The Electromagnetic Field Tensor

Derivation

The Electromagnetic Field Tensor

D-104 Home PU-105 Threads fields · light · symmetry Depends on Lorentz Transformation from the Two Postulates, maxwell-equations-in-vacuum

Statement

Starting from the four-potential A^μ and the four-gradient, the antisymmetric rank-2 tensor F_μν = ∂_μA_ν − ∂_νA_μ is constructed. Its components are shown to be exactly the Cartesian components of E (scaled by 1/c) and B, and its behaviour under a Lorentz boost reproduces the field-mixing transformation laws — establishing that the electric and magnetic fields are the six independent components of a single geometric object, not two separate vector fields.

Why it matters

In pre-relativistic physics E and B are distinct three-vectors governed by four separate Maxwell equations. Once the Lorentz transformation is accepted, a field that is purely electric in one frame acquires a magnetic part in another: the split into "electric" and "magnetic" is frame-dependent. The field tensor is the frame-independent object that both frames agree on; only its components are frame-dependent, exactly as the components of a spatial vector rotate while the vector itself does not.

Collecting the fields into F_μν also compresses Maxwell's four equations into two tensor equations and makes the two field invariants (B² − E²/c² and E·B) manifest, which controls what is physically possible under boosts — for example whether a frame exists in which the field is purely electric.

Assumptions

Special relativity holds: spacetime is flat Minkowski space with metric η_μν = diag(+1,−1,−1,−1).If dropped for a curved metric, partial derivatives must become covariant derivatives; however F_μν survives unchanged because the Christoffel terms cancel by antisymmetry — the construction generalises, but the flat-space component identities below do not.

The fields derive from potentials: E = −∇φ − ∂A/∂t and B = ∇×A.If dropped, F_μν cannot be written as a curl of A_μ; this is equivalent to admitting magnetic monopoles (∇·B ≠ 0), and the homogeneous Maxwell equations would no longer be identities.

The four-potential A^μ = (φ/c, A) transforms as a genuine four-vector.If dropped, the object built from ∂_μA_ν would not transform as a tensor and the boost law derived at the end would be meaningless. This assumption is itself justified by demanding that the wave equation for A^μ (in Lorenz gauge) be Lorentz-covariant.

Gauge freedom A_μ → A_μ + ∂_μχ leaves physics unchanged.If dropped, the antisymmetric combination would not be the natural object — but it is precisely gauge invariance that selects the antisymmetric curl, since ∂_μ∂_νχ − ∂_ν∂_μχ = 0.

Derivation

A^μ = (φ/c, A_x, A_y, A_z) , A_μ = η_μνA^ν = (φ/c, −A_x, −A_y, −A_z)

Definition of the four-potential; the factor 1/c makes the time component share the units of the spatial components. Lowering with η flips the sign of the spatial part. A

E = −∇φ − ∂A/∂t , B = ∇×A

Assumed prior result (potentials of the vacuum Maxwell equations). These are the identities the tensor must reproduce. A

∂_μ = ∂/∂x^μ = ( (1/c)∂_t , ∂_x , ∂_y , ∂_z )

The four-gradient with a lower index. With x⁰ = ct, the time slot carries 1/c; it transforms as a covariant four-vector. B

F_μν ≡ ∂_μA_ν − ∂_νA_μ

The unique gauge-invariant rank-2 object linear in first derivatives of A: under A_μ→A_μ+∂_μχ the extra term ∂_μ∂_νχ−∂_ν∂_μχ vanishes. Antisymmetry F_μν = −F_νμ is built in, leaving 6 independent components. C

F_0i = ∂₀A_i − ∂_iA₀ = (1/c)∂_t(−A_i) − ∂_i(φ/c) = −(1/c)[ ∂_tA_i + ∂_iφ ] = E_i/c

Insert the lowered potential from step 1 and the gradient from step 3, then recognise the bracket as −E_i from step 2. The time-space block of the tensor is the electric field. B

F_ij = ∂_iA_j − ∂_jA_i = −(∂_iA_j^(sp) − ∂_jA_i^(sp)) = −ε_ijkB_k

The lowered spatial components carry a minus sign (step 1); the surviving antisymmetric spatial derivative of A is exactly the curl, i.e. B. So the space-space block of the tensor is the magnetic field. B

