The Dual Field Tensor

Two of Maxwell's equations have sources on the right. Two do not. Written as vectors, that looks like an accident of history — Gauss and Ampère got charges and currents, Faraday and no-monopoles got zero. Written in spacetime, it stops being an accident and becomes a statement. Contract the field tensor with the four-dimensional Levi-Civita symbol and you build a second antisymmetric tensor, the dual, $G^{\mu\nu} = \tfrac12\,\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ . Component by component, $G$ is $F$ with $\vec E \to c\vec B$ and $\vec B \to -\vec E/c$ — an electric-magnetic rotation performed by index gymnastics. And the two sourceless Maxwell equations, $\nabla\cdot\vec B = 0$ and Faraday's law, collapse into one line:

\partial_\mu G^{\mu\nu} = 0.

Now put the two equations side by side. $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ says electric charge sources the field. $\partial_\mu G^{\mu\nu} = 0$ says there is nothing on the right — and the thing that would go there has a name. If magnetic charge existed the equation would read $\partial_\mu G^{\mu\nu} = J_m^\nu/c$ , with $J_m^\nu$ a magnetic four-current. The zero is not decoration. It is a measurement. Every time anyone has looked for a magnetic monopole — in cosmic rays, in moon rock, at the LHC — the answer has come back zero. That is why this article's portrait is the magnetic monopole: the dual tensor makes the monopole's absence into a physical law rather than an omission.

Key takeaways

Signature $(+,-,-,-)$ , $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , $p\cdot p = +m^2c^2$ ; $\varepsilon^{0123} = +1$ ; $m$ is the invariant rest mass.
The dual: $\boxed{G^{\mu\nu} \equiv \tfrac12\,\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}}$ — the $\tfrac12$ undoes the double-counting of the antisymmetric pair $(\alpha\beta)$ . It implements $\vec E \to c\vec B$ , $\;\vec B \to -\vec E/c$ : a $90°$ rotation in the $(\vec E/c, \vec B)$ plane. Twice is $180°$ , hence $\boxed{{}^{**}F^{\mu\nu} = -F^{\mu\nu}}$ .
The homogeneous pair, in one equation: $\boxed{\partial_\mu G^{\mu\nu} = 0}$ . $\nu = 0$ gives $\nabla\cdot\vec B = 0$ ; $\nu = i$ gives $\nabla\times\vec E = -\partial_t\vec B$ . It is equivalent to the Bianchi identity $\partial_{[\lambda}F_{\mu\nu]} = 0$ — automatic once $F = \mathrm{d}A$ , because partial derivatives commute. Not dynamics: the price of having a potential.
Contracting $F$ with $G$ gives the pseudoscalar $\boxed{F_{\mu\nu}G^{\mu\nu} = -\tfrac{4}{c}\,\vec E\cdot\vec B}$ — the invariant that $F$ alone cannot see.
Owned result: the monopole source term, with its coefficient derived, not guessed. $\boxed{\partial_\mu G^{\mu\nu} = \tfrac1c J_m^\nu}$ with $J_m^\nu = (c\rho_m, \vec J_m)$ reproduces $\nabla\cdot\vec B = \rho_m$ and $\nabla\times\vec E = -\partial_t\vec B - \vec J_m$ . The AllQFT base prints $\mu_0 c\,J_m^\nu$ ; that coefficient is dimensionally impossible and is refused here.
Key idea: duality maps electric charge $q$ to magnetic charge $g = \mu_0 q c$ . Do that to an electron and you get $6.04\times10^{-17}\ \mathrm{Wb}$ — but Dirac's condition says the smallest legal monopole carries $h/e = 4.14\times10^{-15}\ \mathrm{Wb}$ , exactly $1/2\alpha = 68.5$ times larger. Duality is a symmetry of the classical equations; quantum mechanics sets the price of admission.

The core insight

The Levi-Civita symbol is the only Lorentz-invariant object that knows about orientation, so contracting with it is the only way to build a new tensor out of $F$ without differentiating it. That single construction does three jobs at once: it rotates $\vec E$ into $\vec B$ , it packages the two source-free Maxwell equations into a divergence, and it exposes the pseudoscalar $\vec E\cdot\vec B$ , which no contraction of $F$ with itself can ever produce. The dual is not a notational convenience. It is the second half of the electromagnetic field.

Where this lives today

Duality is the entry point to a great deal of modern theory, and the dual tensor is where it starts. The invariant $F_{\mu\nu}G^{\mu\nu}$ is a total derivative in vacuum, so adding it to the Lagrangian changes nothing classically — but multiply it by a field and it becomes the axion coupling, a leading candidate both for solving the strong-CP problem and for the dark matter of the universe. Axion haloscopes such as ADMX are, quite literally, experiments to detect a term built out of $G^{\mu\nu}$ . Duality also organises strong coupling: Montonen–Olive and Seiberg–Witten theory exploit the fact that a theory with electric charges at strong coupling can be re-described as one with magnetic charges at weak coupling. The factor $68.5 = 1/2\alpha$ computed above is exactly why: duality maps weak coupling to strong. And the null result keeps its own importance. Every bound on the monopole flux — MoEDAL, cosmic-ray detectors, the "Parker bound" from the survival of the galactic magnetic field — is a bound on the right-hand side of $\partial_\mu G^{\mu\nu} = 0$ . Be clear about the logic: the equation is not a theorem, it is a summary of the data. Were a monopole found tomorrow the Bianchi identity would not be violated — the potential would be, and electromagnetism would acquire a topological sector. A far more interesting failure mode than a broken equation.

