- #1

- 135

- 1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Kolahal Bhattacharya
- Start date

- #1

- 135

- 1

- #2

- 13,033

- 588

[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]

- #3

- 15,393

- 686

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].

Last edited:

- #4

- 135

- 1

However,What about

And also if curl F=0 where F=F(t),or,F=F(v) does it mean the field is conservative?

- #5

- 15,393

- 686

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.

- #6

- 223

- 2

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.

- #7

- 135

- 1

StatMechGuy:I really did not understand:Drag in a constant density fluid satisfies these conditions, and is definitely not conservative.

I really dislike it when classes take the perspective that if the curl is zero, then it has to be conservative

- #8

- 15,393

- 686

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.

- #9

vanesch

Staff Emeritus

Science Advisor

Gold Member

- 5,028

- 17

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).

Share:

- Replies
- 54

- Views
- 1K