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Diffstat (limited to 'mindmap/conservative force.org')
| -rw-r--r-- | mindmap/conservative force.org | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/mindmap/conservative force.org b/mindmap/conservative force.org index 83d1c36..9b01117 100644 --- a/mindmap/conservative force.org +++ b/mindmap/conservative force.org @@ -15,7 +15,7 @@ A conservative force has this property: \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 do the +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} |