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Take $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. The four-current, four-potential and four-gradient areThe first two transform as contravariant four-vectors under an inertial-frame Lorentz transformation, within a consistently transformed electromagnetic gauge. The lowered-index gradient is a covector. Their contraction is a Lorentz scalar:Indeed, $J^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{J}^{\nu }$ and ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\alpha }{}_{\mu }{\partial }_{\alpha }$, so the two matrices cancel in the contraction. Local charge conservation in one inertial frame implies local charge conservation in every inertial frame. Integrating over a region gives the charge change equal to minus its outward current flux; global conserved charge requires no flux through the system boundary.

Use metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and common spacetime origins. A four-vector transforms as $A^{\prime }=\Lambda A$ under a Lorentz transformation. A Lorentz invariant is a scalar unchanged by that transformation. Since ${\Lambda }^T\eta \Lambda =\eta$,The wave phase labels the same stage of a physical oscillation at a spacetime event, independently of the observer's coordinates. Equal-phase wavefronts and interference phase differences are physical; fixing a common phase origin makes the continuous phase a Lorentz scalar. For a plane wave,The gradient of a scalar is a covector; raising its index gives the four-wavevector $K^{\mu }=(\omega /c,\mathbf{k})$. Under an $x$ boost,For vacuum light, $k_x=(\omega /c)\cos \theta$, givingA photon has four-momentum $p^{\mu }=\hslash K^{\mu }$, with $K^2=0$. For two photons separated by angle $\alpha$,In the centre-of-momentum frame the least possible pair energy is $2m_e{c}^2$, when both final particles are at rest. Thus the threshold has $(p_1+p_2{)}^2=4m_e^2{c}^2$; above it, excess energy supplies their kinetic energy. ThereforeThe kinematic equality is a limiting threshold; nonzero production phase space requires energies above it. For parallel photons no finite threshold is available, whereas head-on incidence minimizes the required energy:The blackbody-peak photon pair-production estimate, using the frequency-spectrum blackbody peak, $E_1=2.82144k_{B}T=7.29\times 10^{-4}\mathrm{eV}$ at $3\mathrm{K}$, gives $E_{2,\min }\simeq 3.58\times 10^{14}\mathrm{eV}$, or $358\mathrm{TeV}$. If “peak” means the wavelength-spectrum peak, $E_1=4.96511k_{B}T$ instead gives $2.03\times 10^{14}\mathrm{eV}$. The spectral convention changes the prefactor, not the conclusion that photons of order a few $10^{14}\mathrm{eV}$ are needed for this idealized peak-photon threshold.

wave phase labels the stage of a physical oscillation at a spacetime event. Two observers comparing the same event must identify the same crest, trough or intermediate stage: a change of coordinates cannot change the number of completed cycles between physically specified wavefronts. The continuous wave phase can therefore be treated as a Lorentz scalar, up to a common choice of wave phase origin. This does not assert invariance of the measured angular frequency, which is the rate of change of that wave phase along the observer's trajectory.

For Lorentz invariance of plane-wave phase, writeusing the metric with signature $(+,-,-,-)$. The gradient $K_{\mu }={\partial }_{\mu }\Phi$ transforms as a covector because $\Phi$ is a scalar. Raising its index therefore gives the four-wavevector. Explicitly, substituting the inverse boost $t=\gamma (t^{\prime }+v{x}^{\prime }/c^2)$, $x=\gamma (x^{\prime }+v{t}^{\prime })$ givesThese are exactly the Lorentz transformation laws for $(\omega /c,\mathbf{k})$. In vacuum, $K_{\mu }{K}^{\mu }=0$; the Doppler effect and aberration of light are two consequences of transforming the same four-vector.

Use $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and restore $\hslash$ temporarily. The free Dirac equation is $(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, with $c=1$. Its Hermitian conjugate is $-i\hslash ({\partial }_{\mu }{\psi }^{†})({\gamma }^{\mu }{)}^{†}-m{\psi }^{†}=0$. Multiplying on the right by ${\gamma }^0$ and using $({\gamma }^{\mu }{)}^{†}{\gamma }^0={\gamma }^0{\gamma }^{\mu }$ gives the adjoint equationThe Dirac adjoint and original equations give $({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }\psi =im\bar{\psi }\psi /\hslash$ and $\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi =-im\bar{\psi }\psi /\hslash$. Adding gives the continuity equationThe Dirac current transforms as a four-vector; the positive density is the time component, not a Lorentz scalar. With ordinary units the number current is $c\bar{\psi }{\gamma }^{\mu }\psi$ and the electron charge current is $-e$ times it.

