Parametric and covariant expressions for the acceleration vector I am reading S. Neil Rasband book about Classical Dynamics. In the first chapter, there are two different forms of the acceleration:


*

*What he calls the "intrinsic". Given a trajectory with parameter
$s(t)$, considers $x(s)$ and $\dot{x}(s)$, then:
$$\boldsymbol{a}(t)=
\ddot{s}\hat{\boldsymbol{\tau}}+\frac{v^2}{R}\hat{\boldsymbol{n}}$$
where $\hat{\boldsymbol{\tau}}$ is the tangent vector to the
trajectory, $v=\dot{s}$ the speed along the trajectory, $R$ is the
radius of curvature and $\hat{\boldsymbol{n}}$ is the normal vector.

*And the covariant form:
$$a^{i}=\frac{d^{2}x^{i}}{dt^{2}}+\Gamma^{i}_{jk}\frac{dx^{j}}{dt}\frac{dx^{k}}{dt}$$
where $\Gamma^{i}_{jk}$ is a Christoffel Symbol.
I know these are two ways of describing the same thing, one does not deal with coordinates but a trajectory, and the other one is valid for any coordinate system. 
My question is the next one: Is there a way of going from one description to the other one? What is the explicit relation between them? (if there is any).
 A: The "Instrinsic" acceleration:
\begin{align*}
  \vec{a} & =\ddot{s}\,\vec{\hat{\tau}}+\frac{v^2}{|R|}\vec{\hat{n}}\\
  \text{with:}\quad\\
  \frac{1}{|R|}&=|k|\,\,,\text{$k$  curvature}\,,\\
  \vec{\hat{n}}&=\frac{\vec{k}}{|k|}\,,\\
  v&=\frac{ds}{dt}\,,\\
  \tau&=\tau(s)\,,\Rightarrow\quad |\tau|=1\\\\
\vec{a} & =\ddot{s}\,\vec{{\tau}}+v^2\,\vec{{k}}\\\\
   &\text{with:}\\
  \vec{\tau}&=\frac{d\vec{r}}{ds}\\\vec{k}&=\frac{d^2\vec{r}}{ds^2}\,
  \Rightarrow\\\\
   \quad\vec{a}&=\frac{d\vec{r}}{ds}\,\ddot{s}+\frac{d^2\vec{r}}{ds^2}
  \,\dot{s}^2 \tag{1}
\end{align*}
The position vector to the geodetic line is :
\begin{align*}
  \vec{r}&=
  \begin{bmatrix}
     x(s) \\
     y(s) \\
     z(s) \\
  \end{bmatrix}\,\Rightarrow\\
  \vec{v}&=\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{ds}\,\frac{ds}{dt}\\
   \vec{a}&=\frac{d^2\vec{r}}{dt^2}=\frac{d}{dt}
   \left(\frac{d\vec{r}}{ds}\,\dot{s}\right)= \frac{d\vec{r}}{ds}\ddot{s}+\frac{d^2\vec{r}}{ds^2}
  \,\dot{s}^2 \tag{2}
\end{align*}
So equation (2) is identical to equation (1)   $\checkmark$
But the covariant equation is the ODE for the geodetic line $s(t)$
with:
\begin{align*}
  &\frac{d\vec{r}}{ds}\ddot{s}+\frac{d^2\vec{r}}{ds^2}
  \,\dot{s}^2=0\,\Rightarrow\\
  &\left(\frac{d\vec{r}}{ds}\right)^T\left(\frac{d\vec{r}}{ds}\ddot{s}+\frac{d^2\vec{r}}{ds^2}
  \,\dot{s}^2\right)=0\,,\text{solve for $\ddot{s}$}\\
  &\ddot{s}+\underbrace{g^{-1}\left(\frac{d\vec{r}}{ds}\right)^T\frac{d^2\vec{r}}{ds^2}}_{\Gamma^i_{jk}}
  \,\dot{s}^2=0\quad,g=\left(\frac{d\vec{r}}{ds}\right)^T\,\left(\frac{d\vec{r}}{ds}\right)\,,\text{$g$ is the Metric}
\end{align*}
