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Vector calculus

  1. Nov 10, 2004 #1
    Suppose we have a conservative vector field on a plane. Suppose also that we have a closed curve C on that plane. Then we have:

    [tex] \int_C \mathbf{F}\cdot d\mathbf{r} = 0 [/tex]

    The line integral around C is zero because F is conservative. Here is what I don't understand:

    If you have one or more singularities (points at which F is undefined) within the area bound by C then the line integral around C is no longer zero! How can this be?
  2. jcsd
  3. Nov 10, 2004 #2


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    Why do you say the integral is no longer zero? Certainly it is if the singluarity corresponds to a point charge. If it's not zero then the force is not a conservative force!
  4. Nov 11, 2004 #3
    Maybe I'm getting this wrong...

    My book states that whether a field is conservative or not depends on the area upon which it is defined.

    A at a lecture the proffessor gave this example:

    We have a conservative vector field defined on [tex]\mathbb{R}^2[/tex]. We consider a closed curve C (say, a circle), which encloses the origin (0,0). Because the vector field is conservative we have:

    [tex]\int_C \mathbf{F}\cdot d\mathbf{r} = 0[/tex]

    because the field is conservative.

    Now we remove the point (0,0) so that the vector field is now defined on [tex]\mathbb{R}^2 \backslash (0,0)[/tex]. Now he was able to show (I can' remember how) that the line integral around C is no longer zero, but it measures the number of times that C is traversed during the integration. If C is a circle and traversed only once then:

    [tex]\int_C \mathbf{F}\cdot d\mathbf{r} = 2\pi r[/tex]

    I don't understand this at all!
  5. Nov 11, 2004 #4


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    You surely left out a step.

    Your professor removed {0} from the domain, and then also changed to a different vector field. Otherwise the integral wouldn't change, since it depends only on the value of the field along the path, which doesn't include the origin.

    What he presumably did is use a function which is conservative everywhere except at the origin, but which cannot be extended to a conservative field which includes the origin. For example, let

    [tex]F(x,y) = (\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2})[/tex]

    It's undefined at {0}. Everywhere except {0} its curl is zero (assuming I calculated it right) so an integral around any path which does not enclose {0} will be zero. Integrating along a path which encloses the origin, however, also encloses the point where the curl is undefined (infinite).
    Last edited: Nov 11, 2004
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