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Specify the Minkowski metric convention ${\eta }_{\mu \nu }={\eta }^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, with $x^0=ct$ and ${\partial }_0=c^{-1}{\partial }_t$. For the electromagnetic four-potential, $A_{\mu }={\eta }_{\mu \nu }{A}^{\nu }=(\varphi /c,-\mathbf{A})$. The electromagnetic field tensor isUsing $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, its two component matrices areChanging metric/sign conventions changes the component convention, so stating it is necessary. Under a Lorentz transformation, the tensor law is $F^{\prime \mu \nu }(x^{\prime })={\Lambda }^{\mu }{}_{\rho }{\Lambda }^{\nu }{}_{\sigma }{F}^{\rho \sigma }(x)$. For the stated Lorentz boost, put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$:Multiplication $F^{\prime }=\Lambda F{\Lambda }^T$ gives, for example, $F^{\prime 02}=-\gamma E_y/c+\gamma \beta B_z$ and $F^{\prime 12}=-\gamma B_z+\gamma \beta E_y/c$. Reading off all entries proves Lorentz transformation of electromagnetic fields:For the wire, the lab line charge is zero and the current is $I=nqu+(-nq)(-u)=2nqu$. At points off the wire, write $s^2=y^2+z^2>0$. Gauss's law and Ampère's law giveThe boost leaves $y,z$ unchanged, so the field transformations giveIts physical source is the relativistic charge density of counterstreaming beams. Their boosted number densities are $n_{+}^{\prime }=\gamma n(1-vu/c^2)$ and $n_{-}^{\prime }=\gamma n(1+vu/c^2)$, by transforming each charge-current four-vector. Thus ${\lambda }^{\prime }=q(n_{+}^{\prime }-n_{-}^{\prime })=-2\gamma nquv/c^2=-\gamma vI/c^2$. A line charge has radial field ${\lambda }^{\prime }(0,y,z)/(2\pi {ϵ}_0{s}^2)$; using ${\mu }_0{ϵ}_0{c}^2=1$ reproduces the boxed field. Oppositely moving populations contract differently under the boost, so neutrality in one frame does not imply neutrality in the other. The ideal wire's fields are singular on its axis.

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincare lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.

Under a gauge transformation, the electromagnetic field tensor changes byThe partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the Maxwell Lagrangian depends only on a gauge-invariant tensor:Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.

Antisymmetry gives $-F^{\rho \mu }{\partial }_{\rho }{A}^{\nu }=F^{\mu \rho }{\partial }_{\rho }{A}^{\nu }$. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:This is the electromagnetic stress-energy tensor in the $(+,-,-,-)$ convention. The product $F^{\mu \rho }{F}^{\nu }{}_{\rho }$ is symmetric in $\mu ,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are ${\partial }_{\rho }{F}^{\rho \mu }=0$. Hence the given correction equalson shell. This is an antisymmetric superpotential improvement: $B^{\rho \mu \nu }=-F^{\rho \mu }{A}^{\nu }$ is antisymmetric in $\rho ,\mu$, so ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Conservation of the canonical stress-energy tensor therefore implies ${\partial }_{\mu }{\Theta }^{\mu \nu }=0$. Equivalently, direct differentiation gives ${\partial }_{\mu }{\Theta }^{\mu \nu }=-F^{\nu }{}_{\rho }{\partial }_{\mu }{F}^{\mu \rho }$ after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed $\nu$, with charges $P^{\nu }=\int d^{3}x{\Theta }^{0\nu }$. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,Thus all requested properties are

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

Write the electromagnetic field tensor as $F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }$. Substitution into the Covariant Maxwell equation with the minus-plus-plus-plus metric givesChoose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. Since ${\partial }_{\mu }{\partial }^{\mu }={\nabla }^2-c^{-2}{\partial }_t^2$, the electromagnetic four-potential obeysConvolution with the outgoing retarded Green function gives the retarded electromagnetic potential

For the point particle, ${\tau }_{\ast }(x)$ is its unique retarded proper time, characterized byThus $y({\tau }_{\ast })$ is where the particle's worldline intersects the past light cone of the observation event $x$; the displayed expression is the covariant form of the Lienard-Wiechert potentials.

On the $z$-axis put $D=\sqrt{R^2+z^2}$. The retarded time is $t_{\ast }=t-D/c$. For the circular orbit, the spatial velocity at that time isBecause the displacement from the retarded source point to the observation point is orthogonal to this velocity, $|R^{\nu }{\dot{y}}_{\nu }|=\gamma cD$, while ${\dot{y}}^{\mu }=\gamma (c,\mathbf{v})$. The factors of $\gamma$ cancel, leaving

The operator $ia\bar{\psi }{\sigma }^{\mu \nu }{\gamma }^5\psi F_{\mu \nu }$ is the relativistic fermion electric dipole moment operator. Its nonrelativistic limit contains $\mathbf{S}\cdot \mathbf{E}$: spin angular momentum is an axial vector, whereas the electric field is a polar vector. It is consequently odd under parity symmetry in quantum field theory. Both the pseudotensor fermion bilinear and the electromagnetic field tensor are odd under charge conjugation, so their product is even. ThereforeThe CPT theorem then makes it odd under time-reversal symmetry. Such an interaction can arise only from CP violation. The Cabibbo-Kobayashi-Maskawa matrix therefore induces a nonzero Standard Model electric dipole moment at sufficiently high loop order, but its flavor structure and loop suppressions make the result extraordinarily small.

