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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!

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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.

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  • $\begingroup$ Thank you very much, I would think about this in the afternoon. Could you point out some literatures for further details? Thanks. $\endgroup$ – Witty Viper of Hidden Glen Mar 15 '16 at 3:17
  • $\begingroup$ @Henry Wald, General Relativity (1984) covers some of the same material, but for the most part, Hawking-Ellis was clear enough for me. $\endgroup$ – Ryan Unger Mar 15 '16 at 3:21
  • $\begingroup$ Your reasoning on derivation of equations are very clear! But I cannot understand the physical meaning of $A_{\alpha \beta}$ (I can feel it but not clearly. I mean why did he decompose it into matrix $O$ and $S$, where $O$ is in rotation group, but what about $S$? Why separation?) I don't think (4.9) is a solution to the differential equation, since the matrix $A$ is undetermined itself, only satisfying another differential equation (4.10). $\endgroup$ – Witty Viper of Hidden Glen Mar 15 '16 at 9:07
  • $\begingroup$ What about the determinant of $A$ (or $S$), which later serves as the volume expansion on equation (4.20)? The author mention an analogous to Newtonian hydrodynamics though. Before finally giving rise to the celebrated Raychaudhuri equation, the naming is not so clear (maybe I should take it with a grain of salt). $\endgroup$ – Witty Viper of Hidden Glen Mar 15 '16 at 11:07
  • $\begingroup$ @Henry I updated the question. What "naming" is not so clear? $\endgroup$ – Ryan Unger Mar 15 '16 at 13:40

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