Relativistic Transformation of E and B

Statement

Starting from the covariant transformation law of the electromagnetic field tensor, \( F'^{\mu\nu} = \Lambda^{\mu}{}_{\alpha}\Lambda^{\nu}{}_{\beta}F^{\alpha\beta} \), we derive the explicit mixing of the electric and magnetic fields under a Lorentz boost of speed \( v \) along the \( x \)-axis: the field components parallel to the boost are unchanged, \( E'_{\parallel} = E_{\parallel} \), \( B'_{\parallel} = B_{\parallel} \), while the perpendicular components mix as \( \vec{E}'_{\perp} = \gamma\left(\vec{E} + \vec{v}\times\vec{B}\right)_{\perp} \) and \( \vec{B}'_{\perp} = \gamma\left(\vec{B} - \dfrac{\vec{v}\times\vec{E}}{c^{2}}\right)_{\perp} \). A pure electric field in one frame is therefore an electric and magnetic field in another: magnetism is the relativistic residue of moving charge.

Why it matters

This result dissolves the apparent duality of electromagnetism. "Electric" and "magnetic" are not two forces but two frame-dependent projections of a single object, the field tensor \( F^{\mu\nu} \). A charge at rest beside a current-carrying wire feels a force in one frame that is purely magnetic and in another frame purely electric; the transformation law is what guarantees both observers predict the same physical outcome. Historically this is the puzzle that opens Einstein's 1905 paper — the asymmetry of magnet-and-conductor explanations — and this derivation is its resolution.

Practically, the transformation is the working tool of accelerator physics (beam self-fields), astrophysical plasma theory (ideal MHD's \( \vec{E} + \vec{v}\times\vec{B} = 0 \) is the statement that the plasma rest frame sees no electric field), and the analysis of relativistic beams, where a Coulomb field pancakes into a nearly transverse pulse accompanied by an almost equal magnetic field.

Assumptions

Derivation

Result

Limiting cases

• Galilean limit \( \beta \ll 1 \): \( \gamma \to 1 \), giving \( \vec{E}' \approx \vec{E} + \vec{v}\times\vec{B} \) and \( \vec{B}' \approx \vec{B} \) to first order in \( v/c \) — the field felt in the frame of a slowly drifting charge, the basis of the motional EMF and of ideal MHD.
• Pure electrostatic source (\( \vec{B} = 0 \) in \( S \)): the moving frame sees \( \vec{B}' = -\gamma\,\vec{v}\times\vec{E}/c^{2} = -\vec{v}\times\vec{E}'/c^{2} \) — the magnetic field of a moving charge appears from nothing but Coulomb's law plus relativity.
• Parallel fields, boost along them: if \( \vec{E} \parallel \vec{B} \parallel \vec{v} \), nothing changes at all: \( \vec{E}' = \vec{E},\ \vec{B}' = \vec{B} \).
• Ultra-relativistic boost \( \beta \to 1 \): transverse fields grow without bound as \( \gamma \to \infty \) and approach the radiation condition \( |\vec{E}'| \to c|\vec{B}'| \), \( \vec{E}'\perp\vec{B}' \): the field of an ultra-relativistic charge mimics a pulse of light.
• Crossed fields: if \( \vec{E}\cdot\vec{B} = 0 \), a frame exists where the field is purely electric (if \( E > cB \), drift speed \( \beta = cB/E \)) or purely magnetic (if \( E < cB \), drift speed \( \beta = E/cB \)); at exactly \( E = cB \) neither frame exists — the null field of a plane wave.

