Can a conservative force not conserve mechanical energy because of explicit time dependence?

Question

Let us define a conservative force as being a force whose work is path independent. Then, in particular, a vanishing force is conservative.

If a force acting on a particle can be written from a scalar function, $\vec F=-\vec\nabla U(\vec r,t)$, then the change in mechanical energy is $$\frac{dE}{dt}=\frac{\partial U}{\partial t}.$$ If $U$ depends explicitly on time, then mechanical energy is not conserved.

Now consider a case where the potential energy depends only on time $U=U(t)$. For example, a charged particle inside a charged conducting sphere whose charge changes with time. The force on the particle is zero, therefore it is conservative. On the other hand, the potential change uniformly with time and $$\frac{dE}{dt}=\frac{\partial U}{\partial t}\neq 0.$$

Is it true that a conservative force does not imply energy conservation as this example suggests?

Answer 1

Yes, that is accurate.

Let the position of our particle be given by $\mathbf x(t)$, and let the potential energy function be $U(\mathbf r, t)$. The total energy of the particle at time $t$ is then

$$E(t) = \frac{1}{2}m|\dot{\mathbf{x}}(t)|^2 + U\big(\mathbf x(t) , t\big)$$

and so

$$\dot E(t) = \dot{\mathbf{x}}(t) \cdot \left[m\ddot{\mathbf{x}}(t)+\nabla U \big(\mathbf x(t),t\big)\right] + \frac{\partial U}{\partial t}\big(\mathbf{x}(t),t\big)$$

Because the force on the particle at time $t$ is given by

$$\mathbf F(t) = -\nabla U\big(\mathbf x(t),t\big) \underbrace{=}_{\text{Newton's 2nd}} m\ddot{\mathbf{x}}(t)$$

the first term always vanishes, whether $U$ depends on position or not. This leaves us with

$$\dot E(t) = \frac{\partial U}{\partial t}\big(\mathbf x(t),t\big)$$

Answer 2

I) That's a good question.

1. In this answer$^1$ we would first like to relax the conventional definition of a conservative force ${\bf F}({\bf r},t)$ to include explicit time dependence (but let's not allow velocity dependence$^1$ here for simplicity).

2. Definition. A Force ${\bf F}({\bf r},t)$ is a conservative force if it produces no work for an arbitrary closed loop: $$ W ~\equiv~ \oint_{{\bf r}(t_f)={\bf r}(t_i)} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf F}({\bf r}(t),t) ~=~ 0.\tag{1}$$

3. Let us choose a potential $$U({\bf r}_f,t_f)~:=~-\int_{{\bf r}(t_i)={\bf r}_i}^{{\bf r}(t_f)={\bf r}_f} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf F}({\bf r}(t),t) \tag{2}$$ as minus the work from a reference/fiducial point $({\bf r}_i,t_i)$. The definition (2) does not depend on the curve from $({\bf r}_i,t_i)$ to $({\bf r}_f,t_f)$ by assumption (1).

4. Moreover one can argue that$^2$ $$\frac{U({\bf r}_f,t_f)}{\partial {\bf r}_f} ~\stackrel{(13)}{=}~-{\bf F}({\bf r}_f,t_f),\tag{3}$$ i.e. that $${\bf \nabla}U ~\stackrel{(3)}{=}~-{\bf F},\tag{4}$$ and therefore $$ {\bf \nabla} \times {\bf F}~\stackrel{(4)}{=}~{\bf 0}\tag{5} $$ as usually.

5. Inserting eq. (4) into eq. (1) yields $$\begin{align} 0~\stackrel{(1)}{=}~&W ~\stackrel{(4)}{=}~ -\oint_{{\bf r}(t_f)={\bf r}(t_i)} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf \nabla}U({\bf r}(t),t) \cr ~=~& \int_{{\bf r}(t_i)={\bf r}_{\ast}}^{{\bf r}(t_f)={\bf r}_{\ast}} \!\mathrm{d}t~\left(\frac{\partial}{\partial t}-\frac{d}{dt}\right)U({\bf r}(t),t)\cr ~=~& \int_{{\bf r}(t_i)={\bf r}_{\ast}}^{{\bf r}(t_f)={\bf r}_{\ast}} \!\mathrm{d}t~\frac{\partial U({\bf r}(t),t)}{\partial t} +U({\bf r}_{\ast},t_i) -U({\bf r}_{\ast},t_f) .\end{align}\tag{6}$$ For constant $r_{\ast}, t_f, t_i$ one could consider a curve that stays a finite time at some point $r_m$. By varying $r_m$ in eq. (6), one can argue that $\frac{\partial U({\bf r},t)}{\partial t}$ cannot depend on ${\bf r}$, i.e. that $$ \frac{\partial^2 U({\bf r},t)}{\partial {\bf r} \partial t}~\stackrel{(6)}{=}~0 , \tag{7}$$ i.e. that the potential is separable $$ U({\bf r},t)~=~V({\bf r})+W(t). \tag{8}$$

