# Kerr throat solution derivative

I'm going through this article, since I'll need a part of it for my thesis. And I am trying to derive the Kerr throat solution, from which I should be able, with the change of coordinates get to near-horizon extreme Kerr metric.

I am following the article, and so far I got the $dt^2$, $dr^2$ and $d\theta^2$ part, exactly like in the article, but when I try to get the $\left(d\phi+\frac{r^2}{r_0^2}dt\right)^2$ part, I get stuck.

First I tried by putting already extremal expression for $\omega$, but then I get the correct result without(!) $dt$ part :\ All the mixed terms ($dtd\phi$) and $dt^2$ parts cancel each other out. So I figured, ok, maybe I need to put everything in, before I let $\lambda\to 0$, and see if things will cancel each other out, but in that case my expression becomes infinite! (I either have extra $1/\lambda$ or $1/\lambda^2$ term that don't cancel).

What am I doing wrong here? Is there something I'm missing?

Using : $a = M$ (extremal limit), $\tilde r = M + \lambda r$ $(2.5)$, and the definition of $\omega$ in $(2.3)$, we first get, with some easy algebra, $\omega$ at the first order in $\lambda$:

$$\omega = \frac{1}{2M} (1 - \frac{\lambda r}{M}) + \lambda~ O(\lambda)$$ where $O(\lambda) \rightarrow 0$ when $\lambda \rightarrow 0$

Then we have, at first order in $\lambda$ :

$$(d \tilde \phi - \omega d \tilde t) = (d \phi + \frac{1}{2M \lambda} dt - \frac{1}{2M} (1 - \frac{\lambda r}{M}) \frac{dt}{\lambda}) - O(\lambda) dt$$

When $\lambda \rightarrow 0$, we get :

$$(d \tilde \phi - \omega d \tilde t) = (d \phi + \frac{r}{2M^2} dt)$$

• I tried to do some 'easy algebra' and I get this: $\frac{1+\frac{r\lambda}{M}}{2M\left(1+\frac{2r\lambda}{M}+\frac{2r^2\lambda^2}{M^2}+\frac{r^3\lambda^3}{M^3}+\frac{r^4\lambda^4}{4M^4}-\frac{r^2\lambda^2\sin^2 \theta}{4M^2}\right)}$, now I'm confused a bit, since I have all this 'extra' lambda terms, I can just say that's $O(\lambda)$? Also I don't have - in the first term :\ Commented Jul 17, 2013 at 8:32
• The $\lambda^2,\lambda^3,\lambda^4,$ terms could be written $\lambda ~ O(\lambda)$. Then move the denominator $\lambda$ term, in the numerator, and you got it - because $\frac{1}{1 + 2 (r/M) ~\lambda + \lambda~ O(\lambda)} = 1 - 2 (r/M) ~ \lambda~ + \lambda O'(\lambda)$ - and because $(1 + (r/M)~ \lambda) (1 - 2 (r/M) ~ \lambda~ + \lambda O'(\lambda)) = 1 - (r/M) ~ \lambda~ + \lambda O''(\lambda)$ Commented Jul 17, 2013 at 8:42
• Thanks! I was puzzled how you moved the $\lambda$ in the numerator, and then realized it's just expansion :D Commented Jul 17, 2013 at 13:40