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

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