Fermi - Walker coordinate system In which "differential geometry" book I can find the proof of the following statement?
Let $M$ be a lorentzian manifold. Given a timelike geodesic $γ$ and a point $p∈γ$, there is a neighborhood $U∋p$ equipped with coordinates, $x_0,x_1,x_2,x_3$ such that in the portion of $γ$ included in $U$, exactly along $γ$, the derivatives of the metric vanish in the said coordinates.
Many thanks!
 A: It's probably easier to give a sketch of proof for this than to search for a reference. Assume the geodesic $\gamma: (-a, a) \rightarrow M$ is given and $E_1(t) = \gamma^\prime(t) $ is the tangent vector field along $\gamma$. Assume in addition $\gamma(t)$ is parametrized by arclength, i.e. $g(E_1, E_1) = 1$ ($=-1$ in case of a timelike geodesic in a Lorentz manifold). 
Let $E_i, 2\le i \le n$ an orthonormal extension of $E_1$ to a basis of $T_{\gamma(0)}M$ and let $E_i(t) $ the parallel transport of $E_i$ along $\gamma$. Since the connection is compatible with the metric, the frame $E_i(t)$ remains orthonormal along $\gamma$.
Now define the coordinate system in some neighbourhood of $0$ by
$x: \mathbb{R}^n \rightarrow M$ by
$$(x^1, \ldots, x^n) \mapsto \exp_{\gamma(x^1)}(\sum_{i=2}^n(x^i E_i(\gamma(x^1)))$$
It has to be shown that this is in fact a coordinate system and that it has the desired properties.
Note first, that the image of the $x^1$ -axis is just a portion of the geodesic near the origin, since $(t, 0,\ldots,0)$ maps to $\exp_{\gamma(t)}(0) = \gamma(t)$
Also note that, along $\gamma$, the tangent vector $\frac{\partial}{\partial x^i}$ is just $E_i$, since the differential of the exponential map in $0$ (this is $0_{\gamma(x^1)}$) is just the identity. So the differential of the map has full rank along $\gamma$, and so, by the implicit function theorem, is a local diffeomorphism in a neighbourhood of the $x^1$-axis, in other words it's coordinate system. 
The metric tensor along $\gamma$ is given by $g_{ij} = g(E_i, E_j) = \delta_{ij}$ by construction.
To see that it's derivatives vanish along $\gamma$ note that
 $$(*)\quad E_k g(E_i, E_j) = g(\nabla_{E_k}E_i, E_j) + g(E_i, \nabla_{ E_k} E_j) $$ 
This is $=0$ for $k=1$ since the $E_i$ are parallel along $\gamma$ (including $E_1$, since $\gamma $ is a geodesic).
For $k\ge 2$ consider first $i=k$. Since the curves 
$$t\mapsto \exp_{\gamma(x^1)}(tE_k)$$
are normal geodesics to $\gamma$, $\nabla_{E_k}E_k=0$ along $\gamma$.
Similarly, for $i, j>1$ , $$t\mapsto \exp_{\gamma(x^1)}(t(E_i + E_j))$$
is a geodesic normal to $\gamma$, so
$$\nabla_{E_i+E_j}(E_i+E_j)=\nabla_{E_i}E_i+ \nabla_{E_i}E_j+\nabla_{E_j}E_i+ \nabla_{E_j}E_j=\nabla_{E_i}E_j+\nabla_{E_j}E_i=0$$
so $\nabla_{E_i} E_j = -\nabla_{E_j} E_i$. On the other hand, since the $E_i$ are coordinate vector fields, $[E_i,E_j]=\nabla_{E_i} E_j  -\nabla_{E_j} E_i=0$. Combining these identities we get $\nabla_{E_i} E_j=-\nabla_{E_j} E_i$, so this has to be $=0$ as well. This implies that the left hand side in $(*)$ is $=0$ if $i,j, k >1$
Still missing is the case 
$$E_kg(E_1, E_i) = g(\nabla_{E_k}E_1,E_i)+g(E_1,\nabla_{E_k} E_i) $$
with $k>1$
If $i\neq 1$ then the second term on the right hand side vanishes according to the previous results, so only the first remains. Using again $[E_1, E_k]=0$ we see that $\nabla_{E_k}E_1=\nabla_{E_1}E_k=0$ since $E_k$ is parallel along $\gamma$ by construction. Incidentally this then also cover the case $i=1$.
(A proof in a textbook is, e.g.,  found in Klingenbergs Riemannian Geometry, Proposition 1.12.1 , but you will very likely have difficulties to find your way through the notation you'll find there).
(The 'classical' case of a Riemannian Normal coordinate system where you have the desired properties in one given point can be found in most books on Differential Geometry by searching for said term 'Riemannian Normal coordinates', e.g. in Spivaks' Comprehensive Introduction to Differential Geometry, vol 2 (2nd edition), page 159. The proof has to go through essentially the same steps, the main difference is that the distinction $k=1$ vs $k>1$ is not necessary).
