On pp. 79, it is obvious that equation (4.2) \begin{equation} \frac{D}{\partial s}Z^a = {V^a}_{;\ b}Z^b \end{equation} holds, where $Z$ is the deviation vector and $V$ is the unit tangent vector along the timelike curves.
The author then project the deviation vector to ${}_\bot Z^a = {h^a}_bZ^b$, where ${h^a}_b = {g^a}_b+V^aV_b$ (the author used ${\delta^a}_b$, but I think it should be the metric?) is the tensor which projects a vector into its component in the subspace orthogonal to $V$.
My question is on equation (4.3): \begin{equation} {}_\bot \frac{D}{\partial s}({}_\bot Z^a) = {V^a}_{;\ b} {}_\bot Z^b \end{equation} In components it should be \begin{equation} {h^a}_c({h^c}_d Z^d)_{;\ b} V^b = {V^a}_{;\ b}{h^b}_c Z^c \end{equation} Using equation (4.2), which in component \begin{equation} {Z^a}_{;\ b}V^b = {V^a}_{;\ b}Z^b \end{equation} I just can make equation (4.3) equal on both side. Since the equation looks so simple, the derivation should be rather intuitive?
Also, there is no derivation to equation (4.4). It looks quite formidable though. Any hint on the derivation would be appreciated. There are a lot of detail derivation on geodesic deviation, but they did not project the deviation vector to the orthogonal ones though.
Thanks!