F_μν = [ 0 , E_x/c , E_y/c , E_z/c ]
[ −E_x/c , 0 , −B_z , B_y ]
[ −E_y/c , B_z , 0 , −B_x ]
[ −E_z/c , −B_y , B_x , 0 ]

Collect steps 5–6 (μ,ν = 0,1,2,3) into a single antisymmetric matrix. The six independent entries are precisely the three components of E and the three of B. A

F^μν = η^μαη^νβF_αβ ⇒ F⁰ⁱ = −E_i/c , F^ij = −ε^ijkB_k

Raising both indices flips the sign of every entry that has exactly one time index (the electric block); the purely spatial block is unchanged (two sign flips). The contravariant form differs from step 7 only in the sign of E. B

F′^μν = Λ^μ_α Λ^ν_β F^αβ

A rank-2 tensor transforms with one Lorentz matrix Λ (from lorentz-transformation-from-postulates) per index. Applying this to a boost along x and reading off components yields the field-mixing laws below — the transformation is a rotation in the 4-dimensional space of the tensor's components, not of E and B separately. C

boost +v along x: Λ⁰₀=Λ¹₁=γ , Λ⁰₁=Λ¹₀=−γβ , β=v/c
F′⁰² = Λ⁰₀F⁰² + Λ⁰₁F¹² = γ(−E_y/c) − γβ(−B_z) = −(1/c)γ(E_y − vB_z)

A representative contraction: since F′⁰² = −E′_y/c, this gives E′_y = γ(E_y − vB_z). Repeating for every pair (μ,ν) produces the complete set in the Result box. C

Result

F^μν = ∂^μA^ν − ∂^νA^μ, with F⁰ⁱ = −E_i/c , F^ij = −ε^ijkB_k

Boost along x: E′_x = E_x , B′_x = B_x
E′_y = γ(E_y − vB_z) , E′_z = γ(E_z + vB_y)
B′_y = γ(B_y + (v/c²)E_z) , B′_z = γ(B_z − (v/c²)E_y)

Reading. The six numbers we call E and B are the components of one antisymmetric spacetime tensor. Components parallel to the boost are untouched; components transverse to it get scaled by γ and rotated into each other, so a boost turns electric field into magnetic field and vice versa. There is no absolute distinction between the two — only the tensor is frame-independent.

Units check. Every entry of F^μν must share one unit. The spatial block is B in tesla (T). The time-space block is E/c with units (V·m⁻¹)/(m·s⁻¹) = V·s·m⁻² = T. Both blocks are teslas, so the tensor is dimensionally homogeneous. In the boost laws, vB has units (m·s⁻¹)(T) = V·m⁻¹ matching E, and (v/c²)E has units (s·m⁻¹)(V·m⁻¹) = T matching B. ✓

Limiting cases

v → 0 (γ → 1): E′ = E, B′ = B — the fields do not mix; the Galilean world where electricity and magnetism are separate is recovered.
Non-relativistic boost (β ≪ 1, γ ≈ 1): E′_y ≈ E_y − vB_z, i.e. a charge moving through B feels the magnetic force as an effective electric field in its rest frame — this is the relativistic origin of the motional EMF and of the Lorentz force's v×B term.
Pure magnetic field, B along boost axis: B′_x = B_x, E′ = 0 — a longitudinal magnetostatic field is boost-invariant.
Pure electric field, boosted transversely: a magnetic field of order vE/c² appears — the "magnetism as a relativistic effect" seen from the frame of a moving observer.
c → ∞ formally: the (v/c²) terms vanish and E and B decouple, but E′_y = E_y − vB_z survives — reproducing the inconsistent Galilean electromagnetism whose failure motivated relativity.

