3.1 Spherically symmetric potential - Conservation laws

3.2 Conservation Laws

Lagrangian for a body moving in spherically symmetric potential is given by

$$ L = \frac 1 2 \mu \dot{\vec{r}^2} - V(r) \hspace{5em} (8) $$

where we have used the notation \( r = | \vec x |. \) The Lagrangian does not contain time explicitly, hence we obtain energy conservation

$$ \frac 1 2 \mu \dot{\vec{x}^2} + V(r) = E \hspace{2em} (constant). \hspace{5em} (9) $$

The Lagrangian is also invariant under rotations about any axis and in particular about the coordinate axes. This gives us conservation of angular momentum. Thus we have

$$ \vec L = \mu \vec x \times \vec v = constant \;\; of \;\; motion \hspace{5em} (10) $$

3.3 Orbits lie in a plane

We shall now make use of the conservation laws to give solution of motion in a spherically symmetric potential to quadratures.

Since \( \vec L = \mu \vec x \times \vec v \) is a constant of motion the magnitude as the direction of \( \vec L \) does not change with time. Also \( \vec r \) and \( \vec v \) always perpendicular to \( \vec L \) which points in a fixed direction. Hence \( \vec r \) and \( \vec v \) remain in the plane perpendicular to \( \vec L \). Therefore for a particle in a spherically symmetric potential, the motion is confined to a plane.

If \( \vec L \) is zero, then \( \vec r \times \vec v = 0 \) and \( \vec r \) will always be parallel to \( \vec v \) and the particles moves in a straight line.

3.4 Using Plane Polar coordinates

Since the potential depends only on \( r \), we work with plane polar coordinates. We start with Lagrangian for a particle in two dimensions in plane polar coordinates

$$ L = \frac 1 2 \mu \dot{\vec{r}^2} + \frac 1 2 \mu \vec r^2 \dot{\phi^2} - V(r). \hspace{5em} (11) $$

3.5 \( \phi \) is a cyclic coordinate

Since \( \dot{\phi} \) is a cyclic coordinate we have

$$ \frac {\partial L} {\partial \dot{\phi}} = \mu r^2 \dot{\phi} = constant, say \; L. \hspace{5em} (12) $$

\( \mu r^2 \dot{\phi} \) is in fact seen to be equal to the magnitude of angular momentum.

3.6 Energy conservation

The expression for energy, associated with relative motion, given by

$$ E = \frac 1 2 \mu \dot{\vec{r}^2} + \frac 1 2 \mu \vec r^2 \dot{\phi^2} + V(r). \hspace{5em} (13) $$

The energy is conserved because the Lagrangian (11) is independent of time.

3.7 Effective potential

The angular velocity, \( \dot{\phi} \), can be eliminated using

$$ \dot{\phi} = \frac L {2 \mu r^2} \hspace{5em} (14) $$

and the total energy can be written in terms of \( r \) and \( \dot r \) only. Making use of (13) and (14) we get,

$$ E = \frac 1 2 \mu {\dot r}^2 + \frac {L^2} {2 \mu r^2} + V (r) = \frac 1 2 \mu {\dot r}^2 + V_{eff} (r) \hspace{5em} (15) $$

where we have introduced the notation

$$ \boxed{V_{eff} (r) = V (r) + \frac {L^2} {2 \mu r^2}} \hspace{5em} (16) $$