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BABAR used asymmetric electron-Positron collisions near the Upsilon meson $4S$ resonance to prepare neutral B meson pairs. It reconstructed one decay into charmonium plus a kaon, and used flavour tagging on the other B meson with charged leptons or kaons. Silicon tracking reconstructed decay vertices for the decay-time difference. Self-tagging decays calibrated mistags and time resolution; charged decays provided a control. The tagged time distributions established CP violation, limited by backgrounds, mistags and vertex alignment.^[ref]

The Lorentz boost converts vertex separation to $\Delta t\simeq \Delta z/({\beta }_{\mathrm{boost}}{\gamma }_{\mathrm{boost}}c)$ at leading order. Unequal beam energies thus make a time difference measurable as a distance.^[ref]

For $U={\varphi }^2({\varphi }^2-b^2{)}^2/2$ and $b>0$, the three scalar-field vacua are $-b,0,b$. The static first integral is ${\varphi }_x^2=2U$. In either adjacent interval, $y={\varphi }^2/b^2$ obeys $y_x=\pm 2b^{2}y(1-y)$, giving the displayed four oriented kink profiles with $\sigma ,\eta \in \{1,-1\}$. Each has rest energy $b^4/4$. The intermediate-vacuum obstruction to a kink excludes a direct static connection from $-b$ to $b$: reaching zero would give zero derivative, and uniqueness of the smooth second-order field equation forces the zero solution. A Lorentz boost replaces $x-x_0$ by $(x-vt-x_0)/\sqrt{1-v^2}$.

// Destination: relativistic-quantum-field.bigb

For the coordinate Lorentz boost $t^{\prime }=t\cosh \alpha +x\sinh \alpha$, $x^{\prime }=t\sinh \alpha +x\cosh \alpha$, the spinor convention ${\psi }^{\prime }(x^{\prime })=S\psi (x)$ gives $S=\exp (\alpha {\gamma }^5/2)$. In the Dirac field in two spacetime dimensions basis, this yields the displayed weights. The derivatives ${\partial }_t+{\partial }_x$ and ${\partial }_t-{\partial }_x$ scale by $e^{-\alpha }$ and $e^{\alpha }$, respectively, so the two kinetic terms are invariant. Opposite weights cancel in the Dirac mass term. Reversing the coordinate boost convention reverses both exponents.

Uniform perpendicular fields with $E^2>c^2{B}^2$ admit a subluminal Lorentz boost along $\mathbf{E}\times \mathbf{B}$ that eliminates the magnetic field. The transformed motion has constant transverse momentum and hyperbolic acceleration in the electric direction. An initially stationary laboratory particle generally has nonzero transverse velocity in this adapted frame; dropping it gives the wrong laboratory trajectory.

In units $c=1$, consider spacetime events $O=(0,0)$, $C=(1,2)$ and $D=(1,0)$, with coordinates ordered $(t,x)$. A Lorentz boost of speed $1/2$ assigns $t_O^{\prime }=t_C^{\prime }=0$, while the original inertial frame assigns $t_C=t_D=1$. Nevertheless $O,D$ are timelike separated, so no inertial frame regards them as simultaneous. Hence the relation “simultaneous in at least one inertial frame” is not transitive. This does not contradict transitivity of Minkowski simultaneity relative to an inertial observer for one fixed observer.

For perpendicular electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$, the sign of $E^2-c^2{B}^2$ determines which field can be removed by a Lorentz boost. If positive, $\mathbf{w}=c^2\mathbf{E}\times \mathbf{B}/E^2$ gives ${\mathbf{B}}^{\prime }=0$; if negative, $\mathbf{w}=\mathbf{E}\times \mathbf{B}/B^2$ gives ${\mathbf{E}}^{\prime }=0$. Both results follow from ${\mathbf{E}}_{\perp }^{\prime }=\gamma (\mathbf{E}+\mathbf{w}\times \mathbf{B})$ and ${\mathbf{B}}_{\perp }^{\prime }=\gamma (\mathbf{B}-\mathbf{w}\times \mathbf{E}/c^2)$; the relevant strict inequality ensures $|\mathbf{w}|<c$. A nonzero null crossed field cannot be reduced to a single field by a finite Lorentz boost. A particle initially at rest in the laboratory has initial velocity $-\mathbf{w}$ in the chosen inertial frame, which must be retained when solving its trajectory.

The spinor boost $\mathrm{diag}(e^{\kappa /2},e^{-\kappa /2})$ changes the component-magnitude ratio by $e^{-\kappa }$. The associated active positive-$z$ Lorentz boost therefore changes the polar half-angle as displayed and leaves azimuth unchanged. Its null-vector time scale changes by $\cosh \kappa +\sinh \kappa \cos \theta$, which must be retained separately when normalizing a spinor to encode direction.

