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

  1. Feb 19, 2007 #1
    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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  3. Feb 19, 2007 #2


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    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]
  4. Feb 19, 2007 #3

    D H

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    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: Feb 19, 2007
  5. Feb 19, 2007 #4
    I thank you both.And I was not interested about constant fields.
    However,What about F=F(v) where v=dr/dt
    And also if curl F=0 where F=F(t),or,F=F(v) does it mean the field is conservative?
  6. Feb 19, 2007 #5

    D H

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    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.
  7. Feb 19, 2007 #6
    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.
  8. Feb 19, 2007 #7
    DH:If F=F(v) has curl F=0,then what do you mean by this?
    StatMechGuy:I really did not understand:
  9. Feb 19, 2007 #8

    D H

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    I agree. A zero curl simply means the field is irrotational, period.
  10. Feb 20, 2007 #9


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    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).
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