Covariant Formulation of Electrodynamics

Derivation

Covariant Formulation of Electrodynamics

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Statement

In Minkowski spacetime with metric signature \((+,-,-,-)\), define the antisymmetric electromagnetic field tensor \(F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\) from the four-potential \(A^{\mu}=(\varphi/c,\ \vec{A})\), and the four-current \(J^{\mu}=(c\rho,\ \vec{J})\). Then the two inhomogeneous Maxwell equations (Gauss's law and the Ampère–Maxwell law) are the single covariant statement \(\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\), while the two homogeneous equations (\(\nabla\cdot\vec{B}=0\) and Faraday's law) are the identity \(\partial_{\lambda}F_{\mu\nu}+\partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}=0\), which holds automatically because \(F=dA\). Both equations are manifestly Lorentz covariant, and \(\vec{E}\) and \(\vec{B}\) are unified as components of one rank-2 tensor.

Why it matters

The three-vector form of Maxwell's equations hides the fact that electricity and magnetism are one field seen from different inertial frames. Writing them as \(\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\) makes Lorentz covariance manifest: every index is a spacetime index, so if the equation holds in one frame it holds in all frames, with no separate check required. What one observer calls a pure electric field, another moving observer calls a mixture of \(\vec{E}\) and \(\vec{B}\) — exactly as the tensor transformation law demands.

This formulation is also the gateway to the rest of physics. Charge conservation drops out as an algebraic identity, gauge invariance becomes the statement that \(F^{\mu\nu}\) is unchanged by \(A^{\mu}\to A^{\mu}+\partial^{\mu}\chi\), and the whole theory follows from a Lorentz-invariant action \(S=-\tfrac{1}{4\mu_{0}}\int F_{\mu\nu}F^{\mu\nu}\,d^{4}x-\int A_{\mu}J^{\mu}\,d^{4}x\), the template for every gauge theory that came after.

Assumptions

Flat Minkowski spacetime with a global inertial frame.If dropped (curved spacetime), \(\partial_{\mu}\) must be promoted to the covariant derivative \(\nabla_{\mu}\) and \(\eta_{\mu\nu}\to g_{\mu\nu}(x)\); the divergence becomes \(\tfrac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}\,F^{\mu\nu})=\mu_{0}J^{\nu}\).

Metric signature \((+,-,-,-)\) with \(\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)\).The opposite convention \((-,+,+,+)\) flips the sign of \(F^{0i}\) and of the source term; the physics is identical but every explicit sign below would swap.

The potential \(A^{\mu}\) exists globally and is at least \(C^{2}\).If dropped, \(F=dA\) fails; mixed partials no longer commute, the Bianchi identity is not automatic, and magnetic monopoles or Aharonov–Bohm topology can appear.

SI units with \(c^{2}=1/(\mu_{0}\varepsilon_{0})\).If dropped for Gaussian or Heaviside–Lorentz units, factors of \(c\) and \(4\pi\) redistribute; the tensor structure is unchanged.

Derivation

\[ x^{\mu}=(ct,\ \vec{x}),\qquad \partial_{\mu}=\frac{\partial}{\partial x^{\mu}}=\left(\frac{1}{c}\frac{\partial}{\partial t},\ \nabla\right),\qquad \partial^{\mu}=\eta^{\mu\nu}\partial_{\nu}=\left(\frac{1}{c}\frac{\partial}{\partial t},\ -\nabla\right) \]

Fix the coordinate four-vector and the gradient operators. Raising the index with \(\eta^{\mu\nu}\) flips the sign of the spatial part. A

\[ A^{\mu}=\left(\frac{\varphi}{c},\ \vec{A}\right),\qquad J^{\mu}=(c\rho,\ \vec{J}) \]

Package the scalar and vector potentials into one four-vector, and charge and current densities into another. That these transform as four-vectors is required for covariance and is confirmed a posteriori by the field-strength construction. A

\[ F^{\mu\nu}\equiv\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu} \]

Define the field-strength tensor as the antisymmetrised four-gradient of the potential. Antisymmetry \(F^{\mu\nu}=-F^{\nu\mu}\) is built in, so \(F^{\mu\nu}\) has six independent components — exactly the three of \(\vec{E}\) plus three of \(\vec{B}\). A

\[ F^{0i}=\partial^{0}A^{i}-\partial^{i}A^{0}=\frac{1}{c}\frac{\partial A_{i}}{\partial t}+\frac{1}{c}\frac{\partial\varphi}{\partial x_{i}}=-\frac{E_{i}}{c} \]

Evaluate the time–space components. Using \(\vec{E}=-\nabla\varphi-\partial\vec{A}/\partial t\), the bracket is \(-E_{i}\). This identifies the electric field with the \(F^{0i}\) block. B

\[ F^{ij}=\partial^{i}A^{j}-\partial^{j}A^{i}=-(\partial_{i}A_{j}-\partial_{j}A_{i})=-\varepsilon^{ijk}B_{k} \]

Evaluate the space–space components. Using \(\vec{B}=\nabla\times\vec{A}\), i.e. \(B_{k}=\varepsilon_{kij}\partial_{i}A_{j}\), the space–space block is the magnetic field. Explicitly \(F^{12}=-B_{z}\), \(F^{23}=-B_{x}\), \(F^{31}=-B_{y}\). B

\[ F^{\mu\nu}=\begin{pmatrix} 0 & -E_{x}/c & -E_{y}/c & -E_{z}/c \\ E_{x}/c & 0 & -B_{z} & B_{y} \\ E_{y}/c & B_{z} & 0 & -B_{x} \\ E_{z}/c & -B_{y} & B_{x} & 0 \end{pmatrix} \]

Assemble the full matrix from Steps 4–5. One antisymmetric object now carries both fields; a Lorentz boost mixing rows and columns is precisely what mixes \(\vec{E}\) and \(\vec{B}\). B

\[ \partial_{\mu}F^{\mu 0}=\partial_{i}F^{i0}=\partial_{i}\frac{E_{i}}{c}=\frac{1}{c}\nabla\cdot\vec{E} \;\overset{!}{=}\; \mu_{0}J^{0}=\mu_{0}c\rho \]

Take \(\nu=0\) of \(\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\). The \(\mu=0\) term vanishes by antisymmetry (\(F^{00}=0\)). Multiplying by \(c\) and using \(c^{2}=1/(\mu_{0}\varepsilon_{0})\) gives \(\nabla\cdot\vec{E}=\rho/\varepsilon_{0}\): Gauss's law. B

\[ \partial_{\mu}F^{\mu j}=\partial_{0}F^{0j}+\partial_{i}F^{ij}=-\frac{1}{c^{2}}\frac{\partial E_{j}}{\partial t}+(\nabla\times\vec{B})_{j}\;\overset{!}{=}\;\mu_{0}J^{j} \]

Take \(\nu=j\). The time term gives the displacement current; the space term gives the curl since \(\partial_{i}F^{ij}=-\varepsilon^{ijk}\partial_{i}B_{k}=(\nabla\times\vec{B})_{j}\). Rearranged: \(\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\varepsilon_{0}\,\partial\vec{E}/\partial t\), the Ampère–Maxwell law. B

\[ \partial_{\nu}\partial_{\mu}F^{\mu\nu}=\mu_{0}\partial_{\nu}J^{\nu}\quad\Longrightarrow\quad \partial_{\nu}J^{\nu}=0 \]

Contract the field equation with \(\partial_{\nu}\). The left side is the contraction of the symmetric \(\partial_{\nu}\partial_{\mu}\) with the antisymmetric \(F^{\mu\nu}\), which vanishes identically. Charge conservation \(\partial\rho/\partial t+\nabla\cdot\vec{J}=0\) is therefore a built-in consistency requirement, not a separate postulate. C

\[ \partial_{\lambda}F_{\mu\nu}+\partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}=0 \]

Because \(F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\), each term expands into second derivatives of \(A\) that cancel in pairs (equality of mixed partials). This Bianchi identity is equivalent to \(\partial_{\mu}{}^{\star}\!F^{\mu\nu}=0\) with the dual \({}^{\star}\!F^{\mu\nu}=\tfrac{1}{2}\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}\), and unpacks to \(\nabla\cdot\vec{B}=0\) and \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\). C

Result

\[ \boxed{\ \partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\ ,\qquad \partial_{\lambda}F_{\mu\nu}+\partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda\mu}=0\ } \]

Reading. The first equation is Gauss's law (\(\nu=0\)) and the Ampère–Maxwell law (\(\nu=1,2,3\)) fused into one four-component tensor equation; the source is the four-current. The second is the homogeneous pair (\(\nabla\cdot\vec{B}=0\) and Faraday's law), which holds automatically because the field is derived from a potential. Every symbol carries spacetime indices, so both equations keep their form under any Lorentz transformation — covariance is manifest, not something to be verified frame by frame. \(\vec{E}\) and \(\vec{B}\) are not independent fields but components of the single object \(F^{\mu\nu}\).

Units check. Components of \(F^{\mu\nu}\) are \(E/c\) (in \(\mathrm{V\,s\,m^{-2}}=\mathrm{T}\)) and \(B\) (in \(\mathrm{T}\)), so \(F^{\mu\nu}\) is uniformly in tesla. Then \(\partial_{\mu}F^{\mu\nu}\) has units \(\mathrm{T\,m^{-1}}\). On the right, \(\mu_{0}\) is \(\mathrm{T\,m\,A^{-1}}\) and \(J^{\nu}\) (both \(c\rho\) and \(\vec{J}\)) is \(\mathrm{A\,m^{-2}}\), giving \(\mathrm{T\,m^{-1}}\). Both sides match.

Limiting cases

Electrostatics (\(\partial_{t}\to0\), \(\vec{B}=0\)): only \(F^{0i}\) survives and \(\partial_{i}F^{i0}=\mu_{0}c\rho\) reduces to \(\nabla\cdot\vec{E}=\rho/\varepsilon_{0}\).
Magnetostatics (\(\partial_{t}\to0\), \(\rho=0\)): only \(F^{ij}\) survives and the field equation becomes \(\nabla\times\vec{B}=\mu_{0}\vec{J}\).
Source-free vacuum (\(J^{\nu}=0\)): \(\partial_{\mu}F^{\mu\nu}=0\) together with the Bianchi identity yields \(\Box A^{\nu}=0\) in Lorenz gauge — electromagnetic waves at speed \(c\).
Non-relativistic charge (\(v\ll c\)): the \(F^{0i}\) (electric) block dominates the \(F^{ij}\) (magnetic) block by a factor \(c/v\), recovering the Coulomb picture with magnetism as a small correction.

Breaks when

Spacetime is curved. In a gravitational field the flat metric \(\eta_{\mu\nu}\) and the ordinary divergence \(\partial_{\mu}\) are no longer valid; one must use \(\tfrac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}\,F^{\mu\nu})=\mu_{0}J^{\nu}\), and the Bianchi identity uses covariant derivatives.
Magnetic monopoles exist. The homogeneous equation \(\partial_{\mu}{}^{\star}\!F^{\mu\nu}=0\) fails; it must be replaced by \(\partial_{\mu}{}^{\star}\!F^{\mu\nu}=\mu_{0}J_{m}^{\nu}\), and \(F\) can no longer be written globally as \(dA\).
Fields are strong enough for quantum effects. Near the critical field \(E_{c}=m_{e}^{2}c^{3}/(e\hbar)\approx1.3\times10^{18}\,\mathrm{V\,m^{-1}}\), vacuum polarization makes the equations nonlinear (Euler–Heisenberg); the linear Maxwell tensor equation is only the leading term.
Media with polarization/magnetization. Inside matter one splits into \(F^{\mu\nu}\) and the excitation tensor \(H^{\mu\nu}\) (built from \(\vec{D}\), \(\vec{H}\)); the source equation becomes \(\partial_{\mu}H^{\mu\nu}=J_{\text{free}}^{\nu}\), and constitutive relations are needed.

Failure modes

Sign errors from the metric convention. Mixing \((+,-,-,-)\) and \((-,+,+,+)\) mid-calculation flips \(F^{0i}\) and the source sign; pick one signature and never switch.
Forgetting the factor of \(c\). The electric entries are \(E/c\), not \(E\), and \(J^{0}=c\rho\), not \(\rho\); omitting these breaks the units check and misplaces Gauss's law.
Confusing \(F^{\mu\nu}\) with \(F_{\mu\nu}\). Lowering both indices flips the sign of the \(E/c\) entries (one time index lowered); the magnetic block is unchanged. Track index height carefully.
Treating the Bianchi identity as a dynamical equation. It carries no source and follows from \(F=dA\); it is a constraint, not something to solve for the fields.
Assuming \(\vec{E}\) and \(\vec{B}\) transform as separate three-vectors. They do not — only the combined tensor \(F^{\mu\nu}\) transforms cleanly, which is why a boost mixes them.
Losing antisymmetry. Writing \(F^{\mu\nu}=\partial^{\mu}A^{\nu}\) (dropping the second term) destroys gauge invariance and charge conservation.

Discussion

The deepest content of \(\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\) is that electromagnetism was already relativistic before relativity was discovered. Maxwell's equations are not approximately Lorentz covariant to be fixed at high speed; they are exactly covariant, and it was their incompatibility with Galilean transformations that forced Einstein to abandon absolute time. The tensor form makes this visible in a single line: because every free index is a spacetime index and both sides transform the same way under \(\Lambda^{\mu}{}_{\nu}\), a valid equation in one inertial frame is automatically valid in every other.

The unification of \(\vec{E}\) and \(\vec{B}\) is not a notational convenience but a physical statement. A static point charge produces only \(\vec{E}\); boost to a frame where it moves and the same \(F^{\mu\nu}\), now with rotated components, shows a \(\vec{B}\) as well. The magnetic force between currents is, from this viewpoint, the electric force seen relativistically — a length-contracted charge imbalance. The two Lorentz invariants \(F_{\mu\nu}F^{\mu\nu}=2(B^{2}-E^{2}/c^{2})\) and \({}^{\star}\!F_{\mu\nu}F^{\mu\nu}\propto\vec{E}\cdot\vec{B}\) tell every observer, independent of frame, whether a field is electric-dominated, magnetic-dominated, or null (as for a plane wave, where both invariants vanish).

Gauge invariance also becomes transparent. The substitution \(A^{\mu}\to A^{\mu}+\partial^{\mu}\chi\) leaves \(F^{\mu\nu}\) untouched because \(\partial^{\mu}\partial^{\nu}\chi\) is symmetric and cancels in the antisymmetric combination. This redundancy is not a flaw; it is the seed of the entire Standard Model, where promoting the global phase symmetry \(U(1)\) to a local one forces the existence of the photon and fixes its couplings. The covariant Maxwell equations are the abelian prototype of every non-abelian gauge theory that followed.

At the level of an action principle, both equations descend from \(S=-\tfrac{1}{4\mu_{0}}\int F_{\mu\nu}F^{\mu\nu}\,d^{4}x-\int A_{\mu}J^{\mu}\,d^{4}x\). Varying \(A_{\mu}\) yields \(\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\) as the Euler–Lagrange equation, while the homogeneous equation is a geometric identity requiring no variation at all — it is the statement \(dF=d(dA)=0\) in the language of differential forms, where \(F\) is a 2-form and Maxwell's equations read simply \(dF=0\) and \(d{}^{\star}F=\mu_{0}{}^{\star}J\). Noether's theorem applied to the translation symmetry of this action delivers the symmetric, gauge-invariant stress–energy tensor \(T^{\mu\nu}\), whose components are the energy density, the Poynting vector, and the Maxwell stresses — momentum and energy of the field itself.

Common misconceptions. The covariant formulation does not add new physics beyond the four vector equations — it reorganises them so that their relativistic content is manifest. It is also a common error to think the tensor form "derives" relativity from electromagnetism; rather, it displays a compatibility that was there all along. Finally, \(F^{\mu\nu}=0\) does not imply \(A^{\mu}=0\): the potential can be pure gauge (\(A^{\mu}=\partial^{\mu}\chi\)) yet physically detectable through topology, as in the Aharonov–Bohm effect.

Worked examples

Boost of a pure electric field. In frame \(S\) there is a uniform field \(\vec{E}=(0,\ E_{y},\ 0)\) with \(E_{y}=1000\ \mathrm{V\,m^{-1}}\) and \(\vec{B}=0\). Find \(\vec{E}\,'\) and \(\vec{B}\,'\) in a frame \(S'\) moving at \(v=0.6c\) along \(+x\). B

\[ E_{y}'=\gamma\,(E_{y}-v B_{z}),\qquad B_{z}'=\gamma\left(B_{z}-\frac{v}{c^{2}}E_{y}\right) \]

These transformation laws follow directly from applying \(\Lambda\) to \(F^{\mu\nu}\); the boost mixes the \(F^{0y}\) and \(F^{xy}\) entries. Symbols first, numbers next. B

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

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

\[ E_{y}'=1.25\times1000=1250\ \mathrm{V\,m^{-1}} \]

With \(B_{z}=0\) the electric field is amplified by \(\gamma\). A

\[ B_{z}'=-\gamma\frac{v}{c^{2}}E_{y}=-1.25\times\frac{1.8\times10^{8}}{9\times10^{16}}\times1000 \]

Insert \(v=0.6c=1.8\times10^{8}\ \mathrm{m\,s^{-1}}\) and \(c^{2}=9\times10^{16}\ \mathrm{m^{2}\,s^{-2}}\). A

\[ E_{y}'=1250\ \mathrm{V\,m^{-1}},\qquad B_{z}'=-2.5\times10^{-6}\ \mathrm{T}=-2.5\ \mu\mathrm{T} \]

Reading. A magnetic field appears in \(S'\) even though there was none in \(S\): the moving observer sees the boosted charges as a current. This is the sense in which magnetism is relativistic electricity. The invariant \(B'^{2}-E'^{2}/c^{2}=B_{z}'^{2}-E_{y}'^{2}/c^{2}\) equals \(-E_{y}^{2}/c^{2}\) in both frames, confirming the field stays electric-dominated.

Invariants of a plane electromagnetic wave. A wave in vacuum has \(E=300\ \mathrm{V\,m^{-1}}\) with \(\vec{E}\perp\vec{B}\) and \(B=E/c\). Compute the two Lorentz invariants and interpret. B

\[ I_{1}=F_{\mu\nu}F^{\mu\nu}=2\left(B^{2}-\frac{E^{2}}{c^{2}}\right),\qquad I_{2}={}^{\star}\!F_{\mu\nu}F^{\mu\nu}=-\frac{4}{c}\,\vec{E}\cdot\vec{B} \]

The only two independent scalars formable from \(F^{\mu\nu}\); both are frame-independent by construction. C

\[ B=\frac{E}{c}=\frac{300}{3\times10^{8}}=1.0\times10^{-6}\ \mathrm{T} \]

Compute the magnetic amplitude from the vacuum relation. A

\[ B^{2}=1.0\times10^{-12}\ \mathrm{T^{2}},\qquad \frac{E^{2}}{c^{2}}=\frac{9\times10^{4}}{9\times10^{16}}=1.0\times10^{-12}\ \mathrm{T^{2}} \]

Evaluate each term of \(I_{1}\); they are equal. And \(\vec{E}\cdot\vec{B}=0\) since the fields are orthogonal, so \(I_{2}=0\). A

\[ I_{1}=2\,(1.0\times10^{-12}-1.0\times10^{-12})=0,\qquad I_{2}=0 \]

Reading. Both invariants vanish: this is a null electromagnetic field, the signature of radiation. Because the invariants are the same in every frame, no observer can boost to a frame where the wave is purely electric or purely magnetic — the relation \(E=cB\) and the orthogonality hold universally. Any field with \(I_{1}=I_{2}=0\) (but \(F\neq0\)) is radiation-like.

