Effective potential of a two-body system

Question

This question is with regards to a two body system consisting of two masses that interact via a conservative central force.

In polar coordinates, the Lagrangian can be written as

$$\frac{1}{2}M\dot{\vec{R}}^2+\frac{1}{2}\mu(\dot{r}^2+r^2\dot{\phi}^2)-U(r).\tag{1}$$

The $r$ equation is then

$$\mu \ddot{r}=-\frac{dU_{eff}(r)}{dr},\tag{2}$$

where

$$U_{eff}(r)=\frac{\ell^2}{2\mu r^2}+U(r).\tag{3}$$

($\ell\equiv$ angular momentum)

Simple analysis shows that the first term in $U_{eff}$ is simply $$(1/2) I\omega^2,\tag{4}$$ which is the rotational kinetic energy.

My question is, what is the physical significance / meaning / reason for combining the rotational kinetic energy with the potential energy as opposed to with the translational kinetic energy? That is, why define an effective potential energy instead of an effective kinetic energy?

Answer 1

The motivation for writing the rotational kinetic energy as potential term is to reduce the number of variables needed to describe the system.

Let us forget about the centre of mass degrees of freedom since they are not playing any role here. There are initially three degrees of freedom which can be the spherical coordinates $r$, $\theta$ and $\phi$ of one of the particles relative to the other. As for any central force field, the angular momentum vector is constant so the motion of the system lies on a plane. This reduce the number of degrees of freedom by one. Moreover, the magnitude of the angular momentum $$L=\mu r^2\dot\phi,$$ is also conserved so whenever there is a $\dot\phi$ in the lagrangian (or in the kinetic energy) of the system we can write in terms of a quantity that depends only on $r$. The system is left therefore with only one variable, the coordinate $r$. It is useful to interpret this rotational kinetic energy $$\frac{\mu r^2\dot\phi^2}{2}=\frac{L^2}{2\mu r^2},$$ as a potential term $V_{\mathrm{ce}}$. In fact this is called the centrifugal potential since its gradient gives a centrifugal force, $$F_{r}^{\mathrm{ce}}=-\frac{\partial V(r)_{\mathrm{ce}}}{\partial r}=\frac{L^2}{\mu r^3}=|-\mu\vec\Omega\times(\vec\Omega\times\vec r)|,$$ where $\vec\Omega=\dot\phi \vec e_k$.

Answer 2

For a central force interaction, the angular momentum of the system will be constant. That means that the first term, $$ \frac{\ell^2}{2\mu r^2},$$ depends only on $r$. The differential equation in the $r$ coordinate, consequentially, doesn't depend on any functions of $\theta$.

If you replace that term with $\frac{1}{2}I\omega^2$, it's not as obvious that you have a differential equation purely in $r$ because $\omega = \dot{\theta}$.

Answer 3

It seems worth pointing out that the centrifugal potential (3) and (4) has the opposite sign of what one might naively expect if one keeps in mind

• (i) the original Lagrangian (1) and

• (ii) the fact that kinetic and potential terms appear opposite in a Lagrangian.

This peculiarity is explained in this Phys.SE post. The philosophy of the centrifugal potential is explained in e.g. this Phys.SE post.