Breaks when

Magnetic monopoles exist. The construction assumes B = ∇×A, forcing ∇·B = 0. If monopoles are present, no global four-potential exists, F_μν is no longer an exact curl, and the homogeneous Maxwell equations (∂_[αF_μν] = 0) fail as identities.
Curved spacetime with strong gravity. Partial derivatives must be promoted to covariant derivatives. F_μν itself is unchanged (Christoffel terms cancel by antisymmetry), but the flat-space component identities and the constant-Λ boost law no longer hold — the transformation becomes position-dependent.
Media with polarisation/magnetisation. In matter one must distinguish the field tensor F^μν (built from E, B) from the excitation tensor G^μν (built from D, H). Treating a single tensor as both, with the vacuum relation, breaks inside a dielectric or magnetic medium.
Non-inertial frames / acceleration. The single global Lorentz matrix Λ of step 9 assumes an inertial-to-inertial boost. For an accelerating observer the "boost" varies along the worldline and the simple mixing law is only instantaneously valid.

Failure modes

Dropping the 1/c. Writing F_0i = E_i instead of E_i/c (i.e. forgetting A⁰ = φ/c) makes the tensor dimensionally inhomogeneous and every boost law comes out wrong by a factor of c.
Sign confusion between F_μν and F^μν. Raising indices flips the sign of the electric block only. Students who use the covariant matrix in a formula that expects the contravariant one get the wrong sign of E in the Lorentz force or the boost.
Boosting E and B as independent 3-vectors. Applying a naive length-contraction/γ factor to each field separately misses the cross terms (−vB_z, +(v/c²)E_z) — the whole point is that they mix.
Rotating parallel and transverse components the same way. Only transverse components scale by γ; applying γ to E_x (the boost-axis component) is a common slip.
Assuming the invariants are frame-dependent. Computing B² − E²/c² in one frame and expecting a different number in another; it is a scalar and must be identical.
Metric-signature drift. Silently switching between (+−−−) and (−+++) mid-calculation flips signs in the raising step and corrupts the final matrix.

Discussion

The deepest content of this derivation is that magnetism is a relativistic corollary of electrostatics. Given Coulomb's law and the requirement that physics respect the Lorentz transformation, the magnetic field is not an independent postulate — it is what the electric field of a moving charge looks like from another frame. The boost laws show this quantitatively: start with a pure Coulomb field (all B = 0) and any transverse boost manufactures a magnetic field of order vE/c². Historically Maxwell's equations were found first and relativity extracted from them; logically the arrow runs the other way.

Because F_μν is a tensor, scalars built from it are frame-invariant. There are exactly two independent Lorentz scalars: F_μνF^μν = 2(B² − E²/c²) and ε^μναβF_μνF_αβ ∝ E·B. Their signs are physically decisive. If B² − E²/c² < 0 and E·B = 0, a frame exists where the field is purely electric; if > 0 with E·B = 0, a frame exists where it is purely magnetic. If E·B ≠ 0, then E and B are non-orthogonal in every frame — no boost can eliminate either. A light wave has E ⊥ B and E = cB, so both invariants vanish: a null field that looks like radiation in every inertial frame, which is why light cannot be boosted to rest.

The tensor also compresses Maxwell's equations. The inhomogeneous pair (Gauss + Ampère) becomes ∂_μF^μν = μ₀J^ν, and the homogeneous pair (no monopoles + Faraday) becomes the identity ∂_λF_μν + ∂_μF_νλ + ∂_νF_λμ = 0, which holds automatically because F is a curl of A. Four vector equations collapse to two tensor equations, and Lorentz covariance is manifest rather than hidden.

Geometrically, F is a 2-form, F = dA, and the homogeneous equations are just dF = d²A = 0 — the statement that the exterior derivative squares to zero. The gauge freedom A → A + dχ is the freedom to add an exact form, and physical states are cohomology classes. This viewpoint makes F_μν the prototype for Yang–Mills field strengths, where A becomes a Lie-algebra-valued connection and F = dA + A∧A acquires a self-interaction term. The abelian electromagnetic case is the simplest curvature of a gauge connection.

Common misconceptions. (i) "The field tensor is just bookkeeping." It is not — the fact that six numbers pack into one tensor is a physical claim (they mix under boosts) that would be false if E and B were truly independent. (ii) "A boost can always remove the magnetic field." Only when E·B = 0 and the field is electric-dominated; otherwise never. (iii) "E/c is a made-up factor." It is forced by requiring the time component of a four-vector to share units with the spatial components, i.e. by the geometry of spacetime itself.