Starting from rest, compose $n$ positive collinear Lorentz boosts, each with the same finite rapidity $\alpha >0$. Each step has relative velocity $w=c\tanh \alpha$, measured in the preceding rest inertial frame. The relativistic velocity-addition formula givesThese are equal increments of rapidity, not equal increments of velocity measured in the original inertial frame. They can describe a chain of relatively moving inertial frames or successive idealized changes of a massive particle's motion; no particular acceleration schedule is assumed. ^[ref].

For a relative velocity $w=c/2$ at each step,The first four velocities are $u_0=0$, $u_1=c/2$, $u_2=4c/5$, and $u_3=13c/14$. At step eight, $u_8/c=3280/3281\approx 0.99969521$. The rapidities grow by the same amount at each step, while the successive velocity increments become smaller.

A short derivation applies the hyperbolic-function identity for $\tanh (a+b)$ to the recurrence, giving ${\theta }_{n+1}={\theta }_n+\alpha$. Substituting $\tanh x=(e^{2x}-1)/(e^{2x}+1)$ and $e^{2\alpha }=3$ gives the displayed closed expression. In particular,for every finite nonnegative integer $n$. The limit is $u_n\to c$ as $n\to \infty$, but no finite sequence reaches the speed of light. This illustrates the beauty of rapidity: unbounded additive parameters coexist with bounded physical velocities. Replacing every step by a negative rapidity gives the corresponding approach to $-c$.

The light cone of a spacetime event $O$ consists of $O$ together with the spacetime events having null separation from it. In the coordinates of an inertial frame centered on $O$, it obeysFor $t>0$ it traces an expanding spherical flash of light, with radius $ct$ at time $t$; for $t<0$ it describes possible incoming light rays. A spacetime diagram with one spatial coordinate shows the two lines $x=\pm ct$. The apparent $45^{\circ }$ angle uses equal scales for $x$ and $ct$ and is not a separate physical angle.

The light cone is the boundary, while its interior has timelike separation from $O$. An outgoing light ray follows the boundary; a future-directed massive observer's world line remains inside it. The exterior has spacelike separation: no signal travelling at or below the speed of light can connect its spacetime events directly to $O$. This separates possible causal influence from merely being assigned a later coordinate time.

In the diagram, $A=(4\mathrm{s},3\mathrm{ls})$ is inside the future boundary, $N=(3\mathrm{s},3\mathrm{ls})$ is on it, and $B=(2\mathrm{s},3\mathrm{ls})$ is outside. Their squared Minkowski intervals are respectively $-7,0,5$ in ${\mathrm{ls}}^2$. Thus $B$ occurs later than $O$ in the displayed inertial frame while still being unable to receive a light-speed-or-slower signal sent from $O$.

The usefulness is immediate for an experiment: a detector's reading can depend on a source event only if a causal influence can reach it. A Lorentz boost changes coordinates but preserves this classification. For a boost with $\beta =0.8$ and Lorentz factor $\gamma =5/3$, applying the Lorentz transformation gives, in light-second length units,The corresponding squared Minkowski intervals remain $-7,0,5$. In particular $B$ is now earlier than $O$, which is allowed for spacelike separation; $A$ remains later.

With signature $(-,+,+)$, two Lorentz boost generators and one spatial rotation obey $[J,K_1]=K_2$, $[J,K_2]=-K_1$ and $[K_1,K_2]=-J$. Replacing the boosts by $i{K}_1,i{K}_2$ gives the compact cyclic brackets, proving the complexified isomorphism with the SU(2) Lie algebra. The real algebra has dimension three and rank one. Its finite-dimensional complex irreducible representations have dimensions $m+1$ for nonnegative integer highest weight $m$.

For Minkowski spacetime with unit light speed, $L_i=t{\partial }_i+x_i{\partial }_t$ generates a Lorentz boost. It commutes with the d'Alembert operator and controls derivatives transverse to time slices in the vector field method for wave equations.

For a vacuum plane electromagnetic wave, a Lorentz boost multiplies both transverse Jones vector amplitudes by the same positive Doppler factor when the basis is carried to the aberrated ray. Their ratio and phase difference therefore remain unchanged. All Stokes parameters gain the same squared factor, leaving normalized parameters, ellipse shape, handedness and degree of polarization unchanged. This does not mean the propagation direction or the components in an arbitrarily fixed pair of spatial axes remain unchanged.

