Seeing in shell reference frame in general relativity I'm trying to understand how an observer in a shell reference frame sees an object (star) near a black hole. I'm specifically trying to understand the equation:
$$\tan \theta_{shell} = \left(1-\frac{2M}{r}\right)^{1/2} \tan \theta_{schw}$$
From section 18.6 in chapter 18 in here:
http://www.opensourcephysics.org/document/ServeFile.cfm?ID=7375&DocID=527
My two question are:


*

*What angle does $\theta$ represent? (Some kind picture would be especially nice :) )

*In what reference frame is $\theta_{schw}$ measured?

 A: This shows the situation as viewed by the Schwarzschild observer i.e. an observer far from the black hole:

(Note that the angle $\theta$ is not connected to the Schwarzschild $\theta$ coordinate.)
The angle $\theta$ is (obviously) given by:
$$ \tan\theta = \frac{b}{a} = \frac{b}{r_2 - r_1} $$
But we've calculated the angle using the Schwarzschild $r$ coordinate, and if you're a shell observer sitting at $r_1$, and you align your ruler along the radial direction you will measure a different distance. You will get a larger value for $a$, and the larger value for $a$ will make $\tan\theta$ smaller.
The distance measured with a ruler (the proper distance) is calculated using the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{2M}{r}\right)} + r^2 (d\theta^2 + sin^2\theta d\phi^2) $$
To calculate the proper distance from $r_1$ to $r_2$ we integrate along the radial direction at fixed time so $dt = d\theta = d\phi = 0$, and the metric simplifies to:
$$ ds^2 = \frac{dr^2}{1-\frac{2M}{r}} $$
And the proper distance is:
$$ s = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{2M}{r}}} $$
The integral is a bit messy, so we assume that $r_2 - r_1 \ll r$, in which case we can take $r$ as approximately constant to get:
$$ s \approx \frac{r_2 - r_1}{\sqrt{1-\frac{2M}{r}}} $$
And using the proper distance in our expression for the angle $\theta$ gives us:
$$ \tan\theta_{\text{shell}} = \frac{b}{s} \approx \frac{b}{r_2 - r_1} \sqrt{1-\frac{2M}{r}} $$
or:
$$ \tan\theta_{\text{shell}} \approx \tan\theta \sqrt{1-\frac{2M}{r}} $$
Which is where we came in.
You don't say if you're familiar with GR. If you aren't I should point out that the calculation uses the usual convention that $G = c = 1$. If you're going to actually do calculations we need to put $G$ and $c$ back and the expression looks like:
$$ \tan\theta_{\text{shell}} \approx \tan\theta \sqrt{1-\frac{2GM}{c^2r}} $$
