Multivariable calculus

Curl of the Vortex Field Around the z-Axis

The curl of the vortex field is the zero vector at every point where the field is defined, which is everywhere off the z-axis. Zero curl does not make it conservative: the circulation around a counterclockwise circle enclosing the axis is 2 pi, not 0, because the domain has a line removed and is not simply connected.

P(x,y,z)=−yx2+y2P(x,y,z) = \frac{-y}{x^2 + y^2}
Q(x,y,z)=xx2+y2Q(x,y,z) = \frac{x}{x^2 + y^2}
R(x,y,z)=0R(x,y,z) = 0
∇×F=⟨0, 0, 0⟩for (x,y)≠(0,0)\nabla \times \mathbf{F} = \langle 0,\ 0,\ 0 \rangle \quad \text{for } (x,y) \neq (0,0)

The computation, quotient rule and all

Neither PP nor QQ contains zz, and RR is identically zero, so the first two curl components are zero on sight. Everything happens in the third.

∂Q∂x=(x2+y2)−x(2x)(x2+y2)2=y2−x2(x2+y2)2\frac{\partial Q}{\partial x} = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}
∂P∂y=−(x2+y2)+y(2y)(x2+y2)2=y2−x2(x2+y2)2\frac{\partial P}{\partial y} = \frac{-(x^2+y^2) + y(2y)}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}

The two are identical, so their difference is zero and the curl vanishes at every point where the field is defined. It is undefined on the whole zz-axis, where x=y=0x = y = 0 makes the denominator zero, and that missing line is the entire story of this page.

Zero curl and still not conservative

Take CC to be the unit circle in the plane z=0z = 0, traversed counterclockwise. On it the field has length 11 and points along the direction of travel, so the circulation is the circumference.

∮CF⋅dr=2π\oint_C \mathbf{F} \cdot d\mathbf{r} = 2\pi

A conservative field has zero circulation around every closed loop, so this field is not conservative on the region where it lives. The same 2π2\pi comes out for a circle of radius 22 or radius 12\frac{1}{2}, since the speed 1r\frac{1}{r} falls off exactly as fast as the circumference grows.

Stokes' theorem does not rescue the situation, because it needs a surface spanning CC on which F\mathbf{F} is smooth. Every such surface crosses the zz-axis, where the field does not exist, so the theorem simply does not apply.

Locally the field is still a gradient. On the half space x>0x > 0 it equals ∇arctan⁡(yx)\nabla \arctan\left( \frac{y}{x} \right), the gradient of the polar angle. The angle cannot be defined continuously all the way around the axis, so that potential is local only, and going once around adds 2π2\pi to it.

The mistake: quoting the theorem without its hypothesis

The statement students carry is that zero curl means conservative. The real statement is that zero curl means conservative on a simply connected domain, and this field is the standard proof that the hypothesis is doing work.

The domain here is space with a line removed, and a loop around that line cannot be shrunk to a point without leaving the domain. Removing a single point would not have caused any trouble: space minus a point is still simply connected. It takes a whole missing line, or a missing tube, to break the implication.

There is a picture mistake here too. The flow lines are circles, so it looks like rotation, but a paddle wheel dropped into this flow revolves around the axis without turning about its own centre. The inner edge moves faster and the outer edge slower, in the exact ratio that cancels the turning.

The zero above is checked numerically on every build at sample points away from the axis, where the field is defined.

Frequently asked questions

Why does the flow look like it is rotating if the curl is zero?

Because revolving and rotating are different. Every particle circles the axis, which is revolution, but the 1r\frac{1}{r} speed profile means a small paddle wheel keeps its orientation as it goes around. Curl measures the second thing, the turning about the particle's own centre, and here that is exactly zero.

Where exactly is this field undefined?

On the whole zz-axis, the set x=y=0x = y = 0, where x2+y2x^2 + y^2 vanishes. That missing line is what makes the domain fail to be simply connected. If only the origin had been removed, the domain would still be simply connected and zero curl would have forced a potential to exist.