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conservative force

Table of Contents

1. Definition

A conservative force has this property:

∮f→⋅dl→=0\begin{aligned}\oint\vec{f} \cdot d\vec{l} = 0\end{aligned}

In other words, work done by f→\vec{f} is path independent, because in any closed loop integral, you go from point a→\vec{a} to point b→\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 f→\vec{f} will be the same amount of force no matter what path you take. Using Stokes' theorem:

∫S(∇→×f→)⋅da→=∮f→⋅dl→\begin{aligned}\int_{S}(\vec{\nabla} \times \vec{f}) \cdot d\vec{a} = \oint\vec{f} \cdot d\vec{l}\end{aligned}

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

∫a→b→f→⋅dl→=V(a→)−V(b→)\begin{aligned}\int_{\vec{a}}^{\vec{b}}\vec{f} \cdot d\vec{l} = V(\vec{a}) - V(\vec{b})\end{aligned}

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

f→=−∇→V\begin{aligned}\vec{f} = -\vec{\nabla}V\end{aligned}

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

∇→×f→=0\begin{aligned}\vec{\nabla} \times \vec{f} = 0\end{aligned}

Which is consistent with the result from above.