Transformation of E and B under Boosts

Derivation

Transformation of E and B under Boosts

D-295 Home PU-305 Threads symmetry · fields Depends on Covariant Formulation of Electrodynamics

Statement

Working in Minkowski space with signature \((+,-,-,-)\) and the electromagnetic field tensor \(F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu\), we derive how the fields measured in a frame \(S'\), moving at velocity \(\vec v=v\,\hat x\) relative to \(S\), relate to those in \(S\). Applying the tensor transformation \(F'^{\mu\nu}=\Lambda^\mu{}_\alpha\,\Lambda^\nu{}_\beta\,F^{\alpha\beta}\) component by component yields the field-mixing rules \(E'_\parallel=E_\parallel,\ B'_\parallel=B_\parallel\) and \(\vec E'_\perp=\gamma(\vec E+\vec v\times\vec B)_\perp,\ \vec B'_\perp=\gamma\!\left(\vec B-\tfrac{1}{c^2}\vec v\times\vec E\right)_\perp\), establishing \(\vec E\) and \(\vec B\) as two faces of one rank-2 tensor.

Why it matters

Electricity and magnetism are not separate phenomena that merely happen to appear together in Maxwell's equations. The transformation law shows that a field which is purely electric in one frame carries a magnetic part in another, and vice versa: the split into \(\vec E\) and \(\vec B\) is frame-dependent, while the object \(F^{\mu\nu}\) is not. This resolves the century-old asymmetry that opened Einstein's 1905 paper — a magnet moving past a coil and a coil moving past a magnet must give identical physics.

Practically, these relations are the engine behind motional EMF, the electric field of a fast-moving charge (the "pancake" field), synchrotron and beam optics, and the covariant statement of the Lorentz force. They also expose the two Lorentz invariants \(\vec E\cdot\vec B\) and \(B^2-E^2/c^2\), which classify every field configuration independently of the observer.

Assumptions

Special-relativistic (flat) spacetime with a global inertial frame.In curved spacetime or non-inertial frames the constant boost matrix \(\Lambda\) is replaced by position-dependent tetrads, and the simple algebraic mixing below no longer holds globally.

The boost is a single pure Lorentz boost along \(\hat x\) with constant relative velocity \(\vec v=v\,\hat x\).For an arbitrary boost direction one must first resolve \(\vec E,\vec B\) into components parallel and perpendicular to \(\vec v\); rotations combined with the boost would additionally rotate the field components among the axes.

Fields evaluated at the same spacetime event in both frames.The tensor law is pointwise; comparing fields at different events (e.g. same lab coordinates but different times) mixes in the transformation of the arguments and gives wrong results.

\(F^{\mu\nu}\) is a genuine rank-2 antisymmetric tensor, i.e. Maxwell's equations are Lorentz covariant (prior result).If \(F^{\mu\nu}\) did not transform tensorially, the derived rules would be mere coordinate bookkeeping with no invariant meaning; the antisymmetry \(F^{\mu\nu}=-F^{\nu\mu}\) is what guarantees exactly six independent field components, three electric and three magnetic.

Metric/sign convention: \((+,-,-,-)\), \(A^\mu=(\varphi/c,\vec A)\), so \(F^{0i}=-E^i/c\) and \(F^{ij}=-\varepsilon^{ijk}B_k\).Adopting the opposite metric signature or the Gaussian unit system flips signs and drops the explicit \(c\) factors; the physics is unchanged but every intermediate sign below would need re-checking.

