Why do we consider $L^2/(2mr^2)$ part of effective potential energy?

Question

The energy of a particle under the action of a radial conservative force is given by

$$E = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2+ \frac{L^2}{2mr^2} + U(r),$$

where the last two terms provide the effective potential energy. This is derived using:

$$v^2 = \left(\frac{dr}{dt}\right)^2 + \left(r\frac{d\theta}{dt}\right)^2,$$

and then substituting this expression into $E= \frac{1}{2}mv^2+U(r).$

Given that $L^2/(2mr^2)$ comes from the kinetic energy, why is it considered part of the potential energy of the system? Doesn't it directly relate to the velocity of the particle from the derivation? And isn't $\left(r\frac{d\theta}{dt}\right)^2$ is in the same tangential direction as the velocity?

I'm a first year calculus student, so this may be a simple misunderstanding.

Answer 1

Assume that we have solved the problem of the motion of the particle, so that $\theta=\theta(t)$ is a known function. Put yourself in a non-inertial reference frame $K'$ which rotates with respect to the inertial one $K$ with angular velocity $\vec{\omega}(t)= \dot{\theta}(t) \frac{\vec{L}}{L}$, where we have exploited the fact that, since the potential energy $U$ is radial, the motion in $K$ is in the plane orthogonal to $\vec{L}$, which is constant in $K$.

The kinetic energy in $K'$ is $$T|_{K'} = \frac{m}{2}\dot{r}^2$$ since there is no angular motion there.

In $K'$ some apparent forces take place in addition to the force associated to $U$. They are the centrifugal force, the Coriolis force, and the Euler force (see below).

The potential energy in $K'$ takes the potential energy of the apparent centrifugal force into account, $$U|_{K'}(r) = U(r) + \frac{m}{2}\omega^2r^2 = U(r) + \frac{m}{2}\dot{\theta}^2r^2 = U(r) + \frac{L^2}{2mr^2}\:.$$ Indeed, the centrifugal force reads $$\vec{f} = -m \vec{\omega}\wedge(\vec{\omega} \wedge re_r)) = m \dot{\theta}^2 r e_r = \frac{L^2}{mr^{3}}e_r = - \frac{d}{dr}\left( \frac{L^2}{2mr^2}\right) e_r\:.$$

In $K'$ all (real or apparent) forces are

(a) conservative: the force associated to $U(r)$ and the centrifugal force one,

or

(b) they do not dissipate work because they acts orthogonally to the motion, i.e., orthogonally to $e_r$: the Coriolis force $-2m \vec{\omega}\wedge \dot{r} e_r$ and the Euler force $-m\dot{\vec{\omega}}\wedge re_r $.

Therefore the total mechanical energy is conserved in $K'$: $$E|_{K'}= \frac{1}{2}m\dot{r}^2 + \left(U(r) + \frac{L^2}{2mr^2}\right) = constant$$ In $K'$, differently from $K$, the so-called "effective potential energy" is a true potential energy (though of an apparent force) and it is not part of the kinetic energy as instead it happens in $K$.

Answer 2

The short answer is that it's because it looks like a potential.

A term containing a time derivative (e.g. $\frac12 m\dot{x}^2$) is clearly kinetic. A term without a time derivative is often a potential (e.g. $\frac12 kx^2$).

In a 1-dimensional system the total energy usually has the form $E = \frac12 m\dot{x}^2 + V(x).$ This is a form we meet early in the education and it's relatively easy to get an understanding of it.

For a 3-dimensional system the total energy has the similar form $E = \frac12 m\dot{\mathbf{x}}^2 + U(\mathbf{x}).$ When the potential is spherically symmetric we can write it in spherical coordinates ($r=|\mathbf{x}|$) as $$ E = \frac12 m\dot{r}^2 + \frac{L^2}{2mr^2} + U(r) $$ where it has been used that angular momentum $L$ is constant. This formula is effectively 1-dimensional; we have only one space variable, $r.$ Therefore we can reason about it as a 1-dimensional equation, but then we must treat the $L$-term as part of the potential, i.e. set $$ V(r) = \frac{L^2}{2mr^2} + U(r). $$

Answer 3

When a spherically symmetric potential energy $U(r)$ causes a radial force $-\nabla U(r)=-U^\prime(r)\hat{r}$ on a mass-$m$ test particle, the angular momentum $\vec{L}=mr^2\dot{\theta}\hat{k}$ is conserved, so its motion is confined to a plane orthogonal to $\hat{k}$, and$$E=\frac12m\left(\dot{r}^2+r^2\dot{\theta}^2\right)+U(r)=\frac12m\dot{r}^2+\frac{L^2}{2mr^2}+U(r).$$Although the space is $3$-dimensional, the particle's motion is confined to a $2$-dimensional plane, and comprises a $1$-dimensional path therein. It is therefore convenient to parameterize position along that path by $r$, and compare the above $E$ to the energy of a particle in a $1$-dimensional motion under an arbitrary conservative force, namely $\frac12m\dot{x}^2+V(x)$. This comparison defines $x:=r,\,V(x):=U(x)+\frac{L^2}{2mx^2}$.