Fermi-Propagated Jacobi equation in the book The Large scale structure of space-time On page 81, equation (4.6), the author use the Fermi derivative to write the Jacobi equation
\begin{equation} \tag{4.6}
\frac{{D^2}_\text{F}}{\partial s^2} {}_{\bot}Z^a = -{R^a}_{bcd}{}_{\bot}Z^cV^bV^d + {h^a}_b {\dot{V}^b}_{;c} {}_{\bot}Z^c + \dot{V}^a\dot{V}_b {}_{\bot}Z^b
\end{equation}
and also the equality (4.5)
\begin{equation} \tag{4.5}
\frac{D_{\text{F}}}{\partial s}{}_{\bot}Z^a = {V^a}_{;b}{}_{\bot}Z^b
\end{equation}
where $Z$ is the deviation vector, $V$ is the unit tangent vector along the timeline curves, and ${h^a}_b = {\delta^a}_b+V^aV_b$ is the projection operator (see this).
Using the property of (i) to (iv) of the Fermi derivative (which I can prove), equation (4.6) and (4.5) come naturally from equation (4.3) and (4.4).
The problem is on equation (4.7) and (4.8)
\begin{equation} \tag{4.7}
\frac{\text{d}}{\text{d} s} Z^\alpha = {V^\alpha}_{;\beta} Z^{\beta}
\end{equation}
\begin{equation}\tag{4.8}
\frac{\text{d}^2}{\text{d}s^2} Z^\alpha = (-{R^\alpha}_{4\beta 4} + {\dot{V}^\alpha}_{;\beta} + \dot{V}^\alpha \dot{V}_\beta)Z^\beta
\end{equation}
This is an ordinary differential equation with respect to the component (as a function on the curve), and the Greek indices take the value $1,2,3$, where the time component is on the forth one.
To derive (4.7)
\begin{gather}
\frac{D_{\text{F}}}{\partial s}{}_{\bot}(Z^\alpha\mathbf{E}_\alpha) = {V^\alpha}_{;\beta} \mathbf{E}^\beta \otimes \mathbf{E}_\alpha ({}_{\bot}Z^\gamma \mathbf{E}_{\gamma}) \\
\biggl(\frac{D_{\text{F}}}{\partial s} {}_{\bot}Z^\alpha \biggr) \mathbf{E_\alpha} = \biggl( {V^\alpha}_{;\beta} {}_\bot Z^\beta \biggr) \mathbf{E_\alpha}
\end{gather}
where $\mathbf{E}$ are bases orthogonal to $\mathbf{V}$. Well, I cannot get rid off the $\bot$, but the author wrote on page 82 

As ${}_\bot \mathbf{Z}$ is orthogonal to $\mathbf{V}$ it will have
  components with respect to $\mathbf{E}_1,\mathbf{E}_2,\mathbf{E}_3$
  only. Thus it may be expressed as $Z^\alpha \mathbf{E}_\alpha$.

I guess ${}_{\bot}Z^\alpha = Z^\alpha$ in this notation? 
As for equation (4.8), there are several terms got contracted away. To make the second term of (4.6) equals to (4.8), I think
\begin{align}
{h^a}_b ({V^b}_{;d}V^d)_{;c} {}_\bot Z^c &= {h^a}_b ({V^b}_{;dc}V^d + {V^b}_d {V^d}_{;c}) {}_\bot Z^c \\
& = ({h^a}_b {V^b}_{;d})_{;c}V^d {}_\bot Z^c + {h^a}_b {V^b}_d {V^d}_{;c}{}_\bot Z^c - {h^a}_{b;c} {V^b}_{;d} V^d {}_\bot Z^c \\
& = \dot{V}^a_{;c} - ({V^a}_{;c}V_b + V^a V_{b;c})  V^d {}_\bot Z^c \\
& = \dot{V}^a_{;c} - V^a V_{b;c}V^d {}_\bot Z^c
\end{align}
So there is an extra term in the end, where it should be contracted from the space components of the first term of (4.6)
\begin{align}
0 =& -{R^a}_{bcd} {}_\bot Z^c V^b V^d - V^a V_{b;c}V^d {}_\bot Z^c\\
=& -({V^a}_{dc} - {V^a}_{;cd}){}_\bot Z^cV^d  - V^a V_{b;c}V^d {}_\bot Z^c
\end{align}
I fail to equate this.