Derivation

\[ F^{\mu\nu}=\begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix} \]

Definition: reading off \(F^{0i}=-E^i/c\) and \(F^{ij}=-\varepsilon^{ijk}B_k\) from \(F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu\) with \(A^\mu=(\varphi/c,\vec A)\). The six independent entries are exactly \((\vec E,\vec B)\). A

\[ \Lambda^\mu{}_\nu=\begin{pmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix},\qquad \beta=\frac{v}{c},\ \ \gamma=\frac{1}{\sqrt{1-\beta^2}} \]

The boost of \(S'\) along \(+\hat x\) mixes only the \(0\) and \(1\) coordinates; the transverse axes \(y,z\) are untouched. A

\[ F'^{\mu\nu}=\Lambda^\mu{}_\alpha\,\Lambda^\nu{}_\beta\,F^{\alpha\beta} \]

A rank-2 contravariant tensor carries one \(\Lambda\) per index. This is the content of the prior result (Lorentz covariance of Maxwell): it is legitimate to transform the fields simply by transforming \(F\). B

\[ F'^{01}=\Lambda^0{}_0\Lambda^1{}_1 F^{01}+\Lambda^0{}_1\Lambda^1{}_0 F^{10}=\gamma^2 F^{01}+\gamma^2\beta^2 F^{10}=\gamma^2(1-\beta^2)F^{01}=F^{01} \]

Only \(\alpha,\beta\in\{0,1\}\) survive because \(\Lambda\) is block-diagonal; use \(F^{10}=-F^{01}\) and \(\gamma^2(1-\beta^2)=1\). Since \(F^{01}=-E_x/c\), this gives \(\boxed{E'_x=E_x}\): the longitudinal electric field is unchanged. B

\[ F'^{02}=\Lambda^0{}_0 F^{02}+\Lambda^0{}_1 F^{12}=\gamma\!\left(-\tfrac{E_y}{c}\right)+(-\gamma\beta)(-B_z)=-\tfrac{\gamma}{c}\left(E_y-vB_z\right) \]

The second index is transverse, \(\Lambda^2{}_\beta=\delta^2_\beta\), so only \(\beta=2\) contributes and \(\alpha\in\{0,1\}\). Equating to \(F'^{02}=-E'_y/c\) gives \(\boxed{E'_y=\gamma\left(E_y-vB_z\right)}\). B

\[ F'^{03}=\Lambda^0{}_0 F^{03}+\Lambda^0{}_1 F^{13}=\gamma\!\left(-\tfrac{E_z}{c}\right)+(-\gamma\beta)(B_y)=-\tfrac{\gamma}{c}\left(E_z+vB_y\right) \]

Same structure with the third index transverse; \(F^{13}=B_y\). Equating to \(-E'_z/c\) yields \(\boxed{E'_z=\gamma\left(E_z+vB_y\right)}\). Note the opposite sign to Step 5 — it is the sign of \((\vec v\times\vec B)\). B

\[ F'^{12}=\Lambda^1{}_0 F^{02}+\Lambda^1{}_1 F^{12}=(-\gamma\beta)\!\left(-\tfrac{E_y}{c}\right)+\gamma(-B_z)=-\gamma\!\left(B_z-\tfrac{\beta}{c}E_y\right) \]

Now the first index is boosted and the second is transverse. Since \(F'^{12}=-B'_z\) and \(\beta/c=v/c^2\), we obtain \(\boxed{B'_z=\gamma\!\left(B_z-\tfrac{v}{c^2}E_y\right)}\): a transverse electric field induces a transverse magnetic field. C

\[ F'^{13}=\gamma\!\left(B_y+\tfrac{v}{c^2}E_z\right)\ \Rightarrow\ \boxed{B'_y=\gamma\!\left(B_y+\tfrac{v}{c^2}E_z\right)},\qquad F'^{23}=F^{23}\ \Rightarrow\ \boxed{B'_x=B_x} \]

\(F'^{13}=\Lambda^1{}_0F^{03}+\Lambda^1{}_1F^{13}=(-\gamma\beta)(-E_z/c)+\gamma B_y\) with \(F'^{13}=B'_y\); and \(F'^{23}=F^{23}=-B_x\) because both indices are transverse and untouched by \(\Lambda\). The longitudinal magnetic field is invariant. C

\[ \begin{aligned} \vec E'_\parallel&=\vec E_\parallel, & \vec B'_\parallel&=\vec B_\parallel,\\ \vec E'_\perp&=\gamma\left(\vec E+\vec v\times\vec B\right)_\perp, & \vec B'_\perp&=\gamma\left(\vec B-\tfrac{1}{c^2}\vec v\times\vec E\right)_\perp. \end{aligned} \]