Problems

(A) Components. A field has \(\vec{E}=(E_{x},0,0)\) and \(\vec{B}=(0,0,B_{z})\). Write out all nonzero components of \(F^{\mu\nu}\).
Solution
From the matrix in Step 6, the only nonzero entries involve \(E_{x}\) (the \(0\)–\(1\) block) and \(B_{z}\) (the \(1\)–\(2\) block):
\[ F^{01}=-\frac{E_{x}}{c},\quad F^{10}=+\frac{E_{x}}{c},\quad F^{12}=-B_{z},\quad F^{21}=+B_{z}, \]
and all others vanish. Antisymmetry gives each pair opposite signs; the diagonal is zero.
(B) Gauss from the tensor. Starting from \(\partial_{\mu}F^{\mu 0}=\mu_{0}J^{0}\), recover Gauss's law explicitly, showing where \(c^{2}=1/(\mu_{0}\varepsilon_{0})\) enters.
Solution
The \(\mu=0\) term vanishes (\(F^{00}=0\)), leaving \(\partial_{i}F^{i0}=\mu_{0}J^{0}\). With \(F^{i0}=+E_{i}/c\) and \(J^{0}=c\rho\):
\[ \frac{1}{c}\nabla\cdot\vec{E}=\mu_{0}c\rho\ \Rightarrow\ \nabla\cdot\vec{E}=\mu_{0}c^{2}\rho. \]
Substituting \(\mu_{0}c^{2}=\mu_{0}\cdot\dfrac{1}{\mu_{0}\varepsilon_{0}}=\dfrac{1}{\varepsilon_{0}}\) gives \(\nabla\cdot\vec{E}=\rho/\varepsilon_{0}\).
(B) Field boost. In frame \(S\), \(\vec{E}=(0,\ 500,\ 0)\ \mathrm{V\,m^{-1}}\) and \(\vec{B}=0\). Find \(E_{y}'\) and \(B_{z}'\) for a boost at \(v=0.8c\) along \(+x\).
Solution
\[ \gamma=\frac{1}{\sqrt{1-0.64}}=\frac{1}{0.6}=1.667. \]\[ E_{y}'=\gamma E_{y}=1.667\times500=833\ \mathrm{V\,m^{-1}}. \]\[ B_{z}'=-\gamma\frac{v}{c^{2}}E_{y}=-1.667\times\frac{2.4\times10^{8}}{9\times10^{16}}\times500=-2.2\times10^{-6}\ \mathrm{T}. \]
So \(E_{y}'\approx833\ \mathrm{V\,m^{-1}}\) and \(B_{z}'\approx-2.2\ \mu\mathrm{T}\): a magnetic field is induced by the boost.
(C) Charge conservation. Show that \(\partial_{\mu}F^{\mu\nu}=\mu_{0}J^{\nu}\) forces \(\partial_{\nu}J^{\nu}=0\), and write this out in \(3+1\) form.
Solution
Apply \(\partial_{\nu}\) to both sides: \(\partial_{\nu}\partial_{\mu}F^{\mu\nu}=\mu_{0}\partial_{\nu}J^{\nu}\). The operator \(\partial_{\nu}\partial_{\mu}\) is symmetric in \(\mu\nu\) while \(F^{\mu\nu}\) is antisymmetric, so their double contraction is zero. Hence \(\partial_{\nu}J^{\nu}=0\). Writing \(J^{\nu}=(c\rho,\vec{J})\) and \(\partial_{\nu}=(\tfrac{1}{c}\partial_{t},\nabla)\):
\[ \frac{1}{c}\frac{\partial(c\rho)}{\partial t}+\nabla\cdot\vec{J}=\frac{\partial\rho}{\partial t}+\nabla\cdot\vec{J}=0, \]
the continuity equation.
(C) Invariants and frame choice. A field has \(\vec{E}=(10^{6},0,0)\ \mathrm{V\,m^{-1}}\) and \(\vec{B}=(2\times10^{-3},0,0)\ \mathrm{T}\) (both along \(x\)). Compute both invariants and determine whether a frame exists with (i) pure \(\vec{E}\), (ii) pure \(\vec{B}\), or (iii) \(\vec{E}\parallel\vec{B}\).
Solution
\[ I_{1}=2\left(B^{2}-\frac{E^{2}}{c^{2}}\right)=2\left(4\times10^{-6}-\frac{10^{12}}{9\times10^{16}}\right)=2(4\times10^{-6}-1.11\times10^{-5})=-1.4\times10^{-5}\ \mathrm{T^{2}}. \]\[ I_{2}\propto\vec{E}\cdot\vec{B}=E_{x}B_{x}=(10^{6})(2\times10^{-3})=2\times10^{3}\neq0. \]
Since \(I_{2}=\vec{E}\cdot\vec{B}\neq0\), the fields are not orthogonal in any frame, so neither a pure-\(\vec{E}\) nor a pure-\(\vec{B}\) frame exists (both would require \(\vec{E}\cdot\vec{B}=0\)). Because \(I_{2}\neq0\), there is a frame in which \(\vec{E}\parallel\vec{B}\) — here the fields are already parallel (both along \(x\)) in the given frame. \(I_{1}<0\) tells us the field is electric-dominated, so in the parallel frame the electric part exceeds \(cB\).