Magnetic flux through a superconducting ring

[Lagrange fanboy]

TL;DR

Why isn't the flux 0 since it's equal to a closed line integral of a scalar potential's gradient?

In Feymann's seminar on superconductivity, there was this equation (21.28) $\oint_C \nabla \theta\cdot dl = \frac q \hbar \Phi$. But the gradient theorem demands that $\oint_C \nabla \theta\cdot dl=0$

 

[PeterDonis]

In Feymann's seminar on superconductivity, there was this equation (21.28) $\oint_C \nabla \theta\cdot dl = \frac q \hbar \Phi$. But the gradient theorem demands that $\oint_C \nabla \theta\cdot dl=0$

The answer to your question is already there in the Feynman lecture you linked to. See the discussion between equations 21.28 and 21.29.

 

[Lagrange fanboy]

The answer to your question is already there in the Feynman lecture you linked to. See the discussion between equations 21.28 and 21.29.

Feymann said that it's the case because the region isn't simply connected, but I don't see how that leads to the gradient theorem breaking down, as it only requires the scalar potential to be differentiable along the path.

 

[PeterDonis]

it only requires the scalar potential to be differentiable along the path.

No, it also requires that, for a closed path enclosing an area, the area must be simply connected. (More precisely, it requires that the closed path be continuously reducible to a point, which is equivalent to the area enclosed by the path being simply connected.) Normally this requirement is not mentioned because normal applications of the theorem are for simply connected regions.

 

[PeterDonis]

I don't see how that leads to the gradient theorem breaking down, as it only requires the scalar potential to be differentiable along the path.

Another way of looking at it is that, if the region covered by the potential is not simply connected, the potential cannot be treated as a "scalar function" in the sense the gradient theorem requires, because it is not single-valued. Indeed, in the Feynman lecture you referenced, the potential $\Phi$ is not single-valued.

 

[vanhees71]

It's the famous "potential vortex". The corresponding vector field,
$$\vec{V}=\frac{1}{x^2+y^2} \begin{pmatrix}-y \\x \end{pmatrix}$$
is singular along the entire $z$ axis, i.e., its domain is not simply connected, because you can't contract continuously any closed curve that goes around the $z$ axis.

Neverteless except along the $z$ axis you have $\vec{\nabla} \times \vec{V}=0$.

To define a potential thus you have to choose an arbitrary semi-infinite surface with the $z$ axis as boundary and then calculate the line integral
$$V=\int_{\mathcal{C}} \mathrm{d} \vec{x} \cdot \vec{V},$$
for an arbitrary set of paths all origining from one fixed point $\vec{x}_0$ to any other point $\vec{x}$, not crossing this surface.

The potential has a jump of $2 \pi$ across the arbitrarily chosen surface, which is the value of the line integral for any closed curve encircling the $z$ axis just once.