Collecting Steps 4–8 with \(\vec v=v\hat x\): the unchanged \(x\)-components are the parallel pieces; the \(y,z\)-components reassemble into \(\vec v\times\vec B\) and \(\vec v\times\vec E\) (check: \((\vec v\times\vec B)_y=-vB_z\), \((\vec v\times\vec B)_z=vB_y\)). This coordinate-free form is valid for a boost in any direction. B

Result

\[ \vec E'_\parallel=\vec E_\parallel,\quad \vec B'_\parallel=\vec B_\parallel,\quad \vec E'_\perp=\gamma\!\left(\vec E+\vec v\times\vec B\right)_\perp,\quad \vec B'_\perp=\gamma\!\left(\vec B-\tfrac{1}{c^2}\vec v\times\vec E\right)_\perp \]

Reading. Components of the field along the boost survive untouched; components perpendicular to the boost are amplified by \(\gamma\) and, crucially, mixed — the perpendicular electric field in \(S'\) picks up a \(\vec v\times\vec B\) contribution and the perpendicular magnetic field picks up a \(-\vec v\times\vec E/c^2\) contribution. There is no frame-independent way to say "this is an electric field": \(\vec E\) and \(\vec B\) are the time-space and space-space parts of the single antisymmetric tensor \(F^{\mu\nu}\), and a boost simply re-slices that tensor. The combinations \(\vec E\cdot\vec B\) and \(B^2-E^2/c^2\) are the two quantities the slicing leaves fixed.

Units check. \(E/c\) has units \(\mathrm{(V/m)/(m/s)}=\mathrm{V\,s\,m^{-2}}=\mathrm{T}\), matching \(B\); so \(F^{0i}\) and \(F^{ij}\) share units and the tensor is dimensionally consistent. In \(\vec E'_\perp\), the term \(\vec v\times\vec B\) has units \(\mathrm{(m/s)(T)}=\mathrm{V/m}\), matching \(\vec E\). In \(\vec B'_\perp\), \(\tfrac{1}{c^2}\vec v\times\vec E\) has units \(\mathrm{(s^2 m^{-2})(m/s)(V/m)}=\mathrm{V\,s\,m^{-2}}=\mathrm{T}\), matching \(\vec B\). \(\gamma\) is dimensionless. ✓

Limiting cases

Non-relativistic \((v\ll c,\ \gamma\to1)\): \(\vec E'\approx\vec E+\vec v\times\vec B\) and \(\vec B'\approx\vec B-\tfrac{1}{c^2}\vec v\times\vec E\). The first is the Galilean motional field behind Faraday's law of induction in a moving conductor.
Pure magnetic field in \(S\) \((\vec E=0)\): \(\vec E'_\perp=\gamma\,\vec v\times\vec B\neq0\) — a moving observer sees an electric field. This is the frame-swap that turns the magnetic force on a charge into an electrostatic force.
Pure electric field in \(S\) \((\vec B=0)\): \(\vec B'_\perp=-\tfrac{\gamma}{c^2}\vec v\times\vec E\neq0\) — the magnetic field of a uniformly moving charge, entirely a relativistic view of its Coulomb field.
Null field \((\vec E\perp\vec B,\ E=cB):\) both invariants vanish, so \(E'=cB'\) and \(\vec E'\perp\vec B'\) in every frame — the defining, boost-stable structure of an electromagnetic plane wave.
Fields parallel to boost only: \(\vec E'=\vec E,\ \vec B'=\vec B\); nothing changes, since the mixing lives entirely in the transverse sector.

