conservative force

A conservative force has this property:

\begin{align*} \oint\vec{f} \cdot d\vec{l} = 0 \end{align*}

In other words, work done by \(\vec{f}\) is path independent, because in any closed loop integral, you go from point \(\vec{a}\) to point \(\vec{b}\) and then back. If these forwards and backwards paths end up canceling no matter what path you take, then it is clear that \(\vec{f}\) will be the same amount of force no matter what path you take. Using Stokes' theorem:

\begin{align*} \int_{S}(\vec{\nabla} \times \vec{f}) \cdot d\vec{a} = \oint\vec{f} \cdot d\vec{l} \end{align*}

And therefore, if and only if \(\vec{\nabla} \times \vec{f} = \vec{0}\), this line integral is also \(\vec{0}\). Additionally, if you integrate over \(\vec{f}\), we define \(V(\vec{r})\) such that:

\begin{align*} \int_{\vec{a}}^{\vec{b}}\vec{f} \cdot d\vec{l} = V(\vec{a}) - V(\vec{b}) \end{align*}

because it is path independent, we do not need to consider the infinite paths between \(\vec{a}\) and \(\vec{b}\), which allows us to define this function \(V(\vec{r})\). Then by the fundamental theorem of calculus, using the Gradient:

\begin{align*} \vec{f} = -\vec{\nabla}V \end{align*}

Therefore, conservative forces can be represented by a scalar field. Now taking the Curl of both sides we get:

\begin{align*} \vec{\nabla} \times \vec{f} = 0 \end{align*}

Which is consistent with the result from above.