3
$\begingroup$

I have been reading about the Kerr metric using various sources (Wald's textbook, Visser's The Kerr spacetime: A brief introduction etc.). I could not understand exactly why the singularity structure of the Kerr metric is supposed to be in the form of a ring. Is there an "algebraic derivation" of this in an appropriately chosen coordinate system?

$\endgroup$
2
$\begingroup$

I will not reproduce the derivation, but Ellis and Hawking sketch this in their book "Large Scale Structure of Space-Time" on page 162. One first transforms from the time-radial-angular coordinates to $x,~y,~z, t$. The metric assumes a certain form so the condition $$ r^4~-~r^2(x^2~+~y^2~+~z^2~-~a^2)~-~a^2z^2~=~0 $$ obtains. Here $a~=~J/mc$. Then for $z~=~0$ this reduces to $$ r^2~-~(x^2~+~y^2~-~a^2)~=~0. $$ This defines a family of ellipsoids.

The principal curvature components is $$ R_{rtrt}~=~-\frac{2m}{r^3}, $$ and these hyperboloids give $r~=~\pm\sqrt{x^2~+~y^2~-~a^2}$ and this singularity is seen to be the ring $x^2~+~y^2~=~a^2$. The disk bounded by this ring with $x^2~+~y^2~<~a^2$ is not singular. The two signs for the radius represent to Riemann sheets, where the disk is a form of branch cut. An observer that crosses the disk enters different regions with alternating signs for $r$, in a double covering analytic continuation.

It is not clear whether this singularity actually exists or not. An observer that crosses the inner horizon $r_-$ also crosses the horizon at $\infty$ or ${\cal I}^+$. There will then be a large occurrence of quanta at this region, which might be a Cauchy singularity. So the split $r_-$ might in fact be a type of singularity. The region III with the ring singularity might be a sort of mathematical fiction.

enter image description here

$\endgroup$
  • $\begingroup$ Thanks for the answer. I looked at the page that you referred to. Please correct me if I am wrong. In Kerr, Doran or any other spheriodal coordinates the singularity appears to be the point , $r=0$ . But the use of Kerr-Schild coordinates maps the point singularity to a ring defined by x^2+y^2=a^2 in cartesian coordinates. Am I right ? $\endgroup$ – user91411 Aug 13 '16 at 1:21
  • $\begingroup$ In this metric $r~=~0$ is a condition, which corresponds to the ring. In a sense the spacetime is all "twisted up," and orbiting the singularity results in closed timelike curves and other oddities. $\endgroup$ – Lawrence B. Crowell Aug 14 '16 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.