# Why Kepler problem is equivalent to a free particle on 4 dimensional sphere?

In trying to understand Laplace-Runge-Lenz vector, I read in Wikipedia that the Kepler problem is mathematically equivalent to a particle moving freely on the surface of a four dimensional hypersphere.

The Lagrangian of the free particle on the sphere is

$$\mathcal L=\frac m2R^2(\dot\phi_1^2+\dot\phi_2^2\sin^2\phi_1+\dot\phi_3^2\sin^2\phi_1\sin^2\phi_2)$$

where the spherical coördinates are

$$x_1=r\cos\phi_1,\ x_2=r\sin\phi_1\cos\phi_2,\ x_3=r\sin\phi_1\sin\phi_2\cos\phi_3,\ x_4=r\sin\phi_1\sin\phi_2\sin\phi_3$$

The equation of motion in Kepler problem is $$m\ddot r-mr\omega^2+\frac{2k}{r^3}=0$$

How can I show they are equivalent?

Another small question, they say, "This higher symmetry results from two properties of the Kepler problem: the velocity vector always moves in a perfect circle..."

What do they mean by velocity vector moves in a perfect circle? Velocity vecctor is always perpendicular to the path wich can also be ellipse, parabola and hyperbola.

• Your second question is addressed in the section "Circular momentum hodographs". You have to imagine taking all the momentum vectors at various points along the orbit, moving them to be based at the origin, and seeing the shape their tips trace out. Note that for unbounded orbits you have $A > mk$ and so the circles do not intersect the $p_x$-axis. Jul 27 at 15:06
• @MichaelSeifert Still can't see it. The size of the vectors change so how their tip follows a circle?
– EB97
Jul 27 at 15:24
• The circle isn't centered at the origin. It's centered at $(0, A/L)$ in the $p_xp_y$-plane, as shown in Figured 3 of the Wikipedia article. Jul 27 at 15:26
• John Baez discusses this here Jul 27 at 15:41