Intrinsic derivative of constant vector field along a curve

[harmyder]

Homework Statement

Suppose that $T_i$ is the contravariant component of a vector field $\mathbf{T}$ that is constant along the trajectory $\gamma.$ Show that intrinsic derivative is $0.$

Homework Equations

$$\frac{\delta T_i}{\delta t} = \frac{dT^i}{dt}+V^j\Gamma^i_{jk}T^k$$

The Attempt at a Solution

$$\begin{align}\mathbf{T} = T^i \mathbb{Z}_i\\T^i = \frac{d\mathbf{T}}{dZ_i}\label{ti}\end{align}$$
But from $\ref{ti}$ i see that $T_i=0.$ Probably, $\ref{ti}$ is wrong.

Another attempt:)
$$\begin{align}
\mathbf{T} &= T^i \mathbb{Z}_i\\
\mathbf{T}\cdot\mathbb{Z}^i &= T^i\\
\frac{dT^i}{dt}&= \frac{d\mathbf{T}}{dt}\mathbb{Z}^i + \mathbf{T}\frac{\partial\mathbb{Z}^i}{\partial Z^j}\frac{dZ^j}{dt}\\
&=-\mathbf{T}\Gamma^i_{jk}\mathbb{Z}^k\frac{dZ^j}{dt}\\
&=-\mathbf{T}\mathbb{Z}^k\Gamma^i_{jk} V^j\\
&=-T^k\Gamma^i_{jk}V^j
\end{align}$$

OMH, looks like i have solved it while writing it here. Just need a confirmation.

 

[lippyka]

So, $$\frac{\delta T_i}{\delta t} = \frac{dT^i}{dt}+V^j\Gamma^i_{jk}T^k=-T^k\Gamma^i_{jk}V^j+V^j\Gamma^i_{jk}T^k=0$$