# Proof for condition of a Force being conservative

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?

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

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.

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

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