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The Minkowski inner product of two event-position four-vectors is invariant under a common Lorentz transformation fixing the origin, but changes under a common translation. It is therefore not a criterion for the causal separation of the events. For example $(cT,0)$ and $(0,R)$ are orthogonal while their separation has squared interval $c^2{T}^2-R^2$, which can have any sign. Causal classification uses their difference, not their individual positions.

In the observer's rest frame, $U=(c,0)$ and $P=(E^{\prime }/c,{\mathbf{p}}^{\prime })$. The Lorentz invariance of the Minkowski inner product gives $P\cdot U=E^{\prime }$ andThe relation between four-momentum and velocity is ${\mathbf{p}}^{\prime }=E^{\prime }\mathbf{v}/c^2$, soTaking the nonnegative square root provesFor a future-directed massive particle and observer, $P\cdot U>0$ and the expression lies in $[0,1)$; the massless limit, with nonzero energy, gives $v=c$.

Use the hyperboloid model with Minkowski inner productPut the vertex with angle $\theta$ at $v=(1,0,0)$, and place the endpoints of its adjacent sides of lengths $a$ and $b$ atThe geodesic through $P$ perpendicular to $vP$ has spacelike unit normalwhile the geodesic through $Q$ perpendicular to $vQ$ has spacelike unit normal

These two geodesics form the remaining two sides of the quadrilateral. Their angle equals the angle between their normals in the tangent plane at their intersection. The third right angle therefore givesDividing by $\cosh a\cosh b$ proves the Lambert quadrilateral identity

Assume $m_{\psi }>0$, as required to have a rest frame. Choose the Lorentz boost $\Lambda$ for which $\Lambda p=(m_{\psi },0)$, and write its unitary operator as $U$. The stipulated one-particle state normalization gives $U|\psi ,\mathbf{p}⟩=|\psi ,0⟩$ without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies $U\varphi (0)U^{†}=\varphi (0)$, and the vacuum state satisfies $U|\Omega ⟩=|\Omega ⟩$. Inserting $U^{†}U$ therefore provesWith a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let $q^{\mu }=(E_{\mathbf{q}}^{\chi },\mathbf{q})$ denote the four-momentum of the other energy eigenstate. Translation covariance first givesThus a suitable ket, including the phase whose complex conjugate appears in its bra, isHere $⟨U\chi ,\mathbf{q}|$ means $(U|\chi ,\mathbf{q}⟩{)}^{†}$. The overall phase depends on the fixed insertion point $x$; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum $q^{\prime }=\Lambda q$. Lorentz invariance of the Minkowski inner product givesThere is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.

Now use $(ct,x{)}^T$ and ${\eta }_2=\mathrm{diag}(1,-1)$. WritePreservation of the Minkowski metric says that the first column has squared norm $1$, the second has squared norm $-1$, and the columns are orthogonal for the Minkowski inner product:Orthochronousness gives $a\ge 1$. From the orthogonality equation, $b=de/a$; substituting in the second norm identity gives $e^2=a^2$. Hence $(b,e)=\epsilon (d,a)$ with $\epsilon =\pm 1$. The determinant is then $\epsilon (a^2-d^2)=\epsilon$, so properness forces $\epsilon =1$. Thus $b=d$ and $e=a$.

Set $\beta =d/a$. The first norm identity implies $|\beta |<1$ and $a=(1-{\beta }^2{)}^{-1/2}$. With $u=c\beta$, this givesConversely, the displayed matrix has ${\gamma }^2-(\gamma u/c{)}^2=1$, equal off-diagonal entries, determinant one, and positive time-time entry. Direct multiplication therefore gives $L^T{\eta }_{2}L={\eta }_2$, proving both directions. The plus signs are the Lorentz boost convention used in the paper; reversing the boost direction replaces $u$ by $-u$.

For a massive particle on a timelike world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}=dt/\gamma$. Its four-velocity and four-acceleration are $U=dX/d\tau$ and $A=dU/d\tau$. Differentiating $U=\gamma (c,\mathbf{v})$ givesThe invariant Minkowski inner products areFor the explicitly requested contraction,Thus four-acceleration is orthogonal to four-velocity. Equivalently this follows by differentiating the constant Lorentz scalar $U^2=c^2$ with respect to proper time.

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.

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.

For $p^2=m^2>0$, let a unitary operator $U$ implement a Lorentz boost taking $p$ to $(m,0)$. For a Lorentz scalar field, $⟨q|\varphi (x)|p⟩=⟨\Psi |\varphi (0)|m,0⟩$ with $|\Psi ⟩=e^{-i(q-p)\cdot x}U|q⟩$. The phase follows from translation covariance of a quantum field; its complex conjugate enters the bra. The transformed energy eigenvalue is $(\Lambda q{)}^0=p\cdot q/m$ by invariance of the Minkowski inner product. The reduction requires a massive initial state.

For $J=\mathrm{diag}(1,-1)$, the matrices satisfying $A^{T}JA=J$ and $A_{11}>0$ form the time-orientation-preserving Lorentz group in one spatial dimension. They are exactly $A=H(t)D_{\epsilon }$, where $H(t)=\left(\begin{array}{cc}\cosh t& \sinh t\\ \sinh t& \cosh t\end{array}\right)$, $D_{\epsilon }=\mathrm{diag}(1,\epsilon )$ and $\epsilon =\pm 1$. Preservation of the Minkowski metric makes the first column a future-pointing unit timelike vector; Minkowski inner product orthogonality determines the second up to sign. The identity $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$ gives the multiplication law $(t,\epsilon )(s,\eta )=(t+\epsilon s,\epsilon \eta )$ and inverse $(-\epsilon t,\epsilon )$.

For a two-body W boson decay into a charged lepton $\ell$ and a neutrino, neglect both daughter masses and use natural units $c=1$. Let $\Delta \varphi$ be the angle between the daughter transverse momenta. Identifying the missing transverse momentum with the neutrino's givesThis observable uses no $p_{z\nu }$. The 1983 UA1 experiment paper already used transverse mass as a lower bound on parent mass, before making additional decay-model assumptions for a mass estimate.^[ref]

The bound follows directly from the energy–momentum relation. For each massless daughter write its four-momentum, using rapidity $y$ and azimuth $\varphi$, asUsing the Minkowski inner product with positive time sign, $p^2=E^2-|\mathbf{p}{|}^2$, the two daughter four-momenta give the true parent invariant mass $M$:Since $\cosh u\ge 1$, $m_T\le M$, with equality at equal daughter rapidities when both transverse momenta are nonzero. Thus the unmeasured longitudinal information can enlarge the true invariant mass, but cannot lower it below this ideal transverse mass.

For an original example, stipulate $p_{T\ell }=p_{T\nu }=40\mathrm{GeV}$ and $\Delta \varphi =\pi$. Then $m_T=80\mathrm{GeV}$. Equal rapidities give $M=80\mathrm{GeV}$; choosing $y_{\ell }-y_{\nu }=1$ with exactly the same transverse readout gives $M=80\cosh (1/2)\simeq 90.210\mathrm{GeV}$. One measured transverse configuration therefore permits different true parent masses.