# Understanding conservative forces

I'm trying to better understand conservative forces. I have a decent intuitive idea of what they are, but I've recently learned the mathematical rigor behind it which has made me have some questions. Here's my interpretation of a conservative force -- please do not hesitate to correct me if and where I'm wrong:

• A conservative force is a force that generates $0$ net work when going applying a force from point $A$ to point $B$ and then from point $B$ to point $A$. I don't love this definition, but it's what I can most closely describe it as given my new understanding of the math behind it. Say you have a central force $F(r) \hat r$. The work done from two arbitrary positions would be $$W = \int_{r_1}^{r_2} F \cdot dr$$ For the force to be conservative, the work done from this path integral from $r_2$ to $r_1$ instead must be the exact same. I suppose gravity can be an example although a bit unclear, which is why I don't like the definition exactly. Because, setting the floor to $h = 0$, the work done going up a staircase to $h = h_1$ would be the exact same amount of work done as if one were to fall from $h= h_1$ to $h=0$. The spring force is another example too, and I think it's because, since the work done equals (since the work is only done in one direction): $$W = -k\int_{-x_{max}}^{x_{max}} x \ dx$$ $$-k\int_{-x_{max}}^{x_{max}} x \ dx + k\int_{x_{max}}^{-x_{max}}x = 0$$ So I suppose it's saying if you oscillate a particle from $x_{max}$ it will do the same amount of work as if you did it from $-x_{max}$. However, this doesn't have the central force idea that gravity and electromagnetic forces exhibits.

Some questions of mine:

• If the work is the same irrespective of path, does that mean if the spring made a zig-zag pattern as it oscillated (so there is some orthogonal component of movement in addition), would the net work still be the same as just pulling it back like normal?

• I feel like $\int_{r_1}^{r_2} F \cdot dr = - \int_{r_2}^{r_1} F \cdot dr$ is a mathematical fact, given that (please excuse this horrible abuse of symbols) $\int_{r_1}^{r_2} = -\int_{r_2}^{r_1}$ as a general rule. Why is this not always true?

• I seem to notice that we can draw the conclusion of a conservative force is present if the work done is involving a force that is parallel or anti-parallel to $dr$, like with gravity and the spring force (swap $dr$ with $dx$). If $F$ and displacement are both parallel, can I draw the conclusion of a parallel force in most cases?

The core of OP's question (v2) seems to be resolved by the following fact:

1. On one hand, the statement

Work along any closed loop is zero

is a non-trivial statement. In fact, it is (or is equivalent to) the conventional definition of a conservative force.

2. On the other hand, the statement

Work along a closed path is zero whenever the closed path just retraces the same curve segment twice, back and forth in opposite directions

is a trivial statement true for any force vector field (without explicit time dependence).