I am trying to prove the following theorem.
Let $\gamma=\gamma(s)$ be a regular space curve, parameterised by its arc-length $s$. If $\kappa=0$ is the curvature of $\gamma$, then $\gamma(s)$ is a straight line.
Now I proceeded to use the first Serret-Frenet equation $\mathbf t'=\kappa \mathbf n$ which gives us $\mathbf t'=\mathbf 0$, in other words, $\mathbf t$ is a constant vector. Clearly if a curve has a constant tangent vector, then it must be a straight line.
When I looked up my lecturer's proof, he proved it similarly but performed an additional step:
Now $$\frac{\mathrm d}{\mathrm d s}(\mathbf r\times\mathbf t)=\mathbf t\times\mathbf t+\mathbf r\times\mathbf t'=\mathbf 0,$$ so $\mathbf r\times\mathbf t$ is also a constant vector, i.e. $\mathbf r\times\mathbf t=\mathbf a$ where $\mathbf a$ is some constant vector. This corresponds to the equation of a straight line.
It is this step which I do not understand. How does this correspond to the equation of a straight line? Shouldn't that be $\mathbf r=\mathbf a+t\mathbf b$? And is this additional step necessary to prove the result?
Best Answer
I'm with you, although your sentence starting with "Clearly" could use an explicit proof. Namely, from $\mathbf t=\mathbf c$ constant, you should integrate to get $\boldsymbol\gamma(s)=s\mathbf c + \mathbf d$.
All I see on first principles from your lecturer's proof is that $\mathbf r$ lies in the plane through the origin with normal vector $\mathbf a$.