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I've been reading Wald. The relevant pages are 217 and 46. Page 46 is where he defines the "orthogonal displacement vector". It seems impossible to understand any of this without a thorough understanding of that, so I'll start there. Consider a one-parameter family of timelike curves [itex]\{\gamma_s\}[/itex], parametrized by proper time. I'll write the tangent vector of [itex]t\mapsto\gamma_s(t)[/itex] at t as [itex]\dot\gamma_s(t)[/itex]. I'll write the curve [itex]s\mapsto\gamma_s(t)[/itex] as [itex]\gamma(t)[/itex], and its tangent vector at s as [tex]\dot{\gamma(t)}(s)[/tex]. (I hope this notation isn't too confusing). We define two vector fields T and X by

[tex]T(\gamma_s(t))=\dot\gamma_s(t)[/tex]

[tex]X(\gamma_s(t))=\dot{\gamma(t)}(s)[/tex]

There's an extremely important step in all of this that Wald mentions without proof: It's possible to reparametrize each [itex]\gamma_s[/itex], without destroying the "parametrized by proper time" property, so that X is orthogonal to T at all points that the congruence passes through. If someone wants to see the details, I can type them up.

Why is X a measure of the "distance" to nearby curves in the congruence? The best answer I've been able to come up with uses a coordinate system x:

[tex]x^i(\gamma_{s+r}(t))=x^i(\gamma_s(t))+r\frac{d}{dr}\bigg|_0 x^i(\gamma_{s+r}(t))+\mathcal O(r^2)=\cdots=x^i(\gamma_s(t))+rX^i(\gamma_s(t))+\mathcal O(r^2)[/tex]

[tex]X^i(\gamma_s(t))=\lim_{r\rightarrow 0}\frac{x^i(\gamma_{r+s}(t))+x^i(\gamma_s(t))}{r}[/tex]

So the ith component of X in the coordinate system x is the limit of "coordinate change"/"parameter change" in the direction of X as the parameter change goes to zero.

This result is all we need from page 46. Jump to page 217. I will continue to call the timelike vector field T and the displacement vector X. I will also continue to consider a one-parameter family of curves, because the notation is already complicated enough to be annoying. We have [itex]\mathcal L_T X=0[/itex], because that Lie derivative is equal to the commutator [T,X], which is zero because T and X are the basis vectors associated with the coordinate system [itex]\gamma_s(t)\mapsto(s,t)[/itex].

A crucial step that I haven't done yet: Prove that this really is a coordinate system, given that T is a smooth vector field with integral curves [itex]\gamma_s[/itex], reparametrized to make X orthogonal to T.

It's not hard to show that [T,X]=0 implies [itex]\nabla_X T=\nabla_T X[/itex]. Wald's definition of the tensor B is equivalent to

[tex]B(Y,Z)=g(\nabla_Z T,Y)[/tex]

for all Y,Z. We can then define [tex]B^\sharp=B^i{}_j\partial_i\otimes dx^j[/tex], and use [itex]\nabla_X T=\nabla_T X[/itex] to show that [tex]B^\sharp(\cdot,X)=\nabla_T X[/tex]. This is why [tex]B^a{}_b X^b=B^\sharp(\cdot,X)[/tex] "measures the failure of X to be parallel transported". Note that to get this result, we used[itex]\nabla_X T=\nabla_T X[/itex], which is implied by [X,T]=0.

After that, I don't understand much. I understand that [itex]h^a{}_b=\delta^a_b+T^aT_b[/itex] is a projection operator, but I don't understand why we're looking for a projection operator at all. When I look at the definitions of the tensors that are defined at the end, I see that they don't involve the orthogonal displacement vector X in any way. I have no idea how they can have anything to do with things like rigidity without it. Now I'm also wondering if it was a waste of time to work through the details of the orthogonal displacement vector. It doesn't seem to have anything to do with anything at the end. I really don't get any of this.

I'll give it another shot within the next few days, but if I don't start to get it soon, I'm just going to abandon it. This is taking too much time.