Reparameterization of ODE 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?
 A: 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.
A: 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 $\xi=a  \lambda+b$.
