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
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.
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.
Regarding your PS.
Work can be shown to be frame dependent because velocity is frame dependent. Per the work energy theorem the net work done on an object equals its change in kinetic energy, which in turn depends on the change in the squares of the velocity.
Hope this helps
It is like a kinetic egenrgy, which is frame-dependent. A still body may reach you and hit you if you are in another (moving) reference frame.
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!
