The Electromagnetic Field Tensor

Derivation

The Electromagnetic Field Tensor

D-175 Home PU-204 Threads fields · symmetry Depends on Potentials and Gauge Freedom, Assembly of Maxwell's Equations, lorentz-transformation-derivation

Statement

The scalar and vector potentials are assembled into a four-vector \( A^\mu = (\phi/c,\ \vec{A}) \), and the six field components into the antisymmetric field tensor \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \), whose time–space block is \( \vec{E}/c \) and whose space–space block is \( \vec{B} \). With the four-current \( J^\mu = (c\rho,\ \vec{J}) \), Maxwell's four vector equations collapse into two manifestly Lorentz-covariant tensor equations: the sourced equation \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) (Gauss + Ampère–Maxwell) and the identity \( \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \) (no monopoles + Faraday), with charge conservation \( \partial_\nu J^\nu = 0 \) following automatically from antisymmetry.

Why it matters

Written as four separate vector equations, Maxwell's theory hides its deepest property: it was Lorentz-covariant all along, decades before relativity was formulated. Packaging \( (\phi, \vec{A}) \) into \( A^\mu \) and \( (\vec{E}, \vec{B}) \) into \( F_{\mu\nu} \) makes that symmetry manifest — every index is a spacetime index, so both equations keep their exact form in every inertial frame with no further checking. The frame-dependent split into "electric" and "magnetic" is exposed as bookkeeping: the invariant object is the tensor.

The covariant form is also the launch point for everything downstream: the field transformation laws come from one line of tensor algebra, charge conservation becomes an algebraic identity rather than an extra postulate, the Lorenz-gauge wave equation \( \Box A^\nu = \mu_0 J^\nu \) drops out in one step, and the whole structure is the template copied by every later gauge theory — the weak, strong, and gravitational fields are all described by curvatures of connections modelled on \( F = \mathrm{d}A \).

Assumptions

Spacetime is flat Minkowski space with metric \( \eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1) \) and coordinates \( x^\mu = (ct, \vec{x}) \).If dropped for curved spacetime, every \( \partial_\mu \) acting on a tensor must become a covariant derivative \( \nabla_\mu \); the definition \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \) survives (Christoffel symbols cancel by antisymmetry) but the sourced equation becomes \( \nabla_\mu F^{\mu\nu} = \mu_0 J^\nu \) and picks up metric determinant factors.

The fields derive from potentials: \( \vec{E} = -\nabla\phi - \partial\vec{A}/\partial t \), \( \vec{B} = \nabla\times\vec{A} \) (prior result: scalar and vector potentials).If dropped, \( F_{\mu\nu} \) is not an exact "curl" of \( A_\mu \), the homogeneous equation is no longer an identity, and magnetic monopoles become admissible.

Maxwell's equations in three-vector form hold in every inertial frame (prior result: Maxwell assembly).If dropped, there is nothing to repackage — the covariant equations below are exactly equivalent to the four vector equations, no stronger and no weaker.

Electric charge is a Lorentz scalar, so \( J^\mu = (c\rho,\ \vec{J}) \) is a four-vector.If dropped, \( \rho \) and \( \vec{J} \) would not fit into one four-vector and \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) could not be covariant: a tensor equation needs tensors on both sides. Experimentally, charge quantisation is velocity-independent to better than \( 10^{-20} \) (neutrality of atoms with fast-moving electrons).

\( A^\mu \) transforms as a genuine four-vector under the Lorentz transformation (prior result: Lorentz transformation).If dropped, \( \partial_\mu A_\nu - \partial_\nu A_\mu \) would not be a rank-2 tensor and "manifest covariance" would be an empty notation. The assumption is self-consistent: it is exactly what makes the Lorenz-gauge wave equation frame-independent.

Derivation

\[ A^\mu \equiv \left( \frac{\phi}{c},\ A_x,\ A_y,\ A_z \right), \qquad A_\mu = \eta_{\mu\nu} A^\nu = \left( \frac{\phi}{c},\ -A_x,\ -A_y,\ -A_z \right) \]

Definition of the four-potential. The factor \( 1/c \) gives the time slot the same units as \( \vec{A} \) (\( \mathrm{V\,s\,m^{-1}} = \mathrm{T\,m} \)); lowering with \( \eta \) flips the sign of the spatial part only. A

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

The four-gradient. \( \partial_\mu \) carries a naturally lower index (it transforms with the inverse Lorentz matrix), which is why the gradient of a scalar is a covariant vector. B

\[ \phi \to \phi - \frac{\partial \chi}{\partial t}, \quad \vec{A} \to \vec{A} + \nabla\chi \qquad \Longleftrightarrow \qquad A_\mu \to A_\mu - \partial_\mu \chi \]

