The Dual Field Tensor

The Electromagnetic Field Tensor ended on a deliberate asymmetry. The sourced half of Maxwell's equations — Gauss and Ampère–Maxwell — collapsed into the single divergence $\partial_\mu F^{\mu\nu}=\mu_0J^\nu$ . The sourceless half — no magnetic monopoles, plus Faraday's law — got no such dignity: in terms of $F$ alone it is a cyclic identity with three free indices, correct but ungainly, like a law drafted by a committee. This article builds the object that fixes the aesthetics and, in doing so, exposes real physics: the dual field tensor $G^{\mu\nu}$ , manufactured by contracting $F$ with the four-dimensional Levi-Civita symbol. The construction sounds like bookkeeping and delivers three theorems. First, dualizing rotates the field: every entry of $G$ is an entry of $F$ with $\mathbf{E}/c$ and $\mathbf{B}$ exchanged (one sign flip included) — a $90°$ turn in an abstract field plane, so exact that applying the dual twice returns $-F$ . Second, the homogeneous Maxwell equations become the one-line divergence $\partial_\mu G^{\mu\nu}=0$ , structurally identical to the sourced equation but with an empty source slot — emptiness that is an experimental statement, not a mathematical one: it is where a magnetic four-current would sit if a monopole ever showed up in a detector. Third, because the Levi-Civita symbol flips sign under spatial inversion, $G$ is a pseudotensor — the covariant fingerprint of $\mathbf{B}$ being an axial vector while $\mathbf{E}$ is polar. One object, then: the compact form of half of electrodynamics, the precise statement of electric–magnetic symmetry, and the parity bookkeeping of the fields — with the pseudoscalar invariant it generates handed to Lorentz Invariants of the Field.

Key takeaways

The dual field tensor $G^{\mu\nu}=\tfrac{1}{2}\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$ is built by contracting the field tensor with the four-dimensional Levi-Civita symbol; the $\tfrac12$ compensates the antisymmetric double-count.
Componentwise, dualizing is the substitution $\mathbf{E}/c\to\mathbf{B}$ , $\mathbf{B}\to-\mathbf{E}/c$ — a $90°$ rotation in the $(E,cB)$ plane. Applying it twice gives $\tilde{\tilde F}=-F$ , the signature of a rotation squared.
The homogeneous Maxwell pair — $\nabla\cdot\mathbf{B}=0$ and Faraday's law — collapses to the single covariant statement $\partial_\mu G^{\mu\nu}=0$ , mirroring the sourced $\partial_\mu F^{\mu\nu}=\mu_0J^\nu$ .
That divergence law is equivalent to the cyclic Bianchi identity, and both hold automatically once $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ : half of Maxwell is an identity of the potential formulation, not an independent law.
Index care on $\varepsilon$ is mandatory: with $(+,-,-,-)$ , $\varepsilon^{0123}=+1$ forces $\varepsilon_{0123}=-1$ , and $\varepsilon$ flips sign under parity — making $G$ a pseudotensor, the covariant version of " $\mathbf{B}$ is an axial vector."
If magnetic charge existed, the dual equation would read $\partial_\mu G^{\mu\nu}=\mu_0c\,J_m^\nu$ — the empty right-hand side is a null experimental result, still tested at the LHC today.
Key idea: contracting $F$ with $\varepsilon^{\mu\nu\rho\sigma}$ rotates the field by 90° in the $(E,cB)$ plane and converts the three-index Bianchi identity into the one-line law $\partial_\mu G^{\mu\nu}=0$ — the sourceless half of Maxwell, with a slot reserved for monopoles that nature has so far left empty.

The core insight

Hodge-dualizing $F^{\mu\nu}$ with $\varepsilon^{\mu\nu\alpha\beta}$ rotates the field by $90^\circ$ in the $(E,cB)$ plane, turning the cyclic Bianchi identity into the one-line divergence law $\partial_\mu\tilde{F}^{\mu\nu}=0$ — written $G^{\mu\nu}$ throughout this article.

Where this lives today

The blank right-hand side of $\partial_\mu G^{\mu\nu}=0$ is under active interrogation. MoEDAL — the LHC's dedicated monopole experiment, plastic track detectors and aluminum trapping volumes wrapped around the LHCb interaction point — publishes limits on magnetic-charge production in units of the Dirac charge $q_m=h/e$ , excluding monopoles up to several TeV for charges $1$ – $5\,q_m$ ; a 2022 search in the $\sim10^{16}$ T fields of near-miss heavy-ion collisions set the first bounds free of perturbative-production assumptions. Every null result is a measurement of the zero in equation (6). The pseudoscalar contraction $G_{\mu\nu}F^{\mu\nu}\propto\mathbf{E}\cdot\mathbf{B}$ has meanwhile become one of the busiest objects in physics. Added to the Lagrangian with a coefficient $\theta$ , it is a total derivative classically, yet in quantum chromodynamics it controls CP violation, and the unnaturally small strong-CP angle motivates the axion — a hypothetical field whose coupling is literally $a\,\mathbf{E}\cdot\mathbf{B}$ . Haloscopes like ADMX hunt dark-matter axions by supplying the $\mathbf{B}$ (an 8 T solenoid) and listening for the induced $\mathbf{E}$ ; topological insulators realize the same term as "axion electrodynamics" with quantized $\theta=\pi$ . In the formal direction, the duality $G$ makes manifest seeded the Montonen–Olive conjecture and the S-dualities of gauge and string theory. Downstream in this chapter, $G$ reports for duty immediately: Covariant Form of Maxwell's Equations assembles the full system from the $F$ and $G$ divergences, and Lorentz Invariants of the Field turns the two contractions of equation (4) into the complete classification of electromagnetic fields.

Origins · 1 minTullio Levi-Civita and how this was discovered

Tullio Levi-Civita

1873–1941 · Padua and Rome — whose epsilon symbol is the four lines of machinery this article runs on

Levi-Civita built the calculus this article computes with. With his teacher Gregorio Ricci-Curbastro he created the "absolute differential calculus" — tensor calculus — in a landmark 1900 memoir, systematizing covariant and contravariant indices, contraction, and the totally antisymmetric symbol that now carries his name; the $\varepsilon^{\mu\nu\rho\sigma}$ in the dual's definition is his notation doing his job, a century on. The tools were waiting when physics needed them: Einstein spent years inside Ricci and Levi-Civita's calculus building general relativity, and conducted a famously courteous 1915 correspondence with Levi-Civita, who kept finding — and Einstein kept conceding — errors in the field-equation drafts ("it must be nice," Einstein wrote, "to ride through these fields upon the horse of true mathematics"). His 1917 invention of parallel transport gave curvature its modern geometric meaning. Dismissed from his Rome chair in 1938 by the fascist racial laws, he died in 1941; the calculus he co-invented remains the native language of relativity, and every duality identity checked in this article is, notationally, a sentence in it.

History

1861

— Maxwell's equations appear with their electric–magnetic asymmetry: electric sources exist, magnetic ones do not

1873

— Maxwell's Treatise organizes the theory; the missing-monopole asymmetry is visible but unremarked

1893

— Heaviside, having compressed Maxwell's equations to four, notes their near-symmetry under exchanging electric and magnetic quantities

1900

— Ricci and Levi-Civita publish the absolute differential calculus: tensors, contractions, and the epsilon machinery

1908

— Minkowski writes electrodynamics in four-dimensional form; the field tensor and its dual pair first appear together

1931

— Dirac shows one magnetic monopole anywhere would quantize electric charge, giving the dual equation's empty slot its stakes

1974

— 't Hooft and Polyakov find monopoles as solitons in unified gauge theories: the slot could be filled by theory, not just by hand

1977

— Montonen and Olive conjecture electric–magnetic duality as an exact symmetry of certain gauge theories, seeding modern S-duality

2012

— MoEDAL begins its dedicated monopole hunt at the LHC, converting the blank right-hand side into published exclusion limits

The physical picture

Hold the six numbers $(\mathbf{E}/c,\mathbf{B})$ at one spacetime event and picture them as a single arrow in an abstract plane whose horizontal axis is "electric-ness" and vertical axis is "magnetic-ness." Transformation of Electric and Magnetic Fields showed that boosts slide field content between the axes in a velocity-dependent way. The dual operation is cleaner: a rigid $90°$ rotation of that arrow. Electric content becomes magnetic; magnetic becomes electric with one minus sign — the sign any true rotation needs, since rotating $(x,y)$ by $90°$ gives $(-y,x)$ , not $(y,x)$ . Rotate twice and you must face backwards: the double dual is $-F$ , and the algebra will confirm it to the sign. Why should such a rotation exist at all? Count blocks. The antisymmetric tensor $F^{\mu\nu}$ stores $\mathbf{E}$ in its three time–space slots and $\mathbf{B}$ in its three space–space slots. In four dimensions, every index pair has a complementary pair — $(0,1)$ partners with $(2,3)$ , $(0,2)$ with $(3,1)$ , $(0,3)$ with $(1,2)$ — and the totally antisymmetric $\varepsilon^{\mu\nu\rho\sigma}$ is precisely the machine mapping each pair to its complement. Contracting $F$ with $\varepsilon$ therefore swaps the two blocks: $\mathbf{E}$ and $\mathbf{B}$ trade houses. It is the four-dimensional cousin of a vector-calculus fact — in three dimensions, $\varepsilon_{ijk}$ converts the antisymmetric pair $(i,j)$ into the complementary direction $k$ , which is exactly how $\mathbf{B}$ , secretly an antisymmetric two-index object, gets to masquerade as a vector at all. That masquerade has a price, and the dual makes it visible. Under a spatial inversion, a genuine arrow like $\mathbf{E}$ reverses; $\mathbf{B}$ , really a circulation — an oriented plane, not an arrow — does not. The Levi-Civita symbol carries the orientation of spacetime, flipping sign under improper transformations, and everything built with one factor of $\varepsilon$ inherits the flip: $G$ transforms as a tensor under boosts and rotations but picks up an extra minus under reflection. "Pseudotensor" is not a defect; it is parity bookkeeping, done covariantly. The payoff for all this structure is the cleanest statement in electrodynamics: the two sourceless Maxwell equations become one divergence, $\partial_\mu G^{\mu\nu}=0$ — with a right-hand side one monopole discovery away from filling.

Figure

Dualizing rotates the field 90 degrees in the abstract E-cB plane, and swaps the storage blocks of the tensor

The math: the epsilon machine, the dual, and half of Maxwell for free

Fix conventions first, because every sign in this article flows from them. The field tensor of The Electromagnetic Field Tensor is built from the four-potential as $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ ; with $A^\mu=(V/c,\mathbf{A})$ , metric $(+,-,-,-)$ , and the definitions $\mathbf{E}=-\nabla V-\partial_t\mathbf{A}$ , $\mathbf{B}=\nabla\times\mathbf{A}$ , its components are pinned to

F_{0i}=+\frac{E_i}{c},\qquad F_{ij}=-\varepsilon_{ijk}B_k \quad\Longrightarrow\quad F^{i0}=+\frac{E^i}{c},\qquad F^{ij}=-\varepsilon^{ijk}B_k. \qquad (1)

where: raising the mixed time–space pair with $\eta$ costs one sign ( $F^{0i}=-E^i/c$ : the i-th row, time column carries $+E^i/c$ ), while the purely spatial block is unchanged. Latin $\varepsilon_{ijk}$ is the ordinary three-index symbol with $\varepsilon_{123}=+1$ . Written as a matrix with rows $\mu$ and columns $\nu$ , $F^{\mu\nu}$ has first row $(0,-E_x/c,-E_y/c,-E_z/c)$ and, for example, $F^{12}=-B_z$ . Any sign dispute downstream should be settled by recomputing from (1) — it is the potential definition, not a matrix memorized from a book with a different layout. The new tool is the four-dimensional Levi-Civita symbol,

\varepsilon^{\mu\nu\rho\sigma}=\begin{cases}+1 & \mu\nu\rho\sigma\ \text{an even permutation of}\ 0123\\ -1 & \text{odd permutation}\\ 0 & \text{any repeated index,}\end{cases}\qquad \varepsilon^{0123}=+1\ \Rightarrow\ \varepsilon_{0123}=-1. \qquad (2)

where: the last implication is the trap of the subject. Lowering all four indices multiplies by $\eta_{00}\eta_{11}\eta_{22}\eta_{33}=\det\eta=-1$ , so the all-lower symbol has the opposite sign of the all-upper one — a sign that silently corrupts every duality identity if forgotten. Under a Lorentz transformation the symbol picks up a factor $\det\Lambda$ : invariant under proper transformations, sign-flipped under parity and time reversal — so everything built with one $\varepsilon$ is a pseudotensor. (In curved spacetime the honest tensor carries a $\sqrt{-g}$ ; in special relativity symbol and tensor coincide numerically, which is why general-relativity references decorate their epsilons and we need not.) With the machine in hand, define the dual field tensor:

G^{\mu\nu}\;\equiv\;\tilde F^{\mu\nu}\;=\;\tfrac{1}{2}\,\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}. \qquad (3)

where: the $\tfrac12$ compensates the double-counting of each antisymmetric pair $(\rho,\sigma)$ in the sum, and the contraction is against the lowered $F_{\rho\sigma}$ — feed it $F^{\rho\sigma}$ by mistake and stray signs bloom. Since $\varepsilon$ is antisymmetric in $\mu\leftrightarrow\nu$ , the dual is automatically antisymmetric: six independent components, exactly the budget of another $(\mathbf{E},\mathbf{B})$ -sized object. Which six numbers it holds is the first derivation's business.

Derivation AComponents of the dual and the ninety-degree rotation of the field

The goal is the complete component content of $G^{\mu\nu}$ from definition (3), organized into the duality substitution rule, plus the double-dual theorem $\tilde{\tilde F}=-F$ that certifies "dual = quarter turn." The strategy is block-by-block evaluation: for each index pair of $G$ , the epsilon selects the complementary pair of $F$ , so $G$ 's time–space block reads $F$ 's space–space block and vice versa; the only labor is sign bookkeeping through (1) and (2) — exactly where published treatments diverge, so every sign is shown. Watch for two payoffs: the substitution rule is a rotation (one sign flip, not two, not zero), and the double dual needs the epsilon contraction identity whose leading minus comes from $\det\eta=-1$ .

Step 1. Time–space block. Set $\mu=0$ , $\nu=i$ :

G^{0i}=\tfrac{1}{2}\varepsilon^{0ijk}F_{jk}=\tfrac{1}{2}\varepsilon^{ijk}\left(-\varepsilon_{jkl}B_l\right)=-\tfrac{1}{2}\cdot 2\delta_{il}B_l=-B_i,

using $\varepsilon^{0ijk}=\varepsilon^{ijk}$ (deleting the leading 0 from an even permutation of $0123$ leaves an even permutation of $123$ ), then $F_{jk}$ from (1), then the three-dimensional identity $\varepsilon^{ijk}\varepsilon_{jkl}=2\delta_{il}$ . So the time row of $G$ carries $-\mathbf{B}$ where the time row of $F$ carries $-\mathbf{E}/c$ : the magnetic field has moved into the electric field's slots.

Step 2. Space–space block. Set $\mu=i$ , $\nu=j$ with $i\neq j$ . The complementary pair is $(0,k)$ with $k$ the remaining spatial index, and it enters twice ( $F_{0k}$ and $F_{k0}$ ), killing the $\tfrac12$ :

G^{ij}=\tfrac{1}{2}\left(\varepsilon^{ij0k}F_{0k}+\varepsilon^{ijk0}F_{k0}\right)=\varepsilon^{ij0k}F_{0k}=+\varepsilon^{ijk}\frac{E_k}{c},

where $\varepsilon^{ij0k}=\varepsilon^{0kij}=\varepsilon^{kij}=\varepsilon^{ijk}$ (two transpositions to bring 0 to the front: even) and $F_{0k}=E_k/c$ from (1). The electric field now occupies the magnetic block — with the opposite sign convention to $F$ 's own space block ( $F^{ij}=-\varepsilon^{ijk}B_k$ ): that relative sign is the rotation's minus.

Step 3. Assemble and read the substitution rule. As a matrix,

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},

which is precisely $F^{\mu\nu}$ after the substitution

\frac{\mathbf{E}}{c}\;\longrightarrow\;\mathbf{B},\qquad \mathbf{B}\;\longrightarrow\;-\frac{\mathbf{E}}{c}.

Why one sign flip and not two: a genuine $90°$ rotation of coordinates $(x,y)\to(-y,x)$ negates exactly one slot. Two flips would be a $180°$ rotation; zero flips would be a reflection — and a reflection squared is the identity, which the next step rules out.

Step 4. Double dual. Apply (3) to $G$ itself and reduce the product of epsilons with the contraction identity

\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{\rho\sigma\alpha\beta}=-2\left(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha\right),

whose leading minus sign is $\det\eta=-1$ at work ( $\varepsilon_{0123}=-\varepsilon^{0123}$ ; in Euclidean four-space the identity carries $+2$ ). Then

\tilde{\tilde F}^{\mu\nu}=\tfrac{1}{2}\varepsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}=\tfrac{1}{4}\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{\rho\sigma\alpha\beta}F^{\alpha\beta}=-\tfrac{1}{2}\left(F^{\mu\nu}-F^{\nu\mu}\right)=-F^{\mu\nu},

by antisymmetry of $F$ . Dualizing twice is the $180°$ rotation $(\mathbf{E}/c,\mathbf{B})\to(-\mathbf{E}/c,-\mathbf{B})$ — consistent with running the Step-3 substitution twice: $\mathbf{E}/c\to\mathbf{B}\to-\mathbf{E}/c$ . The dual operation is a genuine quarter turn in field space, and $G$ contains no new information: the same six numbers, filed in the complementary slots. Verified: full contraction (3) executed symbolically (sympy, all 16 components) reproduces the Step-3 matrix exactly; the substitution map applied to F yields G identically; the double dual returns −F for arbitrary fields; the epsilon contraction identity checked over all 256 index combinations. Matrix agrees with Griffiths 4th ed. §12.3.4 (his Eq. 12.120 layout transposes rows and columns; the physics — which field sits in which block, and the single sign flip — is identical) and with Jackson 3rd ed. §11.9.

Figure

The epsilon symbol maps every index pair to its complement — how the contraction swaps the storage blocks

Before the second theorem, note what dualizing does to the contractions of the field. Pairing $F$ with itself gives the scalar $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)$ ; pairing $F$ with its dual gives

G_{\mu\nu}F^{\mu\nu}=-\frac{4}{c}\,\mathbf{E}\cdot\mathbf{B},\qquad G_{\mu\nu}G^{\mu\nu}=-F_{\mu\nu}F^{\mu\nu}=2\left(\frac{E^2}{c^2}-B^2\right). \qquad (4)

where: the first is a new object — a pseudoscalar (one factor of $\varepsilon$ ), proportional to $\mathbf{E}\cdot\mathbf{B}$ , invisible to any contraction of $F$ with itself; the second says $G$ brings no third invariant, only the first one with its sign rotated ( $90°$ in field space turns $B^2-E^2/c^2$ into its negative, exactly as rotating a vector by $90°$ swaps and flips its quadratic forms). The systematic story of these two numbers — the only algebraic invariants the field has, and the classification theorem they generate — is the entire business of Lorentz Invariants of the Field; this article only manufactures the second contraction and certifies its credentials. Now the theorem the dual was invented for. The Electromagnetic Field Tensor delivered the sourced pair of Maxwell's equations as $\partial_\mu F^{\mu\nu}=\mu_0J^\nu$ and left the sourceless pair in cyclic form. Here is the upgrade:

\partial_\mu G^{\mu\nu}=0 \quad\Longleftrightarrow\quad \partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0 \quad\Longleftarrow\quad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu. \qquad (5)

where: the left equation is four component statements ( $\nu=0,x,y,z$ ); the middle (the Bianchi identity) looks like $4^3=64$ statements but, being totally antisymmetric in $\lambda\mu\nu$ , contains only $\binom{4}{3}=4$ independent ones — the same four, as the derivation shows. The right implication is the punchline of the potential formulation: once fields come from a four-potential, the sourceless half of Maxwell is an identity, not a law you could break.

Go deeper · Derivation B · 2 minThe dual divergence is the sourceless half of Maxwell and is automatic from the potential

Derivation BThe dual divergence is the sourceless half of Maxwell and is automatic from the potential

The goal is all three links of (5): unpack $\partial_\mu G^{\mu\nu}=0$ into $\nabla\cdot\mathbf{B}=0$ and Faraday's law; prove the equivalence with the cyclic Bianchi identity in both directions; and show the whole package holds identically once $F=\partial A$ . The strategy is honest componentry for the unpacking (where the physics lives), then a two-line epsilon argument for the equivalence, then a symmetry kill for the potential statement. Watch how four scalar equations, one vector identity, and one theorem about potentials turn out to be the same fact wearing three outfits.

Step 1. Unpack $\nu=0$ . Only spatial $\mu$ contributes ( $G^{00}=0$ ), and Derivation A gives $G^{i0}=-G^{0i}=+B_i$ :

\partial_\mu G^{\mu 0}=\partial_i G^{i0}=\partial_iB_i=\nabla\cdot\mathbf{B}.

Setting it to zero is the no-monopole law — the covariant home of Divergence of B and Absence of Magnetic Monopoles. Why this slot: the time component of the sourced equation $\partial_\mu F^{\mu 0}=\mu_0 J^0$ is Gauss's law with charge density on the right; the dual's time component is Gauss's law for magnetism with nothing on the right. The parallel is exact, and the emptiness is the physics.

Step 2. Unpack $\nu=x$ . Two kinds of terms survive: $\mu=0$ with $G^{0x}=-B_x$ and $\partial_0=\tfrac1c\partial_t$ ; and spatial $\mu$ , where Derivation A gives $G^{yx}=-E_z/c$ and $G^{zx}=+E_y/c$ . Then

\partial_\mu G^{\mu x}=-\frac{1}{c}\frac{\partial B_x}{\partial t}-\frac{1}{c}\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)=-\frac{1}{c}\left[\frac{\partial B_x}{\partial t}+\left(\nabla\times\mathbf{E}\right)_x\right].

Setting all three spatial components to zero gives $\nabla\times\mathbf{E}=-\partial\mathbf{B}/\partial t$ — Faraday's Law of Induction, no flux rule or moving loop in sight, resurfacing as the spatial shadow of one tensor divergence.

Step 3. Equivalence with the Bianchi identity, forward. Insert the definition of $G$ and note that $\partial_\mu\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$ sees only the totally antisymmetric part of $\partial_\mu F_{\rho\sigma}$ in the three indices $\mu\rho\sigma$ (the epsilon projects it out). Therefore

\partial_\mu G^{\mu\nu}=\tfrac{1}{2}\varepsilon^{\nu\sigma\lambda\mu}\,\partial_\lambda F_{\mu\sigma}\big|_{\text{antisym}}=\tfrac{1}{3!}\,\varepsilon^{\nu\sigma\lambda\mu}\left(\partial_\lambda F_{\mu\sigma}+\partial_\mu F_{\sigma\lambda}+\partial_\sigma F_{\lambda\mu}\right),

so if the cyclic sum vanishes, so does $\partial_\mu G^{\mu\nu}$ . Backward: contract the assumed $\partial_\mu G^{\mu\nu}=0$ with $\varepsilon_{\nu\alpha\beta\gamma}$ and apply the contraction identity of Derivation A Step 4; the deltas reassemble exactly the cyclic sum $\partial_\alpha F_{\beta\gamma}+\partial_\beta F_{\gamma\alpha}+\partial_\gamma F_{\alpha\beta}=0$ . Four statements each way — no information lost in either direction: the divergence law and the Bianchi identity are the same four equations.

Step 4. Automatic from the potential. Substitute $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ into the cyclic sum:

\partial_\lambda\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)+\partial_\mu\left(\partial_\nu A_\lambda-\partial_\lambda A_\nu\right)+\partial_\nu\left(\partial_\lambda A_\mu-\partial_\mu A_\lambda\right)=0,

because the six terms cancel in pairs by the symmetry of mixed partials ( $\partial_\lambda\partial_\mu=\partial_\mu\partial_\lambda$ ). Equivalently, in one line: $\partial_\mu G^{\mu\nu}=\tfrac12\varepsilon^{\mu\nu\rho\sigma}\partial_\mu(\partial_\rho A_\sigma-\partial_\sigma A_\rho)=\varepsilon^{\mu\nu\rho\sigma}\partial_\mu\partial_\rho A_\sigma=0$ — a symmetric object ( $\partial_\mu\partial_\rho$ ) contracted against an antisymmetric one ( $\varepsilon$ ) dies identically. The converse is the Poincaré lemma: on any contractible region, $\partial_\mu G^{\mu\nu}=0$ is exactly the integrability condition guaranteeing a four-potential exists — the local existence statement that The Four-Potential trades on. Faraday's law and the absence of monopoles are not extra dynamical input about the world; they are the statement that the world's fields come from potentials. Verified: the ν=0 and all three spatial unpackings recomputed symbolically for arbitrary field functions (sympy): the four components of the divergence equal exactly the divergence of B and minus one-over-c times each Cartesian component of the Faraday combination; the cyclic identity checked to vanish identically for arbitrary A-mu over all 64 index triples. Matches Griffiths §12.3.4 and Jackson §11.9.

The empty right-hand side deserves its own equation. If magnetic charge existed — density $\rho_m$ , current $\mathbf{J}_m$ , packaged as $J_m^\nu=(c\rho_m,\mathbf{J}_m)$ — the symmetric extension of Maxwell's equations would read

\partial_\mu F^{\mu\nu}=\mu_0 J^\nu,\qquad \partial_\mu G^{\mu\nu}=\mu_0 c\,J_m^\nu \qquad (6)

where: the factor $c$ keeps SI units aligned with the Weber convention for magnetic charge. In this extended theory, $\nabla\cdot\mathbf{B}=\mu_0c\rho_m$ , Faraday's law gains a magnetic-current term, and the free theory ( $J=J_m=0$ ) is exactly symmetric under the duality rotation $F\to F\cos\theta+G\sin\theta$ , of which our dual is the quarter turn. Dirac's 1931 observation — one monopole anywhere forces charge quantization via $e\,q_m=nh$ , minimum magnetic charge $q_m=h/e\approx4.14\times10^{-15}$ Wb — is stated here and derived in the quantum-mechanics track, where the required phase machinery lives. Classically, the blank right side is a running experimental scoreboard: every monopole search to date has kept it blank.

Limiting cases

case	result	meaning
pure electric field ( $\mathbf{B}=0$ )	$G^{0i}=0$ , $G^{ij}=\varepsilon^{ijk}E_k/c$	$G$ looks "purely magnetic": all content in the space–space block — the swap in its cleanest form
pure magnetic field ( $\mathbf{E}=0$ )	$G^{0i}=-B_i$ , $G^{ij}=0$	$G$ looks "purely electric": $\mathbf{B}$ occupies the time–space slots
plane wave ( $E=cB$ , $\mathbf{E}\perp\mathbf{B}$ )	$G_{\mu\nu}F^{\mu\nu}=0$ and $F_{\mu\nu}F^{\mu\nu}=0$	null field: dualizing a wave gives the same wave, polarization turned $90°$ (Worked example 2)
double dual	$\tilde{\tilde F}=-F$	the quarter-turn squared; in Euclidean signature the sign would be $+$ — the minus is $\det\eta$ speaking
$J_m^\nu\to0$ in (6)	$\partial_\mu G^{\mu\nu}=0$	standard electrodynamics: the observed world, so far
parity $\mathbf{x}\to-\mathbf{x}$	$F\to$ tensor rule, $G\to-(\text{tensor rule})$	$G$ 's extra sign is exactly $\mathbf{E}$ polar vs $\mathbf{B}$ axial, covariantized

Common trapThe epsilon symbol is not index-gymnastics-safe — three sign traps that corrupt every duality identity

First trap: $\varepsilon_{0123}=-\varepsilon^{0123}$ in Lorentzian signature — lowering all four indices multiplies by $\det\eta=-1$ , and copying an identity between raised and lowered epsilons without the flip silently negates duals, double duals, and the pseudoscalar invariant. Second trap: symbol versus tensor. In special relativity they coincide numerically, but formulas imported from general-relativity texts carry $\sqrt{-g}$ factors that must be stripped consistently. Third trap: convention spread — some texts define the dual with $-\tfrac12$ , some place the $c$ differently, some use $(-,+,+,+)$ ; each choice flips component signs while leaving the physics (which block holds which field, one flip per quarter turn, $\tilde{\tilde F}=-F$ ) intact. The safe procedure is the one used here: fix $\varepsilon^{0123}=+1$ , fix $F$ from the potential as in equation (1), and recompute rather than transcribe. A wrong sign in $G$ is not cosmetic — it reverses Faraday's law.

Worked examples

1 · Dualizing a lab bench field, entry by entry (confidence builder)

Setup. A region contains $\mathbf{E}=(2\times10^5,\,0,\,1\times10^5)$ V/m and $\mathbf{B}=(0,\,0,\,5\times10^{-4})$ T — a tilted electric field over a vertical magnetic one, as in a crossed-field chamber with a misaligned plate. Build $F^{\mu\nu}$ and $G^{\mu\nu}$ numerically, then extract both contractions. Use $c=3.00\times10^8$ m/s.

Step 1. Convert the electric entries to tensor units: $E_x/c=6.67\times10^{-4}$ T, $E_z/c=3.33\times10^{-4}$ T (every slot of $F^{\mu\nu}$ then carries teslas). By (1), the nonzero independent components: $F^{10}=6.67\times10^{-4}$ , $F^{30}=3.33\times10^{-4}$ , $F^{12}=-B_z=-5\times10^{-4}$ .

Step 2. Dualize by substitution: the time–space slots now hold $-\mathbf{B}$ , so $G^{03}=-5\times10^{-4}$ ; the space–space slots hold $+\varepsilon^{ijk}E_k/c$ , so $G^{12}=E_z/c=\mathbf{3.33\times10^{-4}}$ T and $G^{23}=E_x/c=\mathbf{6.67\times10^{-4}}$ T. Every entry of $G$ is an entry of $F$ , relocated — no new numbers, new filing.

Step 3. Contract. The scalar: $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)=2(2.5\times10^{-7}-5.56\times10^{-7})=\mathbf{-6.11\times10^{-7}}$ T² — negative, an electric-dominated field ( $E^2-c^2B^2=+2.75\times10^{10}$ V²/m²). The pseudoscalar: $\mathbf{E}\cdot\mathbf{B}=(1\times10^5)(5\times10^{-4})=50$ V·T/m, so $G_{\mu\nu}F^{\mu\nu}=-4(50)/c=\mathbf{-6.67\times10^{-7}}$ T². Nonzero — by Lorentz Invariants of the Field, no boost anywhere can make these fields perpendicular, let alone eliminate one.

Step 4. Audit with two theorem checks: $G_{\mu\nu}G^{\mu\nu}$ recomputed from $G$ 's entries gives $+6.11\times10^{-7}$ T² $=-F_{\mu\nu}F^{\mu\nu}$ ✓ (equation (4)); and $\det F^{\mu\nu}=(\mathbf{E}\cdot\mathbf{B}/c)^2=(1.67\times10^{-7})^2=2.78\times10^{-14}$ T⁴ ✓ — the determinant is the pseudoscalar in disguise, squared.

The lesson: dualizing is mechanical — relocate six numbers per the substitution rule — and the two contractions fall out in two lines each, with built-in cross-checks.

2 · The dual of a laser beam — Faraday's law as one tensor component (exam level)

Setup. A helium–neon laser beam ( $\lambda=633$ nm, $E_0=300$ V/m, intensity $\approx119$ W/m²) propagates along $+z$ , polarized along $x$ : $\mathbf{E}=E_0\cos(kz-\omega t)\,\hat{\mathbf{x}}$ , $\mathbf{B}=(E_0/c)\cos(kz-\omega t)\,\hat{\mathbf{y}}$ . Verify $\partial_\mu G^{\mu\nu}=0$ component by component, and identify the duality-rotated field.

Step 1. The numbers: $k=2\pi/\lambda=9.93\times10^6$ m⁻¹, $\omega=ck=2.98\times10^{15}$ rad/s, $B_0=E_0/c=1.00\times10^{-6}$ T. The dual's nonzero entries: $G^{02}=-B_0\cos(kz-\omega t)$ and $G^{23}=E_x/c=B_0\cos(kz-\omega t)$ — for a null field the dual's two blocks carry the same profile, another face of $E=cB$ .

Step 2. The $\nu=0$ component: $\partial_\mu G^{\mu0}=\nabla\cdot\mathbf{B}=\partial_yB_y=0$ , since $B_y$ depends only on $z$ and $t$ .

Step 3. The $\nu=y$ component is the live one (Derivation B Step 2 with $x\to y$ ): $\partial_\mu G^{\mu y}=-\tfrac1c[\partial_tB_y+(\nabla\times\mathbf{E})_y]$ . The pieces: $\partial_tB_y=+B_0\,\omega\sin(kz-\omega t)$ and $(\nabla\times\mathbf{E})_y=\partial_zE_x=-E_0k\sin(kz-\omega t)$ , summing to $E_0\sin(kz-\omega t)\left[\omega/c-k\right]$ . With $\omega/c=9.93\times10^6$ m⁻¹ $=k$ exactly, the bracket is zero: Faraday's law holds because the wave rides the dispersion relation $\omega=ck$ ; a hypothetical wave with $\omega\neq ck$ would need magnetic current to exist. ( $\nu=x$ and $\nu=z$ vanish term by term: no $B_x$ , $B_z$ , no $y$ -dependence anywhere.)

Step 4. The duality-rotated field: the substitution rule turns the beam into $\mathbf{E}'=E_0\cos(kz-\omega t)\,\hat{\mathbf{y}}$ , $\mathbf{B}'=-(E_0/c)\cos(kz-\omega t)\,\hat{\mathbf{x}}$ . Check the Poynting direction: $\hat{\mathbf{y}}\times(-\hat{\mathbf{x}})=+\hat{\mathbf{z}}$ — still a beam along $+z$ , same amplitude and frequency, polarization rotated by exactly $90°$ . For free fields, duality rotation is physical: it maps solutions to solutions ( $\partial F=0$ and $\partial G=0$ trade places), and for a linearly polarized wave it is indistinguishable from turning the polarizer a quarter turn.

The lesson: on a plane wave the abstract machinery becomes tactile — the dual equation's only nontrivial component is Faraday's law enforcing $\omega=ck$ , and the $90°$ rotation in field space materializes as a $90°$ rotation of polarization in real space.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §12.3.4–12.3.5 · Jackson, Classical Electrodynamics, 3rd ed., §11.9 and §6.11 · Zangwill, Modern Electrodynamics, §22.5 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 5 (field transformations)

Provenance. Component matrix of $\tilde F$ should be recomputed from $\tilde{F}^{\mu\nu}=\tfrac12\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ with $F^{0i}=E^i/c$ , $F^{ij}=-\varepsilon^{ijk}B_k$ , metric $(+,-,-,-)$ , $\varepsilon^{0123}=+1$ — verify against Griffiths §12.3.4. The two invariants $\tilde F F\propto\mathbf E\cdot\mathbf B$ and $FF=2(B^2-E^2/c^2)$ are standard (Jackson §11.9). The double-dual identity uses the four-index epsilon contraction with the $g=-1$ sign. SI factors of $c$ follow Griffiths/Zangwill conventions; overall signs in the invariant depend on metric signature and should be checked consistently. Addendum for this edition: the recomputation was performed symbolically (all 16 components of $G$ , both contractions, the double dual, the contraction identity over all index combinations), from the potential-pinned convention $F_{0i}=+E_i/c$ , $F_{ij}=-\varepsilon_{ijk}B_k$ — equivalently $F^{i0}=+E^i/c$ — as forced by $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ with $A^\mu=(V/c,\mathbf{A})$ ; the base's provenance line " $F^{0i}=E^i/c$ " carries a sign slip relative to its own (correct) component matrices, resolved here in favor of the potential definition, and every sign the base deferred to convention is fixed explicitly in Derivation A. Division of labor: the invariants' classification theorem belongs to Lorentz Invariants of the Field; the assembled system to Covariant Form of Maxwell's Equations; monopole quantization to the quantum-mechanics track. Worked numbers for this edition ( $\mathbf{E}\cdot\mathbf{B}=50$ V·T/m ⇒ $G\!\cdot\!F=-6.67\times10^{-7}$ T², $F\!\cdot\!F=-6.11\times10^{-7}$ T², $\det F=2.78\times10^{-14}$ T⁴; HeNe wave $k=\omega/c=9.93\times10^6$ m⁻¹ closing Faraday's component) computed and verified numerically for this edition.

