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The two postulates of special relativity are: all nongravitational laws have the same form in every inertial frame, and light in vacuum has the same finite speed of light $c$ in every inertial frame, independently of the motion of its source. The first is Einstein's principle of relativity; the second supplies an experimentally identified invariant speed.

For the stated standard configuration, the Lorentz transformation isTake differentials along the particle's worldline, rather than comparing coordinate displacements at equal times in the two frames. Writing $u_i=d{x}^i/dt$, one obtainsDividing by the transformed time interval proves the velocity-addition formula:For a massive particle, $|\mathbf{u}|<c$ and $|v|<c$, so the denominator $1-v{u}_x/c^2$ is positive.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

The Lorentz transformation is $t^{\prime }=\gamma (t-vx)$, $x^{\prime }=\gamma (x-vt)$ in units $c=1$. Adding and subtracting givesSince $\gamma (1+v)=\sqrt{(1+v)/(1-v)}$ and the other factor is its reciprocal,Here $\eta$ is the boost rapidity. Their product is unchanged, as required by invariance of the Minkowski interval.

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.

For increasing parameter $s$ along a future-directed timelike curve, the elapsed proper time isor $\int \sqrt{1-|\mathbf{v}{|}^2/c^2}dt$ when coordinate time is the parameter. A Lorentz transformation satisfies $L^T\eta L=\eta$, so replacing $\dot{x}$ by $L\dot{\tilde{x}}$ leaves the expression under the square root unchanged. Hence the integral is invariant. A monotone change of path parameter also leaves it unchanged by the ordinary substitution rule.

Special relativity supplies a tenseless mathematical representation: events have spacetime relations, proper-time intervals and causal order whether or not a coordinate labels one event “now.” For spacelike separated events, different inertial frames disagree about simultaneity and temporal order. The invariant structure therefore does not select a global present consisting of all and only events simultaneous with an observer's current event.

That supports an argument for eternalism if one adds that what exists should not depend on an arbitrary inertial frame and that no extra preferred foliation is available. On those assumptions, restricting existence to a frame-dependent simultaneity plane looks unjustified. A spacetime account can instead represent an observer's temporal location as it represents spatial location, without introducing an objective moving present.

The conclusion is not a formal derivation of metaphysics from the Lorentz transformations alone. Relativity establishes the metric and its symmetries, not a complete semantics for existence, becoming or experience. A presentist could add a preferred but empirically hidden slicing, though this adds structure unsupported by ordinary relativistic predictions. Local becoming along a worldline or causal partial orders are other possibilities, and do not reinstate a unique invariant global simultaneity relation.

A tenseless description also does not erase every temporal distinction. Timelike-separated events retain invariant order, light cones distinguish causal possibilities, and clocks measure nonzero proper durations. The absence of a global privileged “now” does not identify past and future dynamically or explain the thermodynamic arrow. Relativity strongly constrains a global present and makes a block representation natural; whether that representation exhausts temporal reality requires additional philosophical premises.

A coordinate transformation considered solely as a relabeling of events has no independent observable outcome: changing labels changes neither a detector record nor a clock comparison. But the Lorentz transformations also express substantive relations between operationally realized inertial coordinates. With clocks and rulers specified, their velocity-dependent time dilation, length relations and velocity composition make testable claims about how material systems and electromagnetic signals behave.

The distinction matters for synchronization. One-way coordinate light speed depends on how distant clocks are synchronized; a conventional alteration changes the displayed transformation law while leaving two-way timing observations unchanged. It would therefore be incorrect to regard every coefficient in an unspecified coordinate formula as a separately measurable physical fact. Testable content comes from the transformation together with the protocol and dynamical behavior of the systems implementing the coordinates.

Tests can compare clock proper times, decay lifetimes or resonance behavior across relative motion and orientation. Agreement supports Lorentz-invariant kinematics and the associated matter dynamics over the range tested; it does not prove exact invariance at every possible scale. A rival theory with an undetectable preferred frame can reproduce the same observations by changing its explanatory commitments. Empirical equivalence of that rival does not make the observable clock relations untestable.

Thus the answer is qualified: transformations as passive mathematical changes are not experiments, while Lorentzian relations among specified measuring procedures have substantial empirical consequences. Separating these roles avoids both treating coordinate convention as a measured force and declaring the entire relativistic structure immune to evidence.