The gauge freedom of the potentials (prior result) compresses into a single four-vector statement: both three-dimensional replacements are the components of one four-gradient shift. B

\[ F_{\mu\nu} \equiv \partial_\mu A_\nu - \partial_\nu A_\mu, \qquad F_{\mu\nu} \to F_{\mu\nu} - \left( \partial_\mu \partial_\nu \chi - \partial_\nu \partial_\mu \chi \right) = F_{\mu\nu} \]

Definition of the field tensor: the antisymmetrised derivative of \( A_\mu \). It is gauge-invariant because mixed partial derivatives commute — gauge invariance is what selects the antisymmetric combination. Antisymmetry \( F_{\mu\nu} = -F_{\nu\mu} \) leaves \( \binom{4}{2} = 6 \) independent components: exactly the count of \( (\vec{E},\vec{B}) \). C

\[ F_{0i} = \partial_0 A_i - \partial_i A_0 = \frac{1}{c}\frac{\partial}{\partial t}(-A_i) - \partial_i\!\left( \frac{\phi}{c} \right) = -\frac{1}{c}\left( \frac{\partial A_i}{\partial t} + \partial_i \phi \right) = \frac{E_i}{c} \]

Insert the lowered components from step 1 and the gradient from step 2, then recognise the bracket as \( -E_i \) from the potential relations. The time–space block of the tensor is the electric field, in units of tesla via the \( 1/c \). B

\[ F_{ij} = \partial_i A_j - \partial_j A_i = -\left( \partial_i A^j - \partial_j A^i \right) = -\varepsilon_{ijk} B_k \]

The lowered spatial components pick up a minus sign (step 1); the surviving antisymmetric derivative is the curl of \( \vec{A} \), i.e. \( \vec{B} \), contracted with the Levi-Civita symbol. The space–space block is the magnetic field. B

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

Raising both indices flips the sign of every entry with exactly one time index (the electric block) and leaves the doubly-spatial block untouched (two sign flips cancel). This contravariant matrix is the one that appears in the sourced equation. B

\[ J^\mu \equiv (c\rho,\ \vec{J}), \qquad \partial_\mu J^\mu = \frac{\partial \rho}{\partial t} + \nabla\cdot\vec{J} = 0 \]

Definition of the four-current: charge density and current density share one four-vector (their transformation into each other under boosts is the relativity of "how much charge is moving"). The continuity equation is its vanishing four-divergence. A

\[ \partial_\mu F^{\mu 0} = \partial_i F^{i0} = \frac{1}{c}\,\nabla\cdot\vec{E} \ \stackrel{!}{=}\ \mu_0 J^0 = \mu_0 c\rho \quad \Longrightarrow \quad \nabla\cdot\vec{E} = \mu_0 c^2 \rho = \frac{\rho}{\varepsilon_0} \]

Propose the covariant sourced equation \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) and check its \( \nu = 0 \) component against the known physics: with \( F^{i0} = E_i/c \) from step 7, it is exactly Gauss's law, using \( c^2 = 1/(\mu_0 \varepsilon_0) \). C

\[ \partial_\mu F^{\mu j} = \partial_0 F^{0j} + \partial_i F^{ij} = -\frac{1}{c^2}\frac{\partial E_j}{\partial t} + \left( \nabla\times\vec{B} \right)_j = \mu_0 J_j \]

The \( \nu = j \) component: \( \partial_0 F^{0j} = \frac{1}{c}\partial_t(-E_j/c) \), and \( \partial_i F^{ij} = -\varepsilon_{ijk}\partial_i B_k = (\nabla\times\vec{B})_j \). Rearranged, this is Ampère–Maxwell, \( \nabla\times\vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\,\partial_t\vec{E} \), displacement current included automatically. One tensor equation therefore reproduces both sourced Maxwell equations. C

\[ \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = \partial_\lambda \partial_\mu A_\nu - \partial_\lambda \partial_\nu A_\mu + \text{cyclic} \equiv 0 \]

The Bianchi identity: substituting \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \), all six second-derivative terms cancel pairwise because partials commute. This is not a new law — it holds identically for any tensor built as the curl of a four-vector. Only the totally antisymmetric part survives, so there are \( \binom{4}{3} = 4 \) independent choices of \( (\lambda\mu\nu) \): four equations. C

\[ (\lambda\mu\nu)=(1,2,3):\ \ \nabla\cdot\vec{B} = 0; \qquad (\lambda\mu\nu)=(0,i,j):\ \ \nabla\times\vec{E} = -\frac{\partial \vec{B}}{\partial t} \]

