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?


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.