6. Then the conservative force $$ {\bf F}({\bf r},t)~\stackrel{(4)}{=}~-{\bf \nabla}U({\bf r},t)~\stackrel{(6)}{=}~-{\bf \nabla}V({\bf r}) \tag{9}$$ cannot depend explicit on time $t$ after all, so we are back to the standard case ${\bf F}({\bf r})$. Moreover the potential choice (2) is then invariant under time-reparametrizations, so it does not depend explicitly on $t_f$.

II) Now we are ready to answer OP's question in the standard case where the conservative force ${\bf F}({\bf r})$ has no explicit time dependence.

1. On one hand, it is well-known that the potential is not a physical quantity; only potential differences are physical. We are allowed to choose a reference energy-level/gauge $W(t)$ that depends explicit on time $t$. E.g. we could be calibrating an experiment in real time.

   From the work-energy theorem $$T+V~=~{\rm const}, \tag{10}$$ the mechanical energy $$\frac{d(T+U)}{d t}~\stackrel{(8)+(10)}{=}~\frac{dW}{d t}\tag{11} $$ would then no longer be conserved, as OP correctly observes.

2. On the other hand, there is a natural gauge where the potential $U$ does not depend explicit on time $t$, so why not use that?

--

$^1$ Velocity-dependent conservative forces are considered in my Phys.SE answer here.

$^2$ One argument could e.g. use an intermediate point $({\bf r}_m,t_m)$, with $$t_m~=~t_f-\epsilon\tag{12}$$ so that

$$\begin{align} U({\bf r}_m,t_m)-U({\bf r}_f,t_f) ~\stackrel{(2)}{=}~&\int_{{\bf r}(t_m)={\bf r}_m}^{{\bf r}(t_f)={\bf r}_f} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf F}({\bf r}(t),t)\cr ~=~&\int_{{\bf r}(t_m)={\bf r}_m}^{{\bf r}(t_f)={\bf r}_f} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf F}({\bf r}(t),t_f) +{\cal O}(\epsilon).\end{align}\tag{13}$$

Answer 3

The idea of potential energy you used above is just a shortcut. What is always going on in situations like this is that we have some larger system, say with a small object and a large object. The potential energy is a function of both their configurations. But often we can treat the large object as fixed in place, in which case the potential energy only depends on the configuration of the small object, and we can think of it as belonging to the small object alone. The small object's energy is then conserved. This of course breaks down if you allow the large, fixed object to start moving.

For example, consider a ball with position $\mathbf{r}$ and the Earth with position $\mathbf{R}$. Let their kinetic energies be $K_b$ and $K_E$. Their potential energy is $$U(\mathbf{r}, \mathbf{R}) = - \frac{G M m}{|\mathbf{r} - \mathbf{R}|}.$$ The total energy is $$E = K_b + K_E + U(\mathbf{r}, \mathbf{R})$$ which is conserved. But for many situations it's good enough to treat the Earth as fixed. In this case, $K_E$ is just zero, and we can plug the fixed position of the Earth into $U(\mathbf{r}, \mathbf{R})$ to get "the potential energy of the ball", $$U(\mathbf{r}) = m g z$$ which only depends on the position of the ball, and is a conservative force. We can think of the remaining energy as the energy of the ball alone, $$E_b = K_b + U(\mathbf{r})$$ and this quantity is conserved. This is the starting point of your analysis.

The problem is that the very next thing you do is start letting the fixed system move. Of course once you do this, the preceding two steps don't make sense anymore, and you have to go back to the full expression for energy, which is conserved.

Answer 4

Can the potential of a conservative force have a dependence on time in the first place? It won't be. And hence, the given example shouldn't work.

Answer 5

The only thing which implies energy conservation of a closed system is the time-independence of its Langrangian according to Noether's theorem.If the potential energy $U$ is only a function of time only the above statement doesnt hold energy doesnt need to be conserved.