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 $c$.

A holonomic constraint can be written as an equation $f(q_1,\dots ,q_n,t)=0$ involving coordinates and possibly time, without an irreducible dependence on velocities. For example a particle confined to a fixed sphere obeys $x^2+y^2+z^2-a^2=0$. A general rolling without slipping constraint in three dimensions need not be integrable into such a coordinate equation.

The number of degrees of freedom is the number of independent coordinates needed to specify an allowed configuration locally. For $N$ unconstrained point particles in three dimensions it is $3N$; $k$ independent regular holonomic constraints reduce it to $3N-k$. Redundant constraints must not be counted twice, and nonholonomic velocity restrictions require separate care.

Generalized coordinates are variables $q_1,\dots ,q_n$ that specify the allowed instantaneous configuration of a mechanical system. They need not be Cartesian positions: a rigid pendulum's angle is a suitable single coordinate. With constraints solved, the particle positions are functions ${\mathbf{r}}_i(q_1,\dots ,q_n,t)$.

A straight diameter at nonzero angle $\chi$ to the rotation axis moves the body toward that axis during the inward half of its trip. With no external axial torque, the total axial angular momentum of the Earth and body is conserved. The body's contribution to the moment of inertia decreases toward the centre, so the Earth spins slightly faster; the reverse happens on the outward half, restoring the original rotation rate at the far surface.

For a body of mass $m$ constrained to follow a tunnel fixed in the Earth, an illustrative angular-momentum balance isHere $I_E$ is the Earth's axial moment of inertia, treated as constant, and $x$ is the signed distance along the diameter. A frictionless guide can still exert normal forces: these supply the transverse acceleration and exchange axial angular momentum with the rotating Earth. Simply allowing an unconstrained body to fall would not keep it inside a narrow nonpolar straight tunnel. The effect vanishes for $\chi =0$ and is largest for an equatorial diameter.

Let $x$ be the signed distance from the Earth's centre along the polar tunnel, with $x(0)=r_0$ and $\dot{x}(0)=0$. The shell theorem leaves only the enclosed mass $4\pi \rho |x{|}^3/3$ contributing to the gravitational field. ConsequentlyThis is simple harmonic motion with $\omega =\sqrt{g/r_0}$ and $x=r_0\cos (\omega t)$. Crossing the centre takes a quarter-period; reaching the opposite surface takes a half-period:The body's mass cancels. The polar diameter lies on the rotation axis, so the tunnel motion has no additional centrifugal force or Coriolis force from the Earth's rotation. A realistic radial density profile would change the harmonic equation.

For a fixed-speed orbital launch-direction error, the initial radius and speed are correct, so the specific orbital energy is still $\epsilon =-GM/(2R)$ and the new semi-major axis is still $a=R$. Resolve the launch velocity into radial and tangential components, taking $\alpha$ as the in-plane deviation from the intended tangent:For a nondegenerate orbit $\cos \alpha \ne 0$, the orbital eccentricity follows from $j^2=GMa(1-e^2)$:The intended circular orbit has radius $R$ everywhere, so the largest radial distance between the orbit curves occurs at either apsis:This is the geometric separation of the two orbit curves, not the distance between two satellites compared at equal elapsed times. The latter also depends on orbital phase. A purely radial launch is a degenerate zero-angular-momentum trajectory through the central body and requires separate treatment of the collision.

Write $\mu =GM$ and let $v_c=\sqrt{\mu /r_0}$ be the original circular orbital speed. An impulse leaves the orbital angular momentum unchanged if it is radial: its angular impulse $\mathbf{r}\times \Delta \mathbf{p}$ is zero. The tangential velocity remains $v_c$, and an added radial component $w$ givesThus kinetic energy increases even though angular momentum does not. For a Radial kick from a circular Kepler orbit, the new specific orbital energy and squared specific angular momentum areThe orbital eccentricity is $e=\sqrt{1+2\epsilon j^2/{\mu }^2}=\sqrt{\beta -1}$, while the semi-latus rectum is $p=j^2/\mu =r_0$. HenceFor $\beta =2$ the trajectory is a parabolic Kepler orbit and has no finite furthest radius; its periapsis is $r_0/2$. For $\beta >2$ it is a hyperbolic orbit, with the same formula for $r_{\min }$ but unbounded radius:One must not use the negative denominator in the bound-orbit expression as a negative physical apocentre. On a departing unbound branch the conic's periapsis may already lie in the past; the future minimum from the kick event is then $r_0$. These are ideal conic orbits; a periapsis inside a real planet would cause impact before the ideal continuation.

