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Transverse metric being 2-dimensional in null case

Physics Asked on January 26, 2021

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?
    In my understanding in timelike and in null case, both reason 1 and 2 are seems equivalent but it is mentioned that in null case 2 reason is independent of 1st.

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

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