Breaks when

Failure modes

• Gamma on the wrong components. Writing \( E'_x = \gamma E_x \). It is the perpendicular components that pick up \( \gamma \); the parallel ones are exactly invariant (steps 4 and 7). Contrast with lengths, where it is the parallel direction that contracts — the patterns are opposite and students conflate them.
• Using the particle's velocity instead of the frame velocity. In \( \vec{E}' = \gamma(\vec{E} + \vec{v}\times\vec{B})_{\perp} \), \( \vec{v} \) is the relative velocity of the two frames. It coincides with a particle's velocity only if you deliberately boost to that particle's instantaneous rest frame.
• Dropping the \( c^{2} \) in the magnetic mixing term. Writing \( B'_z = \gamma(B_z - vE_y) \) — dimensionally wrong in SI. The error typically comes from transplanting Gaussian-unit formulas ( \( \vec{B}' = \gamma(\vec{B} - \vec{\beta}\times\vec{E}) \) ) into SI work.
• Sign errors from the inverse transformation. The fields in \( S \) in terms of those in \( S' \) require \( v \to -v \). Applying the forward formula twice "to go back" multiplies by \( \gamma^{2} \) and doubles the mixing terms.
• Comparing fields at mismatched events. Computing \( \vec{E}'(\vec{r},t) \) with unprimed arguments. The transformed field lives at the transformed event; forgetting to also transform the coordinates gives wrong answers for any non-uniform field.
• Double-counting in the force law. Boosting to the charge's rest frame, computing \( q\vec{E}' \), and then also adding \( q\vec{v}\times\vec{B} \). The \( \vec{v}\times\vec{B} \) force is already inside \( \vec{E}' \); one uses either the lab fields with the full Lorentz force or the rest-frame field alone, never both.

Discussion

The deepest content of this result is ontological: there is no invariant answer to the question "is this field electric or magnetic?" The invariant object is \( F^{\mu\nu} \); \( \vec{E} \) and \( \vec{B} \) are its time–space and space–space slices, and slicing depends on the observer's time axis exactly as the split of spacetime into space and time does. The two quadratic invariants \( E^{2} - c^{2}B^{2} \) and \( \vec{E}\cdot\vec{B} \) are the only frame-independent local statements one can make about the field pair, and they classify fields completely: electric-dominated ( \( E > cB \) ), magnetic-dominated ( \( E < cB \) ), and null ( both invariants zero — radiation ).

The result also explains why magnetism exists at all with everyday materials. In a current-carrying wire the drift speed is of order \( 10^{-4}\ \mathrm{m\,s^{-1}} \), so relativistic corrections of order \( v^{2}/c^{2} \sim 10^{-25} \) seem hopeless. But the electrostatic force between the wire's \( \sim 10^{23} \) mobile electrons per metre and an external charge is astronomically large, and the magnetic force is that huge force multiplied by the tiny relativistic factor: the product is the ordinary, laboratory-scale magnetic force. Magnetism is measurable relativity at walking pace — a length-contraction imbalance of charge densities seen from the frame of a moving test charge.

Common misconceptions. (i) "Magnetic fields are caused by electric fields" — neither causes the other; both are aspects of one tensor sourced by the charge–current four-vector \( J^{\mu} \). (ii) "One can always transform \( \vec{B} \) away" — only if \( E > cB \) and \( \vec{E}\cdot\vec{B} = 0 \); a plane wave or a magnetic-dominated field admits no such frame. (iii) "The transformation applies only to uniform fields" — it applies point by point to any field configuration, provided the coordinates of the evaluation event are transformed too.

Group-theoretically, \( \vec{E} \) and \( \vec{B} \) organize into the complexified combination \( \vec{F} = \vec{E} + ic\vec{B} \), on which a boost of rapidity \( \zeta \) about \( \hat{n} \) acts as a rotation through the imaginary angle \( i\zeta \) about \( \hat{n} \): the Lorentz group \( SO(3,1) \) is locally \( SO(3,\mathbb{C}) \), and the invariant \( \vec{F}\cdot\vec{F} = (E^{2} - c^{2}B^{2}) + 2ic\,\vec{E}\cdot\vec{B} \) packages both scalar invariants at once. This self-dual decomposition, \( F^{\mu\nu} \to F^{\mu\nu} \pm \tfrac{i}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma} \), splits the field into the \( (1,0) \oplus (0,1) \) representation of the Lorentz group and underlies the two photon helicities, spinor-helicity methods, and the Newman–Penrose formalism in general relativity.

