Confusion regarding the scalar potential

[lampCable]

Homework Statement

Consider the following in cylindrical coordinates \rho,\varphi,z. An electric current flows in an infinitely long straight cylindrical wire with the radius R. The magnetic field \mathbf{B} outside of the thread is \textbf{B}(\textbf{r})=\frac{I\mu_0}{2\pi}\frac{\mathbf{e}_{\varphi}}{\rho}, ρ>R. We want to determine wether \mathbf{B} has a scalar potential or not.

Homework Equations

The Attempt at a Solution

Since the domain is not simply connected, there is no use showing that \nabla\times\mathbf{B}=\mathbf{0} for ρ>R. Therefore, we must show that for all points P and Q the line integral of \mathbf{B} from P to Q is independent of path, since this implies that \mathbf{B} has a scalar potential.

Consider a circle, C, that is concentric with the cylindrical wire with radius R_1 such that R_1>R. Line integration of \mathbf{B} over the closed curve C yields \oint_C\mathbf{B}⋅d\mathbf{r} = [d\mathbf{r} = R_1d\varphi\mathbf{e}_{\varphi}] = \frac{I\mu_0}{2\pi}\int^{2\pi}_0\mathbf{e}_{\varphi}⋅\mathbf{e}_{\varphi}d\varphi = I\mu_0 \neq 0. But this says that \mathbf{B} is dependent of path, and therefore \mathbf{B} does not have a scalar potential.

However, if we now consider the function \phi(\varphi) = \frac{I\mu_0}{2\pi}\varphi, we observe that \nabla\phi = \frac{1}{\rho}\frac{I\mu_0}{2\pi}\mathbf{e}_\varphi = \mathbf{B}. But this says that \mathbf{B} is independent of path, and therefore \mathbf{B} has a scalar potential.

And so, we end up with a contradiction. Since I am very confident that the line integral of \mathbf{B} over C is correct, I assume that there is something in the last part that is wrong, but I cannot find the error.

 

[pasmith]

Your problem here is that \varphi lies between 0 and 2\pi, but both \varphi = 0 and \varphi = 2\pi represent the same physical half-plane. Thus if \phi = k\varphi, k \neq 0, then there is a half-plane where \phi is discontinuous: from one side \phi \to 0 and from the other \phi \to 2k\pi. This is a problem for the existence of \nabla \phi *everywhere* in \rho > R.

To avoid this problem, physically acceptable scalar or vector fields are always periodic in \varphi with period 2\pi.

(The other way to avoid this is to admit multivalued potentials, in which case yes, \nabla (k\varphi) = (k/\rho)\mathbf{e}_{\varphi}.)