201911190336 Homework 1, Differential Geometry

Let $\boldsymbol{\alpha} :I\rightarrow \mathbb{R}^3$ be a regular parametrized curve (not necessarily by arc length) and let $\beta :J\rightarrow \mathbb{R}^3$ be a reparametrization of $\boldsymbol{\alpha}$ by the arc length $s=s(t)$ measured from $t_0\in I$ .

Let also $t=t(s)$ be the inverse function of $s$ and denote the derivative of $\boldsymbol{\alpha}$ wrt $t$ by $\boldsymbol{\alpha}'$ . Prove that

i. $\mathrm{d}t/ \mathrm{d}s=1/\|\boldsymbol{\alpha}'\|$ and $\mathrm{d}^2t/ \mathrm{d}s^2=-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}''\rangle /\|\boldsymbol{\alpha}'\|^4$ ;

ii. The curvature of $\boldsymbol{\alpha}$ at $t\in I$ is $\kappa (t)=\displaystyle{\frac{\|\boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|}{\|\boldsymbol{\alpha}'\|^3}}$ ; and

iii. The torsion of $\boldsymbol{\alpha}$ at $t\in I$ is $\tau (t)=\displaystyle{-\frac{\langle \boldsymbol{\alpha}' \wedge \boldsymbol{\alpha}'',\boldsymbol{\alpha}'''\rangle }{\| \boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|^2}}$ .

Solution.

i. By definition $\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}=\boldsymbol{\alpha}'$ .

Using chain rule,

$\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}= \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}$ .

After taking the norm, as a consequence of natural parametrization

(i.e., $\bigg| \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\bigg| =1$ ),

we have

$\| \boldsymbol{\alpha}'\|=\bigg| \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}} \bigg| \bigg| \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}} \bigg|= \bigg| \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}\bigg|$ .

Hence $\mathrm{d}t/\mathrm{d}s=1/ \| \boldsymbol{\alpha}'\|$ .

That said,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}} & =\bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg) \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg) \\ & =\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\displaystyle{\frac{1}{\|\boldsymbol{\alpha}' \|}}\bigg) \\ & =\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\bigg)\\ \end{aligned}$ .

But,

$\begin{aligned} & \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg(\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\bigg)\\ & =-\displaystyle{\frac{1}{2}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\Big( \langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle +\langle \boldsymbol{\alpha}'',\boldsymbol{\alpha}' \rangle \Big) \\ & =-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ \end{aligned}$ .

Thus,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}} & =-\displaystyle{\frac{1}{\| \boldsymbol{\alpha}' \|}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ & =-\displaystyle{\frac{1}{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{1/2}}}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}' \rangle^{-3/2}\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle \\ & = \displaystyle{-\frac{\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}'' \rangle}{\big( \langle \boldsymbol{\alpha}', \boldsymbol{\alpha}'\rangle^{1/2}\big)^4}} \\ & =-\langle \boldsymbol{\alpha}',\boldsymbol{\alpha}''\rangle /\|\boldsymbol{\alpha}'\|^4 \\ \end{aligned}$ .

ii.

$\boldsymbol{\alpha}'= \displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}=\dot{\boldsymbol{\alpha}}s'$ .

Besides,

$\boldsymbol{\alpha}''=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\dot{\boldsymbol{\alpha}}s')=\dot{\boldsymbol{\alpha}}\displaystyle{\frac{\mathrm{d}s'}{\mathrm{d}t}}+s'\displaystyle{\frac{\mathrm{d}\dot{\boldsymbol{\alpha}}}{\mathrm{d}t}}=\dot{\boldsymbol{\alpha}}s''+(s')^2\ddot{\boldsymbol{\alpha}}$ .

Then,

$\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' \rangle =\big( \dot{\boldsymbol{\alpha}}s' \big) \times \big( \dot{\boldsymbol{\alpha}}s''+(s')^2\ddot{\boldsymbol{\alpha}} \big) =(s')^3\langle\dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}} \rangle =\|\boldsymbol{\alpha}'\|^3 \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}} \rangle$ .