Extract components using steps 5–6: the all-spatial choice gives \( -(\partial_x B_x + \partial_y B_y + \partial_z B_z) = 0 \); each mixed choice gives one Cartesian component of Faraday's law. The two homogeneous Maxwell equations are thus consequences of the fields having potentials at all. B

\[ \partial_\nu \partial_\mu F^{\mu\nu} = 0 \ \ (\text{symmetric} \times \text{antisymmetric}) \quad \Longrightarrow \quad \partial_\nu J^\nu = 0 \]

Consistency: take the four-divergence of the sourced equation. The left side vanishes identically because \( \partial_\nu \partial_\mu \) is symmetric under \( \mu \leftrightarrow \nu \) while \( F^{\mu\nu} \) is antisymmetric. Charge conservation is therefore not an extra input — the field equations cannot be solved for a non-conserved current. B

\[ \partial_\mu F^{\mu\nu} = \Box A^\nu - \partial^\nu\!\left( \partial_\mu A^\mu \right) = \mu_0 J^\nu \ \ \xrightarrow{\ \text{Lorenz gauge } \partial_\mu A^\mu = 0\ } \ \ \Box A^\nu = \mu_0 J^\nu \]

Rewrite the sourced equation in terms of the potential, with \( \Box \equiv \partial_\mu \partial^\mu = \frac{1}{c^2}\partial_t^2 - \nabla^2 \). Imposing the Lorenz condition (available by step 3's gauge freedom, and itself a Lorentz-scalar condition) decouples the components into four wave equations — the covariant form of the potential wave equations from the prior result. C

Result

\[ F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu, \qquad \partial_\mu F^{\mu\nu} = \mu_0 J^\nu, \qquad \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \]

Reading. All of electromagnetism is two tensor statements about one antisymmetric object. The first equation couples the field to its sources and contains Gauss's law (\( \nu = 0 \)) and Ampère–Maxwell (\( \nu = 1,2,3 \)). The second is an identity — automatically true once fields come from potentials — and contains \( \nabla\cdot\vec{B} = 0 \) and Faraday's law. Because every index is a spacetime index contracted correctly, both equations hold in identical form in every inertial frame: covariance is visible by inspection, and charge conservation \( \partial_\nu J^\nu = 0 \) is forced by antisymmetry.

Units check. Every entry of \( F^{\mu\nu} \) is in tesla: the magnetic block directly, the electric block as \( E/c \) with \( (\mathrm{V\,m^{-1}})/(\mathrm{m\,s^{-1}}) = \mathrm{V\,s\,m^{-2}} = \mathrm{T} \). Then \( \partial_\mu F^{\mu\nu} \) has units \( \mathrm{T\,m^{-1}} \). On the right, \( \mu_0 J^j \) has units \( (\mathrm{T\,m\,A^{-1}})(\mathrm{A\,m^{-2}}) = \mathrm{T\,m^{-1}} \) ✓, and \( \mu_0 J^0 = \mu_0 c \rho \) gives \( (\mathrm{T\,m\,A^{-1}})(\mathrm{m\,s^{-1}})(\mathrm{C\,m^{-3}}) = \mathrm{T\,m^{-1}} \) since \( \mathrm{C} = \mathrm{A\,s} \) ✓. The homogeneous equation is uniformly \( \mathrm{T\,m^{-1}} = 0 \) ✓.

Limiting cases

Static fields (\( \partial_t = 0 \)): the sourced equation splits cleanly into electrostatics \( \nabla\cdot\vec{E} = \rho/\varepsilon_0 \) and magnetostatics \( \nabla\times\vec{B} = \mu_0\vec{J} \) — the time–space coupling that makes the tensor necessary switches off.
Vacuum (\( J^\nu = 0 \)) in Lorenz gauge: \( \Box A^\nu = 0 \) — four decoupled wave equations propagating at \( c \), recovering electromagnetic waves.
Single charge at rest: \( J^\mu = (c\rho, \vec{0}) \), \( A^\mu = (\phi/c, \vec{0}) \) with \( \phi = q/4\pi\varepsilon_0 r \): only \( F^{i0} \) is non-zero and the tensor reduces to the Coulomb field.
Non-relativistic sources (\( v \ll c \)): \( J^0 = c\rho \) dominates \( |\vec{J}| = \rho v \) by \( c/v \); the field is electric-dominated and magnetism appears as the \( O(v/c) \) correction — consistent with \( B \sim vE/c^2 \) between the blocks of the tensor.
Slowly varying fields (\( \omega \ll c/L \)): the displacement-current term \( c^{-2}\partial_t E \) in step 10 is negligible against \( \nabla\times\vec{B} \), giving quasi-magnetostatics (the regime of circuit theory).

Breaks when

Magnetic monopoles exist. The homogeneous equation is an identity only because \( F = \) curl of \( A \). A monopole density \( \rho_m \) forces \( \partial_\mu \tilde{F}^{\mu\nu} = \mu_0 J_m^\nu \neq 0 \), no global four-potential exists, and the derivation's starting assumption fails; \( A^\mu \) survives only patchwise (Dirac strings / fibre bundles).
Curved spacetime. With gravity, \( \partial_\mu \to \nabla_\mu \) in the sourced equation, \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) becomes \( \frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,F^{\mu\nu}) = \mu_0 J^\nu \), and the global inertial frames used to state "manifest covariance under \( \Lambda \)" no longer exist — only local ones.
Polarisable media. Inside matter the sourced equation must be written for the excitation tensor \( G^{\mu\nu} \) (built from \( \vec{D}, \vec{H} \)) with only free charges on the right: \( \partial_\mu G^{\mu\nu} = J^\nu_{\text{free}} \). Using \( F^{\mu\nu} \) with free charges alone silently drops bound charge and current and gives wrong fields in dielectrics and magnets.
Quantum regime / extreme field strengths. Beyond \( E \sim E_{\text{crit}} = m_e^2 c^3 / e\hbar \approx 1.3\times10^{18}\ \mathrm{V\,m^{-1}} \), vacuum pair creation and Euler–Heisenberg nonlinearities modify the linear relation between field and source; classical linear Maxwell theory, and with it this pair of equations, is only the low-field limit of QED.

Failure modes

Dropping the \( 1/c \) in \( A^0 \). Writing \( A^\mu = (\phi, \vec{A}) \) makes the tensor dimensionally inhomogeneous and every downstream equation wrong by factors of \( c \). The check: all sixteen entries of \( F^{\mu\nu} \) must be in tesla.
Index-position sign errors. \( F^{0i} = -E_i/c \) but \( F_{0i} = +E_i/c \): raising flips the electric block only. Using the covariant matrix where the contravariant one is needed flips the sign of \( \vec{E} \) in the sourced equation.
Treating \( \partial^\mu \) and \( \partial_\mu \) as identical. \( \partial^i = -\partial_i \): forgetting the metric flip on the gradient's spatial part scrambles steps 5 and 14 (a classic source of a wrong-sign d'Alembertian).
Expecting the homogeneous equation to carry dynamics. \( \partial_{[\lambda}F_{\mu\nu]} = 0 \) is an identity given potentials exist — it constrains which field configurations are allowed, but only \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) contains the response to sources.
Counting equations wrongly. Claiming "two equations instead of eight is fewer laws": the tensor equations have \( 4 + 4 \) components — the same eight scalar equations, reorganised so covariance is manifest. Nothing was discarded.
Metric-signature drift. Mixing \( (+,-,-,-) \) and \( (-,+,+,+) \) conventions mid-derivation flips the sign of \( \Box \) and of the electric block; pick one signature and audit every raised index against it.

Discussion

The covariant form settles a question that hung over nineteenth-century physics: Maxwell's equations are not invariant under Galilean transformations, and this was originally read as evidence for a preferred aether frame. Writing them as \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) and the Bianchi identity shows their true symmetry group is the Lorentz group — every symbol transforms as a tensor, so the equations are form-invariant under boosts by construction. Einstein's 1905 move was to accept this symmetry as the symmetry of mechanics too, rather than of electromagnetism alone. In this sense, special relativity was discovered inside Maxwell's equations; the tensor notation merely makes it legible.

The logical asymmetry between the two equations deserves emphasis. The sourced equation is genuine dynamics: it says how charges generate fields, and its structure (a divergence of an antisymmetric tensor) is what forces charge conservation, since \( \partial_\nu \partial_\mu F^{\mu\nu} \) vanishes identically. The homogeneous equation, by contrast, carries no dynamics at all — it is the integrability condition guaranteeing that potentials exist. One can run the logic in either direction: assume potentials and the homogeneous equation is free; or assume the homogeneous equation and (on topologically trivial regions, by the Poincaré lemma) potentials are guaranteed to exist. The pairing of "one identity + one sourced equation" recurs throughout physics, from fluid vorticity to general relativity's Bianchi identities forcing \( \nabla_\mu T^{\mu\nu} = 0 \).

