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The laboratory frame is the reference frame in which an experiment's measuring apparatus is at rest. It need not be the centre-of-momentum frame: a target at rest in the laboratory can be struck by a projectile carrying nonzero total momentum.

Treat the electron and electron antineutrino as effectively massless particles in the centre-of-momentum frame, retain only the charged current, and use the Fermi interaction: $m_e^2\ll s\ll M_W^2$. Take ${\theta }^{\ast }$ between incoming and outgoing electron antineutrino directions. The V-A interaction selects a right-helicity electron antineutrino and a left-helicity electron. Their opposite incoming directions give total angular momentum projection one. The outgoing pair has the same projection along its own axis. The Wigner D-matrix element is $d_{11}^1({\theta }^{\ast })=(1+\cos {\theta }^{\ast })/2$, so the squared helicity amplitude is proportional to $(1+\cos {\theta }^{\ast }{)}^2$. The inactive initial electron helicity is included in the unpolarized normalization supplied by the paper.

The boost from the centre-of-momentum frame to the laboratory frame has $\beta =E/(E+m_e)$, while the incoming electron antineutrino has $E_{\ast }=(s-m_e^2)/(2\sqrt{s})=m_{e}E/\sqrt{s}$ and $s=m_e^2+2m_{e}E$. ThusThe requested identity therefore uses $E\gg m_e$; it is not exact at finite target mass. With $c_{\ast }=\cos {\theta }^{\ast }$, the absolute Jacobian determinant is $|dy/d{c}_{\ast }|=1/2$, givingThe suppression at backward scattering follows from angular momentum conservation: rotating projection $+1$ through $\pi$ cannot leave projection $+1$ along the original axis. Replacing $1+c_{\ast }$ by $2(1-y)$ gives the supplied $(1-y{)}^2$ dependence.

For the relativistic head-on elastic collision with a stationary target, use $c=1$ as stipulated, so masses are measured in energy units. The incoming momentum is $p=\sqrt{E^2-M^2}$ and the invariant squared total four-momentum isThe centre-of-mass frame moves at ${\beta }_{\ast }=p/(E+m)$, with ${\gamma }_{\ast }=(E+m)/\sqrt{s}$. In this frame the incident particle hasA nontrivial head-on elastic collision reverses $p_M^{\ast }$ and preserves $E_M^{\ast }$. Boosting back to the laboratory frame givesThe no-interaction forward branch instead leaves $E^{\prime }=E$; the requested formula describes the actual one-dimensional encounter and recoil.

Let $p$ be the incident momentum and $q_1,q_2$ the outgoing magnitudes in the laboratory frame. Transverse conservation of momentum gives $(q_1-q_2)\sin \theta =0$. For the nondegenerate symmetric scattering, $q_1=q_2=q$, and the two outgoing energies are equal, $ϵ=(E+E_0)/2$. ThenCombining these mass shell relations yieldsHere $E$ includes the incident electron's rest energy. As $E\downarrow E_0$ the limiting angle is $45^{\circ }$, consistent with the nonrelativistic equal-mass elastic collision; at large $E/E_0$ the outgoing tracks become forward directed.

For equal masses, let the incoming laboratory frame velocity be $\mathbf{v}$ and the outgoing velocities be $\mathbf{a},\mathbf{b}$. Conservation of momentum and kinetic energy giveSquaring the first equation and subtracting the second yields $2\mathbf{a}\cdot \mathbf{b}=0$. HenceThe stated oblique collision ensures both outgoing speeds are nonzero. In the degenerate head-on case one particle can stop, and a direction cannot be assigned to its zero velocity.

Use the inverse Lorentz transformation, with ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$:Taking differentials along the object's trajectory givesThis is the collinear relativistic velocity addition law. It concerns the velocity of one object measured in two frames. By contrast the rate of change of the distance between two objects in one laboratory frame can be as large asattained by oppositely directed light signals; for massive objects it is approached but not attained. This is not an object's speed in the other's rest frame and cannot transmit a signal faster than $c$.