For the torsion-free Levi-Civita connection, cyclically summing the displayed coordinate expression and using ${\Gamma }^a{}_{bc}={\Gamma }^a{}_{cb}$ gives the algebraic first Bianchi identityApply the Jacobi identity to three covariant-derivative commutators, or differentiate the coordinate formula in normal coordinates. The third-derivative terms and the derivatives of quadratic connection terms cancel cyclically, giving the differential second Bianchi identityContracting the first and third curvature indices and then contracting once more yieldsor the contracted Bianchi identity

For the metric variation $h_{ab}=\delta g_{ab}$,while the two double-divergence terms in the supplied metric variation of scalar curvature become a boundary term after integration by parts for tensor fields. The Einstein-Hilbert action therefore contributes $G_{ab}+\Lambda g_{ab}$. Varying the Proca action with respect to the metric gives the Proca stress-energy tensor, and the full Einstein-Proca theory equation is

Since $\delta F_{ab}=2{\nabla }_{[a}\delta A_{b]}$, variation with respect to $A_b$ and one integration by parts give the covariant Proca equationTaking its covariant divergence givesbecause the resulting contractions pair the symmetric Ricci tensor with the antisymmetric electromagnetic field tensor. Since $m\ne 0$, the Lorenz constraint in Proca theory follows:Moreover, $F=dA$ and $d^2=0$, so the Electromagnetic Bianchi identity is

Finally, use that identity to differentiate the matter tensor. The result groups naturally asBoth terms vanish by the Proca equation and its Lorenz constraint in Proca theory. Thus ${\nabla }^a{T}_{ab}=0$, exactly as required by the contracted Bianchi identity applied to the Einstein field equations. The vector equation is therefore consistent with the gravitational equation.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a linear system with constant coefficients, so

PutThe initial four-velocity is $(u^0,u^1,u^2,u^3)={\gamma }_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,Taking the initial spacetime point to be the origin and integrating $u^{\mu }=d{x}^{\mu }/d\tau$ gives the relativistic motion in parallel electric and magnetic fieldsThe expressions have their continuous $a\to 0$ or $b\to 0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.

Eliminating proper time between the first two equations yieldsAt early coordinate time,which is uniformly accelerated motion with the transverse relativistic inertia factor ${\gamma }_0$. At late positive time,so the longitudinal speed approaches the speed of light.

The $y$-$z$ projection satisfiesIt is therefore a circle of radius and proper-time period

For the pitch magnitude assume $v_0>0$ and put $\alpha =|q|E/(mc)$, $\Omega =|q|B/m$, andNumber the $n$th proper-time period as $[(n-1)\Delta \tau ,n\Delta \tau ]$. Its longitudinal displacement isConsequently, as $n\to \infty$,where

For $A^{\mu }=(0,0,Bx,0)$, the only independent nonzero component of the electromagnetic field tensor isThe temporal and $z$ components of the Lorentz force show that $u^0$ and $u^3$ are constant. Taking $u^3=0$, the transverse equations arewhere dots now denote proper time derivatives. Witha solution isDirect differentiation verifies both equations. Hence the trajectory is a relativistic cyclotron orbit with proper angular frequency $omega=qB/m$; the sign fixes its orientation.

Use Minkowski spacetime signature $(-,+,+,+)$ and write $dot{x}^{\mu }=d{x}^{\mu }/d\lambda$. The reparametrization-invariant relativistic charged-particle action isThe first term is $-m{c}^2\int d\tau$ because

An electromagnetic gauge transformation isIt leaves the electromagnetic field tensor unchanged because antisymmetrized second derivatives commute. The action changes only byThis endpoint term has zero variation for fixed endpoints, which is the gauge invariance of the charged-particle worldline action.

Withthe electromagnetic field tensor is the antisymmetric tensorUnder a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, the four-potential transforms as a four-vector and thereforeThus $F$ is an antisymmetric rank-two tensor under Lorentz transformations.

Transforming the electromagnetic field tensor by $F^{\prime }=\Lambda F{\Lambda }^T$ and reading off its components givesThese are the parallel and transverse field transformation laws for a boost in the $x$ direction.

The electric components of the electromagnetic field tensor arewhile $F_{ij}$ depends only on the transverse vector. The scalar-vector cross term integrates to zero because ${\partial }_i{A}_i^V=0$, so the scalar action isVarying the nondynamical scalar gives the constraint equation in field theoryFor every nonzero Fourier momentum this fixes ${\Phi }_A=0$, confirming that a source-free massless vector has no propagating scalar polarization.

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

WriteSinceits derivatives areThe Euler-Lagrange equation therefore givesUsing the electromagnetic field tensorand dividing by $d\tau /d\lambda$ givesRaising the first index yields the equivalent requested formThis is the Lorentz-force equation from the worldline action.

A gauge transformation isBecause partial derivatives commute, the added contribution to $F_{\mu \nu }$ isso the electromagnetic field tensor is gauge invariant.

DefineSince ${\partial }_0=c^{-1}{\partial }_t$ and $A_0=-\Phi /c$,

The identityfollows directly from $F=dA$. Its spatial and mixed components areFor the other two equations define the four-currentThen ${\partial }_{\nu }{F}^{0\nu }={\mu }_0{j}^0$ givesand ${\partial }_{\nu }{F}^{i\nu }={\mu }_0{j}^i$ givesThis is the Covariant Maxwell equation with the minus-plus-plus-plus metric.