Two Lorentz scalars can be built from the tensor: \( F_{\mu\nu}F^{\mu\nu} = 2\left( B^2 - E^2/c^2 \right) \) and \( \tilde{F}_{\mu\nu}F^{\mu\nu} = -\frac{4}{c}\,\vec{E}\cdot\vec{B} \), where \( \tilde{F}^{\mu\nu} = \frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma} \) is the dual tensor (in terms of which the homogeneous equation reads \( \partial_\mu \tilde{F}^{\mu\nu} = 0 \), a perfect mirror of the sourced one). These invariants classify fields absolutely: a plane light wave has both zero (a "null" field — it looks like radiation in every frame), while a field with \( \vec{E}\cdot\vec{B} = 0 \) and \( B^2 > E^2/c^2 \) can be boosted to purely magnetic. No boost can change either number.

In the language of differential forms, \( A = A_\mu \mathrm{d}x^\mu \) is a connection 1-form, \( F = \mathrm{d}A \) a curvature 2-form, the homogeneous equation is \( \mathrm{d}F = \mathrm{d}^2 A = 0 \) (the exterior derivative squares to zero), and the sourced equation is \( \mathrm{d}{\star}F = \mu_0\, {\star}J \). Gauge freedom \( A \to A + \mathrm{d}\chi \) is the addition of an exact form, so the physically distinct potentials on a region are classified by its de Rham cohomology — which is precisely why the Aharonov–Bohm effect can detect \( A \) on a non-simply-connected domain even where \( F = 0 \). Promoting the gauge group from \( U(1) \) to a non-abelian group makes \( A \) Lie-algebra-valued and adds a self-interaction, \( F = \mathrm{d}A + A\wedge A \): the Yang–Mills field strengths of the standard model are direct descendants of the object built here.

Common misconceptions. (i) "Covariant form is just compact notation." No — it is a statement of physical content: the components of \( F \) must mix under boosts in the specific tensorial way, which is experimentally testable (and confirmed, e.g. in the fields of ultrarelativistic beams). (ii) "The four equations were unified by relativity." They were already unified; relativity revealed the unification. (iii) "Charge conservation is an independent law." Within this framework it is a theorem — any current that sources \( F^{\mu\nu} \) through the covariant equation is conserved automatically. (iv) "\( \phi \) and \( \vec{A} \) are auxiliary conveniences." At the classical level one may think so, but the covariant structure (and quantum mechanics via Aharonov–Bohm) marks \( A^\mu \) as the fundamental field, with \( F^{\mu\nu} \) its gauge-invariant curvature.

Worked examples

Example 1 — Classifying fields by their invariants. A focused pulsed laser has peak electric field \( E_0 = 1.0\times10^{6}\ \mathrm{V\,m^{-1}} \) with the plane-wave magnetic field \( B_0 = E_0/c \). A separate lab apparatus superposes a capacitor field \( E = 1.0\times10^{6}\ \mathrm{V\,m^{-1}} \) with an independent perpendicular magnet field \( B = 0.50\ \mathrm{T} \). Compute \( F_{\mu\nu}F^{\mu\nu} \) for each and state whether a frame exists in which each field is purely electric or purely magnetic.

\[ F_{\mu\nu}F^{\mu\nu} = 2\left( B^2 - \frac{E^2}{c^2} \right) \]

The scalar invariant, obtained by contracting the matrix of step 7 with its covariant partner; it is the same number in every inertial frame. A

\[ \text{Laser: } B_0 = \frac{E_0}{c} = \frac{1.0\times10^{6}}{2.998\times10^{8}} = 3.34\times10^{-3}\ \mathrm{T} \ \Rightarrow\ F_{\mu\nu}F^{\mu\nu} = 2\left( B_0^2 - B_0^2 \right) = 0 \]

Insert the plane-wave relation \( E = cB \): both terms are \( (3.34\times10^{-3}\ \mathrm{T})^2 = 1.11\times10^{-5}\ \mathrm{T^2} \) and cancel exactly. Since \( \vec{E}\perp\vec{B} \) in a plane wave, the second invariant \( \vec{E}\cdot\vec{B} \) also vanishes. B

\[ \text{Lab: } F_{\mu\nu}F^{\mu\nu} = 2\left[ (0.50)^2 - \left( \frac{1.0\times10^{6}}{2.998\times10^{8}} \right)^{\!2}\, \right] = 2\left[ 0.25 - 1.11\times10^{-5} \right] = 0.50\ \mathrm{T^2} \]