At the threshold found in i, both products are stationary in the centre-of-momentum frame, so they have the same velocity in the laboratory frame. Take the energetic incident photon along $+x$. The total energy is $E+\epsilon$ and total momentum is $(E-\epsilon )/c$, so the centre-of-momentum frame moves atFor the numerical threshold,This is extremely close to, but strictly below, the speed of light. Each product has energy $(E+\epsilon )/2$ in the laboratory frame. Above threshold the same ratio of total momentum to total energy determines the motion of the pair's centre-of-momentum frame, but the individual electron and Positron need not have a common velocity.

Let the incident photon energies be $E$ and $\epsilon$, meeting at angle $\psi$ in the laboratory frame. Add their four-momenta; because each is null, the squared total invariant energy isFor photon-photon pair production, its threshold value is $(2m_e{c}^2{)}^2$: the electron and Positron have zero relative momentum in their centre-of-momentum frame. The minimum possible $E$ occurs for a head-on collision, $\psi =\pi$, soWith $m_e{c}^2\simeq 0.511\mathrm{MeV}$ and $\epsilon =10^{-3}\mathrm{eV}$,The quoted background photon energy is used directly, without substituting a different characteristic value of the cosmic microwave background. A less opposed collision has a higher threshold; parallel photons cannot produce this massive pair.

Choose the initial incident velocity as $u_1\hat{\mathbf{x}}$. The centre-of-mass velocity in the laboratory frame isIn the centre-of-mass frame, the initial velocities are ${\mathbf{u}}_1^{\ast }=m_2{u}_1\hat{\mathbf{x}}/(m_1+m_2)$ and ${\mathbf{u}}_2^{\ast }=-m_1{u}_1\hat{\mathbf{x}}/(m_1+m_2)$. Conservation of momentum and kinetic energy in an elastic collision preserves these two speed magnitudes. Let ${\theta }^{\ast }$ be the outgoing angle of particle 1 in that frame, measured from the initial positive $x$ direction. ThenAdding $\mathbf{V}$ to transform back to the laboratory frame givesThe nonnegative laboratory recoil angle ${\theta }_2$ is measured below the incident axis in this choice of scattering plane. Hence the recoil angle in a two-body elastic collision obeysThe source identity therefore uses ${\theta }^{\ast }$ for the projectile's outgoing direction in the centre-of-mass frame. If instead it denotes particle 2's outgoing direction, the corresponding expression has $1+\cos {\theta }^{\ast }$ in the denominator. A signed laboratory angle for the drawn target trajectory would carry a minus sign. At zero deflection the target remains at rest, so its direction is undefined; ${\theta }_2\to \pi /2$ is a limit.

A Lorentz invariant has the same value when the same physical system is described in any inertial frame. The components of its energy-momentum four-vector change, but their Minkowski norm does not. In particular the relativistic energy-momentum relation iswhere $M$ is the rest mass. Individual total energy and momentum are frame-dependent; the combination on the left is invariant.

For the fixed-target annihilation, the initial total laboratory energy is $E+M{c}^2$ and its total momentum is the incident particle's $\mathbf{p}$, with $c^2{p}^2=E^2-M^2{c}^4$. The centre-of-mass energy available for new-particle creation is thereforeIn the centre-of-mass frame the total momentum vanishes, so $W$ can supply the new particles' total rest energy at threshold. One does not subtract the annihilated particles' original rest energy. The excess laboratory energy beyond $W$ is associated with motion of the whole final system in the laboratory frame.

For the subsequent elastic collision, the equal and opposite transverse components of final momentum imply equal final momentum magnitudes, and hence equal final energies, for the two identical particles. By conservation of relativistic energy, each hasTheir final momentum magnitude $p_f$ and the incident momentum magnitude $p_i$ obeyLongitudinal momentum conservation gives $p_i=2p_f\cos \theta$. It follows that the equal-angle relativistic elastic scattering relation isThis derivation assumes nonzero incident kinetic energy and the stated symmetric outgoing paths. The angle between those paths is $2\theta$, which is evaluated in the two following parts.

In the laboratory frame, the initial target is at rest and the outgoing directions are directly related to detector positions. The centre-of-momentum frame has zero total momentum, so the two outgoing momenta are equal and opposite; it cleanly separates bulk motion from relative kinetic energy. Transforming back is needed to predict laboratory angles and speeds. Neither reference frame removes the need for an appropriate collision law.

Take the incident direction as $\hat{\mathbf{x}}$ and put $S=M_1+M_2$. Momentum conservation fixes the centre-of-momentum frame velocity, and a Galilean transformation gives the initial velocities in that frame:After the collision, $M_1{\mathbf{w}}_1^{\prime }+M_2{\mathbf{w}}_2^{\prime }=0$. Hence both particles have the same final-to-initial speed ratio,The relative kinetic energy is multiplied by $q^2$. This common ratio exists independently of how the outgoing direction is selected.

The meaning of restitution matters here. The usual coefficient of restitution $e_n$ is a ratio of normal separation and approach speeds, not total speeds. In a smooth oblique collision, the tangential relative velocity is unchanged. If the incoming relative velocity makes angle $\chi$ to the contact normal, thenFor example, $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}\ne 1/2$. Therefore the requested identification is not generally true with the standard normal definition. The printed equation is valid if its $e$ denotes the relative speed retention in an oblique collision, namely $e=q$, or if an additional collision model supplies that equality. The elastic cases have $q=e_n=1$ and are unaffected by this qualification.

To derive the scattering formula under that convention, write ${\mathbf{v}}_1^{\prime }=\mathbf{V}+{\mathbf{w}}_1^{\prime }$. Its components give the laboratory scattering circleSolving the resulting quadratic for the nonnegative speed yieldsWith $q=e$, this is the requested formula. A real square root and $v_1\ge 0$ are both necessary: a negative algebraic root is not a second physical speed, and $v_1=0$ does not define a scattering angle. The subsequent limits are taken using this laboratory scattering circle.

In the laboratory frame the incoming particle and the composite move along $+x$, while the target is initially at rest. In the equal-energy two-photon decay of a moving particle, zero total transverse momentum makes the two photon directions symmetric about $+x$. Each photon has energy $2m{c}^2$ and momentum magnitude $2mc$, so longitudinal momentum conservation gives $2(2mc)\cos \theta =2.4mc$. ThereforeTheir momenta can be written $(1.2,\pm 1.6)mc$. The opening angle is not $180^{\circ }$ in this frame, because the parent carried nonzero total momentum.

Maintaining a fixed separation in one inertial laboratory frame is not the same as maintaining a fixed separation measured locally with the accelerating rockets. The attached string is prevented from undergoing the laboratory contraction that an unstrained comoving string would have. It is therefore stretched and develops tensile stress: this is Bell's spaceship paradox, not a failure of Lorentz contraction.

For sufficiently prolonged acceleration and a finite elastic breaking strain, the string breaks. If an acceleration history produces only a small bounded increase in speed, whether it actually breaks depends on its material strength; the kinematic conclusion is stretching, not an unconditional assertion that every arbitrarily weak acceleration must snap every string. Stress also propagates through a real string at a finite speed. The construction enforcing Born rigidity in the following parts is a different set of trajectories chosen to keep the comoving separation fixed.

A particle of total incident energy $E$ and rest energy $E_0$ strikes an identical stationary particle. If the two outgoing paths make equal and opposite nonzero angles $\theta$ to the incoming direction in the laboratory frame, transverse conservation of momentum makes their outgoing energies equal to $(E+E_0)/2$. Longitudinal conservation of momentum and the mass shell relation imply ${\tan }^2\theta =2E_0/(E+E_0)$. The nonrelativistic limit is $\theta =45^{\circ }$; at large incident energy both paths become forward directed.