Transformation of Electric and Magnetic Fields

Derivation

Transformation of Electric and Magnetic Fields

D-105 Home PU-105 Threads fields · symmetry Depends on The Electromagnetic Field Tensor

Statement

For a boost of speed v along the common x-axis relating inertial frames S and S′ (with β = v/c, γ = (1 − β²)^−1/2), the electric and magnetic fields transform by mixing their transverse components. In particular a field that is purely electric in S is measured in S′ as an electric field and a magnetic field, with B′_z = −γ v E_y/c².

Why it matters

Electric and magnetic fields are not separate objects but two faces of one antisymmetric tensor. What one observer calls "the electric field of a static charge" another, moving past, calls a combination of electric and magnetic field. The magnetic force on a moving charge is, in this sense, electrostatics viewed from a moving frame — magnetism is a relativistic corollary of Coulomb's law.

Practically, this transformation is the working tool for relativistic beam physics, synchrotron radiation, motional EMF, and any lab where charges move at appreciable β. It also fixes the sign and size of induced fields without invoking a separate "law", since it follows directly from how F^μν rotates under a boost.

Assumptions

The fields form the antisymmetric tensor F^μν.If E and B were independent Lorentz scalars or three-vectors, they could not mix, and the derivation collapses; the tensor structure is what forces the mixing. The boost is a single-parameter Lorentz transformation along x.Drop linearity/constancy of v and the transformation is no longer a global Lorentz matrix; accelerating frames need the covariant derivative and pick up inertial-force terms, changing the result. Flat spacetime, no gravitation.In curved spacetime the split into "the boost matrix Λ" is only local; tidal terms and connection coefficients enter, so the tidy component formulas hold only in a local Lorentz frame. SI conventions with metric signature (+−−−) and the standard F^μν component map.A different signature or a Gaussian normalisation shifts factors of c; the physics is identical but the literal formulas below change by those factors.

Derivation

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 ]

Adopt the field tensor from the prior result electromagnetic-field-tensor; rows/columns indexed μ,ν = 0,1,2,3 = t,x,y,z. This is a definition, not a step to prove. A

Λ^μ_ν: Λ⁰₀ = Λ¹₁ = γ, Λ⁰₁ = Λ¹₀ = −γβ, Λ²₂ = Λ³₃ = 1

The boost carrying S → S′ (frame S′ moving at +v along x). Transverse axes y,z are untouched; all other entries vanish. A

F′^μν = Λ^μ_α Λ^ν_β F^αβ

F is a rank-2 contravariant tensor, so each index transforms with its own copy of Λ. This single rule contains the entire field transformation. B

F′⁰² = Λ⁰_α Λ²_β F^αβ = Λ⁰_α F^α2 = γ F⁰² − γβ F¹²

Since Λ²_β = δ²_β, only β = 2 survives; the α-sum keeps α = 0,1 because Λ⁰_α is nonzero only there. Pure index bookkeeping. B

−E′_y/c = γ(−E_y/c) − γβ(−B_z) ⇒ E′_y = γ(E_y − v B_z)

Insert F⁰² = −E_y/c, F¹² = −B_z and multiply through by −c, using βc = v. Symbols first; numbers never enter here. B

F′¹² = Λ¹_α F^α2 = γ F¹² − γβ F⁰² ⇒ −B′_z = γ(−B_z) − γβ(−E_y/c)

Same contraction, now with μ = 1: Λ¹_α is nonzero for α = 0,1. This is the partner component that carries the magnetic response. B

B′_z = γ(B_z − v E_y/c²)

Multiply step 6 by −1 and use β/c = v/c². The E_y term is the crucial one: a transverse E feeds a transverse B. B

By the same contraction on the remaining components:
E′_x = E_x, E′_z = γ(E_z + v B_y), B′_x = B_x, B′_y = γ(B_y + v E_z/c²)

Repeat steps 4–7 for μν = 03, 13, 01. The longitudinal (along-x) components are fixed points of the boost; only transverse pairs rotate into each other. C

E² − c²B² and E·B are invariant under Λ

Contracting F_μνF^μν and F_μν(*F)^μν gives frame-independent scalars; one checks the component formulas above preserve them. This certifies the algebra and constrains what field configurations can occur. C

Result

E′_∥ = E_∥, B′_∥ = B_∥; E′_⊥ = γ(E_⊥ + v × B)_⊥, B′_⊥ = γ(B_⊥ − v × E/c²)_⊥

Reading. Components along the boost are unchanged; components perpendicular to it are scaled by γ and mixed. Set B = 0 (pure electric field E_y in S): then E′_y = γ E_y and B′_z = −γ v E_y/c² ≠ 0. The magnetic field appears purely from motion — there was never a current in S.

Units check. B′_z = γ v E_y/c² has units (m s⁻¹)(V m⁻¹)/(m² s⁻²) = V s m⁻² = T. Correct, since 1 T = 1 V s m⁻². And v B has units (m s⁻¹)(T) = V m⁻¹, matching E.

