Magnetic Scalar Potential of Infinite Wire

Question

While working on a question about magnetic scalar potential, I encountered a challenge. The question posits that the magnetic scalar potential, denoted as $\phi$, takes the form $\phi = -\frac{I}{2\pi}\theta$. It then inquires why this expression is not present in Cylindrical Harmonics, the general solution to Laplace's equation in cylindrical coordinates..

I understand that the scalar potential provided here is multi-valued, whereas the solution presented in the form of cylindrical harmonics is a single-valued function. This distinction is crucial. However, my questions are...

1. Why should the solution to Laplace's equation be single-valued? Is it possible for a multi-valued function, such as the one described above, to serve as a solution to the complex form of Laplace's equation? Furthermore, why isn't the complex case of Laplace's equation considered in E&M?

2. I speculate that the multi-valued characteristic arises due to the non-simply connected nature of the considered domain, causing the curl theorem not to hold. Nonetheless, it still satisfies $B = -\mu_0\nabla\phi$ and $\nabla^2\phi = 0$. Is it a general rule that in a non-simply connected region, where $B = -\mu_0\nabla\phi$ and $\nabla^2\phi = 0$ hold, $\phi$ should always be multi-valued, or is this just a fortuitous occurrence?

Answer 1

It depends on the physical interpretation of the quantities. In other contexts, $\phi$ would need to be single valued like if it represented temperature for example. In the context of magnetostatics, only $B$ should be single valued. In particular, $\phi$ need not be single valued (this applies to the vector potential as well). However, for $B$ to be single valued, you'll need the different branches differ by a gauge transformation. In the case of the potential, you'll need the difference to be locally constant. This is the case for your solution, but would not be the case when $\phi\propto \theta^2$ for example.

You usually solve for complex functions in EM. It's just that at the end of the day, the EM fields are real so you keep only the real part by superposition.

For your second question you are on the right track. If you look at the corresponding magnetic field: $$ B = \frac{I}{2\pi \rho}e_\theta $$ If you assume the domain $D$ to be space minus the line $\rho=0$, then it is not simply connected anymore. Even if: $$ \nabla\times B = 0 $$ in $D$, you have $$ \oint B\cdot dx = nI $$ with $n$ the integer winding number of the closed path about the line. You cannot apply the curl theorem for $n\neq 0$ since you cannot construct a surface whose boundary is the loop while still being entirely in $D$. Yes, the potential is multivalued by definition. You define it by: $$ \phi = -\int B\cdot dx $$ and since the RHS, the potential value will depend on the homotopy class of the path, hence be multivalued (does not only depend on the endpoints). In the case of $D$, the homotopy group is $\mathbb Z,+$, which is why the winding number is an integer.

Hope this helps.