A Lorentz transformation preserves the Minkowski metric. For a Lorentz boost, let $S^{\prime }$ move at constant velocity $+v$ along the $x$-axis of $S$, with parallel spatial axes and coincident origins at $t=t^{\prime }=0$. With $\beta =v/c$ and the Lorentz factor $\gamma$, the coordinates of the same spacetime event obey

The inverse Lorentz transformation replaces $v$ by $-v$. In particular, $x^{\prime }=0$ traces the moving origin $x=vt$, fixing the sign convention. The time formula is $t^{\prime }=\gamma (t-vx/c^2)$; dropping its position term would lose the relativity of simultaneity. ^[ref].

For the moving-clock example in time dilation, the second tick has $(t,x)=(5\mu \mathrm{s},3c\mu \mathrm{s})$. At $v=0.6c$, substitution gives $t^{\prime }=1.25(5-0.6\times 3)\mu \mathrm{s}=4\mu \mathrm{s}$ and $x^{\prime }=1.25(3-0.6\times 5)c\mu \mathrm{s}=0$. The clock stays at its own origin, and its coordinate time there equals its proper time.

Uniform perpendicular fields with $E^2<c^2{B}^2$ admit a subluminal Lorentz boost eliminating the electric field. A laboratory particle initially at rest follows a circle in the adapted frame and a stretched cycloid in the laboratory. The mean drift velocity is $\mathbf{E}\times \mathbf{B}/B^2$, while its laboratory kinetic energy is periodic rather than unbounded.

Fix $m>0$, $k=(m,0)$, and a rotationless Lorentz boost $L(p)$ taking $k$ to $p$ on the positive-energy mass shell. The little group on the Lorentz spinor double cover is SU(2). Its spin-$s$ irreducible representation induces a unitary representation on $L^2(d^{3}p/(2p^0))\otimes {\mathbb{C}}^{2s+1}$. Here $W(\Lambda ,p)=L(\Lambda p{)}^{-1}\Lambda L(p)$ is the Wigner rotation, lifted to SU(2) when necessary, and the translation convention is $U(I,a)|p⟩=e^{ip\cdot a}|p⟩$. The little-group cocycle proves the group law, and Wigner's classification gives irreducibility. Every nonnegative integer or half-integer spin is allowed on the double cover; descent to the ordinary proper orthochronous Poincaré group requires integer spin, because its $2\pi$ rotation must act as the identity.

The 1964 report exposed Brookhaven's 80-inch hydrogen bubble chamber to a mass-separated $5.0\mathrm{GeV}/c$ negative-kaon beam. A photographed decay chain, track momenta and bubble density supported the particle identification. Two converted photons helped reconstruct a neutral pion. Alternative assignments were tested kinematically. The authors reported an Omega baryon candidate with $Q=-1$, $S=-3$ and mass $1686\pm 12\mathrm{MeV}/c^2$, reserving a detailed systematic-error discussion for further events.^[ref]

This is an inference from several linked observations. A charged particle can leave a visible bubble-chamber track, while a neutral segment must be reconstructed from its production and decay products. Agreement between possible decay channels, electric charge, momentum conservation and the reconstructed mass is stronger evidence than a suggestive track shape alone. The event tests a predicted hadron pattern; it does not photograph the three strange quarks individually.

For example, an independently constructed two-body kinematic check usesIn a parent rest frame with equal and opposite daughter momenta, the momentum sum vanishes, giving $M{c}^2=E_1+E_2$. In units $c=1$, stipulated daughter masses $m_1=1$, $m_2=1/4$ and momentum magnitude $p=3/4$ give $E_1=5/4$, $E_2=\sqrt{10}/4$ and $M=(5+\sqrt{10})/4\simeq 2.04057$. A common Lorentz boost changes the daughter energies and momenta while preserving this reconstructed invariant mass. These chosen units and numbers are an illustration, not a reanalysis of the 1964 event.

Choose canonical spin states by applying the rotationless Lorentz boost $L(p)$ to rest-frame spin states. Spatial inversion $\Pi =\mathrm{diag}(1,-1,-1,-1)$ commutes with rest-frame rotations, so Schur lemma makes its action on one irreducible rest spin multiplet a phase ${\eta }_P$. Since $\Pi L(p)\Pi =L(\Pi p)$, the formula follows. Canonical spin labels are unchanged; helicity reverses. With $P^2=I$, the intrinsic parity is ${\eta }_P=\pm 1$.

