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?
SR-08 · Electrodynamics §6
Electric and magnetic frame change
Electric and magnetic forces do not exist independently of the coordinate system; a pure electric field in one frame appears as both electric and magnetic fields in another.
Under a boost along x, the parallel field components are unchanged while perpendicular components mix with factor gamma. The field combinations E·B and E² - c²B² are Lorentz invariants.
A test charge experiences force q(E + u × B) in the laboratory frame and q(E' + u' × B') in the moving frame. Transforming the fields and velocities predicts the comoving force exactly according to the relativistic force law.
Section 6's new manner of expression determines the force on a moving charge by transforming the field to the charge's instantaneous rest frame, where the force is purely electric qE''. Electrodynamics becomes kinematic and frame-independent.
Transformation Ledger
| E (Stationary K) | value V/m |
|---|---|
| E′ (Moving k) | value V/m |
| B (Stationary K) | value T |
| B′ (Moving k) | value T |
| Lorentz Factor γ | 1.25 |
| Invariant E² − c²B² | 1 (V/m)² |
| Invariant E · B | 0 T·V/m |
| Laboratory Force F | value N |
| Comoving Force F′ | value N |
Predict: Appearing magnetic field
When a pure electric field in the y direction is described from a frame moving along x at 0.6c, what magnetic field appears?
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 (gamma = 1.25).
Both frames agree exactly on the Lorentz field invariants:
A test charge with charge q at rest in K experiences force Fy = q Ey = 1.602×10⁻¹⁹ N in the laboratory frame. In the moving frame k, the charge has velocity u'x = -0.6c, and experiences the transformed Lorentz force F'y = q(E'y + u'x B'z) = q(1.25 - 0.6×0.75) = 1.0×1.602×10⁻¹⁹ N. The relativistic force transformation law F'y = Fy / gamma gives F'y = 0.8 Fy, matching the kinematics of §6.