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Is there a general way so we can find curvature and torsion of a curve $$\gamma : \mathbb{R} \rightarrow \mathbb{R}^n$$ where $$n=2,3$$ from a coordinate system to a new coordinate system?

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Do you mean a rigid transformation or a complete change of variables? – Pragabhava Oct 24 '12 at 2:14
No rigid transformation, I mean coordinate transformation, for example from cartesian coordinate system to polar coordinate system or from bipolar coordinates to cartesian coordinates. – stavros kassaras Oct 24 '12 at 6:45
up vote 0 down vote accepted

Case $n = 2$ and curvature $\kappa(t)$.

The curve $\gamma(t) = \big(x(t),y(t)\big)^T$, has curvature $$ \kappa(t) = \frac{x'y''-x''y'}{(x'^2 + y'^2)^{3/2}} = \frac{\det(\gamma',\gamma'')}{\|\gamma'\|^3}. $$ Now, suppose $x(t) = x\big(\xi(t),\eta(t)\big)$, $y(t) = y\big(\xi(t),\eta(t)\big)$, then \begin{align} \frac{d x}{d t} &= x_\xi \xi' + x_\eta \eta'\\ \frac{d y}{d t} &= y_\xi \xi' + y_\eta \eta' \end{align} and let $\alpha(t) = \big(\xi(t),\eta(t)\big)^T$, we can write the above relation as $$ \gamma' = \textbf{J} \, \alpha' $$ where $$ \textbf{J} = \begin{pmatrix}x_\xi & x_\eta \\ y_\xi & y_\eta\end{pmatrix} $$ is the Jacobian matrix. Then $$ \gamma'' = \frac{d \textbf{J}}{dt} \alpha' + \textbf{J} \alpha'' $$ and $$ \kappa(t) = \frac{\det(\textbf{J}\,\alpha',\frac{d \textbf{J}}{dt} \alpha' + \textbf{J} \alpha'')}{\|\textbf{J}\,\alpha'\|^3}. $$


Let $x(t) = r(t) \cos \theta(t)$ and $y(t) = r(t) \sin \theta(t)$, then \begin{align} \det(\textbf{J}\alpha',\textbf{J}'\alpha') &= 2 r'^2 \theta' + r^2 \theta'^3\\ \det(\textbf{J}\alpha',\textbf{J}\alpha'') &= -r r'' \theta' + r r' \theta''\\ \|\textbf{J}\alpha'\|^3 &= (r'^2 + r^2 \theta'^2)^{3/2} \end{align}


$$ \kappa(t) = \frac{r^2 \theta'^3 + 2 r'^2 \theta' - r r'' \theta' + r r' \theta''}{(r'^2 + r^2 \theta'^2)^{3/2}}. $$

In the special case $\theta = t$ $$ \kappa(t) = \frac{r^2 + 2r'^2 - r r''}{(r^2 + r'^2)^{3/2}} $$

which, according to eq. (15) of Wolfram Mathworld, is correct.

Now you can do the rest.

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