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

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.

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 $$K$$.

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 $$V(r) = \frac{L^2}{2mr^2} + U(r).$$

• This makes sense, so it is simply out of 'convenience' that we classify those terms as part of the potential energy?
– XXb8
Oct 19, 2021 at 18:04
• Also, given $r=r(t)$, wouldn't the $L^2/smr^2$ part also be function of t since it includes the $r^2$ term? Wouldn't this make it a part of the kinetic energy according to your classification?
– XXb8
Oct 19, 2021 at 18:06
• @XXb8. Yes, one can say that it's out of 'convenience'. For your latter question, note that I write time derivative, not time dependency. We have the same situation in $E=\frac12 m\dot{x}^2 + \frac12 kx^2.$ When $x$ varies over time, both terms will vary. But the form of the terms differ. Only the first depends on the time derivative. That makes it kinetic. Oct 19, 2021 at 18:31

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 $$x:=r,\,V(x):=U(x)+\frac{L^2}{2mx^2}$$.

• Doesn't this answer again imply that $L^2/2mr^2$ is part of the kinetic energy from the above derivation?
– XXb8
Oct 19, 2021 at 8:10
• @XXb8 The kinetic/potential distinction is just an arbitrary convention based on what's convenient for the task at hand (see also the discussion here, especially of the Routhian). The point of an effective potential is to ask which $1$-dimensional analogue correctly predicts the behaviour from $E=\frac12m\dot{x}^2+V$.
– J.G.
Oct 19, 2021 at 8:24
• 'The kinetic/potential distinction is just an arbitrary convention based on what's convenient for the task at hand' This may be true in the present case but does not apply to for example the electromagnetic or the gravitational potential. Oct 19, 2021 at 9:21
• @my2cts Funnily enough, the example at hand can apply to both those interactions. The arbitrariness I refer to is whether to divide based on "effective" definitions or "the other kind". We have three terms, one effectively kinetic, one potential, one part of the kinetic but also part of the effective potential.
– J.G.
Oct 19, 2021 at 9:37
• True, but there is no way that the Coulomb or Newton potential can be effectively seen as 'kinetic', as far as I know. Oct 19, 2021 at 13:22