According to some lecture notes for my new course, the following is "easily seen":
Let $S$ be a curve parameterized by $\lambda$, so that $S = S (\lambda)$, satisfying $\frac{d^2 S}{d \lambda^2}=0$. Now, changing the parameterization $\xi = \xi (\lambda)$ and demanding that $S(\xi)$ satisfy: $\frac{d^2 S}{d \xi^2}=0$, it follows that $\xi = a \lambda +b$, where $a,b$ are constants.
Can anyone explain how they arrive at this?
On
Using the chain rule, one has: $$\frac{\mathrm{d}(S\circ\xi)}{\mathrm{d}\lambda}=\frac{\mathrm{d}\xi}{\mathrm{d}\lambda}\frac{\mathrm{d}S}{\mathrm{d}\xi}\circ\xi.$$ From there, using the chain rule once more, one gets: $$\frac{\mathrm{d}^2(S\circ\xi)}{\mathrm{d}\lambda^2}=\frac{\mathrm{d}^2\xi}{\mathrm{d}\lambda^2}\frac{\mathrm{d}S}{\mathrm{d}\xi}\circ\xi+\left(\frac{\mathrm{d}\xi}{\mathrm{d}\lambda}\right)^2\frac{\mathrm{d}^2S}{\mathrm{d}\xi^2}\circ\xi.$$ Hence, using the assumptions, one gets: $$\frac{\mathrm{d}^2\xi}{\mathrm{d}\lambda^2}=0.$$ Whence the result.
It just a bit of the old chain rule ... \begin{eqnarray*} \color{red}{\frac{d^2 S}{d \lambda ^2}} =\frac{ d \xi}{d \lambda } \frac{d }{d \xi } \left( \frac{ d \xi}{d \lambda } \frac{dS }{d \xi }\right) = \left( \frac{ d \xi}{d \lambda } \right)^2 \color{blue}{\frac{d^2 S}{d \xi ^2}}+\frac{ d \xi}{d \lambda } \frac{d }{d \xi } \left( \frac{ d \xi}{d \lambda } \right) \frac{dS }{d \xi } \end{eqnarray*} Now by hypothesis $ \color{red}{\frac{d^2 S}{d \lambda ^2} =0}$ and $\color{blue}{ \frac{d^2 S}{d \xi ^2}=0}$. So we require \begin{eqnarray*} \frac{d }{d \xi } \left( \frac{ d \xi}{d \lambda } \right) =0. \end{eqnarray*} Which easily integrates to give $\xi=a \lambda+b$.