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For an unpolarized electromagnetic target, Lorentz covariance and parity symmetry in quantum field theory permit a symmetric hadronic tensor built from $g^{\mu \nu }$, $P^{\mu }$ and $q^{\mu }$, where $P=P_H$. Its surviving general form isThe totally antisymmetric tensor structure ${ϵ}^{\mu \nu \rho \sigma }{P}_{\rho }{q}_{\sigma }$ is excluded by electromagnetic parity symmetry in quantum field theory for this unpolarized target. The other possible antisymmetric structure, $P^{\mu }{q}^{\nu }-q^{\mu }{P}^{\nu }$, cannot be transverse for generic scattering kinematics and hence is excluded by current conservation. Current conservation gives the Ward identities $q_{\mu }{W}^{\mu \nu }=q_{\nu }{W}^{\mu \nu }=0$, imposingTwo scalar functions remain. Taking ${\tilde{P}}^{\mu }=P^{\mu }-\nu q^{\mu }/q^2$ gives the transverse basisThese are the deep-inelastic structure functions; $x$ is Bjorken x. At fixed target mass, the two independent Lorentz scalar invariants can equivalently be chosen as $x,q^2$. In this paper $\nu =P\cdot q$ has dimensions of mass squared, rather than the alternative convention $\nu =P\cdot q/M$ used for energy transfer.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

Write a nonzero vacuum plane electromagnetic wave with wavevector $\mathbf{k}=(\omega /c)\mathbf{s}$, where $\mathbf{s}$ is a unit propagation direction. Its propagation Lorentz four-vector is $K^{\mu }=(\omega /c,\mathbf{k})$, or, up to a scalar normalization in any one frame, $n^{\mu }=(1,\mathbf{s})$. It is null because $K_{\mu }{K}^{\mu }=0$.

Choose coordinates $x^{\mu }=(ct,\mathbf{x})$ and Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. With the supplied three-dimensional boost formulas, a consistent convention assigns the printed array to upper-index components,The index conventions for the electromagnetic field tensor then require $F_{0i}=+E_i/c$ and $F_{ij}=F^{ij}$ when both indices are lowered. The printed lower-index label therefore cannot be kept together with its electric signs and a forward propagation vector in this convention.

This is not merely an arbitrary overall sign. For a valid forward wave with $\mathbf{s}=\hat{\mathbf{z}}$, $\mathbf{E}=E_0\hat{\mathbf{x}}$ and $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}$, multiplying the printed array, unchanged, by $n^{\mu }=(1,0,0,1)$ gives an $x$ component $-2E_0/c$, rather than zero. One consistent repair is to retain the printed array as $F^{\mu \nu }$ and contract with $n_{\mu }=(1,-\mathbf{s})$. Equivalently, lower both tensor indices before forming the contraction asked for in the question.

Using that corrected lower-index tensor givesTo establish both electric and magnetic transversality, include the dual electromagnetic field tensor. Choose ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; lowering both indices then givesThe source-free plane-wave Maxwell equations are precisely the pair of covariant conditionsTheir temporal components give $\mathbf{k}\cdot \mathbf{E}=0$ and $\mathbf{k}\cdot \mathbf{B}=0$. Their spatial components give the associated amplitude and orientation relations, $\mathbf{E}/c=-\mathbf{s}\times \mathbf{B}$ and $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$.

To see the Lorentz covariance explicitly, let $V_{\nu }=K^{\mu }{F}_{\mu \nu }$. It is a covector, so if $V_{\nu }=0$ in one inertial frame, its Lorentz transformation is also zero in every other inertial frame. The same is true of $K^{\mu }{\tilde{F}}_{\mu \nu }$. Since the transformed wave still has $K^{\prime 0}>0$ and $|{\mathbf{k}}^{\prime }|={\omega }^{\prime }/c$, their temporal components implyHere $K^{\mu }$, rather than a vector renormalized to have temporal component one in every frame, is the actual Lorentz four-vector; multiplying a zero contraction by a nonzero normalization does not affect the argument.

Finally, the two scalar perpendicularity statements alone are not equivalent to one tensor contraction. They do not enforce the amplitude relation between $\mathbf{E}$ and $\mathbf{B}$. Conversely $n^{\mu }{F}_{\mu \nu }=0$ alone cannot exclude a longitudinal magnetic field: adding $B_{\parallel }\mathbf{s}$ leaves its cross product unchanged. For a genuine source-free plane electromagnetic wave, the remaining Maxwell equations supply that missing constraint, and both tensor contractions vanish. This qualification and the corrected index placement are necessary for the intended invariance argument.

The numerical coordinates assigned to an event can be changed without changing the event. It does not follow that every relation between coordinate systems is devoid of physical content. A standard inertial frame includes a family of resting measuring standards and synchronized clocks. Its coordinates are normalized so that the Minkowski metric has the same standard form. The relation between two such descriptions therefore records properties of the standards and of spacetime, not just an arbitrary choice of names.

Write $x^{\mu }=(ct,x,y,z)$ and use metric signature $(-,+,+,+)$. A Lorentz transformation satisfiesAn origin translation can also be included without changing the Minkowski interval. The displayed condition preserves the interval between events, including the null cone and proper time along a timelike curve. Its physical significance lies in the claim that comparisons made with actual idealized rods and clocks obey this same metric structure.

