# Can a conservative force do not conserve mechanical energy?

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 in energy conservation as this example suggests?

• 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. – Cham Aug 28 at 19:32
• I think you have incorrectly used a partial derivative in your second example as $U$ is only a function of time i.e. $U = U(t)$. – Rumplestillskin Aug 29 at 4:30

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.

• 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. – J. Murray Aug 28 at 19:26
• @J.Murray Sure, it might not be useful, I'm just trying to get at the conceptual point. – knzhou Aug 28 at 19:28
• 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. – Diracology Aug 31 at 13:36
• 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? – Diracology Aug 31 at 13:43

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

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