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I've just learned about the

We should be able to say that $$

J^i = \frac{\delta A^i}{\delta t}

= \frac{\delta^2 V^i}{\delta^2 t}

= \frac{\delta^3 Z^i}{\delta^3 t}

$$ where ##J## is the

\frac{\delta Z^i}{\delta t} = \frac{d Z^i}{dt}

$$ but I can't prove it. This should be equivalent to proving that $$

Z^j \Gamma_{jk}^i V^k = 0

$$ Here's my heroic attempt: $$

\begin{align*}

Z^j \Gamma_{jk}^i V^k &= Z^j V^k \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\

&= Z^j \frac{d Z^k}{dt} \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\

&= Z^j \frac{d \pmb{Z}_j}{dt}\cdot \pmb{Z}^i \\

\end{align*}

$$ Now what? I also tried using the product rule but it doesn't seem to improve the situation...

*covariant derivatives*(##\nabla_i## and ##\delta/\delta t##) and I have a doubt.We should be able to say that $$

J^i = \frac{\delta A^i}{\delta t}

= \frac{\delta^2 V^i}{\delta^2 t}

= \frac{\delta^3 Z^i}{\delta^3 t}

$$ where ##J## is the

*jolt*. This should mean that $$\frac{\delta Z^i}{\delta t} = \frac{d Z^i}{dt}

$$ but I can't prove it. This should be equivalent to proving that $$

Z^j \Gamma_{jk}^i V^k = 0

$$ Here's my heroic attempt: $$

\begin{align*}

Z^j \Gamma_{jk}^i V^k &= Z^j V^k \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\

&= Z^j \frac{d Z^k}{dt} \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\

&= Z^j \frac{d \pmb{Z}_j}{dt}\cdot \pmb{Z}^i \\

\end{align*}

$$ Now what? I also tried using the product rule but it doesn't seem to improve the situation...

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