Breaks when

Non-inertial or gravitational settings. A constant global \(\Lambda\) does not exist for accelerating frames or in curved spacetime; the algebraic rules must be replaced by local tetrad transport, and comparing distant fields becomes ambiguous.
Media with a rest frame (dielectrics, magnetized matter). In matter one must transform \(F^{\mu\nu}\) and the excitation tensor \(G^{\mu\nu}=(\vec D,\vec H)\) separately; the constitutive relations \(\vec D=\varepsilon\vec E,\ \vec B=\mu\vec H\) hold only in the medium's rest frame and generate cross terms (magnetoelectric coupling, Fresnel–Fizeau drag) in other frames.
Quantum / strong-field regime. At field strengths near the Schwinger scale the classical superposition underlying \(F^{\mu\nu}\) fails (vacuum birefringence, photon–photon scattering); the linear transformation still applies to \(F\), but \(\vec E,\vec B\) no longer superpose linearly from sources.

Failure modes

Boosting the whole vector by \(\gamma\). Multiplying every component of \(\vec E\) and \(\vec B\) by \(\gamma\); the parallel components are invariant — only the transverse sector scales.
Dropping the \(1/c^2\) in \(\vec B'\). Writing \(\vec B'_\perp=\gamma(\vec B-\vec v\times\vec E)\) by analogy with the \(\vec E\) rule; the magnetic mixing term carries \(1/c^2\) and is dimensionally forced.
Sign of the cross products. Using \(-\vec v\times\vec B\) in \(\vec E'\) or \(+\vec v\times\vec E\) in \(\vec B'\); the relative sign is fixed by the antisymmetry of \(F^{\mu\nu}\) (Steps 5–8).
Confusing \(\vec v\) of \(S'\) with the source velocity. \(\vec v\) is the relative velocity of the frames, not the velocity of any charge; plugging in a particle speed gives nonsense.
Transforming fields at mismatched events. Reading \(\vec E,\vec B\) at fixed lab coordinates \((t,\vec x)\) and comparing to \(S'\) at the same numbers, forgetting that the event's \(S'\)-coordinates differ.
Assuming \(\vec E\cdot\vec B\) or \(E^2\) alone is invariant. Only \(\vec E\cdot\vec B\) and \(B^2-E^2/c^2\) are; \(E^2\) and \(B^2\) separately are frame-dependent.

Discussion

The deepest lesson is ontological: the electric and magnetic fields are not fundamental. What exists is the tensor \(F^{\mu\nu}\); "electric" and "magnetic" are the labels an observer attaches to its time-space and space-space blocks, and different observers slice it differently. This is the same relationship that ties energy to momentum in the four-vector \(p^\mu\), or space to time in \(x^\mu\). The relativity of the \(\vec E\)/\(\vec B\) split is not a mathematical trick but the literal content of "electromagnetism is a relativistic theory."

The two scalars built from the field, \(F_{\mu\nu}F^{\mu\nu}=2\!\left(B^2-E^2/c^2\right)\) and \({}^\star\! F_{\mu\nu}F^{\mu\nu}\propto\vec E\cdot\vec B\), are boost-invariant and therefore classify fields observer-independently. If \(\vec E\cdot\vec B\neq0\), no frame makes the field purely electric or purely magnetic. If \(\vec E\cdot\vec B=0\) and \(B^2>E^2/c^2\), a frame exists in which the field is purely magnetic; if \(E^2/c^2>B^2\), one in which it is purely electric; the balanced case \(E=cB\) is the radiation (null) field, reachable to no rest frame.

Physically, the transformation unifies apparently distinct laws. The magnetic force \(q\vec v\times\vec B\) on a moving charge becomes, in the charge's instantaneous rest frame, an ordinary electric force \(q\vec E'\) — the two descriptions agree precisely because \(\vec E'=\gamma\,\vec v\times\vec B\) to leading order. Likewise the magnetic field encircling a current-carrying wire is, for a co-moving electron, largely the electrostatic field of a length-contracted, net-charged ion lattice. Relativity is not a small correction here; it is the reason magnetism exists at all despite drift velocities of millimetres per second, the smallness of \(v/c\) being compensated by the enormous charge density.

