = cos µ
(x ¡ x ¤ ) e ¡x dx where x = ¿ sec µ
0
0
·
Z 1
¸
¯1
= cos µ ¡ (x ¡ x¤ ) e¡x ¯0 +
e¡x dx
¯ ¤ 0
£ ¤
¡x ¯1
= cos µ ¡¿ sec µ ¡ e 0
=
cos µ ¡ ¿ ¤
¯
dBº ¯¯
(cos µ ¡ ¿ ¤ ) + ¢ ¢ ¢
d¿ ¯¿ ¤
If we now choose ¿ ¤ = cos µ; we obtain, correct to 2nd order:
Thus
Iv (µ; 0) = Bº (T (¿ ¤ )) +
Iv (µ; 0) = Bº (T (¿ = cos µ))
Then for µ = 0 (ray from the center of the stellar disk)
while from the limb (µ = ¼=2)
Iº (0; 0) ' Bº (T (¿ = 1))
Iº
³¼
´
; 0 ' Bº (T (¿ = 0))
2
Since temperature increases inward (equation 2), the intensity is less at the limb. This is the
phenomenon of |limb darkening}.
3