Parallel Transported Orthonormal Basis

Question

The following argument results in a conclusion that I find strange, and makes me suspect there is something wrong with the reasoning.

First, consider a timelike geodesic $\gamma$ with normalized tangent vector $k^a$. Then consider a set $x^a_{(i)}$ of $(d-1)$ orthonormal spacelike vectors at some point $p$ on $\gamma$ which are also orthogonal to $k^a$ there:

$g_{ab} x^a_{(i)} x^b_{(j)} = \delta_{ij}$ and $g_{ab} x^a_{(i)} k^b = 0$.

The $d$ vectors $(k^a, x^a_{(i)})$ form an orthonormal frame at $p$, and by parallel transporting the $x^a_{(i)}$ along $\gamma$, they yield an orthonormal frame at every point on $\gamma$.

Now, introduce a Fermi coordinate system in a neighborhood of $\gamma$. That is, identify points on $\gamma$ by their proper time $\tau$, and for each $\tau$, introduce the proper distances $y^i$ along spatial geodesics fired off of $\gamma$ in the directions $x^a_{(i)}$. The Fermi coordinates are $(\tau, y^i)$.

$k^a$ and $x^a_{(i)}$ can be thought of as coordinate basis vector fields in a neighborhood of $\gamma$: $k^a = (\partial/\partial \tau)^a$ and $x^a_{(i)} = (\partial/\partial y^i)^a$. They therefore all commute:

$[k,x_{(i)}]^a = 0 = [x_{(i)},x_{(j)}]^a$.

Since the $x^a_{(i)}$ are parallel transported along $\gamma$, we have $k^a \nabla_a x^b_{(i)} = 0$, and therefore the vanishing commutator implies $x^a_{(i)} \nabla_a k^b = 0$. But we also had $k^a \nabla_a k^b = 0$ from the geodesic equation. Since the $(k^a, x^a_{(i)})$ form a basis, these conditions together imply that $\nabla_a k^b = 0$.

I find the result $\nabla_a k^b = 0$ strange and possibly incorrect, but I can't find an error in the reasoning above. Is this argument actually correct, or is there a flaw somewhere?

Answer 1

After some more thought, I think I've realized what's going on. The short answer is that the above result is correct, and is just a specific case of a more general construction.

Here's more explanation: the point is that the original tangent field $k^a$ is only defined on the geodesic $\gamma$, but the expression $\nabla_a k^b$ is only sensible if the vector field $k^a$ is defined in an open neighborhood of $\gamma$. For clarity, let's call the original tangent field (with support only on $\gamma$) $k^a$, and its extension to a neighborhood of $\gamma$ $k^a_\gamma$. Then the sensible object to talk about is $\nabla_a k^b_\gamma$, and not $\nabla_a k^b$.

The key point is that the extension of $k^a$ to $k^a_\gamma$ is not unique. In my original question, I performed this extension by introducing a parallel-transported frame $x^a_{(i)}$ and then using that to define a local coordinate system. The extended vector field $k^a_\gamma$ was just the timelike coordinate basis vector of this coordinate system.

More generally, I could have chosen $k^a \nabla_a x^b_{(i)} = v^b_{(i)}$ for arbitrary $v^b_{(i)}$ with support on $\gamma$. Then proceeding as above, I would have obtained

$x^a_{(i)} \nabla_a k^b_\gamma = v^b_{(i)}$,

and therefore on $\gamma$

$\nabla_a k^b_\gamma = \sum_i (x_{(i)})_a v^b_{(i)}$.

Thus the freedom in choice of $v^a_{(i)}$ gives complete freedom in choosing the extension of $k^a$ to $k^a_\gamma$, and therefore gives complete freedom in choosing $\nabla_a k^b_\gamma$.

In particular, choosing a Fermi coordinate system (i.e. taking $v^a_{(i)} = 0$, or by taking my orthonormal frame to be parallel transported along $\gamma$), I constructed a vector field $k^a_\gamma$ which is covariantly constant on $\gamma$. In fact, this follows directly from the fact that the Christoffel symbols computed in a Fermi coordinate system vanish on $\gamma$. I suspect with some more work, one can use this approach to show that the Christoffel symbols must vanish on $\gamma$ in the first place.