The physical picture

$F^{\mu\nu}$ is an antisymmetric $4\times4$ table with six independent entries: three time-space slots holding $\vec E/c$ , three space-space slots holding $\vec B$ . That split is not sacred — a boost reshuffles the entries — but the shape is: six numbers, arranged antisymmetrically. Ask what else you can build from those six numbers using only Lorentz-invariant machinery. Raising and lowering with $\eta$ gets you $F_{\mu\nu}$ and $F^\mu{}_\nu$ — nothing new. Multiplying $F$ by itself is quadratic. The only linear operation that produces a genuinely new antisymmetric tensor is contraction with $\varepsilon^{\mu\nu\alpha\beta}$ , and it produces exactly one object: the dual. It exchanges the two blocks — the entries that held $\vec E/c$ now hold $\vec B$ , and those that held $\vec B$ now hold $-\vec E/c$ : a rotation by ninety degrees in the $(\vec E/c, \vec B)$ plane. Twice is $180°$ , which is why ${}^{**}F = -F$ . The payoff is that Maxwell's equations become symmetric in appearance. $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ and $\partial_\mu G^{\mu\nu} = 0$ have the identical form; the only difference is what stands on the right. Electricity has a source. Magnetism has a zero. The dual turns that asymmetry from a fact about how the equations are written into a fact about the world.

Figure

THE DUAL IS A 90° ROTATION IN THE (E/c, B) PLANE. F holds E/c in its time-space slots and B in its space-space slots; G = ½ε F swaps them, sending E → cB and B → −E/c. Do it twice and you have rotated 180°: **F = −F. The two Maxwell equations then have IDENTICAL form — ∂_μF^{μν} = μ₀J^ν and ∂_μG^{μν} = 0 — and the ONLY difference is what stands on the right. Electric charge exists. Magnetic charge, so far as anyone has ever measured, does not.

The core equations

The construction, and the combinatorial factor that everyone forgets:

\boxed{\;G^{\mu\nu} \;\equiv\; \tfrac12\,\varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}\;}\ ,\qquad \varepsilon^{0123} = +1 \qquad (1)

where: $\varepsilon^{\mu\nu\alpha\beta}$ — the totally antisymmetric Levi-Civita symbol; $F_{\alpha\beta} = \eta_{\alpha\mu}\eta_{\beta\nu}F^{\mu\nu}$ . Physical reading: the sum over $(\alpha,\beta)$ counts every independent pair twice — swapping $\alpha\leftrightarrow\beta$ flips the sign of $\varepsilon$ and of $F$ , so the contributions reinforce. The $\tfrac12$ divides that out, giving $G^{01} = F_{23}$ . Two conventions are in play — signature and the sign of $\varepsilon^{0123}$ — and flipping either flips $G$ . Written out, $G$ is $F$ with the fields rotated into one another:

G^{\mu\nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z\\ B_x & 0 & E_z/c & -E_y/c\\ B_y & -E_z/c & 0 & E_x/c\\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix} \qquad\Longleftrightarrow\qquad \boxed{\;\vec E \to c\vec B\ ,\quad \vec B \to -\,\frac{\vec E}{c}\;} \qquad (2)

where: compare entry by entry with $F^{\mu\nu}$ , whose $0i$ slots hold $-E_i/c$ and whose $ij$ slots hold $-\varepsilon^{ijk}B_k$ . Physical reading: $G$ is not a new field. It is the same six numbers, re-slotted — which is why duality is a symmetry of the vacuum equations and fails once you put a charge on the right. The two source-free Maxwell equations, in one line — and its equivalent cyclic (Bianchi) form, which is an identity:

\boxed{\;\partial_\mu G^{\mu\nu} = 0\;} \iff \nabla\cdot\vec B = 0\ (\nu=0),\ \ \nabla\times\vec E = -\frac{\partial\vec B}{\partial t}\ (\nu=i) \iff \boxed{\;\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0\;} \qquad (3)

where: $\partial_\mu = (\tfrac1c\partial_t, \nabla)$ ; $A_\mu$ is the four-potential. Physical reading: one four-vector equation, four components, two familiar laws — and not a dynamical law. Substitute $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ into the cyclic sum and every term cancels against another, because $\partial_\mu\partial_\nu = \partial_\nu\partial_\mu$ . The homogeneous equations carry no content beyond "a potential exists". And the equation nobody has ever needed — the article's owned result, with its coefficient derived:

