Taylor's classical mechanics ,chapter 4, states:
A force is conservative,if and only if it satisfies two conditions:
$\vec{F}$ is a function of only the position. i.e $\vec{F}=\vec{F}(\vec{r})$.
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 $\vec{F}=f(r)\hat{r}$".