With $c=1$, the total four-momentum is $P_{\mathrm{tot}}={\sum }_i{P}_i$. A sum of Lorentz four-vectors transforms as a four-vector, sois a Lorentz invariant. Its value is the squared invariant mass of the complete system, irrespective of whether its constituents interact. Isolation makes the total four-momentum constant in time.

For a photon gas, write ${\mathbf{p}}_i=W_i{\mathbf{n}}_i$ with $|{\mathbf{n}}_i|=1$. Its individual null norms do not make the total norm vanish:For an isotropic gas in its momentum rest frame, ${\sum }_i{\mathbf{p}}_i=0$ and $s^2=W_{\mathrm{tot}}^2$. Zero is possible only when all nonzero photons propagate in the same direction. Two equal-energy photons at right angles, for example, give $s^2=2W^2$, rather than zero. Without an angular distribution, the pairwise expression is the general answer; isotropy is an additional assumption.

Let $\Delta t=t_d-t_g$ and $\Delta \mathbf{x}={\mathbf{x}}_d-{\mathbf{x}}_g$. A spacelike separation obeysUnder a boost with velocity $\mathbf{V}$, the time separation is $\Delta t^{\prime }=\Gamma (\Delta t-\mathbf{V}\cdot \Delta \mathbf{x}/c^2)$. ChooseThen $\Delta t^{\prime }=0$, while $|\mathbf{V}|/c=c|\Delta t|/|\Delta \mathbf{x}|<1$ follows from the spacelike inequality. Therefore an inertial simultaneity frame exists. This choice has no transverse boost component and is not claimed to be the unique such frame. If the events are already simultaneous, $\mathbf{V}=0$ suffices.

For a vacuum electromagnetic wave, the four-wavevector is $K^{\mu }=(\omega /c,\mathbf{k})$ with $|\mathbf{k}|=\omega /c$ and $K^2=0$. Its phase can be writtenBecause $K$ and $X$ transform as Lorentz four-vectors, their contraction is a Lorentz scalar. This proves phase invariance, with the same phase convention in both frames.

For a hypothetical free emission $e\to e+\gamma$, four-momentum conservation gives $P=P^{\prime }+Q$. The initial and final electrons have $P^2=P^{\prime 2}=m_e^2{c}^2$, and $Q^2=0$. Squaring gives $P^{\prime }\cdot Q=0$. In the final electron's rest frame, however, $P^{\prime }\cdot Q=m_e{E}_{\gamma }>0$ for a nonzero future-directed photon. ThusAn external field, another body or a medium can exchange momentum and change this conclusion; those are different systems.

The electron moves in the direction for which the electric field does positive work. Its kinetic energy gain is the magnitude $eEL=30\mathrm{MeV}$. HenceThis uses the relativistic energy-momentum relation, rather than the nonrelativistic kinetic-energy formula. Acceleration is opposite the field direction for a negatively charged electron, although the energy gain is positive along its actual displacement.

The equal and opposite incoming momenta make the laboratory the centre-of-mass frame. The initial total four-momentum is timelike. One outgoing photon would have null four-momentum, so an isolated two-body annihilation cannot produce exactly one photon. Two photons are kinematically possible and must have opposite momenta and equal energies:The quoted $0.511\mathrm{MeV}$ is the rest energy $m_e{c}^2$, not a mass in SI units. The energy balance includes both rest energies as well as the $1\mathrm{GeV}$ total kinetic energy.

This minimum concerns the complete final state. A detector with incomplete acceptance could register only one of the two photons, or none; conservation laws do not impose a minimum detected count without an ideal-detection assumption. Additional outgoing photons are also possible.

There is no nonzero, literally constant four-force that is a pure force throughout this motion. Initially $U=(c,0,0,0)$, so purity forces $F^0=0$. If the full four-vector were constant, $U(\tau )=U(0)+F\tau /M_0$; its square would be $c^2+F^2{\tau }^2/M_0^2$, which cannot remain $c^2$ for a nonzero initially spatial $F$. This is a genuine ambiguity in the source wording.

The usual consistent interpretation is constant laboratory relativistic force $f=d{p}_x/dt$, equivalently constant one-dimensional proper acceleration $f/M_0$. Take $f>0$ and let $s=ft/(M_{0}c)$. Integrating momentum conservation with the applied impulse gives $p_x=ft$, soThe corresponding four-force has components $F^{\mu }=\gamma f(\beta ,1,0,0)$, which are not constant, although its invariant magnitude is $F^2=-f^2$. Its temporal component is indeed $F^0=(\gamma /c)dE/dt$. The speed rises from zero and approaches $c$ asymptotically; $\gamma$ starts at one, initially rises quadratically and becomes linear in $t$ at late times.

If instead only the spatial four-force component $F^x=f_x$ is held constant, the distinct consistent answer is most simply parametrized by $u=f_x\tau /(M_{0}c)$:This follows by integrating $d{U}^x/d\tau =f_x/M_0$ and $dt/d\tau =U^0/c$. The temporal component must still vary. These two interpretations must not be conflated with a constant four-vector.

Along a massive particle's world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}$. A pure force preserves its rest mass. Differentiating $P=mU$ and $U^2=c^2$ givesso $\boxed{F\cdot U=0}$ for a pure force. This is a restriction on rest-mass transfer, not a statement that its work in every frame vanishes.

