% Bernd Berg Oct 13 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 8} 

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\item {\bf 27. Motion in 2D plane.}
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We want to fulfill the differential equations
$$ \ddot{r}=r\,\dot{\theta}^2~~{\rm and}~~\ddot{\theta}=-
   \frac{2}{r}\,\dot{r}\,\dot{\theta}\,.$$
The definitions of
$$ r=\sqrt{y_0^2+s^2}~~{\rm and}~~\tan(\theta)=\frac{y_0}{s}\,,$$
imply the additional equations
$$ \sin(\theta)=\frac{y_0}{r}~~{\rm and}~~\cos(\theta)=\frac{s}{r}\,.$$
For the derivatives of $r$ and $\theta$ we find 
$$ \dot{r}=\frac{s}{(y_0^2+s^2)^{1/2}}=\frac{s}{r}=\cos(\theta)\,,
   \qquad\ddot{r}=-\dot{\theta}\,\sin(\theta)\,,$$
$$ \frac{d~}{ds}\,\tan(\theta)=\frac{\dot{\theta}}{\cos^2(\theta)}
   =-\frac{y_0}{s^2} = -\frac{1}{s}\,\tan(\theta)~~\Rightarrow
   ~~~ \dot{\theta} = -\frac{\cos(\theta)}{s}\,\sin(\theta) =
   - \frac{\sin(\theta)}{r}\,. $$
Inserting $\sin(\theta)=-r\,\dot{\theta}$ in the equation for $\ddot{r}$ 
we find the first differential equation $\ddot{r}=r\,\dot{\theta}^2$.

For the second derivative of $\theta$ we have
$$ \ddot{\theta} = \dot{r}\,\frac{\sin(\theta)}{r^2}
   - \frac{\dot{\theta}\,\cos(\theta)}{r}\,. $$
Inserting $\sin(\theta)=-r\,\dot{\theta}$ and $\cos(\theta)=\dot{r}$
we find the second differential equation:
$$ \ddot{\theta}=-\frac{\dot{r}\,\dot{\theta}}{r}-\frac{\dot{r}\,
   \dot{\theta}}{r}=-\frac{2\,\dot{r}\,\dot{\theta}}{r}\,.$$

The motion $x=s$, $y=y_0$ in the $x$-$y$ plane is described by this 
solution.
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