Parametric and covariant expressions for the acceleration vector

Question

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

1. 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.

2. 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).

Answer 1

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*}