Worked examples

Example 1 — the capacitor that grows a magnetic field. A parallel-plate capacitor at rest in \( S \) produces a uniform field \( \vec{E} = E_z\hat{z} \) with \( E_z = 5.0\times10^{4}\ \mathrm{V\,m^{-1}} \) and \( \vec{B} = 0 \). Find the fields in a frame \( S' \) moving at \( v = 0.60c \) along \( +\hat{x} \) (parallel to the plates).

Example 2 — boosting away the magnetic field. In the lab, \( \vec{E} = 3.0\times10^{6}\,\hat{y}\ \mathrm{V\,m^{-1}} \) and \( \vec{B} = 5.0\times10^{-3}\,\hat{z}\ \mathrm{T} \). Show a frame exists in which the field is purely electric, find its velocity, and find the field there.

Problems

1. A uniform magnetic field \( \vec{B} = 0.20\,\hat{z}\ \mathrm{T} \) fills the lab, with \( \vec{E} = 0 \). Find \( \vec{E}' \) and \( \vec{B}' \) in a frame moving at \( v = 0.80c \) along \( +\hat{x} \), and verify \( E'^{2} - c^{2}B'^{2} \) is unchanged.
   Solution

   With \( \beta = 0.80 \), \( \gamma = 1/\sqrt{1-0.64} = 5/3 \). Parallel components: \( E'_x = 0 \), \( B'_x = 0 \). Transverse: \( E'_y = \gamma(E_y - vB_z) = \tfrac{5}{3}\left(0 - 0.80\times3.0\times10^{8}\times0.20\right) = -8.0\times10^{7}\ \mathrm{V\,m^{-1}} \); \( E'_z = \gamma(E_z + vB_y) = 0 \); \( B'_y = \gamma(B_y + vE_z/c^{2}) = 0 \); \( B'_z = \gamma(B_z - vE_y/c^{2}) = \tfrac{5}{3}(0.20) = 0.333\ \mathrm{T} \). So \( \vec{E}' = -8.0\times10^{7}\,\hat{y}\ \mathrm{V\,m^{-1}} \), \( \vec{B}' = 0.333\,\hat{z}\ \mathrm{T} \). Invariant: before, \( E^{2} - c^{2}B^{2} = -\left(3.0\times10^{8}\right)^{2}(0.20)^{2} = -3.6\times10^{15}\ \mathrm{V^{2}\,m^{-2}} \). After, \( (8.0\times10^{7})^{2} - (3.0\times10^{8})^{2}(0.333)^{2} = 6.4\times10^{15} - 1.0\times10^{16} = -3.6\times10^{15}\ \mathrm{V^{2}\,m^{-2}} \). Unchanged, as required; the field remains magnetic-dominated, so no frame can remove \( \vec{B} \) entirely.
2. In the lab, \( \vec{E} = 1.0\times10^{6}\,\hat{y}\ \mathrm{V\,m^{-1}} \) and \( \vec{B} = 0.020\,\hat{z}\ \mathrm{T} \). Decide whether a frame exists with \( \vec{E}' = 0 \) or with \( \vec{B}' = 0 \); find its velocity and the surviving field.
   Solution