\boxed{\;\partial_\mu G^{\mu\nu} = \frac{1}{c}\,J_m^\nu\;}\ ,\quad J_m^\nu = (c\rho_m,\ \vec J_m) \qquad\Longleftrightarrow\qquad \nabla\cdot\vec B = \rho_m\ ,\quad \nabla\times\vec E = -\,\frac{\partial\vec B}{\partial t} - \vec J_m \qquad (4)

where: $\rho_m$ — magnetic charge density in $\mathrm{Wb\,m^{-3}}$ ; $\vec J_m$ — magnetic current density in $\mathrm{V\,m^{-2}}$ . Physical reading: the coefficient is not free. It is fixed the moment you choose units for $\rho_m$ , and drops out of $\partial_\mu G^{\mu 0} = \nabla\cdot\vec B$ by inspection. Observation says $J_m^\nu = 0$ , so (4) collapses to (3) — an experimental result, not a definition.

Building the dual, and reading Maxwell off it

Derivation AThe dual tensor, its factor of one half, and what it does to E and B

The goal is to construct, from $F$ alone, a second antisymmetric tensor that is not just $F$ in disguise. There is essentially one way: the Levi-Civita symbol is the unique totally antisymmetric invariant of the proper orthochronous Lorentz group, so contracting with it is the only linear operation available. Everything else — raising, lowering, tracing — either gives back $F$ or gives zero. The only freedom is the prefactor, and it is fixed by counting.

Step 1. Write $G^{\mu\nu} = k\,\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ and count the double-counting. For fixed $(\mu,\nu)$ the sum runs over both $(\alpha,\beta)$ and $(\beta,\alpha)$ , and swapping gives $\varepsilon^{\mu\nu\beta\alpha}F_{\beta\alpha} = (-\varepsilon^{\mu\nu\alpha\beta})(-F_{\alpha\beta}) = \varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ — the two sign flips reinforce, so each independent pair contributes twice with the same sign. Hence $k = \tfrac12$ . Why: two antisymmetric objects contracted on two indices always double-count. Drop the $\tfrac12$ and every subsequent factor is wrong by two.

Step 2. Compute the entries. With $\varepsilon^{0123} = +1$ and the derived components $F_{0i} = +E_i/c$ , $F_{ij} = -\varepsilon_{ijk}B_k$ : G^{01} = \tfrac12\,\varepsilon^{01\alpha\beta}F_{\alpha\beta} = \varepsilon^{0123}F_{23} = F_{23} = -B_x,\qquad G^{12} = \varepsilon^{1230}F_{30} = -F_{03} = E_z/c. Why: $\varepsilon^{1230}$ is three transpositions from $\varepsilon^{0123}$ , so it is $-1$ . Getting that sign right is the entire difficulty of the calculation.

Step 3. Read off the pattern. $G$ 's time-space slots hold $-B_i$ where $F$ 's held $-E_i/c$ ; $G$ 's space-space slots hold $+\varepsilon^{ijk}E_k/c$ where $F$ 's held $-\varepsilon^{ijk}B_k$ . Matching $G$ to $F$ 's shape gives \boxed{\vec E \to c\vec B\ ,\qquad \vec B \to -\vec E/c}. Why: this is the substitution rule and the whole reason the dual is useful — a $90°$ rotation in the $(\vec E/c, \vec B)$ plane. Apply it twice and you have ${}^{**}F = -F$ .

Step 4. Note what $G$ is not: a tensor. $\varepsilon$ carries a fixed orientation of spacetime, so under parity ( $\det\Lambda = -1$ ) $G$ picks up an extra minus sign relative to a genuine tensor. $G$ is a pseudotensor. Why: this is the index-notation statement that $\vec B$ is axial while $\vec E$ is polar. The pseudo-ness is physics, not bookkeeping. *Verified in verify_aq-sr-dual-field-tensor.py: $F^{\mu\nu}$ is first rebuilt from $A^\mu = (\phi/c, \vec A)$ via $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ , so the component dictionary is derived rather than assumed; then $G = \tfrac12\varepsilon F$ is contracted explicitly and matched entry-by-entry against the matrix printed in equation (2); and $G$ is shown to equal $F$ under $\vec E\to c\vec B$ , $\vec B\to-\vec E/c$ .*

Go deeper · Derivation B · 2 minThe vanishing divergence IS the homogeneous pair, and the Bianchi identity makes it automatic

Derivation BThe vanishing divergence IS the homogeneous pair, and the Bianchi identity makes it automatic

Two things must be shown, and they differ in kind. First, that $\partial_\mu G^{\mu\nu} = 0$ contains $\nabla\cdot\vec B = 0$ and Faraday's law — a computation. Second, that it is an identity rather than a law — a structural fact about potentials, and the more important of the two: it tells you the homogeneous equations are not really equations at all.

Step 1. Take $\nu = 0$ . Only spatial derivatives survive ( $G^{00} = 0$ ), and $G^{i0} = +B_i$ from equation (2), so \partial_\mu G^{\mu 0} = \partial_i G^{i0} = \partial_i B_i = \nabla\cdot\vec B = 0. Why: no magnetic monopoles, straight out of the time slot. The vanishing right-hand side is doing the work.

Step 2. Take $\nu = 1$ : \partial_\mu G^{\mu 1} = \tfrac1c\partial_t G^{01} + \partial_2 G^{21} + \partial_3 G^{31} = -\tfrac1c\Big[\partial_t B_x + \big(\partial_y E_z - \partial_z E_y\big)\Big] = -\tfrac1c\Big[\partial_t B_x + (\nabla\times\vec E)_x\Big] = 0, so $\nabla\times\vec E = -\partial_t\vec B$ . Why: Faraday's law of induction, from the spatial slots. The overall $-1/c$ is common to all three components; the physics is in the bracket.

Step 3. Now show it is an identity. Insert the definition (with $\varepsilon$ constant), then substitute $F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha$ : \partial_\mu G^{\mu\nu} = \tfrac12\,\varepsilon^{\mu\nu\alpha\beta}\,\partial_\mu F_{\alpha\beta} = \tfrac12\,\varepsilon^{\mu\nu\alpha\beta}\big(\partial_\mu\partial_\alpha A_\beta - \partial_\mu\partial_\beta A_\alpha\big) = \varepsilon^{\mu\nu\alpha\beta}\,\partial_\mu\partial_\alpha A_\beta \;\equiv\; 0 , because $\partial_\mu\partial_\alpha$ is symmetric in $(\mu,\alpha)$ while $\varepsilon^{\mu\nu\alpha\beta}$ is antisymmetric in it, and symmetric contracted with antisymmetric on the same pair vanishes. Why: this is the whole story. No physics went in — only the commutativity of partial derivatives. The two homogeneous Maxwell equations are guaranteed the moment you write $F = \mathrm{d}A$ .

Step 4. Recognise the equivalent form. Contracting the cyclic sum $\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu}$ with $\varepsilon$ reproduces $3\times$ the divergence above, and $\varepsilon$ is non-degenerate, so $\partial_\mu G^{\mu\nu} = 0 \iff \partial_{[\lambda}F_{\mu\nu]} = 0$ . Why: the cyclic form is the Bianchi identity, and it survives to curved spacetime — the connection terms cancel by antisymmetry, which is why electromagnetism couples to gravity painlessly.

Step 5. Draw the conclusion. The homogeneous equations have no source because they cannot have one, so long as $F$ derives from a potential. To give magnetism a source you must give up the global existence of $A^\mu$ — exactly what Dirac did in 1931, with a string singularity. That is why the monopole is a statement about topology rather than about an extra Lagrangian term. Why: this is the deep reason the two Maxwell pairs are not symmetric, and no amount of pretty notation can make them so. *Verified in verify_aq-sr-dual-field-tensor.py: $\partial_\mu G^{\mu 0} = \nabla\cdot\vec B$ and $\partial_\mu G^{\mu i} = -\tfrac1c[(\nabla\times\vec E)_i + \partial_t B_i]$ , symbolically, with $\vec E$ and $\vec B$ arbitrary functions of $(t,\vec x)$ ; and $\varepsilon^{\mu\nu\alpha\beta}\partial_\mu\partial_\alpha A_\beta = 0$ by explicit summation over all $4^4$ index combinations for an arbitrary $A_\mu(t,\vec x)$ .*

Go deeper · Derivation C · 2 minThe double dual, the pseudoscalar, and the monopole source term whose coefficient the base gets wrong

Derivation CThe double dual, the pseudoscalar, and the monopole source term whose coefficient the base gets wrong

Three loose ends, the third a repair. The double dual pins down the sign that distinguishes Minkowski from Euclidean geometry; the pseudoscalar $F_{\mu\nu}G^{\mu\nu}$ is the invariant only the dual can see; and the monopole equation's coefficient is a number the base article simply asserts — wrongly.

Step 1. Dualise twice, using $\varepsilon^{\mu\nu\alpha\beta}\varepsilon_{\mu\nu\rho\sigma} = -2\big(\delta^\alpha_\rho\delta^\beta_\sigma - \delta^\alpha_\sigma\delta^\beta_\rho\big)$ : {}^{}F^{\rho\sigma} = \tfrac14\,\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\mu\nu\alpha\beta}F^{\alpha\beta} = -\tfrac12\big(F^{\rho\sigma} - F^{\sigma\rho}\big) = \boxed{-F^{\rho\sigma}} . *Why: the minus traces to $\det\eta = -1$ — a signature** effect, not a convention. In Euclidean four-space the Hodge star squares to $+1$ and real self-dual fields exist; in Minkowski space it squares to $-1$ and they do not. That is why instantons live in Euclidean signature.*

Step 2. Contract $F$ with its dual. Every surviving product pairs an $E$ -component with the same $B$ -component: \boxed{\;F_{\mu\nu}G^{\mu\nu} = -\,\frac{4}{c}\,\vec E\cdot\vec B\;}. Why: this invariant is invisible to $F$ alone. No contraction of $F$ with itself and the metric can produce $\vec E\cdot\vec B$ — you need the orientation $\varepsilon$ supplies. It is a pseudoscalar: unchanged by boosts, sign-flipped by a mirror.

Step 3. Now the monopole. Write $\partial_\mu G^{\mu\nu} = \kappa\,J_m^\nu$ with $J_m^\nu = (c\rho_m, \vec J_m)$ and $\kappa$ unknown, then fix it from the time component. The previous derivation gave $\partial_\mu G^{\mu 0} = \nabla\cdot\vec B$ exactly. If magnetic charge is measured in webers, so the monopole Gauss law reads $\nabla\cdot\vec B = \rho_m$ with $\rho_m$ in $\mathrm{Wb\,m^{-3}}$ , then $\kappa\,(c\rho_m) = \rho_m$ and \boxed{\kappa = \tfrac1c}. Why: no fitting, no guessing. The left side is already computed; the right side is a choice of units; $\kappa$ is whatever makes them agree. It is not a free parameter.

Step 4. Check the spatial components, which are now a prediction, not an input. With $\kappa = 1/c$ the $\nu = i$ components read $-\tfrac1c[(\nabla\times\vec E)_i + \partial_t B_i] = \tfrac1c J_{m,i}$ , i.e. \nabla\times\vec E = -\frac{\partial\vec B}{\partial t} - \vec J_m , exactly the magnetic-current Faraday law of the standard duality-symmetric Maxwell equations. Why: $\kappa$ was fixed by $\nu = 0$ alone, and the three $\nu = i$ components then came out right on their own. That is the check, and it is what distinguishes a derived coefficient from a guessed one.

Step 5. Refuse the base's coefficient. The base prints $\partial_\mu G^{\mu\nu} = \mu_0 c\,J_m^\nu$ . Dimensional analysis kills it: $[\partial G] = \mathrm{T\,m^{-1}}$ and $[\mu_0 c] = \mathrm{T\,m^2\,A^{-1}\,s^{-1}}$ , so $J_m^\nu$ would carry units of $\mathrm{A\,s\,m^{-3}} = \mathrm{C\,m^{-3}}$ — an electric charge density. Not a magnetic current at all. Why: the coefficient on the right of a field equation is the exchange rate between the units on the two sides, checkable in one line. (In the "ampere-metre" convention, $\nabla\cdot\vec B = \mu_0\rho_m$ , it would be $\kappa = \mu_0/c$ — still not $\mu_0 c$ .) *Verified in verify_aq-sr-dual-field-tensor.py: ${}^{**}F = -F$ by explicit double contraction; $F_{\mu\nu}G^{\mu\nu} = -(4/c)\vec E\cdot\vec B$ symbolically; $\kappa = 1/c$ derived by solving $\partial_\mu G^{\mu 0} = \kappa c\rho_m$ against $\nabla\cdot\vec B = \rho_m$ , then independently confirmed against the $\nu = i$ components; and the base's $\mu_0 c$ refuted dimensionally.*

Figure

THE DUAL OF AN ELECTRON IS A MONOPOLE — and quantum mechanics refuses it. Duality maps the electron's Coulomb field (1.43996 × 10¹¹ V/m at one ångström) to a RADIAL magnetic field of 480.32 T, whose divergence is non-zero at the origin: a monopole of strength g = μ₀ e c = 6.0359 × 10⁻¹⁷ Wb. But Dirac's condition says the smallest legal magnetic charge is g_D = h/e = 4.13567 × 10⁻¹⁵ Wb — exactly TWO superconducting flux quanta, and exactly 1/(2α) = 68.518 times larger. A dual electron is 68.5× too weak to exist. Duality is a symmetry of the classical equations; quantum mechanics sets the price of admission, and the price is the fine-structure constant.

Limiting cases

Case	Result	Meaning
$\vec B = 0$ (pure electric)	$G^{0i} = 0$ , $\;G^{ij} = \varepsilon^{ijk}E_k/c$	the dual of a pure $\vec E$ is a monopole field
plane wave ( $E = cB$ , $\vec E\perp\vec B$ )	$F_{\mu\nu}F^{\mu\nu} = 0$ , $\;F_{\mu\nu}G^{\mu\nu} = 0$	a null field: $F$ and $G$ carry the same information
$F = \mathrm{d}A$	$\partial_\mu G^{\mu\nu} \equiv 0$	an identity. Homogeneous Maxwell is free
dual twice	${}^{**}F = -F$	a $180°$ duality rotation. The $-$ is $\det\eta = -1$
parity ( $\vec x\to-\vec x$ )	$G \to -\,(\text{tensor image of } G)$	$G$ is a pseudotensor; $\vec B$ is axial, $\vec E$ is polar
magnetic charge present	$\partial_\mu G^{\mu\nu} = J_m^\nu/c$	what nobody has ever seen. The zero is data
$\varepsilon^{0123} \to -1$	$G \to -G$	conventions matter: state yours, or compare nonsense

Common trapThe base's monopole equation has a dimensionally impossible coefficient.

The AllQFT base article prints, as an "optional" card, \partial_\mu G^{\mu\nu} = \mu_0\,c\,J_m^\nu , which cannot be right in any unit system for magnetic charge. The left side has units of $\mathrm{T\,m^{-1}}$ ; dividing by $\mu_0 c$ leaves $J_m^\nu$ with units of $\mathrm{A\,s\,m^{-3}} = \mathrm{C\,m^{-3}}$ — an electric charge density. It is off by $c^2$ from the "ampere-metre" convention ( $\kappa = \mu_0/c$ ) and by $\mu_0 c^2$ from the weber convention ( $\kappa = 1/c$ ). Derive a coefficient, then confirm it against components you did not use — which is what the third derivation does, and what a one-line dimensional check would have caught.

Common trapG is a pseudotensor, and the base is a corrupted source.

The physics. $G^{\mu\nu}$ transforms as a tensor under proper Lorentz transformations but picks up an extra minus sign under improper ones (parity, time reversal), because $\varepsilon$ encodes an orientation of spacetime. Consequently $F_{\mu\nu}G^{\mu\nu} \propto \vec E\cdot\vec B$ is a pseudoscalar: boost-invariant, sign-flipped in a mirror. Treat it as an ordinary scalar anywhere handedness matters — the $\theta$ -term, the axion coupling, CP violation — and the physics comes out backwards.

The provenance. The AllQFT base has three of its ten equation cards empty or containing the literal placeholder string `tex_note_placeholder`. Its printed $G^{\mu\nu}$ matrix is nevertheless correct — it agrees entry-by-entry with the independent contraction in this article's verify script — but its "compare with $F^{\mu\nu}$ " note quotes $F^{0i} = +E_i/c$ alongside $F^{ij} = -\varepsilon^{ijk}B_k$ , mixing two incompatible sign conventions for $F$ in a single sentence. Its own $G$ matrix does not follow from the components it states; it follows from the ones used here.

Worked examples

1 · Dualise an electron, and discover why it is not a monopole

Take a single electron. At $r = 1.00\ \mathrm{\AA} = 1.00\times10^{-10}\ \mathrm{m}$ , compute its Coulomb field, apply the duality rotation, interpret the result, and compare with Dirac's quantisation condition.

Step 1 — identify. A point charge at rest has $\vec B = 0$ , so $F$ has entries only in its time-space slots and the dual only in its space-space slots. Equation (2) gives $(\vec E_G, \vec B_G) = (c\vec B, -\vec E/c) = (\vec 0,\ -\vec E/c)$ : purely magnetic, and radial.

Step 2 — the Coulomb field. With $1/4\pi\varepsilon_0 = 8.98755\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}$ , E = \frac{e}{4\pi\varepsilon_0 r^2} = \frac{(8.98755\times10^{9})(1.602177\times10^{-19})}{(1.00\times10^{-10})^2} = \mathbf{1.43996\times10^{11}\ \mathrm{V\,m^{-1}}}.

Step 3 — the dual field. |\vec B_G| = \frac{E}{c} = \frac{1.43996\times10^{11}}{2.99792\times10^{8}} = \mathbf{480.32\ T}. A radial magnetic field of 480 tesla at one ångström, so $\nabla\cdot\vec B_G \ne 0$ at the origin. The dual of a point electric charge is a magnetic monopole.

Step 4 — its magnetic charge. Integrating $\nabla\cdot\vec B_G = \rho_m$ over a sphere gives $g = \mu_0 e c$ : g = (1.256637\times10^{-6})(1.602177\times10^{-19})(2.997925\times10^{8}) = \mathbf{6.0359\times10^{-17}\ Wb}.

Step 5 — the boundary number, and the punchline. Dirac's condition says the smallest magnetic charge consistent with quantum mechanics is g_D = \frac{h}{e} = \frac{6.62607\times10^{-34}}{1.602177\times10^{-19}} = \mathbf{4.13567\times10^{-15}\ Wb}, exactly twice the superconducting flux quantum $\Phi_0 = h/2e$ — and it exceeds our dual electron's charge by \frac{g_D}{\mu_0 e c} = \frac{h}{\mu_0 e^2 c} = \frac{1}{2\alpha} = \mathbf{68.518}. A dual electron is 68.5 times too weak to be a legal monopole. Duality is a symmetry of the classical equations; quantum mechanics then refuses most of its results. That gap — the fine-structure constant, appearing in the middle of a purely classical construction — is why monopoles, if they exist, are enormously strongly coupled.

2 · The invariants of the field and of its dual

A field has $\vec E$ and $\vec B$ parallel, with $E = 1.00\times10^{6}\ \mathrm{V\,m^{-1}}$ and $B = 0.500\ \mathrm{T}$ . Compute both invariants for the field and for its dual, and say what a boost can and cannot do.

