# [Physics] Conditions for a force to be conservative

classical-mechanicsconservative-fielddefinitionforceswork

Taylor's classical mechanics ,chapter 4, states:

A force is conservative,if and only if it satisfies two conditions:

1. $$\vec{F}$$ is a function of only the position. i.e $$\vec{F}=\vec{F}(\vec{r})$$.

2. The work done by the force is independent of the path between two points.

Questions:

• Doesn't $$1$$ automatically imply $$2$$? : Since from 1, we can conclude that $$\vec{F}=f(r)\hat{r}$$, for some function $$f$$. Then, if $$A$$ is the antiderivative of $$f$$, we can say that $$\vec{F}=\nabla{A}$$, and therefore the work (line integral) will depend on the final and initial positions only. Or even simply put, $$\vec{F}.d\vec{r}$$ is a simple function of $$r$$ alone, so the integral will only depend on initial and final $$r$$.
• I have seen in many places, only "2" is the definition of a conservative force. In light of this, I cant think of why 1 has to be true: i.e how is it necessary that path independence implies $$\vec{F}=f(r)\hat{r}$$.

It could be that my interpretation of 1 as $$\vec{F}=f(r)\hat{r}$$ is wrong, on which my entire question hinges. Taylor writes $$\vec{F}=\vec{F(\vec{r})}$$ , which I interpreted as : "since F is a function of position vector, F is a function of both the magnitude and direction, and hence $$\vec{F}=f(r)\hat{r}$$".

Your conclusions are not correct. Here is a simple counter-example. Consider this force $$\vec{F}=k(x\hat{y}-y\hat{x})$$ where $$\hat{x}$$ and $$\hat{y}$$ are the unit-vectors in $$x$$ and $$y$$-direction, and $$k$$ is some constant.

From this definition we see, the magnitude of the force is $$F=k\sqrt{x^2+y^2}=kr$$, and its direction is at right angle to $$\vec{r}=x\hat{x}+y\hat{y}$$. So we can visualize this force field like this:

The force circulates the origin in a counter-clockwise sense.

This force clearly satifies your first condition

1. $$\vec{F}$$ is a function of only the position, i.e. $$\vec{F}=\vec{F}(\vec{r})$$

But it is not of the form $$\vec{F}=f(r)\hat{r}$$.

And this force violates your second condition

1. The work done by the force is independent of the path between the two point.

To prove this consider the following two paths:

• Path A (in green): beginning on the right at $$(x=R,y=0)$$, doing a half circle counterclockwise, to the point on the left $$(x=-R,y=0)$$.
• Path B (in red): beginning on the right at $$(x=R,y=0)$$, doing a half circle clockwise, to the point on the left $$(x=-R,y=0)$$.

Then the work for path A is (because here $$\vec{F}$$ is always parallel to $$d\vec{r}$$) $$W_A=\int \vec{F}(\vec{r}) d\vec{r}=kR\cdot\pi R=\pi k R^2.$$

Then the work for path B is (because here $$\vec{F}$$ is always antiparallel to $$d\vec{r}$$) $$W_B=\int \vec{F}(\vec{r}) d\vec{r}=-kR\cdot\pi R=-\pi k R^2.$$

You see, the work is different for the two paths, although the start and end point of the paths are the same.

This is a simple example of a non-conservative force. The non-conservativeness can easily be checked by calculating its curl and finding it is non-zero.

$$\vec{\nabla}\times\vec{F} =\vec{\nabla}\times k(x\hat{y}-y\hat{x}) =2k\hat{z} \ne \vec{0}$$