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RestlessRiver

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**"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 :

7a

^{2}r

^{4}sin

^{4}θ+7a

^{4}r

^{2}sin

^{4}θ-8a

^{2}r

^{2}sin

^{4}θ-16Ma

^{2}r

^{3}sin

^{4}θ+16M

^{2}a

^{2}r

^{2}sin

^{4}θ-2Ma

^{4}rsin

^{4}θ+2a

^{2}r

^{4}sin

^{2}θ+a

^{4}r

^{2}sin

^{2}θ+r

^{6}sin

^{2}θ+2Ma

^{4}rsin

^{2}θ-4M

^{2}a

^{2}r

^{2}sin

^{2}θ-2Mr

^{2}sin

^{2}θ

all devided by r

^{2}+ a

^{2}- 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 (x

^{2}+y

^{2}= 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 =)

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