Condition of a vector field F being conservative is curl F = 0,

By "curl of F=0" we mean

[tex] \nabla_{\vec{r}} \times \vec{F}=0 [/tex]

for an [itex] \vec{F}=\vec{F}\left(\vec{r}\right) [/itex]

If [itex] \vec{F}\neq \vec{F}\left(\vec{r}\right) [/itex] then

[tex] \nabla_{\vec{r}} \times \vec{F}\equiv 0 [/tex]

 

One example: A constant vector field [itex]\vec F(x\hat x + y\hat y + z\hat z) = a\hat x + b\hat y + c\hat z[/itex] has no curl.

The curl of a gradient is necessarily zero:
[tex]\vec F(\vec x) = \nabla \phi(\vec x)[/tex]

So all you need to do is come up with a scalar function [itex]\phi(\vec x)[/itex] that cannot be expressed as a function of [itex]||\vec x||[/itex].

The constant vector field corresponds to [itex]\phi(\vec x) = ax + by + cz[/itex], where [itex]\vec x = x\hat x + y\hat y + z\hat z[/itex]. Then [itex]\nabla \phi(\vec x) = a\hat x + b\hat y + c\hat z[/itex].

 

A picked a constant vector field as a simple counterexample. Any vector field that can be expressed as the gradient of a scalar function has zero curl.

For the latter, [itex]\vec F = \vec F(\vec v), \vec v = d\vec r/dt[/itex], the curl is zero since the partials of [itex]\vec F[/itex] with respect to components of [itex]\vec r[/itex] are zero. Drag in a constant density fluid satisfies these conditions, and is definitely not conservative.

 

Unless you're talking about a viscous fluid and [itex]\mathbf{v} = \mathbf{v}(\mathbf{r}, t)[/itex] is the velocity field. But then things are still more complicated.

Generally velocity/time dependent forcing fields are not conservative. I really dislike it when classes take the perspective that if the curl is zero, then it has to be conservative.

 

I agree. A zero curl simply means the field is irrotational, period.

 

A vector field assigns a vector to each point of the "base space": it is a mapping v(p) from the base space M into a vector space V. In the settings where curl and so on make sense, it can be shown that, if curl_p v = 0 over M, AND IF M IS SIMPLY CONNECTED (no "holes"), that there exists a scalar function f(p) over M, such that v(p) = grad f.

Now, nothing stops you from adding extra parameters to this problem. That is, if you consider a "vector field" which is in fact a *family* of vector fields:
v(p,lambda), with p in M, but lambda any other (set of) parameters, such as time or whatever, well the same theorem holds, for each individual member (indicated by lambda) of the family: if curl_p v(p,lambda) = 0 then v(p,lambda) = grad_p f(p,lambda).