Insert the lab numbers: the magnetic term dwarfs the electric one by a factor \( (cB/E)^2 \approx 2.2\times10^{4} \). The invariant is positive: the field is magnetic-dominated in every frame. B

\[ \text{Laser: } F_{\mu\nu}F^{\mu\nu} = 0 \ \text{(null field)}; \qquad \text{Lab: } F_{\mu\nu}F^{\mu\nu} = +0.50\ \mathrm{T^2} \]

Reading. The laser field is null: no boost can make it purely electric, purely magnetic, or bring it to rest — it is radiation in every frame. The lab field, with \( \vec{E}\cdot\vec{B} = 0 \) and positive invariant, can be boosted to a frame with no electric field at all (the required speed is \( v = E/B = 2.0\times10^{6}\ \mathrm{m\,s^{-1}} \approx 0.0067c \), perpendicular to both fields).

Units check. \( B^2 \) is in \( \mathrm{T^2} \); \( E^2/c^2 \) is \( (\mathrm{V\,m^{-1}})^2/(\mathrm{m\,s^{-1}})^2 = (\mathrm{V\,s\,m^{-2}})^2 = \mathrm{T^2} \) ✓.

Example 2 — The sourced equation applied to a laboratory wire. A long straight wire of cross-section \( 1.0\ \mathrm{mm^2} \) carries a steady current \( I = 10\ \mathrm{A} \) along \( \hat{x} \) and is electrically neutral in the lab frame. Write down \( J^\mu \), verify charge conservation, and use \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) to find \( B \) at perpendicular distance \( r = 5.0\ \mathrm{mm} \).

\[ J = \frac{I}{S} = \frac{10}{1.0\times10^{-6}} = 1.0\times10^{7}\ \mathrm{A\,m^{-2}}, \qquad J^\mu = \left( 0,\ 1.0\times10^{7},\ 0,\ 0 \right)\ \mathrm{A\,m^{-2}} \]

Neutral wire: \( \rho = 0 \) so \( J^0 = c\rho = 0 \); the current density fills the spatial \( x \) slot. (In the frame of the drifting electrons this same four-vector acquires a non-zero \( J^0 \): a boosted neutral wire is charged — the tensor formalism handles this automatically.) A

\[ \partial_\mu J^\mu = \frac{\partial \rho}{\partial t} + \frac{\partial J_x}{\partial x} = 0 + 0 = 0 \]

Steady current, uniform along the wire: the four-divergence vanishes, so this \( J^\mu \) is an admissible source for the field equations (step 13 makes this mandatory, not optional). B

\[ \partial_\mu F^{\mu 1} = \mu_0 J^1, \quad \partial_t \vec{E} = 0 \ \Rightarrow\ \left( \nabla\times\vec{B} \right)_x = \mu_0 J_x \]

The \( \nu = 1 \) component of the sourced equation with static fields (step 10 with the displacement term switched off): covariant Maxwell reduces to Ampère's law for this geometry. B

\[ \oint \vec{B}\cdot d\vec{\ell} = \mu_0 I \ \Rightarrow\ B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(10)}{2\pi (5.0\times10^{-3})} = 4.0\times10^{-4}\ \mathrm{T} \]

Integrate the curl equation over a disc of radius \( r \) and use Stokes' theorem with the circular symmetry; then insert numbers. B

\[ J^\mu = (0,\ 1.0\times10^{7},\ 0,\ 0)\ \mathrm{A\,m^{-2}}, \qquad B(5.0\ \mathrm{mm}) = 4.0\times10^{-4}\ \mathrm{T} \]

Reading. A field of \( 0.4\ \mathrm{mT} \) — about ten times the geomagnetic field — circulates around the wire. In tensor language, the lab-frame field tensor has only a magnetic block; an observer riding with the drift electrons finds a non-zero electric block instead, sourced by the \( J^0 \) their frame assigns to the same four-current.

Units check. \( \mu_0 I / 2\pi r \): \( (\mathrm{T\,m\,A^{-1}})(\mathrm{A})/(\mathrm{m}) = \mathrm{T} \) ✓.

Problems

A uniform electrostatic field \( E_0 = 5.0\times10^{4}\ \mathrm{V\,m^{-1}} \) points along \( \hat{x} \), described by \( \phi = -E_0 x \), \( \vec{A} = 0 \). Write down \( A_\mu \) and compute \( F_{01} \) directly from the definition \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \). Check the sign and units.
Solution
With \( \vec{A} = 0 \): \( A_\mu = (\phi/c, 0, 0, 0) = (-E_0 x / c,\ 0,\ 0,\ 0) \). Then \[ F_{01} = \partial_0 A_1 - \partial_1 A_0 = 0 - \frac{\partial}{\partial x}\!\left( \frac{-E_0 x}{c} \right) = \frac{E_0}{c}. \] This matches the general identity \( F_{0i} = E_i/c \) (step 5), since \( \vec{E} = -\nabla\phi = E_0\hat{x} \). Numerically \( F_{01} = \dfrac{5.0\times10^{4}}{2.998\times10^{8}} = 1.67\times10^{-4}\ \mathrm{T} \). Units: \( (\mathrm{V\,m^{-1}})/(\mathrm{m\,s^{-1}}) = \mathrm{T} \) ✓. The sign is positive because lowering the index on \( A^0 \) leaves the time component unchanged while the spatial gradient supplies the minus sign already present in \( \vec{E} = -\nabla\phi \).
A region contains \( \vec{E} = (3.0\times10^{5},\ 0,\ 0)\ \mathrm{V\,m^{-1}} \) and \( \vec{B} = (0,\ 0,\ 2.0\times10^{-3})\ \mathrm{T} \). Evaluate both invariants \( 2(B^2 - E^2/c^2) \) and \( \vec{E}\cdot\vec{B} \), state whether a purely electric or purely magnetic frame exists, and find the boost speed that reaches it.
Solution
\( \vec{E}\cdot\vec{B} = 0 \) (fields are perpendicular), so a pure-field frame is possible; which kind depends on the sign of the first invariant. \[ \frac{E^2}{c^2} = \left( \frac{3.0\times10^{5}}{2.998\times10^{8}} \right)^{\!2} = (1.0\times10^{-3})^2 = 1.0\times10^{-6}\ \mathrm{T^2}, \qquad B^2 = 4.0\times10^{-6}\ \mathrm{T^2}. \] \[ F_{\mu\nu}F^{\mu\nu} = 2\left( 4.0\times10^{-6} - 1.0\times10^{-6} \right) = +6.0\times10^{-6}\ \mathrm{T^2} > 0. \] Positive invariant with \( \vec{E}\cdot\vec{B} = 0 \): a frame exists in which the field is purely magnetic. Boost with velocity \( \vec{v} = \vec{E}\times\vec{B}/B^2 \), whose magnitude is \[ v = \frac{E}{B} = \frac{3.0\times10^{5}}{2.0\times10^{-3}} = 1.5\times10^{8}\ \mathrm{m\,s^{-1}} = 0.50\,c. \] (Since \( v \lt c \), the boost is physical; had the invariant been negative, \( E/B \gt c \) and the roles reverse: boost at \( v = c^2 B/E \) to a purely electric frame.)
In some region the current density is \( \vec{J} = (\alpha x,\ 0,\ 0) \) with \( \alpha = 2.0\ \mathrm{A\,m^{-3}} \). Use the four-divergence condition \( \partial_\mu J^\mu = 0 \) (forced by the covariant field equation, step 13) to find \( \partial\rho/\partial t \), and compute the charge accumulating per second in a cube of side \( 10\ \mathrm{cm} \).
Solution
\[ \partial_\mu J^\mu = \frac{\partial \rho}{\partial t} + \nabla\cdot\vec{J} = 0 \ \Rightarrow\ \frac{\partial \rho}{\partial t} = -\frac{\partial J_x}{\partial x} = -\alpha = -2.0\ \mathrm{C\,m^{-3}\,s^{-1}}. \] Charge density decreases everywhere: more current leaves each volume element downstream than enters upstream. For the cube, \( V = (0.10)^3 = 1.0\times10^{-3}\ \mathrm{m^3} \): \[ \frac{dQ}{dt} = \frac{\partial \rho}{\partial t}\,V = (-2.0)(1.0\times10^{-3}) = -2.0\times10^{-3}\ \mathrm{C\,s^{-1}} = -2.0\ \mathrm{mA}. \] Cross-check by fluxes: current in through the face at \( x \) is \( \alpha x\,(0.01) \), out at \( x + 0.1 \) is \( \alpha(x+0.1)(0.01) \); net outflow \( = \alpha (0.1)(0.01) = 2.0\times10^{-3}\ \mathrm{A} \) ✓.
