Special relativity · Electrodynamics §6
Fields transform together, not as separate realities.
How do electric and magnetic descriptions change together under a boost, and what does a test charge experience in each frame?
Electrodynamics §6
Electric and magnetic frame change
Static worked example
CurrentThese numbers match the current settings.
Model note
- Primary outputs electricFieldStationary, electricFieldMoving, magneticFieldStationary, magneticFieldMoving: Host calculation (fields.transformSI). Owner fields.transformSI.
- Primary outputs fieldInvariantEDotB, fieldInvariantE2MinusC2B2, fieldInvariantEDotBMoving, fieldInvariantE2MinusC2B2Moving: Host calculation (fields.fieldInvariants). Owner fields.fieldInvariants.
- Primary output lorentzFactor: Host calculation (kinematics.gamma). Owner kinematics.gamma.
- Primary outputs chargeVelocityStationary, chargeVelocityMoving: Host calculation (fields.transformVelocity3D). Owner fields.transformVelocity3D.
- Primary outputs transverseForceLaboratory, transverseForceComoving, electricForceStationary, electricForceMoving, magneticForceStationary, magneticForceMoving: Host calculation (fields.lorentzForce). Owner fields.lorentzForce.
- Primary outputs forceMovingFromFourForce, forceTransformationResidual, particleTimeJacobian: Host calculation (host:three-force-transform-v1). Owner host:three-force-transform-v1.
- Accepted input revision 1.
- Snapshot version 1.
- Not modeled: field sources and currents; radiation and self-fields; radiation reaction; back-reaction on the field; media and polarization; nonuniform fields; accelerated observers.
Whether a field is electric, magnetic or both depends on who describes it. A field that is purely electric for one observer has a magnetic part for an observer moving past, and the two descriptions agree about what a charge in it does.
§6 applies the transformation of §3 to the Maxwell–Hertz equations for empty space and asks what fields must hold in the moving system k if the equations are to keep their form there. For a boost with speed v along x, the components along the motion are unchanged, X′ = X and L′ = L, while the transverse ones mix: Y′ = β(Y − (v/V)N) and N′ = β(N − (v/V)Y), and likewise for Z and M. Einstein's β is the modern γ, and V is the speed of light. The lab's default is a pure electric field of 1 V/m along y and a boost of 0.6c, so γ = 1.25. In k the electric field is 1.25 V/m, and a magnetic field of −2.50 × 10−9 T along z appears where there was none. A charge at rest in K feels 1.60 × 10−19 N; in k it moves at −0.6c through both fields and feels 1.28 × 10−19 N, the value in K divided by γ. §6 then states the result in two ways. The old manner: a charge moving in a field feels, besides the electric force, an electromotive force equal, to first order in v/V, to its velocity crossed with the magnetic force and divided by the speed of light. The new manner: the force on a moving charge is the electric force found by transforming the field to a frame in which the charge is at rest. Let the charge move with k and the lab shows that force, 2.00 × 10−19 N, all of it electric. The electromotive force becomes an auxiliary concept, and the asymmetry between magnet and conductor from the introduction disappears. The ledger also prints two combinations that the boost leaves unchanged, E·B = 0 and E² − c²B² = 1 (V/m)², as a check on the arithmetic; §6 does not use them.
