Jacobi equation in the book The Large scale structure of space-time

Question

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!

Answer 1

We first show that $h=g+V\otimes V$ is a projection operator into the subspace $H_p\mathcal{M}$ orthogonal to $V\in T_p\mathcal{M}$.

Idempotence ($h\circ h=h$). A simple calculation gives: $$h^a{}_bh^b{}_c=(\delta^a{}_b+V^aV_b)(\delta^b{}_c+V^bV_c)=\delta^a{}_b\delta^b{}_c+V^aV_b\delta^b{}_c+\delta^a{}_bV^bV_c+V^aV_bV^bV_c=\delta^a{}_c+V^aV_c+V^aV_c-V^aV_c=h^a{}_c.$$ Identity on orthogonal subspace ($h|_{H_p\mathcal{M}}=\operatorname{id}_{H_p\mathcal{M}}$). This is also not hard. Let $X\in H_p\mathcal{M}$, then $$h^a{}_bX^b=\delta^a{}_bX^b+V^aV_bX^b=\delta^a{}_bX^b=X^a.$$ We note also that $h(V)=0$, since $$h^a{}_bV^b=\delta^a{}_bV^b+V^aV_bV^b=V^a-V^a=0$$

So $h$ is a proper projection operator.

We will now explain the derivation of Eq. (4.3), since there is a little trick involved. The object $${}_\bot \frac{\mathrm{D}}{\partial s}{}_\bot Z^a$$ is understood as follows: project $Z^a\in T_p\mathcal{M}$ into $H_p\mathcal{M}$ and take the covariant derivative of the result along the integral curves of $V$. Then project this into $H_p\mathcal{M}$ again. Thus $${}_\bot \frac{\mathrm{D}}{\partial s}{}_\bot Z^a=h^a{}_b\frac{\mathrm{D}}{\partial s}(h^b{}_cZ^c)$$ Now we begin a long calculation: $$h^a{}_b\frac{\mathrm{D}}{\partial s}(h^b{}_cZ^c)=h^a{}_b\frac{\mathrm{D}h^b{}_c}{\partial s}Z^c+h^a{}_bh^b{}_c\frac{\mathrm{D} Z^c}{\partial s}=h^a{}_bZ^c\frac{\mathrm{D}}{\partial s}(\delta^b{}_c+V^bV_c)+h^a{}_cV^c{}_{;b}Z^b$$ Note that $\delta^a{}_b$ is covariantly constant (page 32), so we have $$\tag{1} h^a{}_cV^c{}_{;b}Z^b+h^a{}_bZ^cV^dV^b{}_{;d}V_c+h^a{}_bZ^cV^bV_{c;d}V^d$$ and the last term vanishes because $h(V)=0$.

Note that since $V^aV_a=-1$, we have $V^a{}_{;b}V_a=0$. This implies $V^a{}_{;b}{}_\bot Z^b\in H_p\mathcal{M}$. Thus $V^a{}_{;b}{}_\bot Z^b=h^a{}_bV^b{}_{;c}{}_\bot Z^c$, and one can show this equals (1) by simply expanding the definitions.