Limiting cases

Low speed (β → 0, γ → 1): E′_y → E_y − v B_z and B′_z → B_z − v E_y/c² — the Galilean/motional-EMF forms, first order in v.
Ultra-relativistic (γ » 1): transverse fields blow up as γ, and the ratio E′_⊥/cB′_⊥ → 1, i.e. the boosted field of a charge collapses toward a "pancake" that looks locally like a plane wave.
Longitudinal fields (E, B along x): completely unchanged, E′_x = E_x, B′_x = B_x.
Plane wave (E = cB, perpendicular): both invariants vanish, so no boost can transform the wave away — a null field stays null in every frame.

Breaks when

Non-inertial frames. For a rotating or accelerating observer the boost parameter varies in space or time, Λ is no longer a constant global matrix, and the component formulas acquire position-dependent and inertial-force corrections.
Curved spacetime / strong gravity. The transformation holds only in a local Lorentz frame; over finite regions tidal effects and the metric's connection modify how fields at different events relate.
Media with a rest frame. Inside a polarizable/magnetizable medium the auxiliary fields D, H and the medium's own four-velocity enter; the vacuum mixing of E, B above no longer captures the measured response (Minkowski/Abraham constitutive relations are needed).

Failure modes

Boosting all three components with γ. The longitudinal component is not scaled; only transverse components carry γ. Multiplying E_x by γ is the most common slip.
Sign of the v× term. Transforming to a frame moving at +v uses the boost with Λ⁰₁ = −γβ; flipping it gives B′_z with the wrong sign, reversing the induced field.
Dropping c factors. Writing B′_z = γ(B_z − v E_y) (missing 1/c²) — a units-check catches it instantly.
Confusing "transforms as a vector". E alone is not a Lorentz three-vector; treating it like the spatial part of a four-vector predicts no E–B mixing and is simply wrong.
Using the field of the wrong charge state. Boosting the static Coulomb field and forgetting that in S′ the source is now a moving charge (so its own field is already the boosted one) — double counting.

Discussion

The single most important lesson is unification: there is one field, the tensor F^μν, and "electric" versus "magnetic" is a frame-dependent slicing of it, exactly as "space" and "time" are frame-dependent slices of spacetime. The transformation rules are not extra postulates bolted onto Maxwell's equations; they are forced the moment you accept that E and B sit in an antisymmetric rank-2 tensor.

This gives the cleanest account of magnetism's origin. A current-carrying wire is neutral in the lab, but a test charge drifting alongside it sees, in its own frame, length-contracted charge densities that no longer cancel — a net electric field. Transform back and that electric force is precisely the magnetic force qv×B. Magnetism is what electrostatics looks like after a boost; the factor v/c² in B′ is the fingerprint of that relativistic origin.

The two invariants E² − c²B² and E·B classify fields into boost-equivalence classes. If E·B = 0 and E > cB there exists a frame with pure E; if E < cB, a frame with pure B. If both invariants vanish (a null field) no frame removes either — this is the algebraic reason light cannot be "boosted away".

Geometrically, a boost acts on F as a hyperbolic rotation in the t–x plane, so the components pair up like the real and imaginary parts of the complex Riemann–Silberstein vector F = E + icB, which transforms by a complex rotation through the rapidity. The invariant F·F = (E² − c²B²) + 2ic E·B packages both scalars at once, and the "boost = complex rotation" picture makes the electric–magnetic duality of vacuum electromagnetism manifest.

Common misconceptions. Students often think the boosted charge "creates" a magnetic field the way a current does; rather, the same field is being re-described in a new frame. Another trap: believing E and B always both grow under a boost — in fact a pure E along the boost direction is entirely unchanged, and a suitable boost can even reduce a transverse field's magnitude relative to its partner.

Worked examples

Pure electric field acquires a magnetic component. In S: E_y = 1.0×10⁶ V m⁻¹, B = 0. Frame S′ moves at v = 0.60c along x. Find E′_y and B′_z.

Set-up: a parallel-plate field seen from a passing frame. A

γ = (1 − β²)^−1/2 = (1 − 0.36)^−1/2 = (0.64)^−1/2 = 1.25

Symbol then number; β = 0.60. A

E′_y = γ(E_y − v B_z) = γ E_y = 1.25 × 1.0×10⁶ = 1.25×10⁶ V m⁻¹

With B_z = 0 the transverse E is just γ-enhanced. A

B′_z = γ(B_z − v E_y/c²) = −γ v E_y/c²; v = 0.60(3.0×10⁸) = 1.8×10⁸ m s⁻¹

Substitute numbers only after the symbolic form is fixed. B

B′_z = −1.25 × (1.8×10⁸)(1.0×10⁶)/(9.0×10¹⁶) = −1.25 × 2.0×10⁻³

Arithmetic: vE_y/c² = 1.8×10¹⁴/9.0×10¹⁶ = 2.0×10⁻³ T. A

E′_y = 1.25×10⁶ V m⁻¹, B′_z = −2.5×10⁻³ T (−2.5 mT)

Reading. A field that was purely electric now has a real, measurable magnetic component of a few milliteslas — produced by nothing but a change of observer.

