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:ID:       6f2aba40-5c9f-406b-a1fa-13018de55648
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#+title: conservative force
#+author: Preston Pan
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* Definition
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 [[id:3587c3b4-c3d8-4ff1-b0ba-8eecb1ef0e4c][Gradient]]:
\begin{align*}
\vec{f} = -\vec{\nabla}V
\end{align*}
Therefore, conservative forces can be represented by a scalar field. Now taking the [[id:b25e0e44-c764-4f0a-a5ad-7f9d79c7660d][Curl]] of both sides we get:
\begin{align*}
\vec{\nabla} \times \vec{f} = 0
\end{align*}
Which is consistent with the result from above.