In this short note, I explain the magnetic monopole and electromagnetic duality. As we know that mathematics and physics are intertwined. Here I show that the electromagnetic duality can be explained very easily using the **differential form** and **gauge transformation**.**Introduction**

It is very well-known that the Maxwell's equation on classical electrodynamic unifies the electric and magnetic forces. In fact this is the first attempt to unify these two forces. In spite of great success of this theory, in 1931, Dirac first notice that these Maxwell's equations are asymmetric with respect to electic and magnetic field. He introduce a magnetic monopole charge and put these equations into symmetric form:

$$\nabla\cdot E = \rho_e \hspace{3cm} \nabla \times E=-{\partial B \over \partial t} - j_m \hspace{4cm}(1)$$$$\nabla\cdot B = \rho_m \hspace{3cm} \nabla \times B= {\partial E \over \partial t} + j_e\hspace{4.4cm}(2)$$

where $\rho_m$ and $j_m$ are the two additional terms, namely, magnetic charge and current density respectively which are not present in Maxwell's equations. The E and B are the electric and magnetic field respectively, $\rho_e$ and $j_e$ are the eletcric charge and current density. By introducing magnetic monopole charge $\rho_m$ , the price one pays is that the magnetic field B can not be represented by the $B = \nabla \times A$ which is very essential and hence it is inconsistence in quantam mechanics [1], [2]. However it is possible to define such vector potential on any contractible region of space that does not contain origin and some line from the origin to infinity. Dirac first showed that in quantam mechanics the magnetic monopole allowed but with a specific amount of magnetic charge.**Differential forms**

Let $x_{\mu}$ be a real variable in d-dimensional space and $A_{\mu}$ is a function of such x in same space. Then the "1-form" is defined as $$\hspace{4.5cm}A = A_{\mu} dx^{\mu}.\hspace{5.5cm}(3)$$ where the differentials $dx^{\mu}$ are treated following Newton and Leibniz. Under the coordinates transformation, it follows standard vector transformation laws i.e. if $x\rightarrow x'$ then $dx^{\mu} = (\partial x^{\mu}/\partial x'^{\mu})dx'^{\nu} $ so that $$A = A'_{\nu}dx'^{\nu}\hspace{3 cm} A'_{\nu} = A_{\mu} (\partial x^{\mu}/\partial x'^{\mu})\hspace{4cm}(4)$$

In a similar manner, one can define "p-form"

$$\hspace{2cm}H = {1\over p!} H_{\mu_1 \mu_2 \cdots \mu_p} dx^{\mu_1} dx^{\mu_2} \cdots dx^{\mu_p}.\hspace{4cm}(5)$$

It is well-known fact that the differentials of area element commute each other (for example $dx\,dy = dy\,dx$ in 2-space). The product of differentials in eqn(5) are not same as area element in $p$-space. In eqn(5) differentials $dx^{\mu}$ are are anti-commuting variables (Mathematicians call it Grass-mann variables) so that $dx^{\mu_1} dx^{\mu_2} = -dx^{\mu_2} dx^{\mu_1}$ and $dx^{\mu_i} dx^{\mu_i} = 0,~ i = 1,\cdots,p$. The physical interpretation of product of differentials in eqn(5)) is the directional area of $p$-space. In other words, it has same magnitude as area of $p$-space with a definite directions. The eqn(5) can be written in terms of wedge product,

$$H = {1\over p!} H_{\mu_1 \mu_2 \cdots \mu_p} dx^{\mu_1}\wedge dx^{\mu_2}\wedge \cdots\wedge dx^{\mu_p}\hspace{4cm}(6)$$where $ dx^{\mu_i}\wedge dx^{\mu_j} = - dx^{\mu_j}\wedge dx^{\mu_i}$. Now define a differential operator "$d$" acting on "$p$"-form gives "$p+1$"-form

$$dH = {1\over p!} \partial_{\nu}H_{\mu_1 \mu_2 \cdots \mu_p} dx^{\nu}\wedge dx^{\mu_1}\wedge dx^{\mu_2}\wedge \cdots\wedge dx^{\mu_p}\hspace{2cm}(7)$$

and hence $$dA = \partial_{\mu}A_{\nu} dx^{\nu}\wedge dx^{\mu} = {1\over 2}(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) dx^{\mu} dx^{\nu}\hspace{2.4cm}(8)$$

