For a regular smooth parametrized curve $\gamma:I\rightarrow \mathbb{R}^3$ of unit speed i.e., with $||\gamma'(s)||=1$ for all $s\in I$ we define curvature of $\gamma$ at $s$ to be $||\gamma''(s)||$.

Suppose $\gamma$ is not of unit speed, there exists a parametrization $\widetilde{\gamma}=\gamma\circ \varphi : J\rightarrow I\rightarrow \mathbb{R}^3$ such that $||\widetilde{\gamma}(s)||=1$ for all $s\in J$. We then have curvature of $\widetilde{\gamma}$ at $s$ to be $||\widetilde{\gamma}''(s)||$.

Let $\gamma(s)=(x_1(s),x_2(s),x_3(s))$ then $\gamma'(s)=(x_1'(s),x_2'(s),x_3'(s))$ and $\gamma''(s)=(x_1''(s),x_2''(s),x_3''(s))$.

I am trying to define curvature of $\gamma$ using curvature of $\widetilde{\gamma}$.

Let $t\in I$ then, it is natural to define curvature of $\gamma$ at $t$ to be curvature of $\widetilde{\gamma}$ at $\varphi^{-1}(t)$ which we expect to be $$\frac{||\gamma'(t)\times\gamma''(t)||}{||\gamma'(t)||^3}.$$

We have $\widetilde{\gamma}=\gamma\circ \varphi$. So, $\widetilde{\gamma}'(\varphi^{-1}(t))=\gamma'(\varphi(\varphi^{-1}(t)))\varphi'(\varphi^{-1}(t))=\gamma'(t)\varphi'(\varphi^{-1}(t))$.

Just to get rid of those inverses, we set $\phi=\varphi^{-1}$ and we have $$\widetilde{\gamma}'(\phi(t))=\gamma'(t)\varphi'(\phi(t))=(x_1'(t)\varphi'(\phi(t)),x_2'(t)\varphi'(\phi(t)),x_3'(t)\varphi'(\phi(t))).$$ Computing its derivative, we have $$\widetilde{\gamma}''(\phi(t))=((x_1'(t)\varphi'(\phi(t)))',(x_2'(t)\varphi'(\phi(t)))',(x_3'(t)\varphi'(\phi(t)))').$$

I have computed $\gamma'(t)\times \gamma''(t)$ as $$(x_2'(t)x_3''(t)-x_2''(t)x_3'(t),x_3'(t)x_1''(t)-x_3''(t)x_1'(t),x_1'(t)x_2''(t)-x_1''(t)x_1'(t)).$$

I do not see that this is going anywhere.

Any help is welcome.

  • $\begingroup$ Do you mean to say that $\gamma$ has a constant speed? $\endgroup$ – Ben Grossmann Mar 13 '17 at 13:56
  • $\begingroup$ No, it is not. @Omnomnomnom $\endgroup$ – user312648 Mar 13 '17 at 14:03

Note that $\phi(t) = \int_0^s \|\gamma'(t)\|\,dt$. In your notation, we must have $\gamma = \tilde \gamma \circ \phi$. Applying the chain rule, we find that $$ \gamma'(t) = \phi'(t) \cdot \tilde \gamma'(\phi(t))\\ \gamma''(t) = \phi''(t) \cdot \tilde \gamma'(\phi(t)) + [\phi'(t)]^2 \cdot \tilde \gamma''(\phi(t)) $$ Note, however, that $\tilde \gamma'$ and $\tilde \gamma''$ are mutually orthogonal (and yield the tangent and normal unit vectors respectively). Verify the cross product formula, noting that $\tilde \gamma ' \times \tilde \gamma ' = 0$, and $\|\tilde \gamma' \times \tilde \gamma ''\| = 1$.

  • $\begingroup$ Note: it is not necessary to write out the cross product in coordinates. In particular, note that the cross product distributes like any other product, so that $a \mathbf u \times (b\mathbf u + c\mathbf v) = ab(\mathbf u \times \mathbf u) + ac (\mathbf u \times \mathbf v)$ $\endgroup$ – Ben Grossmann Mar 13 '17 at 14:13
  • $\begingroup$ I do not understand your first line.. Is your $\varphi$ is same as the function $\varphi:J\rightarrow I$ I have written in question? $\endgroup$ – user312648 Mar 13 '17 at 14:31
  • $\begingroup$ Yes${}{}{}{}{}{}{}$ $\endgroup$ – Ben Grossmann Mar 13 '17 at 14:36
  • $\begingroup$ I have defined $\widetilde{\gamma}$ to be the composition $\gamma\circ \varphi$. How can it happen that $\gamma=\widetilde{\gamma}\circ \varphi$? $\endgroup$ – user312648 Mar 13 '17 at 14:39
  • $\begingroup$ Oh you're right. My $\varphi$ is your $\phi$, it would seem $\endgroup$ – Ben Grossmann Mar 13 '17 at 14:40

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