Ring -singularity (Determinant of the Kerr metric)

"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 :


all devided by r2 + a2 - 2Mr

a=angular momentum/M
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 [Broken])

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

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving