Lorentz Invariants of the Field

Three articles of this chapter have now demonstrated, with increasing machinery, that $\mathbf{E}$ and $\mathbf{B}$ are frame-gossip: Transformation of Electric and Magnetic Fields derived the mixing rules with capacitors, Magnetism as a Relativistic Effect built a magnetic force out of boosted electrostatics, and The Electromagnetic Field Tensor exposed both fields as components of one object $F^{\mu\nu}$ . A fair question after all that demolition: is anything about the local field absolute? Can two observers agree on any number at all? Exactly two numbers, it turns out — and this article computes both, then extracts everything they know. Contracting the field tensor with itself gives $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)$ ; contracting it with the dual of The Dual Field Tensor gives $G_{\mu\nu}F^{\mu\nu}=-4\,\mathbf{E}\cdot\mathbf{B}/c$ . Every inertial observer, at the same event, computes the same two values. And six components minus the freedom of the Lorentz group leaves room for nothing else: every other full contraction is a polynomial in these two. The payoff is a complete classification of electromagnetic fields without performing a single boost. The sign of $E^2-c^2B^2$ decides whether some observer sees a purely electric field, a purely magnetic one, or neither; $\mathbf{E}\cdot\mathbf{B}$ decides whether either field can be eliminated at all; and the doubly-degenerate case — both invariants zero — is the light wave, the field configuration so relativistic that no frame can tame it. The classification theorem is this article's centerpiece, the free-field Lagrangian built from $I_1$ is its forward pointer, and the whole chapter's E–B mixing acquires its bookkeeping: what boosts shuffle, invariants audit.

Key takeaways

From $F^{\mu\nu}$ exactly two independent Lorentz-invariant scalars exist: $I_1=F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)$ and $I_2=F_{\mu\nu}G^{\mu\nu}=-4\,\mathbf{E}\cdot\mathbf{B}/c$ .
Equivalently: $E^2-c^2B^2$ and $\mathbf{E}\cdot\mathbf{B}$ have the same value in every inertial frame, even though $\mathbf{E}$ and $\mathbf{B}$ separately do not. $I_1$ is a true scalar; $I_2$ is a pseudoscalar (parity-odd).
Classification without boosting: if $\mathbf{E}\cdot\mathbf{B}=0$ and $E<cB$ , a frame with pure $\mathbf{B}$ exists (drift $\mathbf{v}=\mathbf{E}\times\mathbf{B}/B^2$ ); if $E>cB$ , a frame with pure $\mathbf{E}$ exists ( $\mathbf{v}=c^2\mathbf{E}\times\mathbf{B}/E^2$ ); the residual field magnitude is fixed by the invariant.
If $\mathbf{E}\cdot\mathbf{B}\neq0$ , neither field can be eliminated in any frame — at best they can be made parallel.
Null fields ( $I_1=I_2=0$ ) keep $E=cB$ and $\mathbf{E}\perp\mathbf{B}$ in every frame: the plane wave has no rest frame, and no boost can purify it.
The invariants build invariant physics: the free-field Lagrangian density is $-I_1/4\mu_0$ , and strong-field QED thresholds (Schwinger pair production) are phrased in $I_1,I_2$ .
Key idea: the field tensor admits exactly two algebraic invariants — $2(B^2-E^2/c^2)$ and $-4\,\mathbf{E}\cdot\mathbf{B}/c$ — and their signs classify every electromagnetic field up to a Lorentz transformation, no boost required.

The core insight

There are precisely two Lorentz invariants of the field: $I_1=F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)$ (a scalar) and $I_2=F_{\mu\nu}\tilde F^{\mu\nu}=-\tfrac{4}{c}\,\mathbf{E}\cdot\mathbf{B}$ (a pseudoscalar). All other contractions reduce to these two.

Where this lives today

Strong-field physics speaks this article's language natively. The Schwinger critical field $E_S=m_e^2c^3/e\hbar=1.32\times10^{18}$ V/m — where the quantum vacuum sparks into electron–positron pairs — is an invariant-phrased threshold: a purely magnetic field, however enormous, produces no pairs, because $I_1>0$ with $I_2=0$ means some frame sees no electric field at all, and the vacuum-decay rate, being frame-independent, must vanish everywhere if it vanishes somewhere. Magnetars ( $B\sim10^{10{-}11}$ G) and petawatt laser facilities probe the regime where the invariants' polynomial corrections — the Euler–Heisenberg Lagrangian, built from $I_1^2$ and $I_2^2$ — predict measurable vacuum birefringence; polarization measurements of X-rays from magnetar surfaces are the current observational frontier. The invariants also police model-building. Any Lorentz-invariant electrodynamics must have a Lagrangian density that is a function of $I_1$ and $I_2$ alone: Maxwell's theory is the linear choice $\mathcal{L}=-I_1/4\mu_0$ (developed in Relativistic Lagrangian and Action for Charged Particles and the field-theory articles beyond), Born–Infeld and Euler–Heisenberg are nonlinear functions of both, and a term linear in $I_2$ — parity-odd — is the $\theta$ -term whose quantum consequences (strong-CP, axions) the dual article previewed. Meanwhile, in the daily physics of accelerators and plasmas: every velocity selector is tuned across the $E=cB$ null boundary, every guiding-center code implements the purifying drift $\mathbf{E}\times\mathbf{B}/B^2$ of Derivation B, and every synchrotron design assumes the storage ring's field is magnetic-dominated in all frames that matter. The two numbers computed here are the quiet bookkeeping behind all of it — and the chapter now has its audit trail: whatever Covariant Form of Maxwell's Equations assembles next, the invariants say what no formulation can change.

Origins · 1 minWolfgang Pauli and how this was discovered

Wolfgang Pauli

1900–1958 · Vienna, Zurich — whose encyclopedia article, written at twenty-one, taught relativity its own invariants

Pauli enters this story absurdly early. In 1921, a twenty-one-year-old doctoral student in Sommerfeld's Munich seminar, he wrote the 237-page relativity article for the Encyklopädie der mathematischen Wissenschaften — the definitive survey of special and general relativity, complete with the systematic treatment of the field invariants and the covariant formulation this chapter builds. Einstein reviewed it in astonishment: "Whoever studies this mature and grandly conceived work would not believe that the author is a twenty-one year old man." The rest of the career kept the standard: the exclusion principle (1925, Nobel 1945) that structures every atom; the prediction of the neutrino (1930) as a desperate bookkeeping device that turned out to exist; the CPT theorem and spin-statistics connection that tie this article's parity discussions to quantum field theory. Famously merciless as a critic — "not even wrong" is his — he was the conscience of quantum physics for three decades, and his encyclopedia article, reprinted as Theory of Relativity, remains in print a century later: the book where the two invariants of the electromagnetic field first got their textbook home.

History

1905

— Einstein's relativity paper transforms the fields between frames; what stays fixed is not yet organized

1908

— Minkowski's spacetime geometry makes invariants the organizing principle of relativistic physics

1910

— Bateman and Cunningham study the invariance structure of Maxwell's equations, including conformal extensions

1921

— Pauli's Encyklopädie article systematizes the two field invariants within the tensor formulation

1935

— Euler and Heisenberg build the effective Lagrangian of the quantum vacuum as a function of the two invariants

1951

— Schwinger computes vacuum pair production: the critical field and its invariant-based selection rules

1970

s — Guiding-center and drift theories embed the purifying boost into plasma-physics practice

2016

— Optical polarimetry of neutron star RX J1856.5−3754 reports evidence of vacuum birefringence, an Euler–Heisenberg effect

The physical picture

Think of an ordinary vector under rotations. Its three components are observer-dependent — tilt your head and they reshuffle — but its length is not, and "length" is the only rotational invariant a single vector has. The field tensor is the four-dimensional, indefinite-metric version of the same story. A boost reshuffles the six components $(\mathbf{E}/c,\mathbf{B})$ according to the mixing rules of Transformation of Electric and Magnetic Fields, but certain quadratic combinations survive untouched, for the reason Four-Vectors and the Metric Tensor industrialized: any honest full contraction of tensors is a scalar. $F$ offers two partners to contract with — itself, and its dual — so the field has exactly two "lengths." Each invariant answers a physical question you can ask without leaving your chair. The scalar $E^2-c^2B^2$ asks: in the tug-of-war between electric and magnetic character, who is winning? — and the answer is frame-proof. A capacitor's field is electric-dominated ( $E^2-c^2B^2>0$ ) for every observer, no matter how fast they fly through it and how much $\mathbf{B}$ they personally measure; a solenoid's is magnetic-dominated for everyone. The pseudoscalar $\mathbf{E}\cdot\mathbf{B}$ asks: are the two fields entangled in a handed way? If it vanishes, the fields are perpendicular for everyone (wherever both exist), and the weaker one is a removable artifact — some observer sees it gone. If it does not vanish, the entanglement is absolute: no observer in the universe sees this field as purely electric or purely magnetic. The degenerate corner of the classification is the most physical of all. Set both invariants to zero and you have a field with $E=cB$ , $\mathbf{E}\perp\mathbf{B}$ — in every frame. That is the electromagnetic wave, and the invariants explain its stubbornness: a boosted observer chasing a light beam sees the same structure at Doppler-shifted strength, never a static field, because there is no invariant scale to purify toward. A null field is the field-space twin of a null four-vector: no rest frame, ever. What the chapter has been dramatizing — fields mutating under boosts — this article disciplines: mutation happens on a leash, and the leash has exactly two knots.

Figure

The classification map — two invariant axes sort every electromagnetic field, with the light wave pinned at the origin

The math: two contractions, and the classification they enforce

The raw material is fixed by the chapter's conventions. From $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ with metric $(+,-,-,-)$ , the field tensor's components are $F_{0i}=+E_i/c$ , $F_{ij}=-\varepsilon_{ijk}B_k$ (so $F^{0i}=-E_i/c$ , $F^{ij}=-\varepsilon^{ijk}B_k$ ), and the dual of The Dual Field Tensor carries $G^{0i}=-B_i$ , $G^{ij}=+\varepsilon^{ijk}E_k/c$ . Contractions of these objects are Lorentz scalars by the up–down cancellation theorem of Four-Vectors and the Metric Tensor — no further proof needed. The only question is their value, and antisymmetry leaves just two candidates:

I_1\equiv F_{\mu\nu}F^{\mu\nu}=2\left(B^2-\frac{E^2}{c^2}\right),\qquad I_2\equiv F_{\mu\nu}G^{\mu\nu}=-\frac{4}{c}\,\mathbf{E}\cdot\mathbf{B}. \qquad (1)

where: $E^2=\mathbf{E}\cdot\mathbf{E}$ and $B^2=\mathbf{B}\cdot\mathbf{B}$ ; both invariants carry units of T². Read them physically before deriving: $I_1$ is a frame-proof verdict on electric-versus-magnetic dominance (negative means electric wins — mind the sign convention), and $I_2$ is a frame-proof measure of E–B alignment. Because $G$ carries one Levi-Civita symbol, $I_2$ is a pseudoscalar: invariant under boosts and rotations, sign-flipped under spatial inversion — which is why it is often quoted through its parity-even square, and why its appearance in a Lagrangian signals CP-sensitive physics. For hand computation, the equivalent "field forms" strip the tensor factors:

E^2-c^2B^2=-\frac{c^2}{2}\,I_1=\text{invariant},\qquad \mathbf{E}\cdot\mathbf{B}=-\frac{c}{4}\,I_2=\text{invariant (pseudo)}. \qquad (2)

where: these are the two numbers every observer extracts identically from their own $(\mathbf{E},\mathbf{B})$ readings at the shared event. A useful third form: the determinant of the field tensor is not independent but the pseudoscalar squared, $\det F^{\mu\nu}=(\mathbf{E}\cdot\mathbf{B})^2/c^2$ — the Pfaffian of an antisymmetric $4\times4$ matrix is linear in $I_2$ , and determinants of $\Lambda F\Lambda^{\mathsf T}$ are automatically invariant since $\det\Lambda=\pm1$ .

Derivation ABoth invariants computed in full, and the invariance double-checked against the boost rules

The goal is equation (1) by explicit component bookkeeping — every sign and factor shown, since the factors of 2 and 4 are the standard casualties of this calculation — followed by a direct verification that the field forms (2) survive the boost rules of Transformation of Electric and Magnetic Fields, so that the tensor argument and the pedestrian algebra certify each other. The strategy for the contraction: split the sixteen-term sum into the time–space block (counted twice by antisymmetry) and the space–space block, then apply the three-dimensional epsilon identity. Nothing else is needed, but the index altitudes must be tracked honestly — that is where the signs live.

Step 1. Split the contraction. For any two antisymmetric tensors, the double sum over $\mu\nu$ decomposes as twice the $(0,i)$ terms plus the $(i,j)$ terms:

F_{\mu\nu}F^{\mu\nu}=2\,F_{0i}F^{0i}+F_{ij}F^{ij}.

Step 2. Evaluate each block of $I_1$ . Time–space: $F_{0i}=+E_i/c$ but $F^{0i}=-E_i/c$ (one metric factor of $-1$ from the spatial index), so $2F_{0i}F^{0i}=-2E^2/c^2$ . Space–space: lowering two spatial indices costs $(-1)^2$ , so $F_{ij}=F^{ij}=-\varepsilon_{ijk}B_k$ , and with the identity $\varepsilon_{ijk}\varepsilon_{ijl}=2\delta_{kl}$ ,

F_{ij}F^{ij}=\varepsilon_{ijk}\varepsilon_{ijl}B_kB_l=2B^2 \quad\Longrightarrow\quad I_1=2B^2-\frac{2E^2}{c^2}.

Why the relative minus is physics: the electric entries live in mixed time–space slots and inherit the metric's indefinite signature; the magnetic entries are purely spatial and do not. $I_1$ is the field-space cousin of the interval $c^2t^2-x^2$ — same minus sign, same origin.

Step 3. Evaluate $I_2$ the same way: $G_{0i}=+B_i$ (lowering $G^{0i}=-B_i$ costs one sign), so $2G_{0i}F^{0i}=2B_i(-E_i/c)=-2\,\mathbf{E}\cdot\mathbf{B}/c$ ; and $G_{ij}=+\varepsilon_{ijk}E_k/c$ against $F^{ij}=-\varepsilon^{ijl}B_l$ gives $-\varepsilon_{ijk}\varepsilon_{ijl}E_kB_l/c=-2\,\mathbf{E}\cdot\mathbf{B}/c$ . The blocks contribute equally:

I_2=-\frac{4}{c}\,\mathbf{E}\cdot\mathbf{B}.

Why equal halves: dualizing swaps the blocks, so the pairing of $F$ 's electric block with $G$ 's magnetic block and vice versa produces the same product twice — a structural check worth remembering when auditing factor conventions across books.

Step 4. Cross-check invariance directly, no tensors. Boost at $v$ along $x$ and apply the six mixing rules: $\bar E_x=E_x$ , $\bar E_y=\gamma(E_y-vB_z)$ , $\bar E_z=\gamma(E_z+vB_y)$ , $\bar B_x=B_x$ , $\bar B_y=\gamma(B_y+vE_z/c^2)$ , $\bar B_z=\gamma(B_z-vE_y/c^2)$ . For the pseudoscalar: $\bar E_x\bar B_x=E_xB_x$ untouched, and

\bar E_y\bar B_y+\bar E_z\bar B_z=\gamma^2\left(1-\frac{v^2}{c^2}\right)\left(E_yB_y+E_zB_z\right)=E_yB_y+E_zB_z,

because the cross terms cancel in pairs ( $+vE_yE_z/c^2$ against $-vE_zE_y/c^2$ , $-vB_zB_y$ against $+vB_yB_z$ ) and $\gamma^2(1-v^2/c^2)=1$ . For the scalar, the same expansion of $\bar E_\perp^2-c^2\bar B_\perp^2$ cancels its cross terms ( $\mp2vE_yB_z$ , $\pm2vE_zB_y$ appear once in each square with opposite signs) and the same $\gamma^2(1-v^2/c^2)=1$ collapses the prefactor. Both field forms survive — the pedestrian route lands exactly where the contraction theorem said it must. Verified: both contractions recomputed symbolically from the full component matrices (sympy, all sixteen terms each): $I_1=2(B^2-E^2/c^2)$ , $I_2=-4\,\mathbf{E}\cdot\mathbf{B}/c$ exactly; $\det F=(\mathbf{E}\cdot\mathbf{B})^2/c^2$ confirmed; the boost cross-check at $\beta=0.8$ with $E=10^5$ V/m reproduces $E^2-c^2B^2=1.00\times10^{10}$ V²/m² in both frames (Worked example 1). Matches Griffiths §12.3.5 and Jackson §11.9.

Why only two? An antisymmetric rank-2 tensor has six independent components; the proper Lorentz group acting at a point has six parameters (three rotations, three boosts) but preserves the pair $(I_1,I_2)$ — and one can show the orbit of a generic field under the group is four-dimensional, leaving exactly $6-4=2$ independent labels. Algebraically the statement is sharper: every fully contracted power of $F$ and $G$ reduces to a polynomial in $I_1$ and $I_2$ . The first nontrivial case is worth recording,

F^{\mu}{}_{\nu}F^{\nu}{}_{\rho}F^{\rho}{}_{\sigma}F^{\sigma}{}_{\mu}=\tfrac{1}{2}I_1^{\,2}+\tfrac{1}{4}I_2^{\,2}, \qquad (3)

where: the quartic trace — the next candidate invariant after the quadratics — collapses onto the two we have (the cubic trace vanishes identically by antisymmetry). There is no third invariant hiding at higher order; the classification below is therefore complete.

Go deeper · Derivation B · 2 minThe classification theorem — which fields can be purified, by which boost, and which cannot

Derivation BThe classification theorem — which fields can be purified, by which boost, and which cannot

The goal is the full field-classification theorem: for $\mathbf{E}\cdot\mathbf{B}=0$ , construct the explicit drift boost that eliminates the weaker field and show the invariant fixes both its existence condition and the survivor's magnitude; for $\mathbf{E}\cdot\mathbf{B}\neq0$ , show no frame eliminates either field; and for the null case, certify that $E=cB$ and $\mathbf{E}\perp\mathbf{B}$ hold in every frame. The strategy is to work directly with the perpendicular-component mixing rules in vector form — $\bar{\mathbf{E}}_\perp=\gamma(\mathbf{E}+\mathbf{v}\times\mathbf{B})_\perp$ , $\bar{\mathbf{B}}_\perp=\gamma(\mathbf{B}-\mathbf{v}\times\mathbf{E}/c^2)_\perp$ — choosing the boost along $\mathbf{E}\times\mathbf{B}$ , the one direction perpendicular to both fields. Watch the invariants act as gatekeepers at every step: they decide what is possible before any algebra runs.

Step 1. Set up the crossed-field case: $\mathbf{E}\cdot\mathbf{B}=0$ , both fields nonzero. Boost with $\mathbf{v}=v\,\hat{\mathbf{n}}$ , $\hat{\mathbf{n}}=\mathbf{E}\times\mathbf{B}/|\mathbf{E}\times\mathbf{B}|$ ; both fields are perpendicular to $\mathbf{v}$ , so the parallel rules are idle and the $\gamma$ -factors multiply everything uniformly.

Step 2. Kill $\mathbf{B}$ (electric-dominated branch). Demand $\mathbf{B}=\mathbf{v}\times\mathbf{E}/c^2$ . With $\mathbf{v}\perp\mathbf{E}$ the magnitudes give $v=c^2B/E$ , and the direction works out exactly along $\hat{\mathbf{n}}$ : using $(\mathbf{E}\times\mathbf{B})\times\mathbf{E}=E^2\mathbf{B}$ for perpendicular fields,

\mathbf{v}=\frac{c^2}{E^2}\,\mathbf{E}\times\mathbf{B} \quad\Longrightarrow\quad \bar{\mathbf{B}}=\gamma\left(\mathbf{B}-\frac{\mathbf{v}\times\mathbf{E}}{c^2}\right)=0.

Subluminality demands $v=c^2B/E<c$ , i.e. $E>cB$ — exactly the electric-dominated condition $E^2-c^2B^2>0$ . The invariant is the gatekeeper: the boost exists precisely when the sign says so.

Step 3. The survivor's strength is fixed by the invariant. With $(\mathbf{E}\times\mathbf{B})\times\mathbf{B}=-B^2\mathbf{E}$ (perpendicular fields),

\bar{\mathbf{E}}=\gamma\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)=\gamma\mathbf{E}\left(1-\frac{c^2B^2}{E^2}\right)=\mathbf{E}\,\sqrt{1-\frac{c^2B^2}{E^2}} \quad\Longrightarrow\quad \bar E=\sqrt{E^2-c^2B^2},

since $\beta=cB/E$ makes $\gamma=(1-c^2B^2/E^2)^{-1/2}$ . No algebra was optional: $\bar E^2-c^2\cdot0^2$ must reproduce the invariant, so the answer was fixed before the calculation started. Why this is the theorem's teeth: the invariant does not merely permit the purification — it dictates the purified field's exact magnitude.

Step 4. Kill $\mathbf{E}$ (magnetic-dominated branch), by the mirror argument: $\mathbf{v}=\mathbf{E}\times\mathbf{B}/B^2$ has magnitude $E/B<c$ exactly when $E<cB$ , annihilates $\bar{\mathbf{E}}$ via $(\mathbf{E}\times\mathbf{B})\times\mathbf{B}=-B^2\mathbf{E}$ , and leaves $\bar B=\sqrt{B^2-E^2/c^2}$ — the cyclotron-clean frame that storage-ring physics quietly assumes. Between the branches sits $E=cB$ : the required boost tends to $c$ , the survivor tends to zero, and the purification fails — the null field.

Step 5. The obstructions, by invariants alone. If $\mathbf{E}\cdot\mathbf{B}\neq0$ : a purely electric (or magnetic) field has $I_2=0$ , and $I_2$ is invariant, so no frame with either field eliminated can exist — no computation required. The best available is alignment: a boost along $\mathbf{E}\times\mathbf{B}$ with $v/(1+v^2/c^2)=|\mathbf{E}\times\mathbf{B}|/(B^2+E^2/c^2)$ renders $\bar{\mathbf{E}}\parallel\bar{\mathbf{B}}$ (the quadratic has one subluminal root, since the roots' product is $c^2$ ; Landau–Lifshitz §25). If $I_1=I_2=0$ : then in every frame $\bar E^2-c^2\bar B^2=0$ and $\bar{\mathbf{E}}\cdot\bar{\mathbf{B}}=0$ — the triad structure $E=cB$ , $\mathbf{E}\perp\mathbf{B}$ that Monochromatic Plane Waves in Vacuum derived from Maxwell's equations is thereby certified frame-independent: every observer chasing or fleeing the wave sees a valid plane-wave geometry, only Doppler-rescaled. Verified: the BAC-CAB reductions $(\mathbf{E}\times\mathbf{B})\times\mathbf{E}=E^2\mathbf{B}$ and $(\mathbf{E}\times\mathbf{B})\times\mathbf{B}=-B^2\mathbf{E}$ hold for $\mathbf{E}\cdot\mathbf{B}=0$ ; both drift formulas checked numerically (Worked example 2: $E=3\times10^6$ V/m, $B=0.02$ T gives $v=0.5c$ , $\bar E=0$ , $\bar B=17.3$ mT $=\sqrt{B^2-E^2/c^2}$ exactly); subluminality conditions match the invariant signs; parallel-frame condition dimensionally audited. Matches Griffiths §12.3.5 problems, Jackson §11.10, Landau–Lifshitz §25.

Figure

The purifying boost — riding the E cross B drift turns a crossed field into a pure one, at the speed the invariant dictates

Limiting cases

case	result	meaning
$E^2-c^2B^2<0$ , $\mathbf{E}\cdot\mathbf{B}=0$	pure- $\mathbf{B}$ frame exists; drift $\mathbf{v}=\mathbf{E}\times\mathbf{B}/B^2$	magnetic-dominated: magnets, storage rings; residual $\bar B=\sqrt{B^2-E^2/c^2}$
$E^2-c^2B^2>0$ , $\mathbf{E}\cdot\mathbf{B}=0$	pure- $\mathbf{E}$ frame exists; drift $\mathbf{v}=c^2\mathbf{E}\times\mathbf{B}/E^2$	electric-dominated: capacitors, point charges; residual $\bar E=\sqrt{E^2-c^2B^2}$
$I_1=I_2=0$	$E=cB$ , $\mathbf{E}\perp\mathbf{B}$ in every frame	null field: the plane wave; no purifying frame — light has no rest frame
$\mathbf{E}\cdot\mathbf{B}\neq0$	neither field removable; a parallel-fields frame exists	generic entangled field; the tilted-field example of the dual article
$\mathbf{B}=0$ in one frame	every other frame: $\bar{\mathbf{B}}=-\mathbf{v}\times\bar{\mathbf{E}}/c^2$ , still $I_2=0$ , $I_1<0$ form preserved	boosted electrostatics stays electric-dominated — the moving-charge pancake
$v\ll c$ purification	$\mathbf{v}\approx\mathbf{E}\times\mathbf{B}/B^2$ is the guiding-center drift	the plasma-physics E-cross-B drift is the nonrelativistic shadow of the purifying boost

Common trapInvariant means frame-independent — not conserved, not everything, and never the fields themselves

Three distinct traps share this article's vocabulary. First: $\mathbf{E}$ and $\mathbf{B}$ are not invariant — only the combinations $E^2-c^2B^2$ and $\mathbf{E}\cdot\mathbf{B}$ are. "The fields are perpendicular in all frames" is true precisely because $\mathbf{E}\cdot\mathbf{B}$ is invariant, not because directions freeze. Second: invariant $\neq$ conserved. An invariant is one event compared across observers; a conserved quantity is one observer compared across times. A pulsed field has time-varying invariants at a fixed point; energy is conserved yet frame-dependent. Third — the exam classic: the energy density $u=\tfrac{\varepsilon_0}{2}(E^2+c^2B^2)$ and the Poynting flux are not invariants (note the plus sign); they are components of the stress-energy tensor built two articles ahead, and every observer measures different ones. If a calculation produces a "frame-independent energy density," a sign error has manufactured it.

Worked examples

1 · Auditing the boosted capacitor with both invariants (confidence builder)

Setup. Transformation of Electric and Magnetic Fields boosted a capacitor's pure field $\mathbf{E}=10^5\,\hat{\mathbf{y}}$ V/m, $\mathbf{B}=0$ through $v=0.8c\,\hat{\mathbf{x}}$ and found $\bar{\mathbf{E}}=1.667\times10^5\,\hat{\mathbf{y}}$ V/m, $\bar{\mathbf{B}}=-4.444\times10^{-4}\,\hat{\mathbf{z}}$ T. Audit that pair of snapshots with both invariants ( $c=3.00\times10^8$ m/s).

Step 1. Pseudoscalar first, in each frame. Lab: $\mathbf{E}\cdot\mathbf{B}=0$ trivially. Boosted: $\bar{\mathbf{E}}\perp\bar{\mathbf{B}}$ ( $\hat{\mathbf{y}}\cdot\hat{\mathbf{z}}=0$ ), so $\bar{\mathbf{E}}\cdot\bar{\mathbf{B}}=0$ . ✓ Invariant — and predictive: since $I_2=0$ was set in the lab, every observer of this capacitor must find perpendicular fields, forever.

Step 2. Scalar, lab frame: $E^2-c^2B^2=(10^5)^2-0=1.000\times10^{10}$ V²/m².

Step 3. Scalar, boosted frame: $\bar E^2=(1.667\times10^5)^2=2.778\times10^{10}$ ; $c^2\bar B^2=(3.00\times10^8)^2(4.444\times10^{-4})^2=1.778\times10^{10}$ . Difference: $\mathbf{1.000\times10^{10}}$ V²/m². ✓ Equal to five digits — the growth of $\bar E^2$ is exactly paid for by the appearance of $c^2\bar B^2$ .

Step 4. Classify: $I_1$ -form positive and $I_2=0$ — electric-dominated. The classification theorem then predicts a frame with $\mathbf{B}$ eliminated and field $\sqrt{E^2-c^2B^2}$ : from the boosted observer's own readings, $\sqrt{2.778\times10^{10}-1.778\times10^{10}}=10^5$ V/m, drift velocity $c^2\bar E\bar B/\bar E^2=0.8c$ backwards — the lab frame itself, recovered from the moving observer's data alone. The invariants let any observer reconstruct the field's "rest frame" without being told the history.

The lesson: invariants are free error-checking on every field-transformation computation — and more: they encode which frame, if any, sees the field in purified form.

2 · Classify, purify, verify — a crossed-field chamber at relativistic drift (exam level)

Setup. A chamber holds $\mathbf{E}=3\times10^6\,\hat{\mathbf{y}}$ V/m and $\mathbf{B}=0.02\,\hat{\mathbf{z}}$ T (strong pulsed fields, laboratory-achievable). Classify the field, construct the purifying frame, and verify the survivor against the invariant.

Step 1. Classify from the chair: $\mathbf{E}\cdot\mathbf{B}=0$ (crossed), and $cB=6\times10^6$ V/m $>E=3\times10^6$ V/m, so $E^2-c^2B^2=9\times10^{12}-3.6\times10^{13}=-2.70\times10^{13}$ V²/m² — magnetic-dominated: a pure- $\mathbf{B}$ frame exists; no frame anywhere sees this field purely electric.

Step 2. The drift: $\mathbf{v}=\mathbf{E}\times\mathbf{B}/B^2$ , direction $\hat{\mathbf{y}}\times\hat{\mathbf{z}}=\hat{\mathbf{x}}$ , magnitude $v=E/B=3\times10^6/0.02=1.5\times10^8$ m/s $=\mathbf{0.5}c$ . Subluminal, as the invariant's sign guaranteed; $\gamma=1/\sqrt{1-0.25}=1.1547$ .

Step 3. Apply the mixing rules at $\beta=0.5$ : $\bar E_y=\gamma(E-vB)=1.1547\,(3\times10^6-1.5\times10^8\times0.02)=1.1547\times0=\mathbf{0}$ ✓. And $\bar B_z=\gamma\left(B-vE/c^2\right)=1.1547\,(0.02-0.005)=\mathbf{1.732\times10^{-2}}$ T.

Step 4. Verify against the invariant: $\sqrt{B^2-E^2/c^2}=\sqrt{4\times10^{-4}-1\times10^{-4}}=\sqrt{3}\times10^{-2}=1.732\times10^{-2}$ T ✓ exactly; and the boosted-frame invariant $0-c^2\bar B^2=-2.70\times10^{13}$ V²/m² ✓ matches Step 1. In the drift frame a charged particle sees pure magnetic field: it circles. Read back in the lab, that circle is the cycloid of $\mathbf{E}\times\mathbf{B}$ drift — the guiding-center motion of plasma physics is the purifying boost seen from the wrong chair. Had the fields been $\mathbf{E}=3\times10^6$ V/m, $\mathbf{B}=5\times10^{-3}$ T instead ( $cB=1.5\times10^6<E$ ), the mirror boost $v=c^2B/E=0.5c$ would kill $\mathbf{B}$ , leaving $\bar E=\sqrt{E^2-c^2B^2}=2.598\times10^6$ V/m.

The lesson: the invariants classify before any boost is computed, select the purifying velocity, and hand over the answer's magnitude as a checksum — three services from two numbers.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §12.3.5 (Invariants of the field) · Jackson, Classical Electrodynamics, 3rd ed., §11.9 · Zangwill, Modern Electrodynamics, §22.4 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 5 · Landau & Lifshitz, Classical Theory of Fields, §25

Provenance. Field-tensor components and conventions follow Griffiths §12.3.5 and Jackson Ch. 11; sign convention $\eta=\mathrm{diag}(+,-,-,-)$ . A verifier should recompute $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)$ and $F_{\mu\nu}\tilde F^{\mu\nu}=-4\mathbf{E}\cdot\mathbf{B}/c$ by explicit summation, confirm the boost-invariance derivation via the field-transformation rules (Griffiths Eq. 12.108), and verify the purifying-boost magnitudes against Landau-Lifshitz §25. Numerical/algebraic factors depend on the SI factors of $c$ ; the dimensionless content (signs, classification) is convention-independent. Addendum for this edition: all contractions recomputed symbolically from the potential-pinned components $F_{0i}=+E_i/c$ , $F_{ij}=-\varepsilon_{ijk}B_k$ shared with The Dual Field Tensor (whose Derivation A supplies the dual's components consumed here); the quartic-trace reduction (3) verified symbolically; the direct boost check and both purifying-drift constructions verified numerically ( $\beta=0.8$ capacitor audit reproducing $1.000\times10^{10}$ V²/m² in both frames; $E=3\times10^6$ V/m, $B=0.02$ T chamber giving $v=0.5c$ , $\bar B=17.32$ mT against $\sqrt{B^2-E^2/c^2}$ exactly). Division of labor: the mixing rules themselves belong to Transformation of Electric and Magnetic Fields (this article consumes them); the plane-wave triad being certified is owned by Monochromatic Plane Waves in Vacuum; the Lagrangian built from $I_1$ is developed in Relativistic Lagrangian and Action for Charged Particles; the stress-energy tensor holding the frame-dependent energy density is The Electromagnetic Stress-Energy Tensor's.

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

If the fields mix under boosts, how can any combination of them possibly be frame-independent?

