astrop
Jan26-10, 09:17 PM
1. The problem statement, all variables and given/known data
Suppose there is an optically thin emitting ring with inner radius r_{in} and outer radius r_{out} seen edge on. Compute the relative surface brightness of the ring as a function of its projected position in the sky.
2. Relevant equations
Radiative transfer:
dE = I_{\nu} dA cos\theta d\nu d\omega dt
Emission:
\frac{dI_{\nu}}{ds}cos\theta = j_{\nu} - k_{\nu}I_{\nu}
3. The attempt at a solution
I'm assuming that the center of the ring is at a distance D>>r_{out}. The angle \theta is the angle between the normal to the ground and the line going to the center of the ring. I think I can drop the extinction coefficient so that:
\frac{dI_{\nu}}{ds}cos\theta = j_{\nu}
I'm not entirely sure how exactly to proceed. Do I just need to figure out what ds is? This would be the area of the ring covered by some angle from the center of the ring?
Suppose there is an optically thin emitting ring with inner radius r_{in} and outer radius r_{out} seen edge on. Compute the relative surface brightness of the ring as a function of its projected position in the sky.
2. Relevant equations
Radiative transfer:
dE = I_{\nu} dA cos\theta d\nu d\omega dt
Emission:
\frac{dI_{\nu}}{ds}cos\theta = j_{\nu} - k_{\nu}I_{\nu}
3. The attempt at a solution
I'm assuming that the center of the ring is at a distance D>>r_{out}. The angle \theta is the angle between the normal to the ground and the line going to the center of the ring. I think I can drop the extinction coefficient so that:
\frac{dI_{\nu}}{ds}cos\theta = j_{\nu}
I'm not entirely sure how exactly to proceed. Do I just need to figure out what ds is? This would be the area of the ring covered by some angle from the center of the ring?