Units check. V m⁻¹ for E, and (m s⁻¹)(V m⁻¹)/(m² s⁻²) = V s m⁻² = T for B. Consistent.

Pure magnetic field acquires an electric component (the dual). In S: B_z = 0.50 T, E = 0. Boost v = 0.80c along x. Find E′_y and B′_z.

Set-up: the field of a magnet viewed from a fast frame. A

γ = (1 − 0.64)^−1/2 = (0.36)^−1/2 = 1/0.60 = 1.667

β = 0.80. A

E′_y = γ(E_y − v B_z) = −γ v B_z; v = 0.80(3.0×10⁸) = 2.4×10⁸ m s⁻¹

With E_y = 0, only the −vB_z term remains. B

E′_y = −1.667 × (2.4×10⁸)(0.50) = −1.667 × 1.2×10⁸ = −2.0×10⁸ V m⁻¹

Numbers substituted last. A

B′_z = γ(B_z − v E_y/c²) = γ B_z = 1.667 × 0.50 = 0.83 T

E_y = 0 leaves the γ-enhanced magnetic field. A

E′_y = −2.0×10⁸ V m⁻¹, B′_z = 0.83 T

Reading. The dual of example 1: a purely magnetic field is seen in the moving frame as an intense electric field plus a stronger magnetic field. This is the field that pushes charges in a moving-magnet dynamo.

Units check. vB = (m s⁻¹)(T) = (m s⁻¹)(V s m⁻²) = V m⁻¹. Correct for E.

Problems

A field in S has E_y = 3.0×10⁵ V m⁻¹, B = 0. For a boost v = 0.50c along x, find E′_y and B′_z.
Solution
γ = (1−0.25)^−1/2 = 1.1547. E′_y = γE_y = 1.1547×3.0×10⁵ = 3.46×10⁵ V m⁻¹. v = 1.5×10⁸ m s⁻¹; B′_z = −γvE_y/c² = −1.1547×(1.5×10⁸)(3.0×10⁵)/(9.0×10¹⁶) = −1.1547×5.0×10⁻⁴ = −5.8×10⁻⁴ T.
Show that the boost with v = 0.60c along x leaves a longitudinal field E_x = 2.0×10⁴ V m⁻¹ unchanged, and explain why in one sentence.
Solution
From the result box, E′_x = E_x = 2.0×10⁴ V m⁻¹ exactly — no γ. The component along the boost lies in the t–x plane's fixed direction for the field tensor: the corresponding tensor component F⁰¹ is antisymmetric and unchanged because Λ⁰_αΛ¹_βF^αβ reduces to (γ²−γ²β²)F⁰¹ = F⁰¹. Physically, motion parallel to E does not length-contract the source distribution transverse to the field.
In S a region has E_y = 2.0×10⁸ V m⁻¹ and B_z = 0.40 T. Find the boost speed v (along x) that makes B′_z = 0.
Solution
Set B′_z = γ(B_z − vE_y/c²) = 0 ⇒ v = c²B_z/E_y = (9.0×10¹⁶)(0.40)/(2.0×10⁸) = 1.8×10⁸ m s⁻¹ = 0.60c. Since E_y = 2.0×10⁸ > cB_z = 1.2×10⁸, the invariant E²−c²B² > 0 confirms a pure-electric frame exists, so a real v < c is guaranteed.
A plane electromagnetic wave in S has E_y = 300 V m⁻¹ and B_z = 1.0×10⁻⁶ T (so E_y = cB_z), propagating along +x. Compute E′_y for a boost v = 0.80c along +x and comment.
Solution
γ = 1.667. E′_y = γ(E_y − vB_z) = 1.667(300 − (2.4×10⁸)(1.0×10⁻⁶)) = 1.667(300 − 240) = 1.667×60 = 100 V m⁻¹. The field is reduced, not enhanced: E′_y = E_y√((1−β)/(1+β)) = 300√(0.2/1.8) = 300/3 = 100 V m⁻¹ — the relativistic Doppler dimming of a wave chased at 0.80c. Note E′_y = cB′_z still holds: a null field stays null.
Using the invariants, decide whether a frame exists in which the field E_y = 1.0×10⁸ V m⁻¹, B_z = 0.50 T is purely magnetic, and if so state the required v (boost along x).
Solution
Invariant E·B: here E ∥ y, B ∥ z, so E·B = 0 — a pure field is possible. Compare magnitudes: cB_z = (3.0×10⁸)(0.50) = 1.5×10⁸ V m⁻¹ > E_y = 1.0×10⁸, so E²−c²B² < 0 — a pure-magnetic frame exists. Set E′_y = γ(E_y−vB_z) = 0 ⇒ v = E_y/B_z = (1.0×10⁸)/0.50 = 2.0×10⁸ m s⁻¹ = 0.667c < c. Valid.