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\begin{document}

\noindent Solution for assignment 14: Problems 1 and 2, 
Landau-Lifshitz p.24. \bigskip

1. The Lagrangians are
\begin{eqnarray} \nonumber
  L &=& \frac{1}{2}\,m\,\left(\frac{d \vec{r}}{d\,t}\right)^2-U\,,\\ 
  \nonumber
  L' &=& \frac{1}{2}\,m'\,\left(\frac{d \vec{r}}{d\,t'}\right)^2-U\,' 
\end{eqnarray}
After substituting $t'=t\,\sqrt{m'/m}$ in $L'$ the Lagrangians agree
and the same paths are obtained for
$$ t'/t\ =\ \sqrt{m'/m}\ .$$

2. The Lagrangians are
\begin{eqnarray} \nonumber
  L &=& \frac{1}{2}\,m\,\left(\frac{d \vec{r}}{d\,t}\right)^2-U\,,\\ 
  \nonumber
  L' &=& \frac{1}{2}\,m\,\left(\frac{d \vec{r}}{d\,t'}\right)^2-U'\,' 
\end{eqnarray}
Assume $U'=\alpha U$. After substituting $t'=t/\sqrt{\alpha}$ we obtain
$$ L'\ =\ \alpha\,L\,. $$
Therefore the same paths are obtained for
$$ t'/t\ =\ \alpha^{-1/2}\ =\ \sqrt{U/U'}\,.$$

\end{document}