Worked examples

Parallel-plate field seen by a moving observer. A capacitor produces a static field E_y = 1.0×10⁵ V/m, B = 0, in the lab frame. An observer moves along x with v = 0.60c. Find E′ and B′.

Boost is transverse to E_y; use the transverse laws. Symbols first, numbers after. A

γ = 1/√(1 − β²) = 1/√(1 − 0.36) = 1/√0.64 = 1/0.80 = 1.25

Lorentz factor at β = 0.60. A

E′_y = γ(E_y − vB_z) = γE_y = 1.25 × 1.0×10⁵ = 1.25×10⁵ V/m

B_z = 0 so only the γ-scaling survives. A

B′_z = γ(B_z − (v/c²)E_y) = −γvE_y/c²
= −1.25 × (1.8×10⁸) × (1.0×10⁵) / (9.0×10¹⁶)

v = 0.60c = 1.8×10⁸ m/s; c² = 9.0×10¹⁶ m²/s². B

E′_y = 1.25×10⁵ V/m , B′_z = −2.5×10⁻⁴ T (all other components 0)

Reading. A purely electric field in the lab acquires a real magnetic field in the moving frame. Check the invariant: B² − E²/c² is (0 − (10⁵/3×10⁸)²) = −1.11×10⁻⁷ in the lab and ((2.5×10⁻⁴)² − (1.25×10⁵/3×10⁸)²) = 6.25×10⁻⁸ − 1.74×10⁻⁷ = −1.11×10⁻⁷ in the moving frame — identical, as required.

Units check. vE/c² = (m/s)(V/m)/(m²/s²) = V·s/m² = T. ✓

Can this field be made purely electric? In a lab frame, E = 3.0×10⁸ V/m along y and B = 0.50 T along z (so E ⊥ B). Determine whether a frame exists with B = 0, and find its velocity.

Test the two invariants, then use the drift-velocity condition. B

E·B = 0 (perpendicular) ✓ ; B² − E²/c² = (0.50)² − (3.0×10⁸/3.0×10⁸)² = 0.25 − 1.0 = −0.75 T² < 0

E·B = 0 permits eliminating one field; the negative sign of B² − E²/c² marks the field as electric-dominated, so B can be removed (not E). C

Boost along x (⊥ both fields). Set B′_z = 0: γ(B_z − (v/c²)E_y) = 0 ⇒ v = c²B_z/E_y

Solve the transverse B′ law for the boost that cancels the magnetic component. Symbols before numbers. B

v = (9.0×10¹⁶)(0.50)/(3.0×10⁸) = 1.5×10⁸ m/s = 0.50c

Well below c, so a physical frame exists (consistent with electric dominance). A

γ = 1/√(1 − 0.25) = 1/√0.75 = 1.1547
E′_y = γ(E_y − vB_z) = 1.1547 × (3.0×10⁸ − 1.5×10⁸×0.50) = 1.1547 × 2.25×10⁸ = 2.60×10⁸ V/m

Evaluate the surviving electric field in the new frame. B

Yes: boost x with v = 0.50c gives B′ = 0 , E′_y = 2.60×10⁸ V/m

Reading. Because the field is electric-dominated and E ⊥ B, a "drift" frame exists in which the magnetic field is completely transformed away, leaving a pure electric field. (Had B² − E²/c² been positive, only a pure-magnetic frame would exist.) Verify: B² − E²/c² in the new frame = 0 − (2.60×10⁸/3.0×10⁸)² = −0.75 T², matching the lab value. ✓