If we consider "1-form", $A = A_{\mu} dx^{\mu}$ is vector potential then $dA$ produces "2-form" electromagnetic field strength $F = (1/2!)F_{\mu\nu} dx^{\mu} dx^{\nu}$. An important property of this differential operator ${\it \bf d}$ is $dd = 0$, which tells that acting with differential operator ${\it d}$ on any * form* twice will give zero. This identity is known as "

Consider a sphere surrounding a magnetic monopole with a magnetic charge g. The 2-form electromagnetic field strength is given by $$F = {g \over 4 \pi} d\cos\theta\,d\phi. \hspace{4.5cm} (9)$$. Therefore the magnetic flux can be calculated by integrating over surface of sphere $S^2$ $$\Phi = \int_{S^2} F = g. \hspace{5cm} (10)$$ Since the magnetic flux $\Phi = \oint_{S}B\cdot dS$, where

$$B= {g\over 4\pi r^2} \hat{r}.\hspace{5.5cm} (11)$$

In quantum mechanics, the vector potential A is the physical. It is easy to guess from eqn(9) that the vector potential would takes the form $A=(g/4 \pi) \cos\theta\,d\phi$. But this is not define in North pole or South pole since $d\phi$ is not defined at these two poles. Instead one can define the vector potential following way $$A_N = -{g \over 4 \pi} (1-\cos\theta)d\phi \hspace{2cm} A_S = {g \over 4 \pi} (1+\cos\theta)d\phi\hspace{2cm}(12)$$ where $A_N(A_S)$ is define locally. The potential $A_N$ is defined on a "coordinate patch" covering the northern hemisphere and extending past the equator as far south as possible except the singular point south pole.

Similarly, the potential $A_S$ is defined on a "coordinate patch" covering the southern hemisphere and extending past the equator as far north as possible except the singular point north pole. In the overlap region (along the equator), the two potentials are not equal:$$\hspace{1cm}A_S-A_N = {g \over 2 \pi}\,d\phi\hspace{4cm}(13)$$ Now consider the gauge transformation of electron field $\Psi \rightarrow e^{i\alpha(x)} \Psi$.

Then the gauge potential $A$ transforms (in $form$ notation) $$A = A + {1\over ie}e^{-i\alpha(x)}\, d \,e^{i\alpha(x)}.\hspace{3cm}(14)$$

On the boundary (the equator of the sphere) of northern and southern hemisphere the two gauge potentials are related by gauge transformation i.e. $$A_S-A_N = {1\over ie}e^{-i\alpha(x)}\, d \,e^{i\alpha(x)}\hspace{3cm}(15)$$

By comparing eqns(14),(15), one gets $\alpha = (eg/2\pi)\phi$. Since $\phi=0$ and $\phi=2\pi$ are the same point (equator of the sphere) and hence hence the wave function must have single valued; in other words,$e^{ieg} = 1$ i.e. $$g = {2\pi \over e} n \hspace{3 cm} n \in \mathbb{Z}\hspace{3cm}(16)$$ This is the famous Dirac's quantization condition for magnetic charge if magnetic monopole exits.

The Dirac's quantization condition eqn(16) has one attractive feature that if $e$ is small then $g$ is large and vice versa. The magnetic charge will be very similar to the electric charge and hence will possess same properties as electric charge (like same magnetic charges repel and opposite attract each other). They too interact with a $1/r$ potential. Therefore one can constract a theory of electromagnetism in terms of magnetic charges, with exchanging the role electric and magnetic field but the theory would be strongly coupled with coupling $g$ rather than $e$. Under duality transformation, a weakly coupled theory is mapped into a strongly coupled theory which is exactly the case electromagnetism if the magnetic charge exit. This concept hase been very widely used in current theoretical physics development (for example string theory (the Theory of Everything!)).

[1] P. A. M. DIRAC, Proceedings of the Royal Society of London A 133, 60 (1931).

[2] J. P. PRESKILL, Annual Review of Nuclear and Particle Science 34, 461 (1984).

[3] Mikio Nakahara, Geometry, Topology and Physics, Institute of Physics Publishing, 2003.

[4] A. Zee, Quantum Field Theory in a Nutshell, University Press, 2010.

Swapan Kumar Majhi (email: tpskm@iacs.res.in) is a Senior Research Associate (Pool Scheme, CSIR) at Indian Association for the Cultivation of Science, Kolkata. He has received Ph.D. from Harish-Chandra Research Institute, Allahabad, India. His research interests includemainly High Energy Physics specially Perturbative QCD, beyond standard model physics and Astroparticle physics. Recently he has also gained interest in mathematical modelling and biophysics.

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