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From book "Introduction to Tensor Analysis and the Calculus of Moving Surfaces" by Pavel Grinfeld, Edition 2013 :

The component $A^{i}(t)$ of acceleration $\mathbf{A}(t)$ of the particle from the preceding exercise is given by \begin{equation} A^{i} = \frac{dV^{i}}{dt}+\Gamma_{jk}^{i}V^{j}V^{k} \tag{5.71} \end{equation}

What is the proof of this equation?

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  • $\begingroup$ Hi Pinton. Welcome to Phys.SE. Note that res. recom. qs are restricted on Phys.SE. Why don't you just ask for the proof directly? $\endgroup$ – Qmechanic Dec 21 '18 at 14:09
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    $\begingroup$ @Qmechanic: 1. Because I thought people rather prefer to refer to a proof instead of writing that here. 2. The question is edited as you said. $\endgroup$ – Pinton Dec 21 '18 at 14:14
  • $\begingroup$ 1. They still can. 2. Better. $\endgroup$ – Qmechanic Dec 21 '18 at 14:18
  • $\begingroup$ The first term here was called relative acceleration: $\nabla_u\nabla_u n+R(n,u)u=0$, is that the same thing as the acceleration you are asking for? $\endgroup$ – Emil Dec 21 '18 at 14:32
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    $\begingroup$ Suggest you learn about the geodesic equation if you have not already done so. You eqn you quoted is basically saying "how much acceleration is there in addition to the part associated with free-fall motion." $\endgroup$ – Andrew Steane Dec 21 '18 at 16:24
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Let $\mathbf{R}(t)$ represent the position of the particle. Then, we have

$$\begin{align} \mathbf{V}(t) = \mathbf{R}'(t) = &\frac{\partial \mathbf{R}}{\partial Z^{i}}\frac{\partial Z^{i}}{\partial t}, \\ =& \mathbf{Z}_{i}\frac{\partial Z^{i}}{\partial t}, \\ =& \mathbf{Z}_{i}V^{i},\end{align}$$

thereby,

$$\begin{align} \mathbf{A}(t) = \mathbf{V}'(t) =& (\mathbf{Z}_{i})'\frac{\partial Z^{i}}{\partial t} + \mathbf{Z}_{i}(\frac{\partial Z^{i}}{\partial t})', \\ =&(\frac{\partial Z_{i}}{\partial Z^{j}}\frac{\partial Z^{j}}{\partial t})\frac{\partial Z^{i}}{\partial t} + \mathbf{Z}_{i}\frac{\partial}{\partial t}(\frac{\partial Z^{i}}{\partial t}), \\ =& \Gamma^{k}_{ij}\mathbf{Z}_{k}\frac{\partial Z^{j}}{\partial t}\frac{\partial Z^{i}}{\partial t} + \mathbf{Z}_{i} \frac{\partial}{\partial t}V^{i}.\end{align} $$

By exchanging $i$, $j$, and $k$ in the first term, we end up with

$$\begin{align} \mathbf{A}(t) = & \Gamma^{i}_{jk}\mathbf{Z}_{i}\frac{\partial Z^{j}}{\partial t}\frac{\partial Z^{k}}{\partial t} + \mathbf{Z}_{i} \frac{\partial}{\partial t}V^{i},\\ = & (\Gamma^{i}_{jk}V^{k}V^{j} + \frac{\partial}{\partial t}V^{i})\mathbf{Z}_{i},\end{align}$$

which concludes

$$A^{i} = \Gamma^{i}_{jk}V^{k}V^{j} + \frac{\partial}{\partial t}V^{i}$$

Q.E.D.

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