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Under normal circumstances, the gravitational force near Earth's surface, $F_g = mg$ is clearly conservative. You can see this either by noting conceptually that the force is constant and work done by gravity is path-independent, or formally by checking that the curl of the field is zero.

What if I changed $g$ so that it was slightly time-dependent, $g(t)$, so the force would vary in magnitude with time, but it would remain spatially homogeneous? Is the new gravitational force conservative?

I have an argument for both yes and no:

YES: this force does not explicitly depend on any spatial coordinates: $F_g(t) = m g(t),$ so the curl of the force must be zero.

NO: Intuitively, imagine picking up an object from $y=0$ to $y=h$, and then setting it back down to $y=0$. Let's imagine we're moving the object up very fast, waiting around for a while, and then moving it down very fast. That way we can approximate $g(t)$ as constant while the object is moving, but significantly different on the way up compared to the way down. In this case, the change in gravitational potential energy on the way up would be $$ \Delta U_{g,up} = mg(t)h $$ and on the way down it would be $$ \Delta U_{g,down} = -mg(t)h. $$ This is a closed loop, so for a conservative gravitational force, $\Delta U_{g}$ should just be zero. However, if $g(t)$ is bigger on the way up than on the way down, the change in potential energy might be non-zero.

So, this force violates both path-independence and it explicitly depends on a non-spatial variable. Thus, it should NOT be conservative.

Which of these explanations is wrong?

Qmechanic
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Bunji
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4 Answers4

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It's hard to define energy conservation for time-depending forcing. In general, for a potential $V(q(t),t)$ with explicit dependence on time, you get a Lagrangian and Hamiltonian functions explicitly dependent on time $\mathscr{L}(\dot q(t), q(t), t)$ and $H(q(t), p(t), t)$, and from Hamiltonian mechanics we know (see as an example link) that

$$\dfrac{d H}{dt} = \frac{\partial H}{\partial t} = \frac{\partial V}{\partial t}\ ,$$

thus energy of the system under investigation is not constant in general. This term describes the power of time-dependent external forces acting on the system.

Remark. If you extend the system under investigation to incorporate the cause of the dependence on time of the potential $V(t)$, you should be able to describe it with a new potential without explicit dependence on time.

basics
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You can make a time dependent gravitational field by moving masses around. This might cause each of the masses to gain or lose gravitational potential energy.

It takes work to move masses toward or away from each other. The total energy, including this work, is conserved.

mmesser314
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It would be path-independent. Just consider we don't even move the mass; force is time-dependent and will change with time, so potential energy, which is dependent on time, would increase without even moving the mass, let alone by moving it. But, just consider two situations: in one, you don't move the mass, and in the other, you do, but in the end, bring it back to the original position in both cases. The mass would have the same energy after a time interval, so it is not dependent on the path, but it is dependent on time. So, even if paths are different, in your argument for "no", the initial and final energy is different because they are different instances in time, not because of the path chosen.

M. A.
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Shaurya Kad
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It seems there are two facets of a force (or a force field) being conservative: one is "spatial" in the sense that the work over a closed path is zero (zero curl), and the second one is "temporal" in the sense that the energy is constant as a function of time.

In the "temporal" sense, energy might not be constant in a time-dependent force field. It is quite easy to construct an example. Imagine all the gravity decaying to zero following an exponential law: after a sufficiently long time, the energy will be arbitrarily close to zero.

In the "spatial" sense, at each fixed instant of time, the gravitational field has all the properties of "the usual" gravitational field, including the force's zero curl.

And R
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