A microwave field is described in Lorenz gauge by \( A^\mu = \left( 0,\ 0,\ A_0\cos(kz - \omega t),\ 0 \right) \) with \( A_0 = 1.0\times10^{-8}\ \mathrm{T\,m} \) and frequency \( f = 1.0\ \mathrm{GHz} \). (a) Verify the Lorenz condition \( \partial_\mu A^\mu = 0 \). (b) Show \( \Box A^\nu = 0 \) requires \( \omega = ck \). (c) Compute the field amplitudes \( E_0 \) and \( B_0 \) and verify \( E_0/B_0 = c \).
Solution
(a) \( \partial_\mu A^\mu = \frac{1}{c}\partial_t A^0 + \nabla\cdot\vec{A} = 0 + \partial_y A_y = 0 \), since \( A_y \) depends only on \( z \) and \( t \) ✓. (b) \( \Box A^2 = \left( \frac{1}{c^2}\partial_t^2 - \partial_z^2 \right) A_0 \cos(kz-\omega t) = A_0\left( -\frac{\omega^2}{c^2} + k^2 \right)\cos(kz-\omega t) \), which vanishes for all \( z, t \) iff \( \omega = ck \). Numerically \( \omega = 2\pi f = 6.28\times10^{9}\ \mathrm{s^{-1}} \), so \( k = \omega/c = 21.0\ \mathrm{rad\,m^{-1}} \) (wavelength \( \lambda = 2\pi/k = 0.30\ \mathrm{m} \) — correct for 1 GHz). (c) \( \vec{E} = -\partial_t \vec{A} = -A_0 \omega \sin(kz - \omega t)\,\hat{y} \Rightarrow E_0 = \omega A_0 = (6.28\times10^{9})(1.0\times10^{-8}) = 63\ \mathrm{V\,m^{-1}}. \) \( \vec{B} = \nabla\times\vec{A} \): only \( B_x = \partial_y A_z - \partial_z A_y = -\partial_z A_y = A_0 k \sin(kz-\omega t) \) survives, so \( B_0 = k A_0 = (21.0)(1.0\times10^{-8}) = 2.1\times10^{-7}\ \mathrm{T}. \) Ratio: \( E_0/B_0 = \omega/k = c = 3.0\times10^{8}\ \mathrm{m\,s^{-1}} \) ✓; also \( \vec{E}\perp\vec{B}\perp\hat{z} \), the full plane-wave structure, from one four-potential.
A parallel-plate capacitor at rest in frame \( S \) produces a pure electric field \( \vec{E} = 1.0\times10^{5}\ \mathrm{V\,m^{-1}}\,\hat{z} \), \( \vec{B} = 0 \). Frame \( S' \) moves at \( v = 0.80\,c \) along \( \hat{x} \). Using the tensor transformation \( F'^{\mu\nu} = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta F^{\alpha\beta} \) (equivalently the standard boost laws), find \( \vec{E}' \) and \( \vec{B}' \), then verify numerically that \( B'^2 - E'^2/c^2 \) equals its frame-\( S \) value.
Solution
\( \gamma = 1/\sqrt{1 - 0.64} = 1/0.60 = 1.667 \). For a boost along \( x \), the transverse laws give \[ E'_z = \gamma\left( E_z + v B_y \right) = \gamma E_z = (1.667)(1.0\times10^{5}) = 1.67\times10^{5}\ \mathrm{V\,m^{-1}}, \] \[ B'_y = \gamma\left( B_y + \frac{v}{c^2}E_z \right) = (1.667)\,\frac{(0.80)(2.998\times10^{8})}{(2.998\times10^{8})^2}\,(1.0\times10^{5}) = (1.667)(2.67\times10^{-4}) = 4.45\times10^{-4}\ \mathrm{T}, \] with all other components zero (\( E_x \) is along the boost but is zero here; \( E'_y = B'_z = 0 \) since \( E_y = B_z = 0 \)). A magnetic field has appeared from a purely electric one. Invariant check. In \( S \): \( B^2 - E^2/c^2 = 0 - \dfrac{(1.0\times10^{5})^2}{(2.998\times10^{8})^2} = -1.11\times10^{-7}\ \mathrm{T^2} \). In \( S' \): \( B'^2 = (4.45\times10^{-4})^2 = 1.98\times10^{-7}\ \mathrm{T^2} \); \( E'^2/c^2 = \dfrac{(1.67\times10^{5})^2}{(2.998\times10^{8})^2} = 3.09\times10^{-7}\ \mathrm{T^2} \). \[ B'^2 - \frac{E'^2}{c^2} = 1.98\times10^{-7} - 3.09\times10^{-7} = -1.11\times10^{-7}\ \mathrm{T^2}\ \checkmark \] The invariant is preserved to rounding; it is negative (electric-dominated) in both frames, as it must be — no boost can make this field purely magnetic.