Take the lab's default. In the stationary system K there is an electric field of 1 V/m pointing along y and no magnetic field. The moving system k travels along x at v = 0.6c. First the factor: γ = 1/√(1 − v²/c²) = 1/√(1 − 0.36) = 1/√0.64 = 1/0.8 = 1.25. Now §6's formulas, written in SI units, where the magnetic field carries a factor of c. The component of E along the motion is unchanged, so E′x = 0. The y component becomes E′y = γ(Ey − vBz) = 1.25 × (1 − 0) = 1.25 V/m. The z component of the magnetic field becomes B′z = γ(Bz − vEy/c²) = 1.25 × (0 − 0.6c × 1/c²) = −0.75/c = −0.75/(2.998 × 108) = −2.50 × 10−9 T. So a field that was purely electric in K is electric and magnetic in k. Now place a charge q = 1.602 × 10−19 C at rest in K. There the force is qE = 1.602 × 10−19 × 1 = 1.60 × 10−19 N along y. In k the charge moves at u′ = −0.6c along x, so the force there is q(E′ + u′ × B′). The y component of u′ × B′ is −u′xB′z = −(−0.6c)(−0.75/c) = −0.45 V/m, so the total is 1.25 − 0.45 = 0.80 V/m, and the force is 0.80q = 1.28 × 10−19 N, smaller than in K by the factor 0.8 = 1/γ. This is not a contradiction. Force is momentum gained per unit time; the transverse momentum is the same in both frames, but the time between two events on the charge is longer in the frame in which it moves, so the force there is smaller. Next, put the charge at rest in k instead. There its velocity is zero, only the electric field acts, and the force is qE′ = 1.25q = 2.00 × 10−19 N: this is §6's new manner, the electric force in the charge's rest frame. In K the same charge moves at +0.6c through a field with no magnetic part, so the force there is qE = 1.60 × 10−19 N, again smaller by γ. Last, the check. E·B is 0 in K, and E′·B′ is also 0 in k, since E′ points along y and B′ along z. E² − c²B² is 1 in K, and in k it is 1.25² − c²(0.75/c)² = 1.5625 − 0.5625 = 1.
In §6 Einstein writes the electric force as (X, Y, Z), the magnetic force as (L, M, N), the speed of light as V and the modern γ as β, in units in which electric and magnetic forces have the same dimension. He fixes an undetermined factor ψ(v) to 1 by requiring the transformation and its inverse to agree and by symmetry. He calls the electromotive force an auxiliary concept, owed to the fact that electric and magnetic forces have no existence independent of the state of motion of the coordinate system, and says that questions about the seat of the electromotive force in unipolar machines lose their point. Lorentz's 1904 theory had already transformed the fields, as mathematical aids in an ether at rest. The invariants E·B and E² − c²B², and the treatment of the field as one object, come from Minkowski in 1908.
Predict before the numbers
When a pure electric field in the y direction is described from a frame moving along x at 0.6c, what magnetic field appears?
The result appears when you choose, say you have one in mind, or skip.
Worked example: seen from the frame moving at 0.6c, the electric field (0, 1, 0) V/m becomes (0, 1.25, 0) V/m and the magnetic field (0, 0, 0) T becomes (0, 0, −2.5 × 10⁻⁹) T.
Laboratory frame K, SI units
Transformation ledger
| E (Stationary K) | (0, 1, 0) V/m |
|---|---|
| E′ (Moving k) | (0, 1.25, 0) V/m |
| B (Stationary K) | (0, 0, 0) T |
| B′ (Moving k) | (0, 0, −2.5017 × 10⁻⁹) T |
| Lorentz Factor γ | 1.25 |
| Invariant E² − c²B² | 1 (V/m)² |
| Invariant E · B | 0 T·V/m |
| Laboratory force F | (0, 1.6022 × 10⁻¹⁹, 0) N |
| Comoving force F′ | (0, 1.2817 × 10⁻¹⁹, 0) N |
Not modeled: field sources and currents; radiation and self-fields; radiation reaction; back-reaction on the field; media and polarization; nonuniform fields; accelerated observers.
Worked case (readable without JavaScript)
Consider a pure electric field in the stationary system K with Ey = 1 V/m and B = 0, viewed from a coordinate system k boosted along the x-axis at speed v = 0.6c (γ = 1.25).
Both frames agree exactly on the Lorentz field invariants:
A test charge q at rest in K feels the force Fy = qEy, which is 1.602×10⁻¹⁹ N for an elementary charge, in the laboratory frame. In the moving frame k the charge has velocity u′x = −0.6c and feels the transformed Lorentz force F′y = q(E′y − u′xB′z) = q(1.25 − 0.45) V/m = 0.8 × 1.602×10⁻¹⁹ N, since u′xB′z = (−0.6c)(−0.75/c) = 0.45 V/m. The force transformation F′y = Fy/γ gives the same 0.8 Fy, matching the kinematics of §6.
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