How to derive the Kerr killing vector?

Question

The Kerr metric have two killing vectors:

$$t^{\mu} \equiv (k_{t})^{\mu} = (1,0,0,0)\hspace{5mm} \mathrm{and}\hspace{5mm} \phi^{\mu} \equiv (k_{\phi})^{\mu} = (0,0,0,1). \tag{1}$$

In general, it is possible to construct another killing vector given by,

$$\chi^{\mu} = a t^{\mu} + b \phi^{\mu}. \tag{2}$$

It is well-known that in Kerr spacetime the general killing vector is when: $a=1$ and $b=\Omega =- \frac{g_{t\phi}}{g_{\phi \phi}}|_{H}$.

So, using the fact that a horizon is a null surface, then we could try to find $a$ and $b$ using:

$$\chi_{\mu}\chi^{\mu}=0. \tag{3}$$

but in the end of the day my calculation resulted in:

$$\chi_{\mu}\chi^{\mu}= g_{tt}a^{2}+g_{\phi\phi}b^{2}+2g_{t\phi}ab=0. \tag{4}$$

So, my question is:

Given the Kerr metric $[1]$, and the general linear combination of the Killing vectors $(2)$, how can I prove that $a=1$ and $b=\Omega =- \frac{g_{t\phi}}{g_{\phi \phi}}|_{H}$?(*)

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$[1]$ Kerr geometry: https://en.wikipedia.org/wiki/Kerr_metric

(*) It is possible to just define the well-known vector field $\chi^{\mu} = t^{\mu} + \Omega\phi^{\mu}$. You take it for granted and derive many important things. But at some point in black hole history someone faced the equation $(2)$ and by some kind of argument (a mathematical proof, say; or not) was stablished that the most interesting killing combination is when $a=1$ and $b=\Omega =- \frac{g_{t\phi}}{g_{\phi \phi}}|_{H}$, and that is the reasoning that I'm not grasping (in both physical and mathematical ways).

Answer 1

I believe you were almost there :)

If you consider that without loss of generality you can set a=1 (or, in alternative, you wanna ask yourself what is b going to look like with a set to 1) you then have a quadratic equation to solve for b. You will soon find out that both solutions coincide and they happen to be exactly $\Omega$:

$$ g_{\phi\phi}b^2+2g_{t\phi}b+g_{tt}=0 $$ then $$b_\pm = \frac{g_{t\phi}\pm\sqrt{g_{t\phi}^2-g_{tt}g_{\phi\phi}}}{g_{\phi\phi}} = \frac{g_{t\phi}\pm\sqrt{\Delta}\text{sin}\theta}{g_{\phi\phi}}$$ where $$\Delta = r^2+\alpha^2-2Mr = (r-r_+)(r-r_-)$$ and $r_+,\,r_-$ are the two horizons and $\alpha=J/M$, angular momentum over mass of the Kerr solution. Being on the horizon, the plus-minus term disappears and we're left with $$b=-\frac{g_{t\phi}}{g_{\phi\phi}}=\Omega_H$$