For my class in classical differential geometry I have to solve the following problem:
Let $\alpha$ be a regular curve in $\mathbb{R}^2$. Proof that:
a) $\quad$ $\alpha$ is a straight line if and only if the curvature vanishes.
b) $\quad$ If $\dot{\alpha}$ and $\ddot{\alpha}$ are linearly dependent, then $\alpha$ is a straight line.
Conisder a). At first this problem seemed easy to me. Because if the curvature is defined via
$$\ddot{\alpha}(t) = \kappa(t) \cdot n(t) \qquad \qquad (1)$$
where $n(t)$ is the unit vector perpendicular to the tangent vector, then it is easy to see that $\kappa = 0$ implies $\ddot{\alpha} = 0$ and therefore $\alpha$ is a straight line. However, I realsied, that the above equation is derived by using the fact
$$\big< \dot{\alpha} , \dot{\alpha} \big> = 1 \quad \Longrightarrow \quad \big< \dot{\alpha}, \ddot{\alpha} \big> = 0$$
in the case of $\alpha$ being parameterised by arc length. Since my given $\alpha$ is not parameterised by arc length, I cannot assume equation (1) to hold and therefore cannot assume my $\ddot{\alpha}$ to be perpendicular to the tangent vector. My conclusion would be wrong then. Any ideas on how to solve this problem for general curves $\alpha$?
b) I have no idea on how to show this. Any tips here?
Best Answer
Bearing in mind that any regular curve can be parametrized by the arc length then
a) if $\alpha$ is a straight line, i.e., $\alpha(t)=\text{p}+\text{v}t$ and so $k(t)=||\alpha''(t)||=||0||=0$.
If $k(t)=0$ then $\alpha''(t)=0$. If you calculate the antiderivative twice you obtain $\alpha'(t)=\text{v}$ and finally $\alpha(t)=\text{p}+\text{v}t$ for some $\text{p,v}\in\mathbb{R^2}$.
b) As you saw $\langle\alpha',\alpha''\rangle=0$ then $\alpha''$ is null vector or $\alpha'$ is perpendicular to $\alpha''$, but as they are linear dependent then $\alpha''=0$ and by a) it can only be a straight line.