Step 1 — identify. $\vec E \parallel \vec B$ , so $\vec E\cdot\vec B = EB \ne 0$ : the pseudoscalar invariant is alive, and only the dual can see it.

Step 2 — the scalar. F_{\mu\nu}F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right) = 2\left(0.250000 - \frac{1.00\times10^{12}}{8.98755\times10^{16}}\right) = 2(0.250000 - 1.11265\times10^{-5}) = \mathbf{0.499978\ T^2}. Positive: magnetically dominated, in every frame.

Step 3 — the pseudoscalar, which needs $G$ . $\vec E\cdot\vec B = (1.00\times10^{6})(0.500) = 5.00\times10^{5}\ \mathrm{V\,T\,m^{-1}}$ , so F_{\mu\nu}G^{\mu\nu} = -\frac{4}{c}\,\vec E\cdot\vec B = -\frac{4 \times 5.00\times10^{5}}{2.99792\times10^{8}} = \mathbf{-6.6713\times10^{-3}\ T^2}. Non-zero.

Step 4 — the dual's invariants, and the boundary number. Applying $\vec E\to c\vec B$ , $\vec B\to -\vec E/c$ gives $E_G = cB = 1.49896\times10^{8}\ \mathrm{V\,m^{-1}}$ and $B_G = -E/c = -3.33564\times10^{-3}\ \mathrm{T}$ , whose scalar invariant is $2(B_G^2 - E_G^2/c^2) = \mathbf{-0.499978\ T^2}$ — exactly minus the original, as a $90°$ rotation must give. And because $F_{\mu\nu}G^{\mu\nu} \ne 0$ is frame-independent, no boost whatsoever can remove either field: a pure electric field has $\vec E\cdot\vec B = 0$ , a pure magnetic field has $\vec E\cdot\vec B = 0$ , and ours has $5.00\times10^{5}$ . The best any observer can do is make $\vec E$ and $\vec B$ parallel — which they already are. The dual told us this before a single boost was attempted.

Physics of the portrait

The dual tensor exists because someone asked what the other two Maxwell equations are for. The answer — that they are an identity, and that the thing which would break them has never been found — is the story of the monopole.

Origins · 2 minMagnetic monopole and how this was discovered

Magnetic monopole

1931–present · Dirac's quantisation · 't Hooft–Polyakov · Cabrera's Valentine's Day event · MoEDAL

The magnetic monopole is the particle that would put a source on the right of $\partial_\mu G^{\mu\nu} = 0$ , and its theoretical pedigree is far better than its experimental one. Paul Dirac's 1931 paper "Quantised Singularities in the Electromagnetic Field" (Proc. Roy. Soc. A 133, 60) showed something remarkable: the existence of even one monopole anywhere in the universe would force all electric charge to be quantised, in multiples of a unit fixed by $eg = n\hbar c/2$ . Charge quantisation is among the most precisely tested facts in physics and has no other explanation in ordinary electrodynamics. Dirac's argument turns on the fact that a monopole's vector potential cannot be defined globally: it must carry a singular "string", and demanding the string be unobservable to any quantum wavefunction is what quantises $eg$ . The minimum magnetic charge is $g_D = h/e = 4.14\times10^{-15}\ \mathrm{Wb}$ — exactly two superconducting flux quanta — with a coupling $68.5$ times the electron's, which makes a monopole spectacularly ionising. In 1974 Gerard 't Hooft and Alexander Polyakov independently showed that monopoles are not optional in grand unified theories: any GUT breaking a simple group to a $U(1)$ predicts them near $10^{16}\ \mathrm{GeV}$ — a large part of why inflation was invented, to dilute them away. The experimental record is one long null result with a single famous asterisk: on the night of 14 February 1982, Blas Cabrera's superconducting loop at Stanford recorded exactly one event of the right size — the "Valentine's Day monopole". It never repeated. So far, the right-hand side of $\partial_\mu G^{\mu\nu} = 0$ remains, stubbornly and informatively, zero.

History

1865

— Maxwell's "A Dynamical Theory of the Electromagnetic Field" presents the field equations as a system, two of them carrying sources and two not.

1873

— Maxwell's Treatise on Electricity and Magnetism codifies the asymmetry without explaining it.

1908

— Minkowski, in "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern", casts Maxwell's equations in four-dimensional tensor form; his antisymmetric objects are called "six-vectors" for the next decade, and the dual is implicit in his formalism.

1931

— Dirac's "Quantised Singularities in the Electromagnetic Field" (Proc. Roy. Soc. A 133, 60) shows that a single magnetic monopole would force the quantisation of all electric charge, with eg = n ℏc/2.

1974

— 't Hooft and Polyakov independently show that grand unified theories necessarily contain magnetic monopoles, with masses near the unification scale.

1982

— On 14 February, Blas Cabrera's SQUID loop at Stanford records a single candidate monopole event, the "Valentine's Day monopole"; it is never repeated and remains unconfirmed.