Units check. c²B/E = (m²/s²)(T)/(V/m) = (m²/s²)(V·s/m²)/(V/m) = m/s. ✓

Problems

Write out the four non-trivial components of F^μν along the first row (μ = 0) for a field E = (E_x, E_y, E_z), and state their common SI unit.
Solution
The μ = 0 row of F^μν is (0, E_x/c·(−1), …). Using F⁰ⁱ = −E_i/c: F⁰⁰ = 0, F⁰¹ = −E_x/c, F⁰² = −E_y/c, F⁰³ = −E_z/c. Each has units (V/m)/(m/s) = V·s/m² = tesla (T), the same unit as the magnetic entries. Note the sign is opposite to the covariant F_0i = +E_i/c because raising the single time index flips the sign.
A field is purely magnetic in the lab: B = 0.20 T along z, E = 0. An observer boosts along x with v = 0.80c. Find E′ and B′.
Solution
γ = 1/√(1−0.64) = 1/√0.36 = 1/0.60 = 1.667. Transverse laws with E = 0: B′_z = γB_z = 1.667×0.20 = 0.333 T. E′_y = γ(E_y − vB_z) = −γvB_z = −1.667 × (0.80×3.0×10⁸) × 0.20 = −1.667 × 2.4×10⁸ × 0.20 = −8.0×10⁷ V/m. All other components zero. Result: B′_z = 0.333 T, E′_y = −8.0×10⁷ V/m. Invariant check: B²−E²/c² lab = 0.04; new = 0.333² − (8.0×10⁷/3.0×10⁸)² = 0.111 − 0.0711 = 0.040 ✓ (positive: magnetic-dominated, cannot be transformed to pure E).
Show that F_μνF^μν = 2(B² − E²/c²) by summing the squares of the tensor components.
Solution
F_μνF^μν = Σ_μν F_μνF^μν. Time-space terms: there are 3 pairs (0i) and their transposes (i0), 6 terms total. F_0i = E_i/c, F⁰ⁱ = −E_i/c, so each product F_0iF⁰ⁱ = −E_i²/c²; summing the 6 (0i)+(i0) terms gives 2×3 averaged… precisely 2·Σ_i(−E_i²/c²) = −2E²/c². Space-space terms: F_ij = F^ij = −ε_ijkB_k; the 6 off-diagonal ij terms give Σ_ij(ε_ijkB_k)² = 2Σ_kB_k² = 2B². Total: −2E²/c² + 2B² = 2(B² − E²/c²). Since it is a contraction of tensors, it is a Lorentz scalar — the same in every frame.
An electromagnetic plane wave has |E| = c|B| with E ⊥ B. Compute both field invariants and explain why no boost can bring the wave to rest.
Solution
Invariant 1: B² − E²/c² = B² − (cB)²/c² = B² − B² = 0. Invariant 2 ∝ E·B = 0 since E ⊥ B. Both invariants vanish — this is a "null" electromagnetic field. Under any boost the transformed fields still satisfy |E′| = c|B′| and E′ ⊥ B′ (the invariants are preserved and both zero), so the field is a wave in every inertial frame; there is no frame in which E and B could be made static or in which one vanishes while the other survives. Physically this is the field-tensor statement that light travels at c in all frames — it cannot be boosted to rest.
In a region E = (0, 4.0×10⁵, 0) V/m and B = (0, 0, 1.0×10⁻³) T. Is there a frame where the field is purely electric? If so, give the required boost velocity (along x).
Solution
Check invariants. E·B = 0 (E along y, B along z) ✓. B² − E²/c² = (1.0×10⁻³)² − (4.0×10⁵/3.0×10⁸)² = 1.0×10⁻⁶ − (1.333×10⁻³)² = 1.0×10⁻⁶ − 1.78×10⁻⁶ = −7.8×10⁻⁷ < 0 → electric-dominated, so a pure-electric frame exists. Cancel B′_z: v = c²B_z/E_y = (9.0×10¹⁶)(1.0×10⁻³)/(4.0×10⁵) = 9.0×10¹³/4.0×10⁵ = 2.25×10⁸ m/s = 0.75c. Since v < c the frame is physical. In it, γ = 1/√(1−0.5625) = 1/√0.4375 = 1.512, and E′_y = γ(E_y − vB_z) = 1.512×(4.0×10⁵ − 2.25×10⁸×1.0×10⁻³) = 1.512×(4.0×10⁵ − 2.25×10⁵) = 1.512×1.75×10⁵ = 2.65×10⁵ V/m, with B′ = 0.