By the same mechanism that makes the interval $c^2t^2-x^2$ invariant while $t$ and $x$ separately are not: the mixing is not arbitrary but metric-preserving. A boost reshuffles $(\mathbf{E}/c,\mathbf{B})$ exactly as it reshuffles $(ct,x)$ — with hyperbolic weights satisfying $\gamma^2(1-\beta^2)=1$ — and the quadratic forms built with the metric's sign pattern feel only that combination. Derivation A's Step 4 shows it concretely: the dangerous cross terms cancel in pairs, and the leftover prefactor is exactly $\gamma^2(1-\beta^2)$ . The tensor formalism packages the same fact more efficiently — contractions are scalars because $\Lambda^{\mathsf T}\eta\Lambda=\eta$ — but nothing is hidden: you can verify both invariants with high-school algebra and the six mixing rules.

Why is the energy density not an invariant when it looks so similar to the first invariant?

One sign — and the sign is the whole story. The invariant is $E^2-c^2B^2$ (a difference, metric-flavored); the energy density is $\tfrac{\varepsilon_0}{2}(E^2+c^2B^2)$ (a sum, Euclidean-flavored). Sums of squares are what rotations preserve; differences are what boosts preserve. Physically, energy density cannot be invariant: energy is the time component of the four-momentum, so field energy must transform — a moving observer sees the capacitor's stored energy changed, along with a nonzero momentum density where the lab sees none. Those pieces organize into the stress-energy tensor $T^{\mu\nu}$ (built in The Electromagnetic Stress-Energy Tensor), whose components are honestly frame-dependent. The invariants are the only pieces of quadratic field information not caught up in that transformation.

Given the two invariants, can I reconstruct the field up to a Lorentz transformation?

Generically yes — that is what "classification" means. For $(I_1,I_2)$ generic, any two field configurations with the same pair are related by some proper Lorentz transformation (plus a rotation): the standard form can be taken as parallel $\mathbf{E}$ and $\mathbf{B}$ along a common axis, with the two magnitudes solving $2(B^2-E^2/c^2)=I_1$ and $-4EB/c=\pm I_2$ . The degenerate cases carry small print: for null fields ( $I_1=I_2=0$ ) the invariants do not distinguish a plane wave's amplitude — a strong beam and a weak beam have the same (vanishing) invariants and are not boost-related; the overall scale is carried by the frame-dependent energy, not by any invariant. So the honest statement: the invariants classify the field's type completely, and its full configuration up to Lorentz transformations except on the null stratum, where an additional scale (and propagation direction) must be supplied.

Does E·B being invariant mean it is conserved in time?

No — the two words answer different questions, and conflating them is a classic error. Invariant compares observers at one spacetime event: every frame computes the same $\mathbf{E}\cdot\mathbf{B}$ there, then. Conserved compares times for one observer: some integral stays constant as the system evolves. A single event has no dynamics; a field can have wildly time-varying invariants (switch on a capacitor: $E^2-c^2B^2$ at a fixed point climbs from zero). There is a related conservation-flavored fact — the spacetime integral of $I_2$ is a topological term, insensitive to local deformations, which is why it drops out of classical equations of motion — but that is a statement about the action, not about a conserved charge. When this chapter needs genuine conservation laws, it builds them from the stress-energy tensor, not from the invariants.

The classification says a frame with parallel E and B exists whenever E·B is nonzero. What is that frame good for?

It is the field's "normal form" — the relativistic analogue of diagonalizing a matrix. In the parallel frame the field configuration is as simple as it can be made: one common axis, two magnitudes, and those two magnitudes are exactly the solutions of the invariant equations (a convenient way to compute the standard form from any frame's data). Theoretical work uses it constantly: strong-field QED rates (pair production in combined fields) are derived in the parallel frame where the algebra collapses, then exported as functions of $I_1$ and $I_2$ ; the classification of charged-particle orbits in constant fields (done covariantly in Relativistic Dynamics of a Charged Particle) splits into cases along exactly these lines. It plays the role for constant fields that the rest frame plays for massive particles — and the null field's refusal to have one is, again, the light-has-no-rest-frame fact in field clothing.

Why do only quadratic invariants appear — could a linear or cubic invariant of F exist?

Linear: the only candidate is the trace $F^\mu{}_\mu$ , which vanishes identically by antisymmetry — no linear invariant. Cubic: the trace of three $F$ 's vanishes too (an odd product of antisymmetric matrices has zero trace; equivalently, transposing flips its sign). Quartic and beyond: nonzero, but never new — equation (3) shows the quartic trace is $\tfrac12I_1^2+\tfrac14I_2^2$ , and the Cayley–Hamilton theorem for $4\times4$ matrices guarantees every higher trace is a polynomial in the traces up to fourth order, hence in $I_1$ and $I_2$ . Adding the dual to the mix helps nothing: $G\!\cdot\!G=-I_1$ and mixed higher contractions reduce the same way. Two is not an accident of low order; it is the complete answer, which is what licenses the strong claim that any Lorentz-invariant electrodynamics has $\mathcal{L}=\mathcal{L}(I_1,I_2)$ .