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So let we have a magnetic field which is $$B_\mu=\frac{q}{2}\frac{x_\mu}{|x|^3}-2\pi q\delta_{3\mu}\theta(x_3)\delta{(x_1)}\delta{(x_2)},\tag{4.65}$$ where $\theta$ is step function and $\delta{(x_\mu)}$ is dirac delta.

So in many places (for example Polyakov's book on gauges and strings at eq 4.64 and eq 4.65), it is said that for this magnetic field we have

$$\nabla\times\vec{B}=0 \tag{4.64}$$

but I can't see how, because the curl of the second term is not zero at least if I take the curl naively it will involve derivatives of Dirac delta and will be non-zero.

Qmechanic
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physshyp
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1 Answers1

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  1. Yes, OP is right: The curl $\vec{\nabla}\times \vec{B}$ of the $\vec{B}$-field (4.65) is non-zero and violates Ampere's Law due to the presence of the Dirac string. However, the Dirac string is a gauge artifact from using a globally defined magnetic vector potential (MVP) $$\begin{align} \vec{A}~=~&\lim_{\epsilon\searrow 0^+}\begin{pmatrix}y\cr -x\cr 0\end{pmatrix}\frac{q/2}{\sqrt{r^2+\epsilon}(\sqrt{r^2+\epsilon}-z)}, \cr \vec{r}~\in~&\mathbb{R}^3.\end{align}\tag{1}$$ The $\vec{A}$-field in eq. (1) and the corresponding magnetic field $\vec{B}=\vec{\nabla}\times \vec{A}$ are generalized functions/distributions. For fixed regularization parameter $\epsilon>0$, they are globally defined smooth $C^{\infty}$-maps that satisfy $\vec{\nabla}\cdot \vec{B}=0$.

  2. The presence of a magnetic monopole $$\begin{align} \vec{B}~=~&\lim_{\epsilon\searrow 0^+}\frac{q/2}{(r^2+\epsilon)^{3/2}}\vec{r}, \cr \vec{\nabla}\cdot \vec{B}~=~&\lim_{\epsilon\searrow 0^+}\frac{3q/2}{(r^2+\epsilon)^{5/2}}~=~2\pi q ~\delta^3(\vec{r}),\cr \vec{r}~\in~&\mathbb{R}^3, \end{align}\tag{2}$$ at the origin $\vec{r}=\vec{0}$ is an obstruction for the existence of a globally defined MVP. The best we can do is to cover the punctured 3-space $\mathbb{R}^3\backslash\{\vec{0}\}=\bigcup_{\alpha}U_{\alpha}$ by locally defined MVPs $\vec{A}_{\alpha}: U_{\alpha}\to\mathbb{R}^3$ in open neighborhoods $U_{\alpha}$ such that each pairs of MVPs in overlaps $U_{\alpha}\cap U_{\beta}$ are connected by gauge transformations, cf. the Wu-Yang bundle construction. This is e.g. worked out explicitly in Refs. 1 & 2.

  3. The magnetic monopole (2) does satisfy Ampere's law $\vec{\nabla}\times \vec{B}=\vec{0}$ globally.

References:

  1. M. Nakahara, Geometry, Topology and Physics, 1989; section 1.3.

  2. M. Nakahara, Geometry, Topology and Physics, 2003; section 1.9.

Qmechanic
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