A more general passive relabeling is allowed, but then the metric components must change with the labels. For example, let $t^{\prime }=2t$ and leave $x,y,z$ fixed. In the new coordinates,Nothing physical has changed, but $t^{\prime }$ no longer has the original clock-normalized meaning of $t$. Retaining the old metric components as well as the new labels would change the mathematical model. This distinction is part of the physical content of a passive Lorentz transformation: the transformation between standard inertial coordinate systems preserves a specified metric form, whereas arbitrary relabelings need not preserve that form.

For an operational example, a clock moving at velocity $v$ between two events on its world line accumulatesin the first frame. In its own rest frame the same result is simply the interval between its local readings. The equality is a physical agreement about one clock, derived using the preserved Minkowski interval. It is not a claim that the two coordinate-time differences are identical. Similarly, an observer measures a light angular frequency through $\omega =-k_{\mu }{u}^{\mu }$, where $k^{\mu }$ is the wave four-vector and $u^{\mu }$ the observer's four-velocity. Transforming both sets of components leaves this scalar unchanged. Comparing two different observers gives the physical relativistic Doppler effect rather than a disagreement over one observer's measurement.

There is a further distinction between passive transformations of the same situation and active changes to a situation. Re-expressing a particular experiment in different coordinates is always possible if all geometric fields are transformed consistently. Preparing a second, moving copy of the apparatus is a different physical arrangement. Its relation to the first is governed by a dynamical symmetry only if the relevant laws and background structure have the appropriate invariance. An active Lorentz transformation of a vector field makes the changed field configuration explicit.

This is why mere Lorentz covariance of notation must not be confused with the Lorentz invariance of the dynamics. A physically preferred timelike vector field can be transformed as a vector under passive relabeling while distinguishing particular states of motion. Holding that preferred field fixed can destroy the active boost symmetry of the matter dynamics. The covariant equation with a preferred timelike vector is a concrete example.

Coordinates are therefore labels, but the Lorentz transformation relates labels tied to shared physical standards and a definite metric structure. Its measurable consequences follow from that structure and the dynamical assumptions, not from the bare act of changing labels.^[ref]

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

A symmetry is a transformation of a physical system that preserves its laws: it maps admissible solutions and consistently transformed data to admissible solutions. The central symmetry of special relativity is Lorentz invariance, expressing equivalence of inertial frames and preservation of the Minkowski interval and the universal speed of light. With spacetime translations included, the spacetime symmetry group is the Poincaré group.

A practical sufficient check is to express an equation as an equality between Lorentz tensors of the same type, with every repeated upper/lower pair correctly contracted and the same free indices on both sides. Every ingredient, including sources and background fields, must have its stated Lorentz transformation rule. For example, $p^{\mu }{p}_{\mu }=-m^2{c}^2$ is a Lorentz scalar equality, and $d{p}^{\mu }/d\tau =q{F}^{\mu }{}_{\nu }{U}^{\nu }$ equates four-vectors. Such manifest Lorentz covariance makes the form of the equation the same in every inertial frame.

This is a sufficient test when the transformation rules are genuine physical ones, not a guarantee from writing index symbols. A fixed preferred four-vector could make an equation formally covariant if it too were transformed, while its chosen physical background breaks Lorentz invariance. Conversely, an equation written in three-dimensional components can be fully consistent with special relativity, as the Lorentz force is, even when the covariance is not manifest.

For $m>0$, the three physical polarization vectors are transverse to $k$ and have norm $-1$ in metric $(+---)$. In the rest frame their sum is $\mathrm{diag}(0,1,1,1)$; Lorentz covariance then gives the displayed expression for $k^2=m^2$. This includes the longitudinal mode. It cannot be specialized to $m=0$ by setting its denominator to zero.

Between spin-zero pseudoscalars, Lorentz covariance decomposes a vector current matrix element into $(p+k{)}^{\mu }{f}_{+}(q^2)+(p-k{)}^{\mu }{f}_{-}(q^2)$, with $q=p-k$. Strong interaction parity symmetry in quantum field theory makes the corresponding axial current matrix element vanish.

Define ${\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }]$ using the gamma matrix Clifford algebra. Moving ${\gamma }^{\rho }$ through the two factors with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$ proves the displayed operator commutator. For real antisymmetric ${\omega }_{\mu \nu }$ and $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$, it gives $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$. This is the infinitesimal Lorentz covariance identity required by the Dirac equation.

In the minimal Super-Poincaré algebra, Lorentz covariance allows $[P^{\mu },Q_{\alpha }]=A{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{Q}}^{\dot{\beta }}$. The graded Jacobi identity with $P^{\mu },Q_{\alpha },Q_{\beta }$ forces $A=0$. Thus each supercharge preserves four-momentum and its mass Casimir, giving supersymmetric mass degeneracy in an unbroken physical supermultiplet.

Lorentz covariance allows only a multiple of $p^{\mu }$ for a vacuum-to-spin-zero one-particle current. In its rest frame only the temporal component can remain. A parity-even temporal vector current between a parity-even vacuum and parity-odd particle changes sign, forcing that component to vanish. A temporal axial current is parity-odd and is allowed. This explains why the pion decay constant enters the axial, rather than vector, part of the weak charged current.