Is there any motion hidden within potential energy?

[Bob Enyart]

Of potential and kinetic energy in their various forms, in their own reference frames, which involve motion? Heat, light, nuclear, kinetic, etc., seem to involve motion. Does potential energy, in any way whatsoever, involve motion? Thermal does. Does nuclear energy involve motion? Seems to because the nucleus is in motion and within the nucleons so are their quarks. So how about gravitational energy, including as potential energy? Do these involve motion in any way?

 

[etotheipi]

What you say is mostly nonsense; what is the 'own reference frame' of energy, for instance?

That said, there is one point to make. In standard conservative systems the potential energy functions depend only on the co-ordinates, $V = V(\boldsymbol{q})$, and not the derivatives of the co-ordinates $\dot{\boldsymbol{q}}$.

However, there are some special non-conservative systems for which, if the generalised forces $Q_j$ can be written $Q_j = \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial U / \partial \dot{q}^j \right) - \partial U / \partial q^j$ for some function $U = U(\boldsymbol{q}, \dot{\boldsymbol{q}}, t)$, then this function $U$ is called a velocity dependent potential. For example, the Lorentz force can be described by a velocity dependent potential $U = q \phi - q \dot{\boldsymbol{x}} \cdot \mathbf{A}$, where $\mathbf{A}$ is the vector potential. It is a good exercise to show that this potential does indeed yield the correct generalised forces $q(\mathbf{E} + \dot{\boldsymbol{x}} \times \mathbf{B})_j$!

 

[russ_watters]

Of potential and kinetic energy in their various forms, in their own reference frames, which involve motion? Heat, light, nuclear, kinetic, etc., seem to involve motion. Does potential energy, in any way whatsoever, involve motion? Thermal does. Does nuclear energy involve motion? Seems to because the nucleus is in motion and within the nucleons so are their quarks. So how about gravitational energy, including as potential energy? Do these involve motion in any way?

No, none of those except kinetic energy and certain types of thermal energy/heat involve motion.

 

[kuruman]

It's not clear to me what you mean when you ask whether potential energy "involves" motion. In your chosen example of gravitational potential energy, the expression for it is not an explicit function of some velocity or speed, but a function of position only. When a ball is held at some height above the Earth, the potential energy of the Earth-ball system does not change. When the ball is released and allowed to fall, the potential energy changes as the relative position changes but it is still a function of position only. Arguably then, a change in potential energy implies relative motion between Earth and ball. Conversely, if there is no relative motion between Earth and ball, the potential energy does not change. So relative motion is a necessary and sufficient condition for potential energy change. Is that "involved" enough?

 

[Ibix]

Does potential energy, in any way whatsoever, involve motion?

If the underlying question here is "is all energy secretly kinetic energy" then I would say no. The rest energy (aka rest mass) of a body can include kinetic energy, but if it were solely due to kinetic energy then any isolated particle would have to have a rest energy of zero. This is not the case, including for particles that appear to be fundamental indivisible pointlike objects.

 

[Dale]

Of potential and kinetic energy in their various forms, in their own reference frames, which involve motion? Heat, light, nuclear, kinetic, etc., seem to involve motion. Does potential energy, in any way whatsoever, involve motion? Thermal does. Does nuclear energy involve motion? Seems to because the nucleus is in motion and within the nucleons so are their quarks. So how about gravitational energy, including as potential energy? Do these involve motion in any way?

"Involve motion" is a little vague. So let me give some background that may be helpful. We usually write the laws of physics in terms of the Lagrangian, $L(q,\dot q, t)$ where $q$ are the generalized positions, $\dot q$ are the generalized velocities, and $t$ is time. The Lagrangian is used with the principle of least action to determine the equations of motion, which are the description of how the system evolves over time.

Now, in the end the Lagrangian is determined by looking at the behavior of actual physical systems in real experiments and figuring out what the Lagrangian must have been to produce that behavior. For many systems (a surprising number) the Lagrangian can be written as $L(q, \dot q, t) = T(\dot q, q)-V(q)$. In other words, the experimental observations imply a Lagrangian that has several terms. The terms that depend on the velocity, $\dot q$, we collect and call "kinetic energy" and the terms that depend only on the position, $q$ we collect and call "potential energy".

Now, physics doesn't know anything about that split. The Lagrangian just has all of the terms that it has. But we arbitrarily separate those terms into kinetic and potential terms. So the potential energy does not depend on motion, by definition. That is precisely the defining property that we used to decide which terms were kinetic and which terms were potential. And as @etotheipi mentioned sometimes the Lagrangian is not so neatly separable and we wind up terms that involve both $q$ and $\dot q$. In those cases there really isn't a normal "potential energy" to speak of anyway.

Note: edited due to correction below

 

[etotheipi]

It's worth to note that even for a conservative holonomic system the kinetic energy is a function of both the co-ordinates and generalised velocities, i.e. the kinetic energy is a homogenous quadratic form with position dependent coefficients,$$T(\boldsymbol{q}, \dot{\boldsymbol{q}}) = a_{jk}(\boldsymbol{q})\dot{q}^j \dot{q}^k \quad \mathrm{where} \quad a_{jk}(\boldsymbol{q}) = \frac{1}{2}\sum_a m_a \left( \frac{\partial \boldsymbol{x}_a}{\partial q^j} \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^k}\right)$$in which case the Lagrangian has the form $L(\boldsymbol{q}, \dot{\boldsymbol{q}}) = T(\boldsymbol{q}, \dot{\boldsymbol{q}}) - V(\boldsymbol{q})$. Further, Lagrange's equation for any holonomic system $\frac{\mathrm{d}}{\mathrm{d}t} \left( \partial T / \partial \dot{q}^j \right) - \partial T / \partial q^j = Q_j$ is formulated only in terms of $T(\boldsymbol{q}, \dot{\boldsymbol{q}})$.

For more complex holonomic systems with moving constraints [$\boldsymbol{x}_a = \boldsymbol{x}_a(\boldsymbol{q}, t)$], the kinetic energy may also contain a functional time dependence $T(\boldsymbol{q}, \dot{\boldsymbol{q}}, t) = a_{jk}(\boldsymbol{q}, t) \dot{q}^j \dot{q}^k + b_j(\boldsymbol{q}, t) \dot{q}^j + c(\boldsymbol{q}, t)$. Also, with moving constraints the potential energy now depends on time, $V = V(\boldsymbol{q}, t)$.

[Edit: and also, as mentioned above, it's possible to have even weirder systems where the potential now also depends on $\dot{\boldsymbol{q}}$, in which case $L(\boldsymbol{q}, \dot{\boldsymbol{q}}, t) = T(\boldsymbol{q}, \dot{\boldsymbol{q}}, t) - U(\boldsymbol{q}, \dot{\boldsymbol{q}}, t)$!]

 

[Dale]

Oops, yes, you are right. Particularly with generalized coordinates that are not the same as normal Cartesian coordinates.