Builds on: The Electromagnetic Field Tensor · Four-Vectors and the Metric Tensor · Lorentz Transformations · Transformation of Electric and Magnetic Fields Leads to: Lorentz Invariants of the Field · Magnetism as a Relativistic Effect

Does the dual tensor contain any physics that F does not?

No new information — the same six numbers $(\mathbf{E},\mathbf{B})$ populate $G$ , filed in complementary slots — but genuinely new access to it. Three things only $G$ delivers cleanly: the sourceless half of Maxwell as a one-line divergence (in terms of $F$ alone it needs three free indices); the pseudoscalar invariant $\mathbf{E}\cdot\mathbf{B}$ (no contraction of $F$ with itself can produce it); and the covariant statement of parity, since $G$ 's pseudotensor character is exactly the axial-vector nature of $\mathbf{B}$ . Think of $F$ and $G$ as the same ledger sorted two ways: no entry changes, but each sorting makes a different total trivial to read.

Is the duality rotation an actual symmetry of electromagnetism or just notation?

For the free field it is an exact symmetry: with $J^\nu=0$ the divergences $\partial F=0$ and $\partial G=0$ trade places under $F\to F\cos\theta+G\sin\theta$ , so duality rotations map vacuum solutions to vacuum solutions — Worked example 2's quarter turn rotated a laser's polarization by $90°$ and nothing else. With sources the symmetry breaks: nature supplies electric four-currents but (so far) no magnetic ones, and the rotation would shuffle charge into magnetic charge that does not exist. That breakage is empirical, not structural — equations (6) restore full symmetry at the cost of one hypothetical particle, which is why monopole searches are, in covariant language, tests of whether duality is fundamental or accidental.

If B is really an antisymmetric tensor in disguise, why do we get away with treating it as a vector everywhere?

Because three dimensions is a coincidence-rich place: an antisymmetric $3\times3$ object has exactly three independent components, and the three-index epsilon converts it to a vector that transforms correctly under rotations. The disguise slips in two situations. Under reflections: a true vector reverses, but $\mathbf{B}$ — really an oriented circulation — does not, the "axial vector" caveat every magnetostatics course carries as a footnote. And under boosts: no three-vector rule can describe how $\mathbf{B}$ mixes with $\mathbf{E}$ , because the mixing is the rank-2 tensor transformation of $F$ . The dual tensor is where the disguise comes off covariantly: one factor of the four-epsilon, and $\mathbf{B}$ 's parity character is carried by the formalism instead of by a footnote.

The homogeneous equations are identities once potentials exist — so are they laws of nature or not?

They are the statement that potentials exist, which is a law of nature. Given $A^\mu$ , the sourceless pair follows identically (Derivation B Step 4) — no experiment could falsify it without first falsifying the potential description. Given only fields, $\partial_\mu G^{\mu\nu}=0$ is precisely the integrability condition (the Poincaré lemma's hypothesis) under which a four-potential can be constructed on any contractible region — so its experimental status is exactly that of " $\mathbf{E}$ and $\mathbf{B}$ admit potentials." A single verified monopole would break the global statement: around it $\nabla\cdot\mathbf{B}\neq0$ , no single-valued $\mathbf{A}$ exists, and one is forced into Dirac strings or fiber-bundle patches — machinery the quantum-mechanics track builds. The empty right-hand side of the dual equation is the classical fingerprint of a deep structural choice nature seems to have made.

Different books give me different signs for the dual and the pseudoscalar invariant — who is right?

Everyone, within their own conventions; almost no one, when formulas are transplanted. Four independent choices scatter the signs: metric signature, the value of $\varepsilon^{0123}$ versus $\varepsilon_{0123}$ (some texts fix the lower one to $+1$ ), the sign convention inside $F$ itself, and SI-versus-Gaussian factors of $c$ . Each flip propagates into the dual's matrix, the sign of $G\!\cdot\!F$ , even the epsilon contraction identity ( $-2$ Lorentzian, $+2$ Euclidean). The convention-independent content — what to check when reconciling books — is structural: which field occupies which block, exactly one sign flip per quarter turn, $\tilde{\tilde F}=-F$ , the pseudoscalar proportional to $\mathbf{E}\cdot\mathbf{B}$ , and the equivalence of the dual divergence with the sourceless Maxwell pair. This article's signs all descend from three pinned choices: $(+,-,-,-)$ , $\varepsilon^{0123}=+1$ , and $F=\partial A$ .

What would actually change in this article if a monopole were discovered tomorrow?

Remarkably little machinery, and remarkably much meaning. The dual equation gains a source, $\partial_\mu G^{\mu\nu}=\mu_0c\,J_m^\nu$ : Gauss's law for magnetism becomes $\nabla\cdot\mathbf{B}=\mu_0c\rho_m$ , Faraday's law gains a magnetic-current term, and the two Maxwell columns become perfectly symmetric. The potential formulation takes the real damage: with $\nabla\cdot\mathbf{B}\neq0$ somewhere, no globally single-valued $A^\mu$ exists, so Derivation B's "automatic" theorem survives only patchwise, with Dirac-string seams between patches. The consistency condition of that patching is Dirac's quantization $e\,q_m=nh$ — one monopole anywhere would explain why every measured electric charge is an integer multiple of $e$ , a fact standard electrodynamics simply postulates. That trade — one exotic particle for an explanation of charge quantization — is why the empty slot in equation (6) has been hunted for ninety years.