Radiative Transfer - Emitting Ring

[astrop]

Homework Statement

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.

Homework Equations

Radiative transfer:
$$dE = I_{\nu} dA cos\theta d\nu d\omega dt$$

Emission:
$$\frac{dI_{\nu}}{ds}cos\theta = j_{\nu} - k_{\nu}I_{\nu}$$

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?

 

[Jhero]

Or do I need to integrate over the entire ring?

First, let's define some variables:

D = distance to the ring
r_{in} = inner radius of the ring
r_{out} = outer radius of the ring
\theta = angle between the normal to the ground and the line going to the center of the ring
dA = area element of the ring
ds = path length along the ring
d\nu = frequency element
d\omega = solid angle element
dt = time element

Next, let's consider the radiative transfer equation:

dE = I_{\nu} dA cos\theta d\nu d\omega dt

This equation describes the change in energy (dE) as it travels through the ring. We can rewrite this equation in terms of the surface brightness (I_{\nu}):

dE = I_{\nu} dA cos\theta d\nu d\omega dt
I_{\nu} = \frac{dE}{dA cos\theta d\nu d\omega dt}

Next, we can use the emission equation to relate the surface brightness to the emission coefficient (j_{\nu}):

\frac{dI_{\nu}}{ds}cos\theta = j_{\nu}

Substituting this into our previous equation, we get:

I_{\nu} = \frac{j_{\nu} ds}{cos\theta d\nu d\omega dt}

Now, we need to determine the value of ds. This is the path length along the ring, which can be calculated using basic geometry:

ds = r d\theta

Where r is the radius of the ring at a given angle \theta. In this case, we can define r as a function of \theta:

r = \frac{r_{out}+r_{in}}{2} + \frac{r_{out}-r_{in}}{2} cos\theta

Substituting this into our equation for ds, we get:

ds = (\frac{r_{out}+r_{in}}{2} + \frac{r_{out}-r_{in}}{2} cos\theta) d\theta

Now, we can plug this back into our equation for the surface brightness:

I_{\nu} = \frac{j_{\nu} (\frac{r_{out}+r_{in}}{2