realistic More line profiles, using Eddington Approximation for T structure Te :=
5800 K
4
model the solar spectrum
T ( ) := Te
3 2 + 4 3
Eddington Approximation for T structure
1.5 .10
4
T
(0) = 4877 K
T
2 = 5800 K 3
T 0 i
( )
1 .10
4
T
(1) = 6133 K
2 Gaussian line profile
5000
0
0
i
5
10
0 line ( , 0 , 0) := 0 e
line 0
is the ratio of line to continuum optical depth is the optical depth in the line down to the continuum optical depth = 1 level. It is proportional to the absorber's column density above the continuum optical depth = 1 level
I ( ) = B T =
( (
)
1) 1 1
Intensity = source function at optical depth = 1 in general, we see to total optical depth, line+ continuum = 1 continuum optical depth when total optical depth = 1
= c + c line = c , 0 , 0 :=
(
1 + line ( , 0 , 0)
I , 0 , 0 := B , T c , 0 , 0
(
)
(
( (
)))
Simplified radiative transfer, we "see" to optical depth = 1.
0 := i :=
500 nm
line near peak of Planck function i := 0
0 .. 100
5 +
14
i
10
4 .10
( ) I( i , 0 , 2) 2 .1014 I( i , 0 , 20)
I i , 0 , .2
3 .10
14
1 .10
14
0 499.7 499.8 499.9
500 500.1 500.2 500.3
i nm
0 := i :=
5000 nm
IR line in Rayleigh-Jeans tail of Planck function i := 0
0 .. 100
5 +
11
i
10
8 .10
( ) I( i , 0 , 2) 4 .1011 I( i , 0 , 20)
I i , 0 , .2
6 .10
11
2 .10
11
0 4999.7 4999.8 4999.9 5000 5000.1 5000.2 5000.3
i nm
0 := i :=
200 nm
UV line in Wien tail i := 0
0 .. 100
5 +
13
i
10
3 .10
( ) I( i , 0 , 2) I( , i 0 , 20)
. I i , 0 , .2 2 10
13
1 .10
13
0 199.7 199.8 199.9
200
i nm
200.1 200.2
----------------------------------------------------------------------------------------------------------------------------The above simulations correspond to theta = 0, i.e. lines from the center of the disk of the sun. How would these lines change for a distant star, where the intensity I is integrated over all theta's to give the flux F? Hint, consider an absorption line from the limb of the sun, i.e. theta = 90 degrees. Hint, what will the flux be at the bottom of an optically thick line be? Hint, what will the flux in the continuum be?
We "see" to optical depth 1 at the center of the disk, but only 2/3 when averaged over the disk. Compare 2/3 to 1 at different wavelengths
B
(500 nm , T (.667)) B ( 500 nm , T ( 1) )
=
0.762
line near peak of Planck fct.
B
(5000 nm , T (.667)) B ( 5000 nm , T ( 1) )
=
=
0.932
IR, Rayleig...