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

## Main Question or Discussion Point

When we say condition of a vector field F being conservative is curl F=0,does it mean that F=F(r)?.I know normally it does not look so.Please,then site an example where F is not a function of r,but still curl F=0.

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dextercioby
Homework Helper
By "curl of F=0" we mean

$$\nabla_{\vec{r}} \times \vec{F}=0$$

for an $\vec{F}=\vec{F}\left(\vec{r}\right)$

If $\vec{F}\neq \vec{F}\left(\vec{r}\right)$ then

$$\nabla_{\vec{r}} \times \vec{F}\equiv 0$$

D H
Staff Emeritus
When we say condition of a vector field F being conservative is curl F=0,does it mean that F=F(r)?.I know normally it does not look so.Please,then site an example where F is not a function of r,but still curl F=0.
One example: A constant vector field $\vec F(x\hat x + y\hat y + z\hat z) = a\hat x + b\hat y + c\hat z$ has no curl.

The curl of a gradient is necessarily zero:
$$\vec F(\vec x) = \nabla \phi(\vec x)$$

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

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

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I thank you both.And I was not interested about constant fields.
And also if curl F=0 where F=F(t),or,F=F(v) does it mean the field is conservative?

D H
Staff Emeritus
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, $\vec F = \vec F(\vec v), \vec v = d\vec r/dt$, the curl is zero since the partials of $\vec F$ with respect to components of $\vec r$ are zero. Drag in a constant density fluid satisfies these conditions, and is definitely not conservative.

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, $\vec F = \vec F(\vec v), \vec v = d\vec r/dt$, the curl is zero since the partials of $\vec F$ with respect to components of $\vec r$ 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 $\mathbf{v} = \mathbf{v}(\mathbf{r}, t)$ 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.

DH:If F=F(v) has curl F=0,then what do you mean by this?
Drag in a constant density fluid satisfies these conditions, and is definitely not conservative.
StatMechGuy:I really did not understand:
I really dislike it when classes take the perspective that if the curl is zero, then it has to be conservative

D H
Staff Emeritus
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.

vanesch
Staff Emeritus