Proper distance and embedding diagrams? I'm trying to understand proper distance equation in Schwarzschild spacetime.
$d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}$.
I'm sure I'm missing something really obvious here, but how do I use this to find the coordinate distance $r$
  for a particular proper distance $\sigma$
 . For example, if I found the proper circumference of circle going round the Sun that roughly coincides with the Earth's orbit. Then I move radially inwards one proper mile, how would I then find the circumference of the circle I now find myself on. This example is also in the context of trying to understand the spacing of concentric circles in embedding diagrams.
Thank you
 A: You have the Schwarzschild metric
$ds^2=(1-R_s/r)c^2dt^2-(1-R_s/r)^{-1}dr^2-r^2(d\theta^2+sin^2\theta d\phi^2)$
For an equatorial orbit, put $\theta=\pi/2; d\theta=0$.
The proper distance between two events is defined as the integral of $ds$ along a spacelike path between them.  I'm guessing the two events we're interested in are the (identical) start and end point of an elliptical path, where the path is traversed in zero coordinate time (so $dt=0$).  The only things that vary on the path are $r$ and $\phi$.  This path will be a function like
$r=a(1-e^2)/(1-e.cos\phi)$
where a and e are a couple of fixed parameters.
Since spacelike separations are negative in this signature, we apply an extra minus sign and get
$d\sigma = \sqrt{(1-R_s/r)^{-1}dr^2-r^2d\phi^2}$
Using the formula for the ellipse, you can get $dr$ in terms of $d\phi$ and take the square root, leaving just a $d\phi$ on the RHS.  Integrating from 0 to 2$\pi$ should then give you your proper distance.
When you want to compare this with the value "one mile in", you need to decide what your measure of the radius of the elliptical path is (maybe average r), and adjust the parameters accordingly.
Edit: I just noticed you're using circular orbits.  That simplifies it a bit !  In fact, doesn't it make it trivial, since $dr=0$ on your orbit ?
