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?

  • 4
    $\begingroup$ If $U$ depends on $t$, then the system is not closed: there's some external agent which is bringing/removing something from outside. No wonder that the system's energy isn't conserved in this case. $\endgroup$
    – Cham
    Commented Aug 28, 2019 at 19:32
  • 2
    $\begingroup$ I think you have incorrectly used a partial derivative in your second example as $U$ is only a function of time i.e. $U = U(t)$. $\endgroup$ Commented Aug 29, 2019 at 4:30

5 Answers 5


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: $$\begin{align} W ~\equiv~& \oint_{{\bf r}(t_f)={\bf r}(t_i)} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf F}({\bf r}(t),t) \cr ~=~&\oint^{t(s_f)=t_f}_{t(s_i)=t_i,~{\bf r}(s_f)={\bf r}(s_i)} \!\mathrm{d}s~ \dot{\bf r}(s)\cdot {\bf F}({\bf X}(s)) ~=~ 0,\end{align}\tag{1}$$ where we have introduced a 4-vector notation ${\bf X}\equiv\begin{pmatrix} {\bf r}\cr t\end{pmatrix}$ and a worldline parameter $s$ with $\frac{dt}{ds}>0$.

  3. Let us choose a potential $$\begin{align} U({\bf X}_f)~\equiv~U({\bf r}_f,t_f)~:=~&-\int_{{\bf r}(t_i)={\bf r}_i}^{{\bf r}(t_f)={\bf r}_f} \!\mathrm{d}t~ \dot{\bf r}(t)\cdot {\bf F}({\bf r}(t),t)\cr ~=~&-\int_{{\bf X}(s_i)={\bf X}_i}^{{\bf X}(s_f)={\bf X}_f} \!\mathrm{d}s~ \dot{\bf r}(s)\cdot {\bf F}({\bf X}(s))\cr ~=~&-\int_{{\bf X}(s_i)={\bf X}_i}^{{\bf X}(s_f)={\bf X}_f} \!\mathrm{d}s~ \dot{\bf X}(s)\cdot \begin{pmatrix} {\bf F}({\bf X}(s))\cr 0\end{pmatrix} \end{align}\tag{2}$$ as minus the work from a reference/fiducial point ${\bf X}_i\equiv({\bf r}_i,t_i)$. The definition (2) does not depend on the curve from ${\bf X}_i\equiv({\bf r}_i,t_i)$ to ${\bf X}_f\equiv({\bf r}_f,t_f)$ 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 intuitive argument for eq. (3) could e.g. use an intermediate point ${\bf X}_m\equiv({\bf r}_m,t_m)$ close to the final point ${\bf X}_f\equiv({\bf r}_f,t_f)$ with time almost frozen

$$\forall s\in[s_m,s_f]:~~ 0~<~\frac{dt}{ds}~<~\epsilon, \tag{12}$$

so that

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

  • $\begingroup$ mathematically (2) doesn't make sense because you are integrating respect to time, so the limits of integration must be times, not points in the space. From there is really hard to follow what you really mean by (3) or (13) $\endgroup$
    – Masacroso
    Commented May 9 at 21:02
  • $\begingroup$ instead of (2) I would write something like $-\int_{(\mathbf{r}_i,t_i)}^{(\mathbf{r}_f,t_f)}\mathbf{F}(\mathbf{r},t)\cdot d(\mathbf{r},t)$, where $\mathbf{F}(\mathbf{r},t)\in \mathbb{R}^3 \times \{0\}$. Then we find (3) immediately $\endgroup$
    – Masacroso
    Commented May 9 at 22:21
  • $\begingroup$ Hi @Masacroso. Thanks for the feedback. I updated the answer. $\endgroup$
    – Qmechanic
    Commented May 10 at 10:12

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

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


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.

  • 1
    $\begingroup$ I agree that this is true, but question the general utility of expanding a model until energy conservation holds. After all, if we want to model a charged particle in an oscillating field, I think we'd have to either (a) expand our model to include the fusion reactions in the sun, or (b) just write down a potential with explicit time-dependence. $\endgroup$
    – J. Murray
    Commented Aug 28, 2019 at 19:26
  • $\begingroup$ @J.Murray Sure, it might not be useful, I'm just trying to get at the conceptual point. $\endgroup$
    – knzhou
    Commented Aug 28, 2019 at 19:28
  • $\begingroup$ I agree with you. But what bothers me is the fact that the definition of a conservative force depends only on the field of force. It does not matter what generates it or how the system is defined. On the other hand, conservation of energy depends on particles and the definition of the system. $\endgroup$
    – Diracology
    Commented Aug 31, 2019 at 13:36
  • $\begingroup$ By the way, if I include the change in the energy of the sphere (my example) then the change in mechanical energy of the system (sphere+particle) should vanish, right? $\endgroup$
    – Diracology
    Commented Aug 31, 2019 at 13:43

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.


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.


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