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

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1 Answer

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 ?

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– Peter4075 Dec 21 '11 at 9:31
– Peter4075 Dec 21 '11 at 9:37
thought I should point out that at my level nothing is trivial. – Peter4075 Dec 21 '11 at 9:45
Ah, OK, looking at that reference, I think the point he was trying to make was that if, instead of the coordinate radius r, you use the proper radius defined as $\sigma$ in the formula in your question, then the relation between circumference and radius is no longer $C=2\pi. radius$.  Proper radius is the distance between two events for which $t, \theta and \phi$ are the same. – twistor59 Dec 21 '11 at 10:31
Thanks. I quite like that picture of the two concentric circles because I can then easily see how it builds into the kind of upside down witch's hat shape of an embedding diagram. But how does he calculate the circumference of the inner circle to be 0.99999999x2pi miles less than the outer circle? Does it involve some nasty integral of the proper distance equation I gave in my question? My definition of "nasty integral" is pretty all encompassing. – Peter4075 Dec 21 '11 at 11:00
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