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I apologize for the ignorance and the rough English in advance, I have an issue understanding how to match both what happens in physics and what I am seeing in calculus.

We learned that if a vector field be defined on an open simply-connected region and the partial derivatives of its components are continuous and equal over said region, then the field is conservative. This I also then learned is not necessary for a field to be conservative, as in, it may not be defined on a simply connected region and still be conservative, but in general I was told it is simpler to prove it to be conservative if we can apply the theorem.

I don't quite understand however, how can one prove that a vector field, such as the electric field created by a charged particle in a plane, is conservative when by definition it is not defined for that particular point in space (or here in the plane). I know from physics that it is conservative but I don't understand how one can prove that to be the case, since the other theorem I learned to prove that a field is conservative requires the curl of the field to be zero, but it also said that it is true so long as the field is defined on a simply-connected region.

I have the same issue if I take an electric field created in space by an infinitely large rod of charged conducting material, which is the case where the field would not be defined on a line in space.

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  • $\begingroup$ Does Jesse Madnick's answer here help? math.stackexchange.com/questions/30186/… $\endgroup$
    – Andrew D. Hwang
    Commented Jun 1 at 16:42
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    $\begingroup$ Yes quite appreciated friend :D I do however still struggle to comprehend how one could go about proving that a vector field such as $\vec{F}(x;y) = \frac{x}{x^2 + y^2)}\vec{i} + \frac{y}{x^2 + y^2)}\vec{j}$ is conservative, as although it is not defined over a simply connected region, intuitively I suppose I can tell it should be conservative. $\endgroup$
    – Some random guy
    Commented Jun 2 at 3:21
  • $\begingroup$ For that, it's enough to exhibit a function $f$ whose gradient is that field. Since $F$ is radially symmetric, it's enough to look at functions of the form $f(x, y) = \phi(r)$ for some function $\phi$ of one variable (and $r = \sqrt{x^2+y^2}$). <> When you find a potential for $F$, please feel free to write up your own answer! $\endgroup$
    – Andrew D. Hwang
    Commented Jun 2 at 14:23

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