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Consider the resistive force modelled by the function $\vec{F} = -b\vec{v}(t)$.

The curl of this function, $\nabla \times \vec{F}$, is

$$[\frac{\partial}{\partial y} (\frac{dz}{dt}) - \frac{\partial}{\partial z} (\frac{dy}{dt})] \hat{i} \ + \ ...$$

I wrote the $x$-component only because it is too time-consuming to write all of them.

But, for example, $\frac{dz(t)}{dt}$ looks like it's completely unrelated to $y$ so the partial derivative should be zero. Applying this logic to the latter term and to the other components of the curl, we can conclude that $\nabla \times \vec{F} = \vec{0}$.

As far as I know, this is a necessary and sufficient condition for the force to be conservative, however, the friction force is obviously not a conservative force. There should be something wrong with my calculation.

abouttostart
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1 Answers1

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Given a force field $\vec{F}(\vec{x})$, the curl being zero everywhere is a necessary and sufficient condition for $\vec{F}$ to be conservative. But the frictional force $\vec{F}=-b\vec{v}$ is not a field at all, so this does not apply.

Writing $\vec{\nabla}\times\vec{F}$ does not even make sense for this force, since curl is only defined for vector fields, and the force you've given is not a field at all.

Chris
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  • So, F should be a function of the position vector? But we can still calculate the work done by taking the inner product of -bv and v dt.. – abouttostart Jun 27 '20 at 07:29
  • @curious But what you can't do, for the reasons Chris gave, is rewrite $-\int bv^2dt$ as $\int F\cdot dx=\int(\nabla\times F)\cdot dS$. – J.G. Jun 27 '20 at 07:40
  • @curious You can calculate the work, yes. It doesn't come out to zero. – Chris Jun 27 '20 at 07:41