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.


  • $\begingroup$ I'm sorry. I'm a little confused what the question exactly is. Are you asking how one should derive (4.3)? $\endgroup$
    – Prahar
    Commented Mar 13, 2016 at 12:54
  • $\begingroup$ I though it should be the Lorentz metric c.f. section 2.7 $\endgroup$ Commented Mar 13, 2016 at 17:23
  • $\begingroup$ From equation (1) of your post the Lie derivative of the projected separation vector along the congruence vanishes. So here on equation there should not be an extra projection operator on the left side? $\endgroup$ Commented Mar 13, 2016 at 17:33
  • $\begingroup$ Thanks a lot! I'm self studying though. Before this book I've finished the book Geometrical Methods of Mathematical Physics by Schutz. I'm still not very fluent in tensor gymnastics. $\endgroup$ Commented Mar 13, 2016 at 17:35

1 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.

  • $\begingroup$ Hopefully OP can use these clarifications to understand (4.4), which was not addressed here. $\endgroup$
    – Ryan Unger
    Commented Mar 13, 2016 at 18:09
  • $\begingroup$ In the first sentence, do you mean 'orthogonal' to? I skip section 2.9 so I'm not sure what is horizontal bundle found there. On the second property you've contracted away $V_b X^b$ though. $\endgroup$ Commented Mar 14, 2016 at 1:00
  • $\begingroup$ @Henry Whoops, yes I meant orthogonal. My $H_p\mathcal{M}$ is the $H_q$ in HE. $\endgroup$
    – Ryan Unger
    Commented Mar 14, 2016 at 1:19
  • $\begingroup$ @Henry Note that $g^a{}_b=\delta^a{}_b$ because if you raise the first index in $g_{ab}$ you end up with $g^{ac}g_{cb}=\delta^a{}_b$, cf. page 37 in HE. My first comment on the OP was correct, the second was wrong. Thus $h_{ab}=g_{ab}+V_aV_b$. $\endgroup$
    – Ryan Unger
    Commented Mar 14, 2016 at 1:24
  • $\begingroup$ Yes! I'm about to say I just realize that. I think the trick is on the last part where you've shown ${V^a}_{;b}\in H_p \mathcal{M}$. I have the same result on equation (1). I want to ask: how do we interpret $DZ^a/\partial s$? I would think that we first take covariant derivative of vector $Z$ along curve $s$, and then take the component denoted $a$. Or can we just think of the component as a function defined on the curve and then operate on it? $\endgroup$ Commented Mar 14, 2016 at 1:30

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