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Consider a force of form $\vec F=N\hat v$, where $\hat v$ is a unit vector in direction of velocity or equivalently, in direction of change of displacement and N is a constant. To prove that this is a path-dependent force we will consider the integral $$\oint\vec F\cdot d\vec r$$ Where $d\vec r$ is displacement in any general direction. Since the direction of force and displacement are the same we get that $$\oint\vert\vec F\vert\vert d\vec r\vert$$ Making it clear that this indeed is path-dependent.

The question is can we prove this using the curl form of the integral, that is in general $$\vec\nabla\times\vec F\neq0?$$

Also, can we generalize to this where N is some scalar function instead of a constant?

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  • $\begingroup$ Velocity and displacement are not equivalent. Do you mean $\vec{F} = N \vec{r}/|\vec{r}|$? Ignoring the fact this is undefined at the origin, this is a conservative force field, as $\vec{F} = N \vec{\nabla} |\vec{r}|$, and $\vec{\nabla} \times \vec{F} = 0$. $\endgroup$ Dec 29, 2022 at 23:45
  • $\begingroup$ @RobertIsrael oops I meant change In displacement, now corrected in original post. Also $d\vec r$ here is not a radial vector, but change in displacement vector. $\endgroup$ Dec 30, 2022 at 9:30
  • $\begingroup$ So $\vec{F}$ is dependent on the path your particle is taking, not just on the position? Then it is not a vector field, and there is no such thing as $\vec{\nabla} \times \vec{F}$. $\endgroup$ Dec 30, 2022 at 14:19
  • $\begingroup$ @RobertIsrael Ah I see, I have made this mistake before too, that is I just know what I am doing is wrong but not why it is wrong. For example, the same issue was with this question physics.stackexchange.com/q/723224/330899. Can you guide me where to look further on this? $\endgroup$ Dec 30, 2022 at 14:26

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