Common misconceptions. (i) "Magnetism is a relativistic correction to electricity, so it's tiny." — The mixing coefficient is \(O(v/c)\), yet ordinary magnets are strong because the invariant combination and the source charge densities are large; the effect is not perturbatively negligible. (ii) "A boost can always remove the magnetic field." — Only when \(\vec E\cdot\vec B=0\) and \(E^2/c^2<B^2\); the invariants forbid it otherwise. (iii) "\(\vec E'\) and \(\vec B'\) are just \(\vec E,\vec B\) rotated." — They are not related by a spatial rotation (which preserves \(E^2\) and \(B^2\) separately) but by a boost, which trades \(E^2\) against \(B^2\) while holding \(B^2-E^2/c^2\) fixed. (iv) "The \(\gamma\) applies because the fields are lengths that contract." — \(\gamma\) here comes from the tensor index structure, not from length contraction of the fields themselves.

Worked examples

Example 1 — A purely electric field acquires a magnetic field.

\[ \vec E=(0,\,E_0,\,0),\quad \vec B=0,\quad E_0=1.0\times10^{6}\ \mathrm{V/m},\quad \vec v=0.6c\,\hat x \]

Set-up: transverse electric field, boost along \(\hat x\). A

\[ \gamma=\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{0.64}}=1.25 \]

Lorentz factor for \(\beta=0.6\). A

\[ E'_y=\gamma\left(E_y-vB_z\right)=\gamma E_0=1.25\times10^{6}\ \mathrm{V/m} \]

Transverse \(E\) rule with \(B_z=0\); symbols first, then numbers. B

\[ B'_z=\gamma\!\left(B_z-\frac{v}{c^2}E_y\right)=-\gamma\frac{v}{c^2}E_0=-1.25\cdot\frac{0.6\,(3\times10^{8})}{(3\times10^{8})^2}\,(10^{6}) \]

Transverse \(B\) rule with \(B_z=0\); the field mixes in. B

\[ B'_z=-\frac{1.25\times0.6\times10^{6}}{3\times10^{8}}=-2.5\times10^{-3}\ \mathrm{T} \]

Arithmetic; all other components zero (\(E'_x=B'_x=0\)). A

\[ \vec E'=(0,\ 1.25\times10^{6},\ 0)\ \mathrm{V/m},\qquad \vec B'=(0,\ 0,\ -2.5\times10^{-3})\ \mathrm{T} \]

Reading. A field seen as purely electric in \(S\) has a real magnetic field in \(S'\). Invariant check: in \(S\), \(B^2-E^2/c^2=-\left(10^{6}/3{\times}10^{8}\right)^2=-1.11\times10^{-5}\,\mathrm{T^2}\); in \(S'\), \(B'^2-E'^2/c^2=(2.5{\times}10^{-3})^2-(1.25{\times}10^{6}/3{\times}10^{8})^2=6.25\times10^{-6}-1.74\times10^{-5}=-1.11\times10^{-5}\,\mathrm{T^2}\). ✓ The invariant is preserved.

Example 2 — A purely magnetic field acquires an electric field.

\[ \vec B=(0,\,0,\,B_0),\quad \vec E=0,\quad B_0=2.0\ \mathrm{T},\quad \vec v=0.8c\,\hat x \]

Set-up: transverse magnetic field, boost along \(\hat x\). A

\[ \gamma=\frac{1}{\sqrt{1-0.8^2}}=\frac{1}{\sqrt{0.36}}=1.667 \]

Lorentz factor for \(\beta=0.8\). A

\[ E'_y=\gamma\left(E_y-vB_z\right)=-\gamma v B_0=-1.667\,(0.8)(3\times10^{8})(2.0) \]