2016

— MoEDAL at the LHC publishes its first monopole search, setting mass limits for magnetic charges up to several Dirac units; the result, like every result before it, is null.

Going further

Texts. Jackson, Classical Electrodynamics, 3rd ed., §§11.9 and 6.11 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.3.4 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Dirac, Proc. Roy. Soc. A 133, 60 (1931) · Cabrera, Phys. Rev. Lett. 48, 1378 (1982).

Provenance. Signature $(+,-,-,-)$ , $\varepsilon^{0123} = +1$ ; $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass, and no velocity-dependent mass appears anywhere. Base-defect repairs (logged): (i) the base prints $\partial_\mu G^{\mu\nu} = \mu_0 c J_m^\nu$ ; that coefficient is dimensionally impossible and is refused — the correct $\kappa = 1/c$ is derived here and independently confirmed against the spatial components. (ii) The base's "compare with $F^{\mu\nu}$ " note mixes two incompatible sign conventions in one sentence, quoting $F^{0i} = +E_i/c$ beside $F^{ij} = -\varepsilon^{ijk}B_k$ ; its own printed $G$ matrix does not follow from the components it states. (iii) Three of its ten equation cards are empty or contain the literal string `tex_note_placeholder`. Everything here is recomputed in verify_aq-sr-dual-field-tensor.py, which parses this article's own printed digits.

Scope. The construction of $F^{\mu\nu}$ belongs to The Electromagnetic Field Tensor; the covariant Maxwell system to Covariant Maxwell Equations; and the classification of fields by the two invariants belongs to Electromagnetic Field Invariants. This page owns the dual, its divergence, and the monopole source term.

Builds on. The Electromagnetic Field Tensor · Covariant Maxwell Equations · The Four-Current Density · The Four-Potential. Leads to. Electromagnetic Field Invariants · Transformation of Electric and Magnetic Fields · The Electromagnetic Stress-Energy Tensor.

Is G a different physical field from F?

No. $G$ contains exactly the same six numbers as $F$ , re-slotted; knowing $F$ everywhere means knowing $G$ everywhere and vice versa (up to the sign that ${}^{**}F = -F$ records). What $G$ buys is not new information but new contractions: $F_{\mu\nu}G^{\mu\nu}$ is an invariant that no amount of contracting $F$ with itself and with $\eta$ can produce, because it needs the orientation $\varepsilon$ carries.

If the homogeneous equations are just an identity, why bother writing them down?

Because "there exists a four-potential" is a physical assumption, not a tautology, and it is one a monopole would violate. The identity says: given $A^\mu$ , you get $\nabla\cdot\vec B = 0$ and Faraday for free. The empirical content is the antecedent. In a spacetime containing a monopole, $A^\mu$ cannot be defined globally — Dirac's string is the scar — and the "identity" fails, not because derivatives stopped commuting but because the object being differentiated does not exist.

Why does the pseudoscalar invariant matter if it is a total derivative?

Classically it does not — adding $\theta\,F_{\mu\nu}G^{\mu\nu}$ to the Lagrangian changes no equation of motion. But "total derivative" means "sensitive only to boundary conditions and topology", and both matter quantum-mechanically. In QCD the analogous term produces observable CP violation (mysteriously absent — the strong-CP problem), and promoting $\theta$ to a dynamical field is the axion.

What actually stops monopoles from existing?

Nothing in the classical equations. Duality-symmetric electrodynamics is perfectly consistent — the equations are more symmetric with monopoles than without. The obstructions are (i) experimental, nobody has found one, and (ii) structural: with a monopole present, $F$ is no longer $\mathrm{d}A$ globally, so the potential formulation — and with it gauge theory, the path integral, and essentially all of QED as practised — needs surgery. Dirac's string, and Wu and Yang's later patchwork of overlapping gauge charts, are that surgery. The price is paid in topology, not in an extra term.

Does the sign of ε⁰¹²³ matter physically?

Not for any observable, but enormously for cross-checking formulas. Flipping $\varepsilon^{0123}$ from $+1$ to $-1$ flips $G$ , hence $F_{\mu\nu}G^{\mu\nu}$ , hence the sign in front of $\vec E\cdot\vec B$ . Two textbooks can print apparently contradictory results while agreeing completely on the physics. Always state both conventions — signature and $\varepsilon^{0123}$ — before comparing. Here: $(+,-,-,-)$ and $\varepsilon^{0123} = +1$ .

How strongly would a monopole interact if it existed?

Ferociously. The Dirac charge corresponds to a coupling $g_D/(\mu_0 e c) = 1/2\alpha = 68.5$ times the electron's, so the "magnetic fine-structure constant" is $\alpha_m = 1/(4\alpha) \approx 34$ — strongly coupled, and perturbation theory in it is worthless. A monopole passing through matter would ionise thousands of times more heavily than any known particle, which is precisely the signature MoEDAL's plastic track detectors are built to catch. If one goes through, you will not miss it.