# Metric diameter of a ring singularity

In the Kerr metric the ring singularity is located at the coordinate radius $$r=0$$, which corresponds to a ring with the cartesian radius $$R=a$$.

So the center of the ring singularity in cartesian coordinates is at $$r=-a, \ \theta=\pi/2$$.

But the center in cartesian coordinates is also at $$r=0, \ \theta=0$$ (at $$r=0$$ all $$\theta$$ are in the equatorial plane, at least in Boyer Lindquist and also in Kerr Schild coordinates).

To calculate the physical diameter to see how much fits through the ring (in one example it is a tiger, in another one Alice & Bob), would I integrate

$$(1) \ \ \ \ \theta=\pi/2 , \ \ d =2 \int_{-a}^0 \sqrt{|g_{rr}|} \ \ {\rm d}r = 2 \sqrt{(2-a) a}+4 \arcsin \left(\sqrt{\frac{a}{2}}\right)$$

in the equatorial plane, or is it rather

$$(2) \ \ \ \ r=0 , \ \ d =\int_{-\pi/2}^{\pi/2} \sqrt{|g_{\theta \theta}|} \ \ {\rm d}\theta = 2a$$

since that should also cover the distance from one side of the ring to the opposite.

Approach $$(2)$$ gives exactly the diameter in cartesian coordinates, but I don't know if that's supposed to be so, or only a coincidence, since otherwise the metric distance is not nescessarily the same as the coordinate or cartesian distance.

So which one is it, $$(1)$$ or $$(2)$$? Or is it done in a completely different way?

The coordinates I used are Kerr Schild coordinates, which should cover the inside with the relevant components

$$g_{r r}=-\frac{2 r}{a^2 \cos ^2 \theta +r^2}-1 \ , \ \ g_{\theta \theta }= -r^2 - a^2 \cos^2 \theta$$

I guess it is approach $$(2)$$ since no one can enter the ring singularity from the equatorial plane, but I'd like to hear a 2nd opinion on that

• – user4552
Apr 9 '19 at 0:48
• It seems unlikely to me that you're going to be able to formulate a notion of diameter that makes sense here. Putting aside all questions of the metric's misbehavior at the ring singularity, there is the question of what spacelike path you want to integrate along. For the notion of a diameter to make sense, there would have to be some preferred path. Outside the horizon of a Schwarzschild black hole, we have a preferred stationary observer at any given point, and therefore there is a preferred radial direction that is orthogonal to that observer's world-line. But this doesn't work here.
– user4552
Apr 9 '19 at 23:36
• Therefore I'm using Kerr Schild coordinates (those are like Finkelstein, but for the rotating case), which let me integrate the distance in a free falling frame (the g_tr crossterm is the square of the free fall velocity). I'm pretty sure now that it is method 2 (the g_θθ integral from -90° to +90° along r=0) which gives me the local diameter of an object that just fits through the ring. Apr 10 '19 at 0:15

The ring singularity has coordinates $$r = 0, \theta = \pi/2$$. The radius of the ring is described by $$r = 0, \theta =[0, \pi/2]$$. That means we can integrate along a path defined by those coordinates with $$dt = dr = d\phi = 0$$.
$$ds^2 = g_{\theta \theta} d\theta^2$$
$$g_{\theta \theta} = r^2 + a^2 \cos^2 \theta = a^2 \cos^2 \theta, (r = 0)$$
$$R_{ring} = a \int^{\pi/2}_0 \cos \theta d\theta = a$$
Note that $$g_{\theta \theta}$$ is positive, so the path is spacelike.
Instead the approach (1), that is the integration over the radial coordinate $$r$$ is not viable as $$r$$ is null, i.e. constant along the path meaning $$dr = 0$$.