Ring -singularity (Determinant of the Kerr metric)

[RestlessRiver]

"Ring"-singularity (Determinant of the Kerr metric)

My problem is as follows:
"Calculate the determinant of the Kerr metric. Locate the plac where it is infinite. (In fact, this gives the "ring"-singularity och the Kerr black hole, which is the only one)

I got the determinant to :

7a²r⁴sin⁴θ+7a⁴r²sin⁴θ-8a²r²sin⁴θ-16Ma²r³sin⁴θ+16M²a²r²sin⁴θ-2Ma⁴rsin⁴θ+2a²r⁴sin²θ+a⁴r²sin²θ+r⁶sin²θ+2Ma⁴rsin²θ-4M²a²r²sin²θ-2Mr²sin²θ

all devided by r² + a² - 2Mr

where
a=angular momentum/M
M=Mass
r= radius (of the "ring")

and I talked to my prefessor and he told me that the answer should be the equation of a ring (x²+y² = constant) in spherical coordinates, I have all this in Boyer-Lindquist coordinates I believe, and according to wikipedia

{x} = \sqrt {r^2 + a^2} \sin\theta\cos\phi
{y} = \sqrt {r^2 + a^2} \sin\theta\sin\phi
{z} = r \cos\theta

(http://en.wikipedia.org/wiki/Boyer-L...st_coordinates )


I don't get it to be an eq of a ring (or circle) .. please help =)

 

[Jhero]

I can understand your confusion and frustration with this problem. The Kerr metric is a complex mathematical equation that describes the spacetime around a rotating black hole. Its determinant is an important quantity that can tell us about the properties of the black hole, including the location of its singularity.

To start, let's define the Kerr metric in Boyer-Lindquist coordinates as:

ds^2 = -\left(1-\frac{2Mr}{\Sigma}\right)dt^2 - \frac{4Mar\sin^2\theta}{\Sigma}dtd\phi + \frac{\Sigma}{\Delta}dr^2 + \Sigma d\theta^2 + \left(r^2+a^2+\frac{2Ma^2r\sin^2\theta}{\Sigma}\right)\sin^2\theta d\phi^2

where M is the mass of the black hole, a is its angular momentum, and \Sigma = r^2 + a^2\cos^2\theta and \Delta = r^2 - 2Mr + a^2. Now, the determinant of this metric is given by:

det(g) = -\Sigma^2\sin^2\theta

This is a negative quantity, which means that the metric is not positive definite and therefore cannot be interpreted as a distance or a measure of length. However, we can still analyze the behavior of this determinant to understand the properties of the Kerr metric.

First, let's look at the behavior of the determinant as r approaches the event horizon, which is defined by \Delta = 0. This gives us:

det(g) = -\frac{4Ma^2r\sin^4\theta}{\Sigma^2}

As r gets closer to the event horizon, this determinant approaches infinity, indicating a singularity at the horizon. This is the "ring"-singularity that you mentioned in your problem.

Next, let's consider the behavior of the determinant as r approaches infinity. In this case, we have:

det(g) = -r^2\sin^2\theta

This determinant goes to zero as r goes to infinity, indicating that the metric becomes flat in this limit. This makes sense, as the spacetime around a black hole should approach flat spacetime at large distances.

Now, let's examine the behavior of the determinant as \theta approaches 0 or \pi (