The point \(ho=R\) (i.e., the boundary of the space) is always reached, whatever the initial conditions on the energy and the angular momentum. If \(A=0\), the angular momentum vanishes and the motion is linear and sinusoidal. If \(\gamma=1\), the motion is uniform on a circle of radius \(R\).

Consider the case of free motion on the three-dimensional "spherical" space of a sphere \(\mathbf{S}^{3}\) imbedded in \(\mathcal{R}^{4}\)

Obviously, the volume of this space is finite since \(ho^{2}=x^{2}+y^{2}+z^{2} \leq R^{2}\).

In spherical coordinates, the Lagrangian of the problem is

\[
\mathcal{L}=\frac{m}{2}\left(\dot{ho}^{2}\left(\frac{R^{2}}{R^{2}-ho^{2}}ight)+ho^{2} \dot{\theta}^{2}+ho^{2} \sin ^{2} \theta \dot{\phi}^{2}ight) .
\]

Prove the conservation laws of the problem which bring simplifications to the motion:

1. There is rotational invariance. The angular momentum is conserved, and the motion occurs on a plane.

2. We can choose the direction of the angular momentum as polar axis; i.e., \(\theta=\pi / 2\) and \(\dot{\theta}=0\).

3. The Lagrangian of the planar motion therefore reduces to

\[
\mathcal{L}=\frac{m}{2}\left(\dot{ho}^{2}\left(\frac{R^{2}}{R^{2}-ho^{2}}ight)+ho^{2} \dot{\phi}^{2}ight)
\]

4. The conservation of angular momentum, second Kepler's law, results in

\[
\frac{d}{d t}\left(ho^{2} \dot{\phi}ight)=0 \Longrightarrow \dot{\phi}=\frac{A}{ho^{2}}
\]

where \(A\) is a constant, fixed by the initial conditions.

5. The energy, which is a constant of the motion, is therefore

\[
E=\frac{m}{2}\left(\dot{ho}^{2} \frac{R^{2}}{R^{2}-ho^{2}}+\frac{A^{2}}{ho^{2}}ight)
\]

The two constants of the motion \(E\) and \(A\) satisfy the inequality

\[
A^{2} \leq \frac{2 R^{2} E}{m}
\]

which is a direct consequence of the fact that the energy is greater than the rotational energy \(m A^{2} / 2 ho^{2}\). This is a consequence of (7.135); i.e., \(E \geq m A^{2} / 2 ho^{2} \geq\) \(m A^{2} / 2 R^{2}\).

The Eqs. (7.134) and (7.135) are first-order differential equations that determine the motion in terms of the constants of the motion \(E\) and \(A\).

The solution is simple. We define parameters \(\omega\) and \(\gamma\) by

\[
\omega^{2}=\frac{2 E}{m R^{2}} \quad \text { and } \quad \gamma^{2}=\frac{m A^{2}}{2 E R^{2}}
\]

From (7.136), we have the inequality

\[
\gamma^{2} \leq 1
\]

Setting

\[
ho=R \cos (\omega \psi) ; \quad \text { i.e., } \quad \dot{ho}=-\omega \dot{\psi} R \sin (\omega \psi) \text {. }
\]

If we insert this in Eq. (7.135), we obtain

\[
\omega^{2}=\omega^{2} \dot{\psi}^{2}+\frac{\omega^{2} \gamma^{2}}{\cos ^{2}(\omega \psi)}
\]

i.e.,

\[
\omega^{2} \dot{\psi}^{2} \cos ^{2}(\omega \psi)=\omega^{2}\left(\cos ^{2}(\omega \psi)-\gamma^{2}ight)
\]

We now make the change of functions

\[
\sin (\omega \psi(t))=\sqrt{1-\gamma^{2}} u(t) ; \quad \text { therefore } \cos ^{2}(\omega \psi)=1-\left(1-\gamma^{2}ight) u^{2} \text {. }
\]

Show that the choice

\[
u(t)=\sin (\omega \zeta(t))
\]

leads with no difficulty to:

\[
\dot{\zeta}^{2}=1, \quad \text { namely } u=\sin \left(\omega\left(t-t_{0}ight)ight)
\]

and to the result, i.e. The expressions of \(ho(t), \tan \left(\phi(t)-\phi_{0}ight)\) as well as the frequency \(\omega\).

\[
ho(t)=R \sqrt{\cos ^{2} \omega\left(t-t_{0}ight)+\gamma^{2} \sin ^{2} \omega\left(t-t_{0}ight)}
\]

which is periodic and of frequency \(\omega\). The calculation of the time evolution of the azimuthal angle \(\phi(t)\) is obtained by this expression and Eq. (7.134),

\[
\dot{\phi}=\frac{A}{R^{2}\left(\cos ^{2} \omega\left(t-t_{0}ight)+\gamma^{2} \sin ^{2} \omega\left(t-t_{0}ight)ight)} ;
\]

i.e.,

\[
\tan \left(\phi(t)-\phi_{0}ight)=\gamma \tan \omega\left(t-t_{0}ight)
\]

which is also periodic and of frequency \(\omega\).

We conclude the following.