I seem to be quite confused with Killing vector fields, Killing horizons and null horizons, especially in the context of the Kerr metric. I have a couple of questions regarding this, and the text I am using is Gravitation - Foundations and Frontiers by T. Padmanabhan.
In discussing the event horizon, he looks at two surfaces given by $g_{tt} = 0$ and $g^{rr} = 0$, where $g_{ab}$ is the Kerr Metric. Considering $g^{rr} = 0$ he says that "the normal to a general $r$ = constant surface will be proportional to $\partial_a r$. Thus, if the normal to the surface must become null, then we must have: $g^{ab}\partial_a r\partial_b r = 0 = g^{rr}$.
- I do not understand why the normal to a $r =$ constant surface must be proportional to $\partial_a r$. I have a very simple analogy (it's simplicity is why I think I might be wrong). In 3D, if we want to describe an $r =$ constant surface, then we get a sphere. But the normal to the sphere is along the radial direction. Not perpendicular to it. However, in the Kerr metric the coordinate $r$ does not refer to the standard spherical polar radius. So I am unable to visualise/imagine how the normal to an $r=$constant surface would be, or in which direction it would point in.
Secondly, to understand surfaces given by $g_{tt} = 0$, he considers observers whose four velocities are along the timelike Killing vector $\xi$ and hence $u = R\xi$, with $R = -\xi \cdot \xi$. Now, for any photon to be propagating in this region with a four momentum $p$, the "conserved energy of the photon" is given to be $E = -p \cdot \xi $. I have a few questions with this discussion.
I understand that given an arbitrary Killing vector $K$, then if $v^a$ is a tangent vector along the geodesic, then the quantity $K_av^a$ is conserved along the geodesic. In this case however, there is no mention of a geodesic. So my understanding as to the origin of this 'conservation' is as follows. If, the Killing vector is null, then the vector $\xi$ satisfies the geodesic equation (as given here). Therefore, in this case, for $\xi = \partial_t$ to be null, we must demand that $g_{tt} = 0$. Hence, on this surface alone, the Killing vectors also generate a null geodesic along which the energy as defined above is conserved. Is my reasoning correct?
As an aside, why do we consider $g_{tt} = 0$ and $g^{rr} = 0$ as the surfaces of interest? At least for $g_{tt} = 0$ I am able to see some motivation via the null Killing vectors (if I understand correctly), but what is the reason for interest in $g^{rr}$ (as opposed to $g_{rr}$ even.
