We now turn to the solutions of those radiative transfer problems at the end of the last solar radiative transfer instalment. We look at the first two to start with:
1) A star has a gray atmosphere for which the Eddington approximation:
T^4 = ¾ T_e^4(τ + 2/3)
is valid, where T_e denotes the effective temperature. Use this approximation to obtain the fraction of outward intensity escaping from the star’s surface.
We begin by assuming: a) radiative equilibrium, and b) "LTE" or local thermodynamic equilibrium.
First, find the intensity radially coming out in all directions, from:
I (Θ=0)= INT (0 to oo) B(T) exp(-τ) dτ = σ /π INT (0 to oo) (T)^4 exp(-τ) dτ
I (Θ=0)= 3σ /4π T_e^4 INT(0 to oo) (τ + 2/3) exp(-τ) dτ = 5σ /4π (T_e^4)
Now, the part of the above which originates above the critical layer- call it layer τ_o is:
I (τ < τ_o) = 3σ /4π T_e^4 INT(0 to oo) (τ + 2/3) exp(-τ) dτ
I (τ < τ_o) = 3σ /4π T_e^4 [5/3 - (τ_o + 2/3) exp(-τ_o)
The fraction originating above any given optical depth is then:
f(I(τ < τ_o) ) = 1 - (0.6τ_o + 1) exp (τ_o)
The fraction escaping from the star's surface is then that for which τ_o = 0, so:
1 - (0.6τ_o + 1) exp (τ_o) = 1 - (0.6 (0) + 1) exp (0) = 1 - (0 + 1) (1) = 1 -1 = 0
2) The star Suhail has a (B – V) color index of +1.7. Use this to obtain the net flux (H) passing through the Suhail’s surface. How might you estimate the intensity I from this and the mean intensity J?
First, we see from the previous link,
that a (B – V) color index of +1.7 corresponds to log T_e = 3.52
Then: antilog (3.52) = 3, 300 K
which is the effective temp.
Then the net flux H passing through Suhail's surface is obtained from:
σT_o^ 4/ π = 2H
where the boundary temp. T_o = T_eff /1.189 = 3300 K/ 1.189 = 2770 K
so: H = σT_o ^4/ 2 π = [(5.67 x 10^-8 W m^-2 K^-4)(2770K)^4/2 π
H = 5.3 x 10^5 W
The intensity I and the mean intensity J can be estimate from the above using (see, e.g. http://brane-space.blogspot.com/2012/04/simple-solar-radiative-transfer-1.html:
a) J = 2H/ 3 = 2(5.3 x 10^5 W) / 3 = 3.5 x 10^5 W
b) I1 (τ) = 3Hτ (where I1 is the forward flux)
and setting τ = 2/3:
I1 (τ) = 3H(2/3) = 2H = 2(5.3 x 10^5 W) = 1.1 x 10^6 W
Remaining solutions next time!