   \( \vec{E}\cdot\vec{B} = 0 \). Compare \( E = 1.0\times10^{6}\ \mathrm{V\,m^{-1}} \) with \( cB = 3.0\times10^{8}\times0.020 = 6.0\times10^{6}\ \mathrm{V\,m^{-1}} \): since \( cB > E \), the field is magnetic-dominated, so only \( \vec{E}' = 0 \) is achievable. Setting \( \vec{E}'_{\perp} = \gamma(\vec{E} + \vec{v}\times\vec{B}) = 0 \) gives \( \vec{v} = \vec{E}\times\vec{B}/B^{2} \) (the \( E\times B \) drift velocity), pointing along \( \hat{y}\times\hat{z} = \hat{x} \) with magnitude \( v = E/B = 1.0\times10^{6}/0.020 = 5.0\times10^{7}\ \mathrm{m\,s^{-1}} \), i.e. \( \beta = 1/6 \), comfortably subluminal. Then \( \gamma = 1/\sqrt{1 - 1/36} = 6/\sqrt{35} = 1.0142 \) and \( B'_z = \gamma\left(B_z - vE_y/c^{2}\right) = 1.0142\left(0.020 - \frac{5.0\times10^{7}\times1.0\times10^{6}}{9.0\times10^{16}}\right) = 1.0142\left(0.020 - 5.56\times10^{-4}\right) = 0.0197\ \mathrm{T} \). Check against the invariant: \( B' = \sqrt{B^{2} - E^{2}/c^{2}} = \sqrt{4.0\times10^{-4} - 1.11\times10^{-5}} = 0.0197\ \mathrm{T} \). Agreed: \( \vec{v} = c/6\ \hat{x} \), \( \vec{B}' = 0.0197\,\hat{z}\ \mathrm{T} \), \( \vec{E}' = 0 \).
3. A proton moves at \( v = 0.50c \) along \( +\hat{x} \) in the lab. Using the transformation of its rest-frame Coulomb field, find \( \vec{B} \) in the lab at the point of closest approach a perpendicular distance \( b = 1.0\ \mathrm{mm} \) from its track.
   Solution

   In the proton rest frame \( S' \) the field at perpendicular displacement \( b\,\hat{y} \) is purely electric, \( E'_{y} = \frac{q}{4\pi\varepsilon_{0}b^{2}} \) (the perpendicular distance is the same in both frames since it is transverse to the boost). Transforming to the lab (inverse transformation, \( v \to -v \)): \( E_y = \gamma E'_y \) and \( B_z = \gamma\frac{v}{c^{2}}E'_y = \frac{v}{c^{2}}E_y \), i.e. \( \vec{B} = \vec{v}\times\vec{E}/c^{2} \). Numbers: \( \frac{q}{4\pi\varepsilon_{0}b^{2}} = \frac{(8.99\times10^{9})(1.60\times10^{-19})}{(1.0\times10^{-3})^{2}} = 1.44\times10^{-3}\ \mathrm{V\,m^{-1}} \). With \( \gamma = 1/\sqrt{1-0.25} = 1.155 \): \( E_y = 1.155\times1.44\times10^{-3} = 1.66\times10^{-3}\ \mathrm{V\,m^{-1}} \). Then \( B_z = \frac{vE_y}{c^{2}} = \frac{(1.5\times10^{8})(1.66\times10^{-3})}{9.0\times10^{16}} = 2.8\times10^{-12}\ \mathrm{T} \), directed along \( +\hat{z} \) (circling the track in the right-hand sense of the current). This is exactly the Biot–Savart-like field of a single moving charge, obtained with no magnetism postulated — only Coulomb's law and the boost.
4. Prove algebraically, from the six component formulas of the Result box, that \( \vec{E}'\cdot\vec{B}' = \vec{E}\cdot\vec{B} \), and evaluate both sides for \( \vec{E} = (2.0, 3.0, 0)\times10^{5}\ \mathrm{V\,m^{-1}} \), \( \vec{B} = (0.010, 0, 0.020)\ \mathrm{T} \), \( \beta = 0.60 \).
   Solution

