Advanced Astrophysics (4) Solutions

1) The specific intensity is defined from:

p( F_o ) = 2 p (I(cos (q))

And for q = p/4, then cos (p/4) =  Ö2/2 and:

I = p( F_o ) / 2 p (Ö2/2) =  p( F_o ) / p (Ö2)

Therefore:

I = (6.3 x 10 ⁷ Jm^-2 s^-1) / p(Ö2)

I » 1.4 x  10 ⁷ Jm^-2 s^-1

 

²⁾  The effective temperature is obtained using:

p( F_o ) = s(T_eff)⁴

So:  T_eff  = [p( F_o ) / s] ^1/4

Where s = 5.67 x 10^-8 W m^-2 K^-4

Then:

 

T_eff  = [6.3 x 10 ⁷ Jm^-2 s^-1/ 5.67 x 10^-8 W m^-2 K^-4] ^1/4

T_eff  »  5800 K

The boundary temperature is found from:

T_eff  = (2)^1/4 T_o   = 1.189 T_o

Or:   T_o  =   T_eff  /1.189  = 5800K/ 1.189  » 4800 K

The boundary temperature differs because of being referenced to a different optical depth. The boundary temperature (T_o) approaches the value of the effective (or surface) temperature when t = 0, but this still exhibits a difference in layers so will not be exactly the same!