Covariant derivative of contracted tensor: why is it not obvious In Wald's GR book (1984), he writes on page 221,

In the timelike case, we restricted consideration to deviation vectors $\eta^a$ orthogonal to $\xi^a$ [$\xi^a$ is the normalized vector field of tangents of a smooth congruence of timelike geodesics]. There actually were two independent (though) related reasons for doing so. (1) We have $\xi^a \nabla_a(\xi_b \eta^b) = \xi^a \xi_b \nabla_a \eta^b = \xi_b \mathcal{L}_\xi \eta^b + \eta^a \xi_b \nabla_a \xi^b = 0$ provided $\xi_a\xi^a$ is normalized to be constant. Thus, $\xi_a\eta^a$ is constant along each geodesic, and the behavior of the "nonorthogonal" part of $\eta^a$ is uninteresting. (2) Deviation vectors which differ only by a multiple of $\xi^a$ represent a displacement to the same nearby geodesic. Orthogonality fixes a natural "gauge condition" on $\eta^a$.

(I added the statement in brackets for clarity. The Lie derivative is denoted with $\mathcal{L}$.)
My questions are:

*

*If $\eta^b$ is orthogonal to $\xi^b$, which I presume means $\eta^b \xi_b=0$, why does he go through the work of taking the covariant derivative of this in (1) only to conclude that it's $0$? Wouldn't it be 0 because $\eta^b \xi_b$ is $0$ by assumption?


*Why does he conclude after the calculation in (1) that $\xi_a\eta^a$ is constant along each geodesic, i.e. why does he say it's constant instead of $0$ (which it is by assumption)?
 A: Suppose we have a geodesic congruence.  To describe the relative behavior of nearby geodesics, we "pick out" a one-parameter family of geodesics $\gamma_s(t)$ from it such that for every fixed $s$, $\gamma_s$ is a geodesic with affine parameter $t$, and such that $s$ and $t$ are "good coordinates" on the 2-D submanifold spanned by this family.
We can then define the coordinate vector fields $\xi^a = (\partial/\partial t)^a$ and $\eta^a = (\partial/\partial s)^a$.
The vector fields $\xi^a$ and $\eta^a$ so constructed will not necessarily satisfy $\xi^a \xi_a = -1$, $\xi^a \eta_a = 0$, or even $\xi^a \eta_a = $ constant everywhere on the 2-D submanifold.  It is true that $\xi^a \xi_a$ will be constant along each geodesic, since $t$ is an affine parameter on each geodesic.  But even then, this quantity could be different for different geodesics (i.e., it could vary with $s$.)
However, by using the reparametrization freedom of the geodesics, we can find an equivalent $\xi^a$ & $\eta^a$ on the 2-D manifold spanned by $\gamma_s(t)$ such that $\xi^a \xi_a = -1$ for all geodesics, and that $\xi^a \eta_a = 0$.  The construction which allows one to do so is laid out in more detail in Section 3.3 (p. 46).  This is a very convenient choice to make, but it's still a choice.
The direct answers to your questions are therefore basically "he is not assuming that $\xi^a \eta_a = 0$ at this point, he's showing that we can choose it to be so."  He then rehashes the reasons why, even though $\xi^a \eta_a \neq 0$ for a general one-parameter family of geodesics, we have the freedom to choose $\xi^a \eta_a = 0$.  In particular, all that he has shown directly via the calculation in (1) is that $\xi^a \nabla_a(\xi_b \eta^b) = 0$, which means that the quantity $\xi_b \eta^b$ is constant along geodesics.  He still has in the back of his mind the idea of a general congruence, and is not yet assuming that we have chosen a family of geodesics for which $\xi_a \eta^a = 0$.