   Algebra: \( E'_xB'_x = E_xB_x \). Next, \( E'_yB'_y + E'_zB'_z = \gamma^{2}\left[(E_y - vB_z)\left(B_y + \tfrac{v}{c^{2}}E_z\right) + (E_z + vB_y)\left(B_z - \tfrac{v}{c^{2}}E_y\right)\right] \). Expanding, the cross terms \( \tfrac{v}{c^{2}}E_yE_z - \tfrac{v}{c^{2}}E_zE_y \) and \( -vB_zB_y + vB_yB_z \) cancel, leaving \( \gamma^{2}\left[(E_yB_y + E_zB_z)\left(1 - \tfrac{v^{2}}{c^{2}}\right)\right] = E_yB_y + E_zB_z \) since \( \gamma^{2}(1-\beta^{2}) = 1 \). Hence \( \vec{E}'\cdot\vec{B}' = \vec{E}\cdot\vec{B} \). Numerically: \( \vec{E}\cdot\vec{B} = (2.0\times10^{5})(0.010) + (3.0\times10^{5})(0) + 0 = 2.0\times10^{3}\ \mathrm{V\,T\,m^{-1}} \). Primed frame ( \( \gamma = 1.25 \), \( v = 1.8\times10^{8}\ \mathrm{m\,s^{-1}} \) ): \( E'_x = 2.0\times10^{5} \), \( B'_x = 0.010 \); \( E'_y = 1.25(3.0\times10^{5} - 1.8\times10^{8}\times0.020) = 1.25(3.0\times10^{5} - 3.6\times10^{6}) = -4.125\times10^{6} \); \( B'_y = 1.25(0 + 0) = 0 \); \( E'_z = 1.25(0 + 1.8\times10^{8}\times0) = 0 \); \( B'_z = 1.25\left(0.020 - \frac{1.8\times10^{8}\times3.0\times10^{5}}{9.0\times10^{16}}\right) = 1.25(0.020 - 6.0\times10^{-4}) = 0.02425 \). Then \( \vec{E}'\cdot\vec{B}' = (2.0\times10^{5})(0.010) + (-4.125\times10^{6})(0) + (0)(0.02425) = 2.0\times10^{3}\ \mathrm{V\,T\,m^{-1}} \). Equal, as proved.
5. In the lab, \( \vec{E} = 6.0\times10^{6}\,\hat{y}\ \mathrm{V\,m^{-1}} \) and \( \vec{B} = 0.010\,\hat{z}\ \mathrm{T} \). (a) Show no frame exists where the field is purely magnetic. (b) Find the frame where it is purely electric. (c) A proton ( \( m_p = 1.67\times10^{-27}\ \mathrm{kg} \) ) is instantaneously at rest in that frame; find its proper acceleration.
   Solution

   (a) \( cB = 3.0\times10^{8}\times0.010 = 3.0\times10^{6}\ \mathrm{V\,m^{-1}} \), so \( E^{2} - c^{2}B^{2} = (36 - 9)\times10^{12} = 2.7\times10^{13}\ \mathrm{V^{2}\,m^{-2}} > 0 \). This invariant is positive in every frame; a purely magnetic field would make it negative, so no such frame exists. (b) Boost along \( \vec{E}\times\vec{B} \propto \hat{x} \) with \( \beta = cB/E = 3.0\times10^{6}/6.0\times10^{6} = 0.50 \), i.e. \( v = 1.5\times10^{8}\ \mathrm{m\,s^{-1}} \), \( \gamma = 1.155 \). Check: \( B'_z = \gamma(B_z - vE_y/c^{2}) = 1.155\left(0.010 - \frac{1.5\times10^{8}\times6.0\times10^{6}}{9.0\times10^{16}}\right) = 1.155(0.010 - 0.010) = 0 \). Surviving field: \( E' = \sqrt{E^{2} - c^{2}B^{2}} = \sqrt{2.7\times10^{13}} = 5.20\times10^{6}\ \mathrm{V\,m^{-1}} \) along \( \hat{y} \) (equivalently \( E'_y = \gamma(E_y - vB_z) = 1.155\times4.5\times10^{6} = 5.20\times10^{6}\ \mathrm{V\,m^{-1}} \)). (c) At rest, the force is purely electric: \( a = \frac{eE'}{m_p} = \frac{(1.60\times10^{-19})(5.20\times10^{6})}{1.67\times10^{-27}} = 4.98\times10^{14}\ \mathrm{m\,s^{-2}} \). Because the proton is momentarily at rest in this frame, this is the proper acceleration, and it is the invariant that any frame would compute for the same event.