Solutions to Radiative Transfer Problems (Part 1)

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.

Solution:

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?

Solution:

First, we see from the previous link,

http://brane-space.blogspot.com/2011/09/tackling-intermediate-astronomy_22.html

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

and

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!