# 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:

1. 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?

2. 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)?

## 1 Answer

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 $$\xi_a \eta^a = 0$$.