Write $\mathbf{a}=d\mathbf{v}/dt$. Since $\dot{\gamma }={\gamma }^3\mathbf{v}\cdot \mathbf{a}/c^2$, the four-acceleration isIf $\mathbf{v}\cdot \mathbf{a}=0$, this reduces to $A=(0,{\gamma }^2\mathbf{a})$, whenceHere $\alpha$ is the proper acceleration. The positive expression printed in the source is $A^2$ in the opposite metric signature, or the squared acceleration magnitude in ours. For example, at rest with nonzero acceleration, $A=(0,\mathbf{a})$ has $A^2=-a^2$ in our convention. Thus one must not combine the printed positive sign with a mostly-minus norm.

For a massive particle on a timelike world line, the four-velocity is $U^{\mu }=d{X}^{\mu }/d\tau =\gamma (c,\mathbf{v})$, with $U^2=c^2$. Its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$ and its four-force is $F^{\mu }=d{P}^{\mu }/d\tau$. Here $\tau$ is proper time. A massless trajectory does not admit this massive-particle proper-time parametrization.

The asserted classification of arbitrary orthogonal vectors is false. With the chosen Minkowski metric, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both spacelike vectors. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal null vectors. The valid statement is that a nonzero vector orthogonal to a timelike vector is spacelike. To prove it, choose the rest frame of the timelike vector, $D=(d,0,0,0)$; then $D\cdot G=d{G}^0=0$ forces $G^0=0$, so $G^2=-|\mathbf{G}{|}^2<0$.

The relativistic energy-momentum relation givesTherefore $\boxed{P^2=0⟹m=0}$. A photon is an example: $E=c|\mathbf{p}|$. A collection of photons need not itself have zero invariant mass.

In the massive particle's rest frame, $P=(m_{0}c,0)$ and the physical Pauli-Lubanski pseudovector is $W=(0,m_{0}c{\mathbf{s}}_{\mathrm{rest}})$, up to a common overall sign. Their Minkowski inner product is therefore zero. Since it is a Lorentz scalar,in every inertial frame, for any nonzero scalar normalization of $W$. This proves the requested orthogonality without a boosted component calculation.

The printed assertion that its particular normalization is always a unit vector needs qualification. For the displayed components $W=(\mathbf{s}\cdot \mathbf{p},E\mathbf{s}/c)$,Thus dividing by $m_{0}c|\mathbf{s}|$ is unit spacelike only when $\mathbf{s}\times \mathbf{p}=0$, or when $\mathbf{s}$ in that denominator is explicitly replaced by the proper rest-spin magnitude with correctly boosted components of $W$. As a counterexample to the literal displayed normalization, choose $\mathbf{s}\perp \mathbf{p}$ with nonzero momentum: it gives $U_W^2=-{\gamma }^2$, not $-1$. Orthogonality remains true, but it does not establish normalization.

Take $A^{\mu }=(\phi /c,\mathbf{A})$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and $J^{\mu }=(c\rho ,\mathbf{j})$. ThenExpanding the cyclic derivative of $F^{ab}={\partial }^a{A}^b-{\partial }^b{A}^a$ produces six second derivatives, which cancel in pairs because partial derivatives commute. HenceThis Bianchi identity contains $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. The inhomogeneous Maxwell equations areThe temporal component is $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, since ${ϵ}_0{\mu }_0{c}^2=1$; the spatial components give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{j}$. The source suppresses the coupling factor, as in units with ${\mu }_0=1$ or a rescaled current.

For the physical Lorentz force density on matter, the consistent tensor formula isIts temporal component is power transferred to matter per unit volume divided by $c$. The printed expression $J^{\nu }{F}_{\nu \mu }$ has the opposite sign for the stated definition of $F$ in this convention. A stationary positive charge in a positive $x$ electric field is a concrete check: the physical upper spatial force is $+\rho E_x$, whereas the printed expression, raised to an upper index, gives $-\rho E_x$. It can represent the opposite momentum transfer to the electromagnetic field, but must not be identified with the force on the charge without changing a convention.

For a forward wave, the Faraday's law of induction fixes the relative amplitudes and signs:Thus the two scalar amplitudes in the source cannot both be positive when the stated propagation direction and axes are used. The supplied product magnitude fixes only an overall amplitude in the chosen field units.

One convenient potential representative isIndeed $-{\partial }_t\mathbf{A}$ gives the electric field and $\nabla \times \mathbf{A}$ gives the signed magnetic field above. Moreover $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\phi =0$, the Lorenz gauge condition. The potentials are not unique because residual gauge transformations may preserve this condition.

Since $n>1$, the electromagnetic electromagnetic field invariants satisfy $\mathbf{E}\cdot \mathbf{B}=0$ and $B^2-E^2/c^2=(n^2-1)E^2/c^2>0$ away from field nodes. A boost along positive $z$ with speed $u=c/n$ givesHere ${\omega }^{\prime }={\gamma }_u(\omega -uk)=0$ and $k^{\prime }=(\omega /c)\sqrt{n^2-1}$. The wave becomes a static spatially varying magnetic pattern in this frame. This is possible because its medium wave four-vector is spacelike, $K^2=(1-n^2){\omega }^2/c^2<0$; it is not a vacuum-photon rest frame. The medium moves in the new frame, so its original rest-frame constitutive relation must not be reused there unchanged.

No inertial frame makes the magnetic field vanish throughout a nontrivial wave: that would turn the positive invariant $B^2-E^2/c^2$ into a nonpositive one. Both fields vanish at isolated nodes already, which is distinct from removing a field everywhere.