The (uncharged) Kerr metric for a black hole of mass $M$ and angular momentum $Ma$ takes the form

$$ds^{2} = \Sigma\Big(\frac{dr^{2}}{\Delta} + d\theta^{2}\Big) + (r^{2} + a^{2})\text{sin}^{2}\theta d\phi^{2} + \frac{2Mr}{\Sigma}\Big(a\text{sin}^{2}\theta d\phi - dt\Big)^{2} - dt^{2}$$

where

$$ \Delta = r^{2} - 2Mr + a^{2} \text{ and } \Sigma = r^{2} = a^{2}cos^{2}\theta .$$

In my notes we introduce the coordinates $\chi$ and $r_{*}$ such that

$$dr_{*} = \frac{r^{2} + a^{2}}{\Delta} dr \text{ and } d\chi = d\phi + \frac{a}{\Delta}dr$$

I was just wondering if I could get some help understanding these coordinates. Intuitively I imagine the submanifold of constant $\chi$ to be some sort of spacelike surface that spirals inwards and contains all null geodesics starting at some initial fixed $\phi$. I assume then that the transformation to $r_{*}$ stretches this "sheet" so that the null geodesics that reside in it are "straight." Is this intuition correct? If so, do timelike geodesics starting at the same $\phi$ stay at a fixed $\chi$ during the entire trajectory?

(This is a cross post from Physics Stack Exchange)