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While studying Mechanics, I came to know about a way to test whether a force is conservative. Check whether the expression for the Work done is solvable without the path of the object that is $\int \overline{F}\cdot \mathrm d\overline{r}$ is solvable without dependence on the path the object follows.

Recently I came to know about another method for judging the same. That is to check whether the following system of equations hold true

$$\dfrac{\partial F_x}{\partial y}=\dfrac{\partial F_y}{\partial x}\\\dfrac{\partial F_z}{\partial y}=\dfrac{\partial F_y}{\partial z}\\\dfrac{\partial F_z}{\partial x}=\dfrac{\partial F_x}{\partial z}$$

Is there an intuitive as to why this is true and does this come straight out of the integral $\int\overline{F}\cdot \mathrm d\overline{r}$ being solvable without a trajectory in the first place?

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    $\begingroup$ The shorthand way of writing your relationships is curl $\vec F =0$ as described here. $\endgroup$ – Farcher Sep 30 '19 at 6:05
  • $\begingroup$ An additional condition for a conservative force is that it does not change with time. $\endgroup$ – R.W. Bird Sep 30 '19 at 18:36
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The three PDEs that you give are the component representation of the condition

$\nabla \times\vec{F}=0$

i.e. the curl of $\vec F$ is zero.

If this condition is true everywhere in a connected region $\mathbf R$ then the Kelvin-Stokes theorem (which relates the integral of $\nabla \times \vec F$ over a surface to the line integral of $\vec F$ around the boundary of the surface) tells us that the integral of $\vec F$ around any closed curve inside $\mathbf R$ is zero. This is not intuitive, but it is a well known result in vector calculus.

This in turn tells us that the line integral of $\vec F$ between any two points in $\mathbf R$ is independent of the path taken (as long as that path stays inside $\mathbf R$) - and this means that $\vec F$ is conservative.

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By definition, $$\vec{\nabla}×\vec{F_{cons}}=0$$ Solve this equation and you will reach the desired result.

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  • $\begingroup$ Maybe you want to add why this condition makes the work along a path dependent only on the initial and final points.. $\endgroup$ – pp.ch.te Sep 30 '19 at 6:48

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