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!
Answer
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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