// Target: lie-theory.bigb

Differentiate the defining relation at $\Lambda =I+tX+O(t^2)$. The linear term gives the Lie algebraSolving this equation between matrices yieldswhere a convenient basis isThe first two generate Lorentz boosts and the last a spatial rotation. Direct commutators giveThere are three independent real coefficients, so ${\dim }_{\mathbb{R}}\mathfrak{g}=3$. A Cartan subalgebra is $\mathbb{R}J$, and the rank of a semisimple Lie algebra is one. Explicitly, if $X=a{K}_1+b{K}_2+cJ$ and $Y=a^{\prime }{K}_1+b^{\prime }{K}_2+c^{\prime }J$, their commutator vanishes exactly whenFor nonzero $X$, these equations force $Y$ to be proportional to $X$, so there is no two-dimensional commuting subalgebra. The split real rank is also one, realized by the semisimple Lorentz boost generator $K_1$. ThusFor this Lorentz algebra in three spacetime dimensions, disconnected components of the full Lorentz group do not change its tangent Lie algebra at the identity.

Adopt the Minkowski metric $\eta =\mathrm{diag}(-1,+1)$ and the Clifford algebra convention $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\eta }_{\mu \nu }I$. The source's Pauli matrices giveThey satisfy ${\gamma }_0^2=-I$, ${\gamma }_1^2=I$ and ${\gamma }_0{\gamma }_1=-{\gamma }_1{\gamma }_0=-{\sigma }_3$, which verifies the required Clifford algebra. Raising the indices gives ${\gamma }^0=-{\gamma }_0$ and ${\gamma }^1={\gamma }_1$.

Fix the sign of the rapidity $\theta$ by taking the Lorentz boostThe corresponding Spinor representation of the Lorentz group isFor example $S^{-1}{\gamma }^{0}S={\gamma }^0\cosh \theta -{\gamma }^1\sinh \theta$ and $S^{-1}{\gamma }^{1}S={\gamma }^1\cosh \theta -{\gamma }^0\sinh \theta$, establishing Lorentz covariance by $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$.

Thus the Dirac spinor transforms as ${\psi }^{\prime }(x^{\prime })=S(\theta )\psi (x)$, with infinitesimal lawThese variations compare the fields at corresponding spacetime points. At a fixed coordinate one adds the usual orbital term $-\delta x^{\mu }{\partial }_{\mu }\psi$. Reversing the rapidity convention reverses both boost signs consistently. Infinitesimal transformations describe the proper orthochronous Lorentz group.

Use Minkowski spacetime signature $(+,-)$ and put $\kappa =|\lambda |>0$. The Euler-Lagrange field equation for the real scalar field isIndeed, differentiating the potential $U(\varphi )={\kappa }^2({\varphi }^2-1{)}^2/2$ gives $U^{\prime }=2{\kappa }^2\varphi ({\varphi }^2-1)$, while the kinetic term contributes ${\partial }_{\mu }{\partial }^{\mu }\varphi$.

Translation invariance gives the symmetric stress-energy tensorIts divergence is $({\partial }_{\mu }{\partial }^{\mu }\varphi +U^{\prime }){\partial }^{\nu }\varphi$, which vanishes by the Euler-Lagrange field equation. Consequently the conserved energy and physical spatial momentum areThe minus sign in $P=\int T^{01}dx$ follows from ${\partial }^1=-{\partial }_x$. Conservation assumes the associated stress-energy fluxes vanish at spatial infinity, as they do for the localized fields below.

A static finite-energy field configuration approaches vacua $\varphi =\pm 1$. Multiplying ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$ by ${\varphi }^{\prime }$ giveswhere the vacuum limits fix the constant. For a kink increasing from $-1$ to $1$, ${\varphi }^{\prime }=\kappa (1-{\varphi }^2)$. Integration gives $\mathrm{artanh}\varphi =\kappa (x-X)$ and henceIts negative is the antikink. The arbitrary center $X$ is a translational collective coordinate. The static kink mass is its rest energy; using ${\varphi }^{\prime 2}=2U$ and $y=\kappa (x-X)$ givesEquivalently, completing the static energy into a square gives $E=\frac{1}{2}\int [{\varphi }^{\prime }-\kappa (1-{\varphi }^2){]}^{2}dx+\kappa [\varphi -{\varphi }^3/3{]}_{-\infty }^{\infty }$, so this kink saturates its Bogomolny bound.

A Lorentz boost with velocity $v$ givesWriting $z=\gamma (x-X-vt)$, the derivatives are ${\varphi }_t=-\gamma v{\varphi }_{\mathrm{K}}^{\prime }(z)$ and ${\varphi }_x=\gamma {\varphi }_{\mathrm{K}}^{\prime }(z)$. The momentum integral therefore becomesThe boosted energy isbecause ${\gamma }^{-1}=\gamma (1-v^2)$. Thus $E^2-P^2=M^2$ and $P/E=v$, exactly the energy–momentum relation of a relativistic particle of rest mass $4\kappa /3$. This is the relativistic energy and momentum of a phi-four kink. If $\lambda =0$, the double-well potential disappears and there is no localized static kink connecting these vacua.