Is this case a failure of Stokes' theorem?

Question

In the presence of a hypothetical magnetic point charge at the origin of coordinates, it turns out that an irremovable physical singularity of the vector potential ${\bf A}({\bf r})$ exists for any choice of ${\bf A}({\bf r})$, extending from the origin to infinity in a radially outward direction. This is called the Dirac string. Let us consider a particular choice of ${\bf A}({\bf r})$ such that the Dirac string lies along the positive $z$ axis. This choice is $${\bf A}({\bf r}) = \frac{g}{r(r-z)}(y\hat{{ x}}-x\hat{{y}}),\nonumber\\ ~~~~= -\frac{g(1+\cos\theta)}{r\sin\theta}\hat{{\phi}}. $$ There is a genuine singularity at $\theta=0$ i.e. on the entirety of positive $z$-axis!

Now consider the following steps. First, consider a punctured sphere with its center at the origin and the punctured disc pierced by the positive $z$ axis such that the Dirac string passes through the disc. Next, consider a closed loop $C$ and along the circumference of the disc enclosing the Dirac string.

Is Stokes' theorem valid here? I recall from memory that Stokes' theorem is valid if both the surface and the loop enclosing surface are in a region where the vector field is free from singularities. In this particular situation, the loop $C$, as well as the curved surface of the sphere enclosed by the loop, lie in a singularity-free region of ${\bf A}({\bf r})$. However, if we consider the circular puncture enclosed by the loop, this disc not free from the singularity. Because it is pierced by the positive $z$-axis.

Answer 1

In this kind of cohomology-related things, physicists tend to say that “Stokes theorem always holds,” while allowing the possibility of delta-function sources. On the other hand, mathematicians normally exclude the delta-like sources from the domain of their spaces and say that “you cannot apply Stokes theorem when the ‘inner region’ (the inverse-boundary operation) of the submanifold of integration is not well-defined.“

Take a current $I$ flowing through an infinite line, for example. The magnetic field is $(\mu_0 I/2\pi) (1/s)\hat{\phi}$. Physicists say that Stokes’ theorem $\mu_0\int_\mathcal{S} \vec{J}\cdot d^2\vec{a} = \oint_{\partial\mathcal{S}} \vec{B}\cdot d\vec{l}$ holds perfectly well, as the current density is given by $\vec{J}(x,y,z) = I\delta(x)\delta(y) \hat{z}$. But, mathematicians say that “you cannot call a loop enclosing the current line ‘$\partial\mathcal{S}$’!” It is because, the surface $\mathcal{S}$ filling the loop “$\partial\mathcal{S}$” passes through the line, where the magnetic field diverges. In mathematician’s standards, the “space” is not $\mathbb{R}^3$ but $\mathbb{R}^3\setminus\{x=0,y=0\}$, the current line being removed. Then it has a nontrivial topology and a nontrivial cohomology class… If we imagine $\mathcal{S}$ as a soap membrane, then it will “burst” when it touches the “spiky” delta-like $\nabla\times\vec{B}$, screaming like “ouch”!!

Your original question about magnetic monopole can be also understood in this way. For physicists, applying the Stokes theorem is never a problem at all because they will always push their “black magic with delta distributions” to the end. For instance see Nakahara’s “Geometry, Topology, and Physics” for the current density of the Dirac string: an infinitesimally thin half-infinite solenoid. It is expressed in terms of Dirac delta and Heaviside step function. On the other hand, mathematicians will say that the “space” is not the entire Euclidean space but $\mathbb{R}^3\setminus\{0\}$ so that you cannot call the Gauss surface enclosing the magnetic monopole “$\partial\mathcal{V}$” because there is no such thing as “$\mathcal{V}$” that can “survive” after passing through the spiky singularity at the origin. They just exclude the origin from the domain of the magnetic field, because it blows up there. (Note that physicists include the origin as the domain of the magnetic field and even write down equations like $\nabla\cdot\vec{B}=g\delta(x)\delta(y)\delta(z)$.) Then they will work in the manner precisely you have described in your question: introducing Dirac string, coordinate patches, and so on.

Answer 2

The Stokes' theorem $$ \oint_{C(S)} \vec{A}(\vec{r}) \cdot d\vec{\ell} = \int\int_S \vec\nabla\times\vec{A}\cdot d\vec{S}. $$

Therefore, any closed contour integral vanishes only when $\vec\nabla\times\vec{A} = 0$ (which then implys that the integral between two points will not depend on the path of integral).

In your case, the $\vec\nabla\times\vec{A}$ is:

$$ \vec\nabla\times\vec{A} = \frac{1}{r^2\sin\theta} \left\vert \begin{matrix} \hat{r} & r\hat{\theta} & r\sin\theta\hat{\phi}\\ \frac{\partial}{\partial r} &\frac{\partial}{\partial \theta} &\frac{\partial}{\partial \phi} \\ 0 & 0& r\sin\theta \frac{g(1+\cos\theta)}{r\sin\theta} \end{matrix} \right\vert = \frac{1}{r^2\sin\theta} \hat{r}\frac{\partial}{\partial \theta} g(1+\cos\theta) = -\hat{r}\frac{g}{r^2} $$

The $\vec\nabla\times\vec{A} \ne 0$, therefore the integral between two points will be depended on path of integral.

Answer 3

You don't really have to be worried about the non-analyticity of the gauge potential on the Dirac string, when deducing Dirac's quantization:

With your choice of potential ${\bf{A}} = -g\frac{(1+\cos\theta)}{r\sin\theta}\,\hat{\phi} = -\frac{g}{r}\cot\left(\frac{\theta}{2}\right)\,\hat{\phi}$, there is a singular line (Dirac string) on the positive $z$ axis as you correctly mentioned. In order to get to Dirac's quantization result, consider a pinched sphere S containing the pole (very similar to yours), and has a tiny hole around $\theta = 0$ somewhere on the $z$ axis, from within which the Dirac string goes through. You can, without any issues, use the Stokes' theorem for this surface S having the tiny circular hole C as its boundary. (Yes you are right that you cannot use Stoke's theorem if instead we were trying to work with the tiny cap surface on the upper hemisphere of S, which we have discarded).

Consider a charged particle (electric charge e) and let us move it around this ever so tiny loop C. The total phase accumulated in its wave function would be equal to $e\oint_c \mathrm{d}{\bf{l}}\cdot{\bf{A}} = \int\mathrm{d}{\bf{S}}\cdot{\bf{B}} = 4\pi e g$. The first equality comes about on account of using Stoke's theorem, which as we said above holds just fine for our surface. (To convince yourself you can explicitly calculate the line integral for your choice of potential and see it is the same as the radius of C becomes ever so tiny.) Now this phase accumulated cannot be anything but some integer multiple of one full rotation, $2\pi n$, since the wave function has to be well behaved (continuous). This gives $2eg = n$.

We can repeat this exercise for any other choice of $\bf{A}$ for which the Dirac string was lying somewhere else. For instance you can perform a $\textit{large}$ gauge transformation on your original $\bf{A}$: $\bf{A} \rightarrow \bf{A} + \nabla\alpha$ with $\alpha = g\phi$, to now get ${\bf{A}} = -g\cot(\theta)\hat{\phi}$. This time you have two Dirac strings! One on the positive $z$ axis and another on the negative $z$ axis. We can play the same game as above, this time having two punctures on both the northern and southern hemisphere. We will get the same quantization condition.