Halbebene/Poincaré/Riemannsche Geometrie/Geodätische/Halbkreislösungen/Aufgabe/Lösung

Es ist

${}{\begin{aligned}\gamma '(t)&={\begin{pmatrix}s+r{\frac {\sinh t}{\cosh t}}\\{\frac {r}{\cosh t}}\end{pmatrix}}'\\&=r{\begin{pmatrix}{\frac {\cosh t-\sinh t}{\cosh t}}\\-{\frac {\sinh t}{\cosh t}}\end{pmatrix}}\\&=r{\begin{pmatrix}{\frac {1}{\cosh t}}\\-{\frac {\sinh t}{\cosh t}}\end{pmatrix}}\\&={\frac {r}{\cosh t}}{\begin{pmatrix}1\\-\sinh t\end{pmatrix}}\end{aligned}}$

und

${}{\begin{aligned}\gamma ^{\prime \prime }(t)&=r{\begin{pmatrix}{\frac {1}{\cosh t}}\\-{\frac {\sinh t}{\cosh t}}\end{pmatrix}}'\\&=r{\begin{pmatrix}{\frac {-2\sinh t}{\cosh t}}\\{\frac {-\cosh t+2\sinh t\cosh t}{\cosh t}}\end{pmatrix}}\\&=r{\begin{pmatrix}{\frac {-2\sinh t}{\cosh t}}\\{\frac {-\cosh t+2\sinh t}{\cosh t}}\end{pmatrix}}.\end{aligned}}$

Wegen

${}{\frac {2}{\frac {r}{\cosh t}}}\cdot {\frac {r}{\cosh t}}\cdot (-1){\frac {r\sinh t}{\cosh t}}=-{\frac {2r\sinh t}{\cosh t}}=\gamma _{1}^{\prime \prime }(t)\,$

stimmt die erste Gleichung. Wegen

${}{\begin{aligned}-{\frac {1}{\gamma _{2}(t)}}\gamma _{1}'(t)\gamma _{1}'(t)+{\frac {1}{\gamma _{2}(t)}}\gamma _{2}'(t)\gamma _{2}'(t)&={\frac {\cosh t}{r}}{\left(-{\left({\frac {r}{\cosh t}}\right)}^{2}+{\left({\frac {r\sinh t}{\cosh t}}\right)}^{2}\right)}\\&={\frac {r{\left(-1+\sinh t\right)}}{\cosh t}}\\&={\frac {r{\left(-\cosh t+2\sinh t\right)}}{\cosh t}}\\&=\gamma _{2}^{\prime \prime }(t)\end{aligned}}$

ist auch die zweite Gleichung erfüllt.

Zur gelösten Aufgabe