(For $\mathrm{d}s/\mathrm{d}t=\| \boldsymbol{\alpha}'\|$ of part i. is used.)

It follows that

$\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'' \rangle =\|\boldsymbol{\alpha}\|^3 \| \dot{\boldsymbol{\alpha}} \|\| \ddot{\boldsymbol{\alpha}}\|\sin \measuredangle (\dot{\boldsymbol{\alpha}},\ddot{\boldsymbol{\alpha}})$ .

Note that $\dot{\boldsymbol{\alpha}}=\mathbf{t}$ and $\ddot{\boldsymbol{\alpha}}=\dot{t}$ are orthogonal,

$\| \boldsymbol{\alpha}\|=1$ and $\| \ddot{\boldsymbol{\alpha}}\|=\| \dot{\mathbf{t}}\|=\| \kappa\|$ .

We obtain

$\kappa (t)=\displaystyle{\frac{\|\boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|}{\|\boldsymbol{\alpha}'\|^3}}$ .

iii.

First,

$\dot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}s}}=\displaystyle{\frac{\mathrm{d}\boldsymbol{\alpha}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}=\boldsymbol{\alpha}'\dot{t}$ ;

secondly,

$\ddot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}(\boldsymbol{\alpha}'\dot{t})=\boldsymbol{\alpha}'\ddot{t}+\bigg( \displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\boldsymbol{\alpha}' \bigg)\dot{t}=\boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2$ ;

thirdly,

$\dddot{\boldsymbol{\alpha}}=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\big( \boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2 \big) =\boldsymbol{\alpha}'\dddot{t}+\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}''2\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3=\boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3$ .

Compute $\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle$ as follows:

$\begin{aligned} \langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle & = \big( \boldsymbol{\alpha}'\dot{t} \big) \wedge \big( \boldsymbol{\alpha}'\ddot{t}+\boldsymbol{\alpha}''\dot{t}^2 \big) \cdot \big( \boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3 \big) \\ & =\big( \boldsymbol{\alpha}'\dot{t}\wedge \boldsymbol{\alpha}''\dot{t}^2\big) \cdot \big( \boldsymbol{\alpha}'\dddot{t}+2\boldsymbol{\alpha}''\dot{t}\ddot{t}+\boldsymbol{\alpha}'''\dot{t}^3 \big)\\ \end{aligned}$

Now that the cross product $\boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}''$ is orthogonal to both $\boldsymbol{\alpha}'$ and $\boldsymbol{\alpha}''$ , we can ignore the dot product among them and what remains is

$(\boldsymbol{\alpha}'\dot{t}\wedge \boldsymbol{\alpha}''\dot{t}^2)\cdot \boldsymbol{\alpha}'''\dot{t}^3$ , or,

$\dot{t}^6\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle$ .

Using the result of part i., substitute $1/\| \boldsymbol{\alpha}'\|$ for $\dot{t}$ , we obtain

$\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}}, \dddot{\boldsymbol{\alpha}} \rangle=\displaystyle{\frac{\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle}{\|\boldsymbol{\alpha}'\|^6}}$ .

Using the formula for curvature in part ii. and put it into

$\tau =\displaystyle{\frac{\langle \dot{\boldsymbol{\alpha}}\wedge \ddot{\boldsymbol{\alpha}},\dddot{\boldsymbol{\alpha}} \rangle}{\kappa^2}}=\displaystyle{\frac{\langle \boldsymbol{\alpha}'\wedge \boldsymbol{\alpha}'', \boldsymbol{\alpha}'''\rangle}{\kappa^2 \|\boldsymbol{\alpha}'\|^6}}$ ,

i.e.,

$\tau (t)=\displaystyle{-\frac{\langle \boldsymbol{\alpha}' \wedge \boldsymbol{\alpha}'',\boldsymbol{\alpha}'''\rangle }{\| \boldsymbol{\alpha}' \wedge\boldsymbol{\alpha}''\|^2}}$ .

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