Transverse \(E\) rule with \(E_y=0\). B

\[ E'_y=-8.0\times10^{8}\ \mathrm{V/m},\qquad B'_z=\gamma\left(B_z-\frac{v}{c^2}E_y\right)=\gamma B_0=3.33\ \mathrm{T} \]

Arithmetic for \(E'_y\); \(B'_z\) uses \(E_y=0\). B

\[ \vec E'=(0,\ -8.0\times10^{8},\ 0)\ \mathrm{V/m},\qquad \vec B'=(0,\ 0,\ 3.33)\ \mathrm{T} \]

Reading. The moving observer sees a strong transverse electric field — the seed of motional EMF. Invariant check: in \(S\), \(B^2-E^2/c^2=(2.0)^2=4.0\,\mathrm{T^2}\); in \(S'\), \((3.33)^2-(8.0{\times}10^{8}/3{\times}10^{8})^2=11.1-7.11=4.0\,\mathrm{T^2}\). ✓ Also \(\vec E'\cdot\vec B'=0\), matching \(\vec E\cdot\vec B=0\).

Problems

(A) Warm-up. In \(S\), \(\vec E=(0,0,E_0)\) with \(E_0=300\ \mathrm{V/m}\) and \(\vec B=0\). Boost along \(\hat x\) at \(v=0.5c\). Find \(\vec E'\) and \(\vec B'\).
Solution
\(\gamma=1/\sqrt{1-0.25}=1.1547\). Longitudinal parts vanish. \(E'_z=\gamma(E_z+vB_y)=\gamma E_0=1.1547\times300=346.4\ \mathrm{V/m}\). \(B'_y=\gamma\!\left(B_y+\tfrac{v}{c^2}E_z\right)=\gamma\tfrac{v}{c^2}E_0=1.1547\cdot\tfrac{0.5(3\times10^8)}{(3\times10^8)^2}\cdot300=1.1547\cdot\tfrac{150}{3\times10^8}=5.77\times10^{-7}\ \mathrm{T}\). All other components zero: \(\vec E'=(0,0,346.4)\ \mathrm{V/m}\), \(\vec B'=(0,\,5.77\times10^{-7},\,0)\ \mathrm{T}\).
(B) Crossed fields, invariant \(\vec E\cdot\vec B\). In \(S\), \(\vec E=(0,\,3\times10^{4},\,0)\ \mathrm{V/m}\) and \(\vec B=(0,0,10^{-4})\ \mathrm{T}\). Boost along \(\hat x\) at \(v=0.6c\). Find \(\vec E',\vec B'\) and verify \(\vec E'\cdot\vec B'=\vec E\cdot\vec B\).
Solution
\(\gamma=1.25\). \(E'_y=\gamma(E_y-vB_z)=1.25\!\left(3\times10^4-0.6(3\times10^8)(10^{-4})\right)=1.25(3\times10^4-1.8\times10^4)=1.5\times10^4\ \mathrm{V/m}\). \(B'_z=\gamma\!\left(B_z-\tfrac{v}{c^2}E_y\right)=1.25\!\left(10^{-4}-\tfrac{0.6(3\times10^8)}{9\times10^{16}}(3\times10^4)\right)=1.25(10^{-4}-6\times10^{-5})=5\times10^{-5}\ \mathrm{T}\). So \(\vec E'=(0,1.5\times10^4,0)\), \(\vec B'=(0,0,5\times10^{-5})\). Both fields stay along \(\hat y\) and \(\hat z\) respectively, so \(\vec E'\cdot\vec B'=0=\vec E\cdot\vec B\). ✓
(B/C) Transforming away the electric field. In \(S\), \(\vec E=(0,E_0,0)\), \(\vec B=(0,0,B_0)\) with \(E_0=1.5\times10^{8}\ \mathrm{V/m}\), \(B_0=1.0\ \mathrm{T}\). Find the boost velocity along \(\hat x\) that makes \(\vec E'=0\), and the resulting \(\vec B'\). When is this possible?
