[NOTES/CM-05006] Effective Potential for Spherically Symmetric Problems

Angular momentum conservation
Lagrangian for a body moving in spherically symmetric potential is given by
\begin{equation}\label{EQ05}{
\mathcal L} = \frac{1}{2} \mu \dot{\vec{r}}^{\,2} - V(r)
\end{equation}
where we have used the notation \(r=|\vec{x}|\). The Lagrangian is invariant under rotations about any axis and in particular about the coordinate axes. This gives us conservation of angular momentum. Thus we have
\begin{equation}\label{EQ07}
\vec{L} = \mu \vec{x} \times \vec{v} = \text{constant of motion}
\end{equation}

Orbits lie in a plane
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 particle moves in a straight line. 

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

\begin{equation}\label{EQ08}
\mathcal  L= \frac{1}{2} \mu \dot{\vec{r}}^{\,2} + \frac{1}{2} \mu r^{\,2}
\dot{\phi}^{2} -
V(r).
\end{equation}

\(\phi\) is a cyclic coordinate
Since $\dot{\phi}$ is a cyclic coordinate we have
\begin{equation}\label{EQ10} \frac{\partial {\mathcal L} }{\partial \dot{\phi}} = \mu r^{2}\dot{\phi} = \text{constant, say} L . \end{equation} $\mu r^{2}\dot{\phi}$ is in fact seen to be equal to the magnitude of angular momentum.

Energy conservation
The expression for energy  given by
\begin{equation}\label{EQ09}
E = \frac{1}{2}\mu \dot{r}^{2} + \frac{1}{2} \mu r^{2} \dot{\phi}^{2} +V(r).
\end{equation}
The energy is conserved because the Lagrangian \eqref{EQ08} is independent of time.

The Effective potential
The angular velocity, \(\dot{\phi}\), can be eliminated using
\begin{equation}\label{EQ10A} \dot{\phi} = \frac{L}{2\mu r^2} \end{equation}
and the total energy can be written in terms of \(r\) and \(\dot{r}\) only.
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Making use of \eqref{EQ09} and \eqref{EQ10A} we get,
\begin{equation}\label{EQ11}
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_\text{eff}(r)
\end{equation}
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where we have introduced the notation
\begin{equation}\label{EQ12}
\boxed{V_\text{eff}(r) = V(r) + \frac{L^{2}}{2\mu r^{2}}}.
\end{equation}
The radial motion is like motion of a particle in one dimension in effective potential \(V_\text{eff}(r)\). Solution to this problem can be reduced to a quadrature using energy conservation.