# Conservative or non-conservative? Frame dependent?

Can a force which is conservative in one frame become non-conservative in another frame. Why/Why not?

Basically what does it mean for work to be zero in closed loop? If I am thinking of coordinates of starting and ending point and they can be changed by translation of frame. But I think this would muddle up a lot of things. Please clarify.

P.S.- Also how would the work done by the force vary with frames, since work done by force is frame dependent

• Can we assume by “frame” you mean inertial frame? Mar 8, 2020 at 8:56
• Not necessarily, what would be the difference for the 'another' frame to be inertial or non-inertial? Mar 9, 2020 at 3:25

In classical mechanics, forces are frame invariant. Work is not, in general, because trajectories are not frame invariant.

However, definition of conservative forces requires only that in each reference frame the work depends only on the initial and final point. Variation of this value with the change of reference frame does not make the definition of conservative field frame dependent.

Even if the null work of a conservative force for a given closed trajectory in one frame could correspond to a non-zero work for the same motion as observed in a different frame, this does not change the property of being conservative, because in all these cases the original closed trajectory is mapped into an open curve and all the paths having the same initial and final point do correspond to the same value for the work.

• What about forces in rotating frames? You can create (fictitious) forces if you transform to a rotating frame. Mar 8, 2020 at 13:46
• Also what about electromagnetic forces. Electrostatic field is conservative in an inertial frame with charge at rest. As we move with respect to the charge we encounter non conservative electric fields and magnetic fields which are also non conservative. Do we still say that the net effect would be conservative if we form a closed loop in the new reference frame? Mar 8, 2020 at 17:46
• It is a definition of acceleration who is frame invariant, not the force! See my answer. Mar 8, 2020 at 18:31
• @AccidentalTaylorExpansion I am excluding fictitious forces. I'll add an extended explanation why I exclude them later. Here, I would just anticipate that they are not forces between physical systems. Mar 8, 2020 at 18:58
• @VladimirKalitvianski It depends on what is the underlying definition of force. A very old and never definitely settled problem of classical mechanics. I'll try to touch this point too in a next editing. Unfortunately I cannot do it immediately. Mar 8, 2020 at 19:08

No mathematics from my side, but here is what my argument is.

One everyday life conservative force is gravity. Suppose you and your friend are diving side by side in your cars (which can fly). For the time being neglect friction and assume earth to be flat. From the perspective of a observer standing on ground bothering of you have the same velocity vector, $$\mathbf v = v_x \hat i$$. Now suppose that your friend starts to fly up vertically (with respect to you) and then returns back vertically down to the same point (in your reference frame).

So what do you observe?

You observe that neither of his kinetic or potential energy increase/decrease and hence the gravitational force is conservative.

What does that standing observer observe?

He/she sees that their is no difference in kinetic and potential energy of you and your friend. And hence concludes that gravitational force is conservative. Why so? Because, here, you reach your friend reaches his vertically initial position and gravity acts vertically downwards(hence cannot change the kinetic energy into any other perpendicular directions).

The generalization of this to an arbitrary inertial frame and an arbitrary conservative force requires math (which I can't provide you for the time being).

The electric field of a point charge is conservative in the charge's rest frame. In any other frame, this field transforms into a mixture of electric and magnetic fields, in which the electric field is nonconservative (i.e., its curl is nonzero).

Force does not depend on the inertial frame of the observer. That includes conservative forces. However, the work done by a force does depend on the frame.

The amount of work done by a conservative force between two points may be different in different frames, but for any given frame the work done between two points is independent of path taken.

Worse, in a moving RF the total energy is not censerved any more! Just because the potential becomes time-dependent: $$U(r)\to U(\vec{r}\pm \vec{V}t)$$, so the term $$\partial U(\vec{r},t)/\partial t \ne 0$$. Surprise!