If the curl of the gradient is always zero why isn't it in vorticity definition? Kosterlitz - Thouless - Berezinsky topological transition

Question

Is a well estabilished property that the curl of a gradient is always zero (i.e. $\nabla\times\nabla\Phi=0$) and it's possible to prove it in many ways. e.g.

If $(\nabla\times\nabla\Phi)_i = \epsilon_{ijk}\partial_j\partial_k\Phi$, where Einstein summation is being used to find the $i$th component...

Using Clairaut's theorem $\partial_{i}\partial_{j}\Phi = \partial_{j}\partial_{i}\Phi$, so $$\epsilon_{ijk}\partial_j\partial_k\Phi = \epsilon_{ijk}\partial_k\partial_j\Phi = -\epsilon_{ikj}\partial_k\partial_j\Phi$$

Thus $$\epsilon_{ikj}\partial_k\partial_j\Phi=0 \longrightarrow (\nabla\times\nabla\Phi)=0$$

Now, the definition of vorticity (in continuum mechanics) is $(\vec{\nabla}\times \vec{u})$ and through the Stokes theorem $$\oint_{\partial S} d\vec{l}\cdot\vec{u}=\int_Sd\vec{s}\cdot(\vec{\nabla}\times \vec{u})$$ Now, I'm in dealing with vorticity in the context of the Kosterlitz - Thouless - Berezinsky transition, where $\vec {u}=\vec{\nabla}{\theta}$. Thus, the vorticity $n$, working with 1 vortex or antivortex is defined through $$\oint_{\partial S} d\vec{l}\cdot\vec{\nabla}{\theta}=\int_{ S} d\vec{s}\cdot(\vec{\nabla}\times\vec{\nabla}{\theta})=2\pi n$$ While working with N vortexes or antivortexes is defined through $$\oint_{\partial S} d\vec{l}\cdot\vec{\nabla}{\theta}=\int_{ S} d\vec{s}\cdot(\vec{\nabla}\times\vec{\nabla}{\theta})=2\pi \sum_j^Nn_j$$

Why in this case is not zero? Because in this context we're dealing with complex scalar fields instead of real scalar fields? How i can prove that is not zero using simple arguments like the first classical used in the beginning?

P.S.: $\theta$ is an angular coordinate defined ad $\theta=\arctan\frac{y}{x}$, singular in the center of the vortex/antivortex and the definition of vorticity in this context comes from the fact that $\sum_{closed loop}(\theta_{i+1}-\theta_i)\neq0$ due to the existence of vortexes or antivortexes. Then, going to the continuum limit one get the definition in the question above.

Answer 1

The statement that $\nabla\times\nabla f = 0$, makes a tacit assumption about $f$. It is an "ordinary" function $f: R^2\rightarrow R$.

In contrast, one can have $\nabla\times\nabla\theta = T \neq 0$, where $\theta$ is a phase function. Unlike ordinary functions, a phase function is $\theta: R^2\rightarrow S$, where $S$ is the unit circle on the two-dimensional plane. The difference here is that $S$ has nontrivial topological properties. It is said to have a nontrivial homotopy group.

To see how this works, one can consider a mapping of any closed contour on the two-dimensional plane to the unit circle, and then allow that contour to contract to a point. Sometimes, the mapping would wrap around the circle, in which case it cannot be contracted continuously to a point on the unit circle. This implies that the closed contour on the two-dimensional plane encloses a topological defect where the phase becomes singular. This topological defect is called a vortex. It is an essential singularity with the phase going through all possible values around it.

When we now consider the vector differential operation at the point of this topological defect we get $$ \nabla\times\nabla\theta = \pm 2\pi\delta(\mathbf{x}-\mathbf{x}_0) . $$ This relationship can be proven with the aid of a limit process taking infinitesimal step around the topological defect. The sign comes from the direction around the defect in which the phase increases, and is referred to as the topological charge.

When we integrate the above expression over a region $$ \int_S \nabla\times\nabla\theta\ \text{d}^2x = 2\pi n = 2\pi\sum_n \sigma_n , $$ it adds up the topological charges $\sigma_n$ of all the topological defects in that region and produce $2\pi$ times the net number $n$ of topological defects.

Although, the above discussion assumes a phase function on a two-dimensional plane, one can generalize it to higher dimensions. In three dimensions the topological defects become lines, and the topological charge becomes charge flow along these lines.

Answer 2

As you mentioned, taking $\vec{u}=\vec{\nabla}\theta$ is equivalent to placing a line vortex at $r=0$ along $z$ . In Cartesian coordinates $$\vec{u}=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2},0\right)$$ Now it is easy to check that $\vec{\nabla}\times \vec{u}$ is zero everywhere except $x=y=0$. in other words,

$$\textbf{$\vec{\nabla}\times \vec{u}$ is singular at the origin}$$

Therefore, we cannot use Stokes' theorem when the origin is included. However, Stokes' theorem holds for any region that is punctuated at the origin. For example, let us consider the region between two circles with radii $r_1$ and $r_2$ about the origin ($r_1<r_2$).

$\bullet$ region $r_1<r<r_2$ is not simply connected (https://en.wikipedia.org/wiki/Simply_connected_space). The boundary of $S$ has two parts: one part is located at $r= r_1$ and the other one is located at position $r=r_2$. Let's call them $\partial S_1$ and $\partial S_2$. Then Stokes' theorem tells us that $$\oint_{\partial S_1+\partial S_2} d \vec{l}\cdot \vec{\nabla} \theta=\int _{S} d\vec{s}\cdot \vec{0}=0$$

$$\textbf{You can take $r_1$ as small as you want: $r_1\rightarrow 0$; but not zero. }$$

$\bullet$ For the region $r<r_1\rightarrow 0$ curl is singular and the Stokes theorem doesn't apply. You can find the result by directly computing the line integral (LHS of Stokes' theorem). This gives $2\pi$.

Hope this helps.