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Potentiality of the field

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Given field

    [itex]\vec{F}=(y^2-x^2)\vec{e}_x+(3xy)\vec{e}_y[/itex]


    2. Relevant equations

    It potentially?

    3. The attempt at a solution

    When I try to calculate [itex]\partial_y F_x=\partial_x F_y[/itex] I find [itex]2y=3y[/itex] which is equal only at the origin.

    If I try to calculate [itex]\oint {\vec Fd\vec r = 0} [/itex] along any closed loop which covers the origin, I get the 0.

    That's the question arose, whether this field is a potential field?
     
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The physics terminology would be that [itex]\vec{F}[/itex] is a "conservative field" if and only if it has a "
    "potential function". I have never seen "potential field".
    (The mathematics terminology would be that [itex]\vec{F}[/itex] is "exact" if and only it has an "antiderivative".)

    In any case, why does the question arise? Knowing that [itex]\partial_yF_x\ne \partial_xF_y[/itex], you know immediatly that this is NOT a "conservative field"- there is NO "potential" function.

    You say "When I try to calculate [itex]\oint \vec{F}d\vec{r}[/itex] along any closed loop that covers the origin, I get the 0".

    Really? there are an infinite number of such loops. How in the world did you integrate on all of them?

    In any case, "any closed loop which covers the origin" is irrelevant. To have a potential, the integral along every closed loop must be 0, not just those "which cover the origin". For example, integrating around the boundary of the square with vertices at (0, 0), (1, 0), (1, 1), and (0, 1), I get 5/3.
     
  4. Nov 13, 2013 #3
    Ok, I integrated around the boundary of the square with vertices (-1,1), (1,1), (1,-1), (-1,-1) and also, I integrated around the boundary of the circle [itex]x^2+y^2=1[/itex] and received zero again.

    Let's take the boundary of the square with specified vertices. Then
    [itex] \oint {\vec Fd\vec r = \int\limits_{ - 1}^1 {{F_x}dx + \int\limits_1^{ - 1} {{F_y}dy} + } } \int\limits_1^{ - 1} {{F_x}dx + \int\limits_{ - 1}^1 {{F_y}dy} = \frac{4}{3}} + 0 - \frac{4}{3} + 0 = 0 [/itex]

    In russian literature the "potential field"="conservative field", so I do not argue here, I did not know this nuance.
     
    Last edited: Nov 13, 2013
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