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From the theorem of open simply-connected regions we know that if $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ then $\vec F(x,y)=P\vec i+Q\vec j$ is conservative, well I have two examples related to this that I'm not understanding.

Example 1:

$\vec F(x,y)=\left(\frac{x}{(x^2+y^2)^{3/2}},\frac{y}{(x^2+y^2)^{3/2}}\right)$

$\text{Domain of $\vec F(x,y)$:} \mathbb{R}^2-\{(0,0)\}$

$\text{potential:} f(x,y)=\frac{-1}{\sqrt{x^2+y^2}}$

$\text{Domain of $\vec f(x,y)$:} \mathbb{R}^2-\{(0,0)\}$

Example 2:

$\vec F(x,y)=\left(-\frac{y}{(4x^2+y^2)},\frac{x}{(4x^2+y^2)}\right)$

$\text{Domain of $\vec F(x,y)$:} \mathbb{R}^2-\{(0,0)\}$

$\text{potential:} f(x,y)=\frac{\tan^{-1}\left(\frac{y}{2x}\right)}{2}$

$\text{Domain of $\vec f(x,y)$:} \{(x,y), x\neq 0\}$

The argument for the example 1 one being a conservative field it's because the function of the field has the same domain of the potential function, and the argument for the second example not being a conservative field it's because the domain of the potential function is greater than the domain of the field and because it's line integral along a closed curve isn't zero, but if we didn't used this arguments, based on the theorem above why wouldn't both fields not be conservative, because they have restrictions on their domain so they have a "hole" on the region and wouldn't they fail to be open and simply-connected?

Kayodê Vital
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  • The domain on which the potential in the second example is continuous and differentiable is smaller than that of the vector field. See also this answer. – Kurt G. Mar 12 '23 at 20:53
  • I meant to write smaller but got confused there, about that answer, it's a bit advanced for me, I don't understand how it relates to my question. – Kayodê Vital Mar 12 '23 at 21:15
  • If you change $\vec{F}$ a little bit to be $(\frac{y}{x^2+y^2},\frac{-x}{x^2+y^2})$ it is exactly your question. I am using 2d polar coordinates to find the potential which becomes the polar angle $-\theta,.$ Does the 3d graph there tell you anything? – Kurt G. Mar 12 '23 at 21:20
  • You mean it overlaps in the $xy$-plane so isn't a simple curve? – Kayodê Vital Mar 12 '23 at 21:26
  • The potential is not continuous at the positive $x$-axis which restricts its domain to exclude that part and makes it (as we know) smaller than $\mathbb R^2\setminus {0},.$ – Kurt G. Mar 12 '23 at 21:31
  • So the conclusion we can draw from this is that just because a field has a restriction on the domain, it doesn't mean that it isn't conservative and we have to check the domain of the potential and compare to the field domain to prove if the field is conservative or not? and if a field has no restriction and $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ we can tell straight away that it's conservative without even knowing it's potential? – Kayodê Vital Mar 12 '23 at 21:38
  • I elaborate on this in the answer. – Kurt G. Mar 12 '23 at 22:38

1 Answers1

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  • The second example is essentially the same as here. In short: it is not conservative because the potential in polar coordinates is the polar angle which has a discontinuity at the positive $x$-axis by which a integral over a closed loop is not zero.

  • The most rigorous way to show conservativity of the first example is I think the following: It is the restriction to the $xy$-plane of the vector field $$\tag{1} \vec{F}=\frac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}} $$ which has potential $f=\frac{-1}{\sqrt{x^2+y^2+z^2}}\,.$ Both are defined on the simply connected domain $\mathbb R^3\setminus\{0\}\,.$ (Note that $\mathbb R^\color{red}{2}\setminus\{0\}$ is not simply connected.) The conservativity of (1) in 3d carries over trivially to the two-dimensional case because we just have to consider loops in the $xy$-plane.

  • Another way of looking at it is this: Every closed loop is the boundary of a surface that does not contain the origin and the integral over that surface is zero because of ${\rm curl}\,\vec{F}=0\,.$ Hence, by Stokes' theorem $\oint \vec{F}\cdot \,d\vec{\gamma}=0\,.$ In particular this holds for loops in the $xy$-plane.

Kurt G.
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