Additionally I do not know how to solve this component differential equation, where the author has given a answer to (4.7)
\begin{equation} \tag{4.9}
Z^\alpha (s) = A_{\alpha \beta}(s) Z^\beta|_q
\end{equation}
where $A_{\alpha \beta}(s)$ is a $3\times 3$ matrix which is the unit matrix at $q$ and satisfies
\begin{equation} \tag{4.10}
\frac{\text{d}}{\text{d} s} A_{\alpha\beta}(s) = V_{\alpha ;\gamma} A_{\gamma \beta}(s)
\end{equation}
Equation (4.9) does not even contract to the correct index, and it does not equate (4.7) when plugging back in. And then there is equation (4.11)
\begin{equation}\tag{4.11}
A_{\alpha \beta} = O_{\alpha \delta} S_{\delta \beta}
\end{equation}
where $O_{\alpha \delta}$ is an orthogonal matrix with positive determinant and $S_{\delta \beta}$ is a symmetric matrix. I cannot figure it out the physics behind this.
Any advice would be greatly appreciated, as I'm trying to clarify these new ideas and equations! Thanks!
 A: Let us first argue that ${}_\bot Z^\alpha={}_\bot Z^a$. Now, ${}_\bot\mathbf{Z}$ does not contain the component of $\mathbf{Z}$ along $\mathbf{V}$. Suppose we expand $\mathbf{Z}$ in terms of the the basis $\{\mathbf{E}_1,\mathbf{E}_2,\mathbf{E}_3,\mathbf{V}\}$, then it is clear that the projection of $\mathbf{Z}$ into the subspace orthogonal to $\mathbf{V}$ (which is exactly ${}_\bot\mathbf{Z}$) contains only the components of $\mathbf{Z}$ in the $\mathbf{E}_1,\mathbf{E}_2,\mathbf{E}_3$ directions. But these components are just $Z^\alpha$, which we have shown equal ${}_\bot Z^\alpha$.
We know that ${}_\bot\mathbf{Z}=Z^\alpha\mathbf{E}_\alpha$, but since the frame $\{\mathbf{E}_\alpha\}$ is Fermi-transported, we have 
$$\frac{\mathrm{D}_\mathrm{F}}{\partial s}{}_\bot \mathbf{Z}=\frac{\mathrm{D}_\mathrm{F}}{\partial s}(Z^\alpha\mathbf{E}_\alpha)=\frac{\mathrm{D}_\mathrm{F}Z^\alpha}{\partial s}\mathbf{E}_\alpha+Z^\alpha\frac{\mathrm{D}_\mathrm{F}\mathbf{E}_\alpha}{\partial s}=\frac{\mathrm{d}Z^\alpha}{\mathrm{d}s}\mathbf{E}_\alpha$$
(We used property (iii) on page 81.) This is how the ordinary derivatives appear.
For $\mathbf{X}$ any vector, it is clear that $X^a{}_\bot Z_a=X^\alpha Z_\alpha$ because ${}_\bot Z^4=0$ and ${}_\bot Z^\alpha=Z^\alpha$, as was shown above. (4.7) should now be clear. 
For $\boldsymbol{\omega}$ any covector, we have $\omega_aV^a=\omega_4$ because $\mathbf{V}$ is the fourth basis vector. This explains the presence of the $4$s in (4.8). The $\alpha$s on the other vectors appear because we set $a=\alpha$ on the LHS of (4.6) anyway and are using $h^a{}_b$ to project into $H_p\mathcal{M}$. This should explain (4.8).
The equations (4.9) and (4.10) are the standard solution method for a linear first order differential system like (4.7). To verify this, take the derivative of (4.9) and insert (4.10):
$$\frac{\mathrm{d}Z^\alpha}{\mathrm{d}s}=\frac{\mathrm{d}A_{\alpha\beta}}{\mathrm{d}s}Z^\beta\rvert_q=V_{\alpha;\gamma}A_{\gamma\beta}Z^\beta\rvert_q=V_{\alpha;\gamma}Z^\gamma$$
Since $A$ is a real matrix, at the points where $\det A\ne0$, its polar decomposition is of the form $OS$, where $O\in\mathrm{SO}(3)$ and $S=S^t$. $O$ represents the rotation of the curves because it is an element of $\mathrm{SO}(3)$, the rotation group. $S$ is interpreted as telling us about the separations because it is symmetric. The distance between flow lines in the $\alpha\beta$ direction is the same as in the $\beta\alpha$ direction.