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.

The same Lorentz transformation maps the past light cone to itself. In the primed axes the causal past isthe orange region in the upper-right spacetime diagram of part c. Future and past are not interchanged by the standard boost.

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

The scalar contraction of the electromagnetic field tensor isWith orientation ${ϵ}^{0123}=+1$, define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. Direct substitution givesThe first is a Lorentz scalar; the second is a pseudoscalar, invariant under proper Lorentz transformations such as boosts and rotations. Under spatial inversion $\mathbf{E}$ changes sign while $\mathbf{B}$ does not, so $\mathbf{E}\cdot \mathbf{B}$ reverses sign. For example, $\mathbf{E}=\mathbf{B}=\hat{\mathbf{z}}$ has product $+1$ before inversion and $-1$ afterwards. Thus the unrestricted statement including improper transformations needs this qualification.

For the spacetime integral, assume a globally regular electromagnetic four-potential on ordinary Minkowski space and sufficient boundary decay. Using $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$,ConsequentlyThe magnetic helicity boundary term controls the integral. If its surface flux vanishes and the magnetic helicity $H_B(t)=\int \mathbf{A}\cdot \mathbf{B}{d}^{3}x$ vanishes at both temporal ends, thenCompactly supported smooth potentials are one sufficient version of the intended falloff assumption. The local identity is exact, but vanishing fields at spatial infinity alone is insufficient. For a concrete counterexample, take $\varphi =0$ and, in dimensionless coordinates,where $a$ changes smoothly from zero to one and $\dot{a}$ has compact temporal support. Both electric fields and magnetic fields vanish rapidly at spatial infinity, butThe counterexample leaves a final magnetic field and therefore fails the extra temporal boundary condition. It shows exactly why the printed integral cannot follow from a merely spatial reading of “at infinity”; vanishing of the full boundary term is the needed hypothesis.

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 starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

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.

The four-gradient rules from a giveThe Lorentz transformation coefficients are constant, so the partial derivatives do not differentiate them. This proves the Lorentz invariance of the d'Alembert operator:For example, the two longitudinal terms transform as $-{\gamma }^2({\partial }_0+\beta {\partial }_1{)}^2+{\gamma }^2({\partial }_1+\beta {\partial }_0{)}^2=-{\partial }_0^2+{\partial }_1^2$, using ${\gamma }^2(1-{\beta }^2)=1$ and commuting partial derivatives. The transverse terms are unchanged. Applied to a scalar field, $\square$ therefore gives a Lorentz scalar.

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.

In the center-of-momentum frame, four-momentum conservation gives two opposite photon momenta, each of energy $M_{\pi }/2$. Let $z=\cos {\theta }_{\ast }$, with the axis along the Lorentz boost from that frame to the laboratory. Reversing the axis specified in the source changes $z$ to $-z$ and has no effect on the following product. The Lorentz transformation of four-momentum givesWriting $s=\sin (A/2)$, b now impliesThe isotropy assumption is $dN=a|dz|$ on $-1\le z\le 1$. Each interior opening angle has two inverse branches, $z$ and $-z$, both of which must enter the change of variables for probability densities. The positive branch has derivativeConsequently the diphoton opening-angle distribution isIts support is $2\arcsin (1/\gamma )<A<\pi$ for $0<\beta <1$, and its integral is $2a$, the integral of the original angular probability density before normalization. Thus the normalized probability density is $(2a{)}^{-1}dN/dA$. Omitting one inverse branch would incorrectly halve the total count.

The continuous internal symmetry is local $U(1)$ gauge invariance. With the source's gauge covariant derivative convention it acts asso $D_{\mu }\phi$ transforms like $\phi$ and $F_{\mu \nu }$ is unchanged. Constant phases are its global subgroup, not an additional independent internal rotation group.

In flat four-dimensional space-time the continuous external symmetry is the Poincaré group: four translations and six Lorentz transformations. Noether's theorem associates translations with conserved energy and momentum, spatial rotations with angular momentum, and boosts with the relativistic center-of-energy/boost charges. In tensor notation the latter two are the six components of the conserved total angular-momentum tensor $M^{\mu \nu }$, including field spin contributions where appropriate. The nonzero dimensionful parameter ${\mu }^2$ prevents exact scale invariance. Discrete transformations, such as parity, are outside the requested continuous-symmetry list.