Geometry of Null hypersurfaces

Question

In Wald section 9.2 page 221 he says that

We turn our attention; now , to null geodesic congruences. Again, we parameterize the geodesics by an affine parameter $\lambda$, but , unlike the timelike case, we now have no natural way of normalizing the tangent field $K^\alpha$ and thereby adjusting the scale of $\lambda$ on different geodesics . In the timelike case, we restricted consideration to deviation vectors $\eta^\alpha$ orthogonal to $\xi^\alpha$ . There actually were two independent (though related) reasons for doing so. (1) We have $\xi^\alpha \nabla_\alpha (\xi_\beta \eta^\beta)=0$ provided $\xi^\alpha \xi_\alpha $ is normalized to be constant. Thus, $\xi_\alpha \eta^\alpha$ is constant along each geodesic, and the behavior of the "non orthogonal" partof $\eta^\alpha$ is uninteresting. (2) Deviation vectors which differ only by a multiple of $\xi^\alpha$ represent a displacement to the same nearby geodesic. Orthogonality fixes a natural "gauge condition" on $\eta^\alpha$.

In the case of a null geodesic congruence, the above reasons for restricting the choice of deviation vector still apply, but now they lead to two independent restrictions. First, for any deviation vector $\eta^\alpha$, we again have $k^\alpha \nabla_\alpha (k_\beta \eta^\beta)=0$, so $k^\alpha \eta_\alpha $ does not vary along each geodesic. This implies that an arbitrary deviation vector $\eta^\alpha$ may be written as the sum of a vector not orthogonal to $k^\alpha$ which is parallelly propagated along the geodesic, plus a vector perpendicular to $k^\alpha$ .(Note, however, that there is no natural, unique way of decomposing $\eta^\alpha$ in this manner.) Thus, the behavior of the "nonorthogonal" part of $\eta^\alpha$ again is uninteresting, and we may restrict consideration to deviation vectors satisfying $\eta^\alpha k_\alpha=0$. Second, deviation vectors which differ only by a multiple of $k^\alpha$ again represent a displacement to the same nearby geodesic . Thus, the physically interesting quantity is really the equivalence class of deviation vectors, where two deviation vectors are considered equivalent if their difference is a multiple of $k^\alpha$.Since $k^\alpha$ is null and thus is orthogonal to itself, this second restriction is independent of the first restriction, and it reduces the physically interesting class of deviation vectors to a two-dimensional subspace.

1. I am not able to understand the second reason in timelike and null case case i.e what does he mean that deviation vectors which differ by multiple of $\xi^\alpha$ in timelike or $k^\alpha$ for null case will represent displacement to the same nearby geodesic?

2. How does in null case this reasoning reduces deviation vectors to 2 dimensional subspace?

Answer 1

The heart of this is that, for a null surface, the metric is no longer an invertible matrix.

This means, that you no longer have a natural relationship between $\gamma_{ab}$ and $\gamma^{ab}$. This statement may seem a bit academic, but it actually matters, because, thanks to the null surface being a boundary between the spacelike surface and the timelike surface, if you work out the tangent space and the cotangent space, you'll find that the tangent space is spanned by (outgoing null vector) x (2-geometry), while the co-tangent space is spanned by (ingoing null vector) x (2-geometry).

Another way to see that this has to be true is because the fundamental definition of a basis one form is that that one-form $\theta_{a}$ must have some vector $v^{a}$ such that $\theta_{a}v^{a} = \pm 1$, which is impossible of both the cotangent and tangent space are spanned by the same vector, raised and lowered by the enveloping 4-metric.

The other thing to realize is that your null metric must have some basis where a whole row and column must be zero, and the rest spacelike. Otherwise, there is some way to construct a timelike direction on your null surface.

Once you have these two realizations worked out, I think the rest of Wald's commentary becomes significantly less mysterious.