In order for a force to be derived from the gradient of a potential energy, does the work done by such a force need to be invariant of the path?

Question

Suppose a force $\mathbf{F} = \mathbf{F}(\mathbf{r}, t)$ where $\mathbf{r}$ is a three dimensional space vector and $t$ is time.

I understand that in order to a force be conservative two conditions must be satisfied:

1. the force must be distance dependent only, i.e. $\mathbf{F} = \mathbf{F}(\mathbf{r})$;
2. the work done by such force between two points, say, 1 and 2, must be independent of the path taken.

Firstly, we may automatically say that the force $\mathbf{F}$ that I am talking about is not conservative since it has time dependence.

However, it can be a force such that its curl is $\mathbf{0}$, i.e. $\mathbf{\nabla \times F} = \mathbf{0}$. If that's the case, it's sufficient to argue that this force is such that the work done by this force between two points is path invariant.

I wonder whether $\mathbf{F}$ may be derived from a potential energy even though it does not satisfy condition 1, but satisfy condition 2.

All of this being said, I want to ask: One is always able to derive a force from a potential energy or are there any conditions to do so? Does it need to satisfy any of the conditions mentioned above? Does it need to satisfy both (be conservative)? Even further, if not in the force, are there any conditions that the potential energy must satisfy in order to this relation hold?

Answer 1

Your question reduces essentially to what conditions are necessary for one to be able to write $\mathbf{F}(\mathbf{r}, t) = - \nabla V(\mathbf{r},t)$ for some function $V(\mathbf{r},t)$. I'm assuming we are working in Euclidean space for simplicity, but in curved spaces the answer gets much more complicated.

There are conditions necessary for this to be possible. Firstly, notice that $\mathbf{\nabla\times\nabla}V = \mathbf{0}$, since the curl of a gradient always vanishes. Hence, not all forces can be derived from a potential, they necessarily must have zero curl. If they do, can we do it?

Assuming a few other conditions (the force is smooth enough, it decays sufficiently fast as you go close to infinity), it follows from Hemholtz's theorem that there is such a potential function. The time dependence is irrelevant, for it is essentially a parameter in the expressions. In this case it is no more relevant than, for example, the constant $k$ in the expression for the force of a spring.

Notice, however, that in this situation energy will depend on time, since the potential energy explicitly does it. This can be understood by thinking, for example, of a pendulum whose length varies with time. The energy of the pendulum changes, the reason being that something is doing work to push and pull on the string.

As an extra remark, it is worth mentioning that Helmholtz's theorem goes beyond the case we are dealing with and holds for more general situations. Namely, it states that if $\mathbf{F}(\mathbf{r})$ is sufficiently behaved (smooth, decays fast enough), then one can find a scalar function $V(\mathbf{r})$ and a vector function $\mathbf{A}(\mathbf{r})$ such that $$\mathbf{F}(\mathbf{r}) = - \nabla V(\mathbf{r}) + \nabla\times \mathbf{A}(\mathbf{r}).$$ This fact is quite explored in the potential formulation of Electrodynamics (see essentially any book in Electrodynamics).

Answer 2

The problem with a Force that depends on time and not just on position, is that regarding time, you can always change the work in a path because you can wait until the force changes