Solution
\(E'_y=\gamma(E_0-vB_0)=0\Rightarrow v=E_0/B_0=1.5\times10^8\ \mathrm{m/s}=0.5c\) (possible because \(E_0<cB_0=3\times10^8\)). Then \(\gamma=1/\sqrt{1-0.25}=1.1547\) and \(B'_z=\gamma\!\left(B_0-\tfrac{v}{c^2}E_0\right)=1.1547\!\left(1.0-\tfrac{(1.5\times10^8)(1.5\times10^8)}{9\times10^{16}}\right)=1.1547(1-0.25)=0.866\ \mathrm{T}\). Check invariant: \(B_0^2-E_0^2/c^2=1-0.25=0.75\), and \(B'^2=0.866^2=0.75\). ✓ Possible precisely when \(\vec E\cdot\vec B=0\) and \(E<cB\).
(C) Field with a longitudinal component. In \(S\), \(\vec E=(2\times10^{3},\,10^{3},\,0)\ \mathrm{V/m}\), \(\vec B=(0,0,10^{-5})\ \mathrm{T}\). Boost along \(\hat x\) at \(v=0.6c\). Find \(\vec E'\) and \(\vec B'\).
Solution
\(\gamma=1.25\). Longitudinal: \(E'_x=E_x=2\times10^3\ \mathrm{V/m}\), \(B'_x=0\). Transverse \(E\): \(E'_y=\gamma(E_y-vB_z)=1.25(10^3-0.6(3\times10^8)(10^{-5}))=1.25(10^3-1.8\times10^3)=-10^3\ \mathrm{V/m}\); \(E'_z=\gamma(E_z+vB_y)=0\). Transverse \(B\): \(B'_z=\gamma(B_z-\tfrac{v}{c^2}E_y)=1.25(10^{-5}-\tfrac{1.8\times10^8}{9\times10^{16}}(10^3))=1.25(10^{-5}-2\times10^{-6})=10^{-5}\ \mathrm{T}\); \(B'_y=\gamma(B_y+\tfrac{v}{c^2}E_z)=0\). Result: \(\vec E'=(2\times10^3,-10^3,0)\ \mathrm{V/m}\), \(\vec B'=(0,0,10^{-5})\ \mathrm{T}\). The longitudinal \(E_x\) is untouched while the transverse part is mixed and reduced.
(C) Null (radiation) field is boost-stable. In \(S\), \(\vec E=(0,cB_0,0)\), \(\vec B=(0,0,B_0)\) with \(B_0=10^{-6}\ \mathrm{T}\) (so \(E_0=cB_0=300\ \mathrm{V/m}\), \(\vec E\perp\vec B\), \(E=cB\)). Boost along \(\hat x\) at \(v=0.8c\). Show \(E'=cB'\) still holds.
Solution
\(\gamma=1/\sqrt{1-0.64}=1.667\). \(E'_y=\gamma(E_0-vB_0)=1.667(300-0.8(3\times10^8)(10^{-6}))=1.667(300-240)=100\ \mathrm{V/m}\). \(B'_z=\gamma\!\left(B_0-\tfrac{v}{c^2}E_0\right)=1.667\!\left(10^{-6}-\tfrac{0.8}{3\times10^8}(300)\right)=1.667(10^{-6}-8\times10^{-7})=3.33\times10^{-7}\ \mathrm{T}\). Then \(cB'_z=(3\times10^8)(3.33\times10^{-7})=100\ \mathrm{V/m}=E'_y\). ✓ Both invariants \(\vec E\cdot\vec B=0\) and \(B^2-E^2/c^2=0\) are preserved, so the null structure \(E'=cB'\), \(\vec E'\perp\vec B'\) survives in every frame — as it must for a plane wave.