The three on-shell spatial momenta have nine components. Four conservation constraints leave five, and an unobserved overall spatial rotation removes three for the scalar decay, leaving two internal kinematic variables. The pair invariants obeyThus two determine the third. They also determine each rest-frame daughter energy,and hence all momentum magnitudes and pairwise angles. The momentum triangle is fixed up to overall orientation. The Dalitz plot describes this internal kinematics, rather than claiming to record three independently oriented Four-vectors. Polarized parents or additional measured spin information can require orientation variables beyond the two invariant axes.

Use coordinates $X=(ct,x,y,z{)}^T$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real $4\times 4$ matrix $L$ is a Lorentz transformation precisely when it preserves the Minkowski interval for every $X$, equivalentlyIt is proper when $\boxed{\det L=+1}$, and orthochronous when $\boxed{L^0{}_0>0}$, equivalently $L^0{}_0\ge 1$. The latter selects the orthochronous Lorentz group, which preserves future-directed timelike four-vectors. Indeed the time-time entry of the preservation identity gives $(L^0{}_0{)}^2-{\sum }_{i=1}^3(L^i{}_0{)}^2=1$, so $|L^0{}_0|\ge 1$. Imposing both conditions gives a proper orthochronous Lorentz transformation.

The covariance of a selected particle momentum sum follows because, for a fixed set of particle four-momenta, their sum $P_A^{\mu }={\sum }_{i\in A}{p}_i^{\mu }$ is a four-vector by linearity of the Lorentz transformation. Thereforeis invariant. The subset need not comprise the entire plasma, and its particles need not move together. This expression includes cross terms between different particles; it is not the sum of their individual rest-mass squares in a.

The comparison must use the same subset and the transformed four-momenta at the same chosen events. Selecting “the lowest-energy half” anew in each frame need not do that. For example, take four photons along $\pm x$, with energies $E_0,2E_0$ in each direction. The two lowest in the original frame are an opposite pair and have invariant energy squared $4E_0^2$. After a boost at $c/2$ along $+x$, both forward photons are redshifted enough to become the lowest pair; that new collinear pair has zero invariant mass. This does not contradict covariance, because the subset changed.

If energies are resampled on a different simultaneity slice in an interacting plasma, they can also refer to different events. A physical conserved subsystem total must then include exchanged field energy and any boundary flux as appropriate. Such operational conservation requirements are distinct from the exact algebraic covariance of a sum of fixed four-vectors.

The event separation is timelike precisely whenA simultaneous pair in an inertial frame would instead have $(D-B{)}^2=-|\Delta {\mathbf{x}}^{\prime }{|}^2\le 0$. Since a Lorentz transformation preserves this interval, no such simultaneous frame exists. A timelike pair can be brought to the same spatial position: the required velocity is $\mathbf{u}=\Delta \mathbf{x}/\Delta t$, whose magnitude is less than $c$.

The separate condition $D\cdot B=0$ says that the two event-position four-vectors, measured from the chosen spacetime origin, are orthogonal in the Minkowski inner product. It is invariant under Lorentz transformations about that origin but not under arbitrary translations. It is not a general criterion for their causal separation.

For a concrete example, put $B=(cT,0,0,0)$ and $D=(0,R,0,0)$. Their inner product vanishes, but their separation has squared interval $c^2{T}^2-R^2$, which can be positive, zero or negative. If $B$ is timelike, a rest frame for $B$ makes $D^0=0$; this means that $D$ and the chosen origin are simultaneous in that frame, not that $D$ and $B$ are simultaneous.

An isolated free electron in vacuum cannot spontaneously emit a nonzero real photon. If $P=P_f+Q$, where $Q$ is the emitted photon four-momentum, $P^2=P_f^2=m^2{c}^2$ and $Q^2=0$, conservation would require $P_f\cdot Q=0$. A future-pointing timelike four-vector has strictly positive inner product with every nonzero future-pointing null four-vector, so this is impossible.

The same conclusion follows from the classical Cherenkov radiation condition $v>c/n$: in vacuum $n=1$, while the electron with $\gamma =100$ has $\beta =\sqrt{1-10^{-4}}<1$. Hence there is no vacuum emission angle for the specified free particle. Acceleration by an external field or scattering from another body would change the physical problem and can permit radiation.

Write $\Gamma =(1-v^2/c^2{)}^{-1/2}$ for the Lorentz factor of the inertial frame $S^{\prime }$, distinguishing it from the particles' changing Lorentz factors. Apply the Lorentz transformationto each world line in a. Both particles initially have velocity $-v$ in $S^{\prime }$. The switching events have coordinatesThus the Positron starts accelerating earlier in $S^{\prime }$. Its velocity first becomes less negative, vanishes when its laboratory velocity reaches $v$, and then becomes positive. The electron continues at $-v$ until $t^{\prime }=0$, then accelerates toward $-c$. The relativistic velocity-addition formula makes this precise:The electric field parallel to the Lorentz boost has unchanged magnitude, but its switching surface is $t^{\prime }+v{x}^{\prime }/c^2=0$, rather than $t^{\prime }=0$ everywhere.

In the laboratory, the total three-momentum is zero for all $t$: the two Lorentz forces cancel at equal laboratory times. In $S^{\prime }$, each active force is still $qE$, since this is one-dimensional motion with a parallel electric field. Define $t_p^{\prime }=-\Gamma vd/c^2$. For $v>0$ the total three-momentum, evaluated at equal $t^{\prime }$, isThis follows by integrating $d{p}_p^{\prime }/d{t}^{\prime }=F$ from its switching event and $d{p}_e^{\prime }/d{t}^{\prime }=-F$ from its later switching event. During the interval between them only the Positron feels the prescribed electric field, so momentum conservation for the particle pair fails. Once both forces act, the total three-momentum is constant again.

The equal-time momentum of an externally driven particle pair illustrates a distinction. The sum of particle four-momenta at a fixed pair of events is always a four-vector. However, a fresh sum at equal time in each inertial frame generally uses different events. Here transforming an equal-$t$ laboratory sum would give $-2\Gamma v\sqrt{m^2{c}^4+F^2{c}^2{t}^2}/c^2$, which varies with $t$; this differs from the equal-$t^{\prime }$ answer above. For example, with $c=m=F=d=1$ and $v=0.6$, the late-time equal-$t^{\prime }$ sum is $-0.75$, whereas transforming the equal-$t=0$ sum gives $-1.5$. This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent four-vector.

A closed system's total four-momentum, including the electromagnetic field and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange energy and momentum with an external system, so neither that independence nor universal particle-only momentum conservation is guaranteed.

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.

Write the Lorentz transformation as $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$. Acting on a scalar field, the chain rule gives the four-gradientFor the specified Lorentz boost, the inverse is obtained by changing $\beta$ to $-\beta$, soThese are precisely the covector component rules for the lowered coordinates. Raising the index with the Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ reverses the time component, yieldingIndeed ${\partial }^{\prime 0}=\gamma ({\partial }^0-\beta {\partial }^1)$ and ${\partial }^{\prime 1}=\gamma ({\partial }^1-\beta {\partial }^0)$, the contravariant four-vector rules. Thus the index on a partial derivative with respect to an upper coordinate is lower.

Take the divergence of the Ampère-Maxwell equation and use that the divergence of a curl is zero:The last step uses Gauss's law. Thus the charge continuity equation is exactlyFor a direct justification that this four-current is a four-vector, take one locally comoving population of electric charges, with rest charge density ${\rho }_0$ and four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$. Invariance of electric charge and length contraction give $\rho =\gamma {\rho }_0$ and $\mathbf{J}=\rho \mathbf{v}$, hence $J^{\mu }={\rho }_0{U}^{\mu }$. Summing populations preserves the four-vector rule. In particular, $c{\rho }^{\prime }=\gamma (c\rho -\beta J_x)$ and $J_x^{\prime }=\gamma (J_x-\beta c\rho )$. The Lorentz invariance of its four-divergence then explains why charge conservation holds in every inertial frame. A single zero-divergence condition alone would not determine a field's transformation law; the physical charge density and current density provide the additional content.

The Lorenz gauge condition is $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$. Substitute the electric potential and magnetic vector potential into Gauss's law:Next the Ampère-Maxwell equation, the curl of the curl identity, and the Lorenz gauge condition giveThe gradient terms cancel. Since ${\mu }_0{ϵ}_0{c}^2=1$, the time component obeys $\square (\varphi /c)=-{\mu }_{0}c\rho$, so togetherThe four-potential can be chosen to transform as a four-vector. To make the argument rigorous, choose the retarded potential with no freely added homogeneous solution. If $\square G_{\mathrm{ret}}(x)={\delta }^{(4)}(x)$, this choice isThe retarded Green function is a Lorentz scalar under proper orthochronous Lorentz transformations, and $d^{4}y$ is invariant. The four-vector rule for $J^{\mu }$ therefore proves $A^{\prime \mu }(x^{\prime })={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }(x)$. More generally, transform any homogeneous initial data by this same rule. Both the wave equation and the Lorenz gauge condition are then preserved.

Qualification. The wave equation by itself cannot prove the transformation rule for every independently chosen gauge potential: the d'Alembert operator has a nontrivial kernel of a linear map. For a concrete counterexample, zero sources and zero fields admit $A^{\mu }=0$ in one inertial frame and $A^{\prime \mu }=(0,C,0,0)$ in another. Both satisfy the wave equation and the Lorenz gauge condition, but the latter is not the Lorentz transformation of zero. It is a residual Lorenz gauge symmetry, generated by $\chi =C{x}^{\prime }$. Thus the usual assertion is a covariant choice of four-potential, with a consistent gauge transformation and homogeneous data, rather than a uniqueness result from the displayed equation.

For a massive particle, the proper time is determined by the Minkowski interval along its worldline:Choose future-directed time increments. With the Lorentz factor ${\gamma }_u=(1-|\mathbf{u}{|}^2/c^2{)}^{-1/2}$,We use $U^{\alpha }$ for the source's $u^{\alpha }$ to distinguish four-velocity from velocity $\mathbf{u}$. Since proper time is invariant and $d{x}^{\alpha }$ is a four-vector, $U^{\alpha }$ is a four-vector, with $U_{\alpha }{U}^{\alpha }=-c^2$. The rest-frame result $U^{\alpha }=(c,0)$ follows immediately.

The association between four real coordinates and Hermitian two-by-two matrices is a real-linear bijection. Congruence by $S$ preserves Hermiticity, so $X^{\prime }=SX{S}^{†}$ defines another real four-vector depending linearly on the original. Its determinant isThus the induced invertible linear map preserves the Minkowski quadratic form:The maps obtained continuously from the identity are proper and orthochronous. For example, a future timelike vector gives a positive-definite $X$, which remains positive definite under invertible congruence; its time direction is preserved. The matrices $S$ and $-S$ induce the same four-vector transformation, illustrating the Lorentz spinor double cover.

The source uses $x+iy$ in the upper-right entry, corresponding to $X=ctI+x{\sigma }_x-y{\sigma }_y+z{\sigma }_z$ for the standard Pauli matrices. This is a permissible basis convention, but its minus sign in the $y$ basis must also be retained in the spinor identification.

The free complex Proca field has Poincaré symmetry, with $B_{\mu }$ transforming as a spacetime four-vector, and the global internal U(1) symmetryThe free theory is also invariant under charge conjugation $B_{\mu }\leftrightarrow B_{\mu }^{\ast }$ and the usual parity and time reversal transformations of a vector field. The internal real-field description consists of two equal-mass real Proca fields, with an $O(2)$ symmetry containing the phase rotations and charge conjugation.

A gradient shift of $B_{\mu }$ leaves $G_{\mu \nu }$ unchanged, because partial derivatives commute. It changes the mass term byThis is generally nonzero: for example, $B_{\mu }=0$ and real $\lambda =at$ produce $\Delta \mathcal{L}=M^2{a}^2$. Thus a massive Proca field has no such gauge symmetry. The massless kinetic term alone would have it.

The spacetime coordinates of an event combine into a four-vector $X^a=(ct,x,y,z)$ under Lorentz transformations between inertial frames sharing an origin. For a massive particle, differentiation with respect to proper time gives its four-velocity. Under a spacetime translation, positions acquire an additive offset, while displacement four-vectors transform linearly under the Lorentz part.

A Lorentz transformation with determinant $+1$ that preserves future-directed timelike four-vectors. Equivalently, it belongs to the identity component of the Lorentz group. For the Minkowski metric $(-,+,+,+)$, its matrix satisfies ${\Lambda }^T\eta \Lambda =\eta$ and ${\Lambda }^0{}_0\ge 1$. Such transformations preserve the future light cone, so the retarded Green function of the d'Alembert operator transforms as a Lorentz scalar.