Accelerated coordinates in general relativity I'm self-studying Rindler coordinates in general relativity and I'm rather confused with the meaning of the coordinates. I found an excercise in Misner, Taylor and Wheeler's gravitation book which might help me understand.
The rindler coordinates are $$ds^2 = -(1+g\xi^1)^2 (d\xi^0)^2 + (d\xi^1)^2 + (d\xi^2)^2 + (d\xi^3)^2$$
In excercise 6.6 of this book a clock is attached to every grid point $(\xi^1,\xi^2,\xi^3 )$ and the observer is in hyperbolic motion.  It is asked to show that proper time  as measured by a lattice clock differs from coordinate time at its lattice point as
$$\frac{d\tau}{d\xi^0} = 1 + g\xi^1 $$
What I did was to impose that 
$$-d\tau^2 =   -(1+g\xi^1)^2 (d\xi^0)^2$$
this is because the proper time of the clock only moves in the time direction and the other $d\xi^i$ are zero because the clock is not moving in the grid. Thus with simple algebra one is done.
I'm confused with the question following this one: An accelerometer is attached to each grid point of the local coordinates of an observer in hyperbolic motion. It is asked to calculate the acceration measured by the accelerometer at $(\xi^1,\xi^2,\xi^3 )$.
Here I'm confused because the accelerometers are attached to the grid, thus again I could impose:
$$-d\tau^2 =   -(1+g\xi^1)^2 (d\xi^0)^2$$
(the other differentials are zero for the same reason), thus I would only measure a 4-velocity in the time direction. Thus there is no acceleration in the space directions so an accelerometer measuring in the space direction wouldn't measure anything. Is this reasoning correct? If not any help with understanding and completing this problem would be appreciated.
 A: For the first part, you need to consider a timelike observer moving on a path parametrised by proper time $\xi^{\mu}(\tau)$. The tangent vector to this curve is given by $X=\dot{\xi^{\mu}}e_{\mu}$, where $e_{\mu}=\frac{\partial}{\partial \xi^{\mu}}$. The observer is in hyperbolic motion, which in this coordinate system means $\xi^{i}={\rm const}$ and hence $X=\frac{d \xi^{0}}{d\tau}e_{0}$. For a timelike observer (parametrised by proper time) $g(X,X)=-1$ and hence $$ -1=g(X,X)=g_{00}\left(\frac{d \xi^{0}}{d\tau}\right)^{2}$$
from which you can deduce that (I renamed the constant $g \rightarrow \gamma$ to avoid confusion with the metric) $$\frac{d\tau}{d \xi^{0}}=(1+\gamma\xi^{1}).$$
Note that the choice of positive sign when taking the square root corresponds to looking at a future-directed curve.
For the second part, you need to look at $A=\nabla_{X} X$. Remember that for $A=0$ this is the geodesic equation and geodesics are curves of no acceleration. $A$ measures how much the curve fails to be a geodesic, i.e. how much it is accelerating. 
\begin{align}
\nabla_{X}X &=X^{0}\nabla_{e_0}\left(X^{0}e_{0}\right)\\
& = X^{0}\left(\frac{\partial X^{0}}{\partial \xi^{0}}\right)e_{0}+(X^{0})^2\nabla_{e_{0}}e_{0}\\
& = 0+\frac{\gamma}{1+\gamma\xi^{1}} e_{1}
\end{align}
So we found that the observers accelerate in the $\xi^{1}$ direction
$$\nabla_{X}X=\frac{\gamma}{1+\gamma\xi^{1}} e_{1}$$
To compute $\nabla_{e_{0}}e_{0}$:
$$g(\nabla_{e_{0}}e_{0},e_{0})=\frac{1}{2}\nabla_{e_{0}}g(e_{0},e_{0})=0$$
\begin{align}
g(\nabla_{e_{0}}e_{0},e_{1})&=\nabla_{e_{0}}g(e_{0},e_{1})-g(e_{0},\nabla_{e_{0}}e_{1})\\
&=0-g(e_{0},\nabla_{e_{1}}e_{0})\\
&=-\frac{1}{2}\nabla_{e_{1}}g(e_{0},e_{0})\\
&=(1+\gamma\xi^{1})\gamma
\end{align}
where between the first and second line we used the fact that coordinate basis vectors commute. Finally,
$$g(\nabla_{e_{0}}e_{0},e_{2})=g(\nabla_{e_{0}}e_{0},e_{3})=0$$
Hence
$$\nabla_{e_{0}}e_{0}=(1+\gamma\xi^{1})\gamma e_{1}$$
