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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 $\vec{F}=f(r)\hat{r}$".

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    $\begingroup$ Does this help: tok.fandom.com/wiki/Central_force $\endgroup$
    – R. Emery
    Commented Dec 20, 2020 at 9:03
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    $\begingroup$ A function of the position only has not to be a radial function (central force). A simple counterexample is the 2D harmonic oscillator with two different force constants. It is a conservative but a non-central force. $\endgroup$ Commented Dec 20, 2020 at 9:11
  • $\begingroup$ 1 means that it isnt a function of time. It doesnt change (unless the distances change) $\endgroup$
    – R. Emery
    Commented Dec 20, 2020 at 9:16
  • $\begingroup$ Mathematically the curl of the gradient of any scalar potential is always zero. $\endgroup$
    – R. Emery
    Commented Dec 20, 2020 at 9:22

3 Answers 3

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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:
enter image description here
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 $(x=-R,y=0)$.

enter image description here

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.

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

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    $\begingroup$ I know it's a bit late to ask but does the 2nd condition in op's question (in short, the curl being zero) implies the first condition( that $F$ is a function of position only)? $\endgroup$ Commented Mar 25, 2021 at 16:34
  • $\begingroup$ @GaurangAgrawal Yep I have the same question... Any answers? $\endgroup$
    – user266637
    Commented Sep 16, 2023 at 12:43
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If the function is (as you assumed) one of distance, then you are right. But there are many functions of position coordinates whose curl is not zero, hence non-conservative.

Edit: Try for example $$F = (xy,-xy,0)$$

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The central forces with spherical symmetry are also conservative forces.
You can show this proving that the work does not depend on the path.
HP: $$\vec{F}=F(r)\vec{r}$$ where $\vec{r}$ is the unitary vector to the position vector.
So you have: $$W_{AB}=\int_A^BF(r)\vec{r}\cdot ds$$ Since in polar coordinates elementar displacement in $d\vec{s}=d\vec{r}=dr\vec{r}+rd\theta\vec{\theta}$, have $$W_{AB}=\int_{r_A}^{r_B}F(r)dr$$ So the work done by the force depends only by the initial and final distance.
You can also show that a central force with spherical symmetry is irrotational calculating the curl of the force but in polar coordinates.
Remember that the fact that the curl of the force is $0$ isn't enough for the conservation but the function must to be defined on a simply connetted domain, like gravitational and Coulomb force.

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