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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!

raegakj
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  • I'm sorry. I'm a little confused what the question exactly is. Are you asking how one should derive (4.3)? – Prahar Mar 13 '16 at 12:54
  • I though it should be the Lorentz metric c.f. section 2.7 – raegakj Mar 13 '16 at 17:23
  • 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? – raegakj Mar 13 '16 at 17:33
  • 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. – raegakj Mar 13 '16 at 17:35

1 Answers1

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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.

Ryan Unger
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  • Hopefully OP can use these clarifications to understand (4.4), which was not addressed here. – Ryan Unger Mar 13 '16 at 18:09
  • 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. – raegakj Mar 14 '16 at 01:00
  • @Henry Whoops, yes I meant orthogonal. My $H_p\mathcal{M}$ is the $H_q$ in HE. – Ryan Unger Mar 14 '16 at 01:19
  • @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 $h{ab}=g_{ab}+V_aV_b$. – Ryan Unger Mar 14 '16 at 01:24
  • 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? – raegakj Mar 14 '16 at 01:30
  • @Henry $\mathrm{D}Z^a/\partial s$ is the $a$th component of the vector $\mathrm{D}\mathbf{Z}/\partial s$. The $a$ can also be understood abstractly, but HE does not explicitly follow this convention. – Ryan Unger Mar 14 '16 at 01:34
  • OK, so on the part where you've used the Leibniz rule on $\frac{D} { \partial s} ({h^b}c Z^c)$, the indices of the tensors inside the square bracket are understood abstractly. In tensor notation it should be $\frac{D}{\partial s} h(\,Z) = (\frac{D}{\partial s}h)(_,Z) + h(_,\frac{D}{\partial s}Z)$, right? I'm not sure of a good practice to do these computation. – raegakj Mar 14 '16 at 01:52
  • @Henry Yes, that's correct, but not very enlightening. It's best to use indices for such computations. – Ryan Unger Mar 14 '16 at 02:44
  • I think there is some subtlety on your last part ${V^a}{;b}{}{\bot} Z^b\in H_p\mathcal{M}$. You mean only the contravariant part of the ${V^{a}}_{;b}$ tensor lives in $H_p\mathcal{M}$, right? Also I can equate equation (4.4) now, but I can't see how it is `reduced' from the one above (4.4) in a natural ways without messy algebra. Can you give me some hints? – raegakj Mar 14 '16 at 08:44
  • @Henry Because $V^a{}{;b}V_a=0$ (do you see why this is?), contract with ${}\bot Z^b$ to get $(V^a{}{;b}{}\bot Z^b)V_a=0$. This shows that the vector $V^a{}{;b}{}\bot Z^b$ is orthogonal to $V^a$. – Ryan Unger Mar 14 '16 at 15:00
  • @Henry I don't have a hint for (4.4), a problem like that involving two covarariant derivatives is bound to get very messy. The best advice I have is to plug and chug. – Ryan Unger Mar 14 '16 at 15:06
  • I'm sorry, but equation (4.8) really gives me a headache. I cannot see why something is contracted away. The explanation is really scarce. It seems like something from the second term got cancelled by the first term. What can I do if I keep bumping these situations? – raegakj Mar 14 '16 at 15:40
  • I felt like there are some mathematical tricks on the hypersurfaces. – raegakj Mar 14 '16 at 15:50
  • @Henry Maybe ask another question about (4.8). This chat tower is already really long. – Ryan Unger Mar 14 '16 at 20:12
  • I would really appreciate it if you could answer this. Thanks!! – raegakj Mar 15 '16 at 01:47
  • @ocelo7 Hi, I would appreciate if if you could know the question and answer here. (I'm afraid I did not state the question clear but I feel like there is one.) Thanks!! – raegakj Apr 30 '16 at 03:10