Here $G$ is the gravitational constant, $r$ is the separation of the orbiting object from the central body's centre, and $J$ is the magnitude of the object's conserved orbital angular momentum, not angular momentum per unit mass. Approximate the much heavier central body as fixed, neglect other forces, and use the Newtonian gravitational potential energy $U=-GM{m}_0/r$, with zero at infinity.

A central force confines the motion to a fixed orbital plane. In plane polar coordinates its kinetic energy is $T=\frac{1}{2}{m}_0({\dot{r}}^2+r^2{\dot{\varphi }}^2)$. Conservation of angular momentum gives $J=m_0{r}^2\dot{\varphi }$; eliminating $\dot{\varphi }$ yieldsThe centrifugal term belongs to the radial effective potential; it is the eliminated transverse kinetic energy, not an additional physical interaction. If recoil of the central body mattered, the relative-coordinate reduction would use the reduced mass.

Treat the atmospheric proton as stationary in the Earth frame, the incident antiproton as vertically downward, and neglect atmospheric energy loss, scattering and the Earth's curvature initially. Take $m_p{c}^2=0.9383\mathrm{GeV}$. Conservation of energy and equal outgoing energies giveEqual energies imply equal momentum magnitudes. Their transverse momenta cancel, so the two muons have equal downward components $p_z=p_{\bar{p}}/2$, whereEach muon therefore makes an angle $\theta$ with the downward vertical satisfyingThe paths are not both vertical: replacing their path lengths by $20\mathrm{km}$ would miss this two-body kinematic correction. Since $v_z=p_z{c}^2/E_{\mu }$, the proper flight time over vertical height $H$ isFor equal-energy atmospheric muon-pair survival, independent exponential particle decays with mean proper muon lifetime ${\tau }_0=2.2\mu \mathrm{s}$, both survive with probabilityEquivalently each travels about $20.55\mathrm{km}$, with proper flight time about $0.404\mu \mathrm{s}$, and survives with probability about $0.832$. The result is decay-only survival to an ideal surface, not a prediction including air interactions or detector efficiency. For a spherical Earth of radius $R_E$, the exact path length is $\ell =(R_E+H)\cos \theta -\sqrt{R_E^2-(R_E+H{)}^2{\sin }^2\theta }$; inserting $P_{\mathrm{both}}=\exp [-2\ell /({\gamma }_{\mu }{v}_{\mu }{\tau }_0)]$ gives a small curvature correction.

For fixed-target electron-Positron formation, set $m_1=m_2=m_e$ in a and solve for the incident total energy:The incident kinetic energy is smaller by another $m_e{c}^2$, a negligible relative change here. Most of the fixed-target energy becomes motion of the final rho meson in the laboratory.

With equal and opposite incoming momenta, the centre-of-momentum frame is the laboratory. Each incoming particle has the same total energy, and there is no final translational kinetic energy at formation:This large contrast illustrates the invariant-energy advantage of colliding beams. The given $768\mathrm{MeV}/c^2$ resonance mass is used rather than replacing it with a modern tabulated value.

The incident particle obeys the energy–momentum relation $E^2=p^2{c}^2+m_1^2{c}^4$. The stationary target contributes energy $m_2{c}^2$ and zero momentum. A single product inherits the total four-momentum, so its invariant mass satisfiesHenceHere $E$ is the incident total energy, including rest energy. In a one-particle final state, exact production of a specified mass fixes the incident energy; there is no adjustable final relative motion to absorb excess energy. A resonance with a finite decay width relaxes that ideal exact equality over its line shape.

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 nonrelativistic mechanics and let $\mathbf{V}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/(m_1+m_2)$ be the constant centre-of-mass velocity. In the centre-of-mass frame, conservation of momentum gives $m_1{\mathbf{u}}_1+m_2{\mathbf{u}}_2=0$ before and after the elastic collision. Thus ${\mathbf{u}}_2=-(m_1/m_2){\mathbf{u}}_1$, and the kinetic energy isThe decomposition $T_{\mathrm{lab}}=T_{\ast }+\frac{1}{2}(m_1+m_2)V^2$ and constant centre-of-mass velocity show that the elastic collision preserves $T_{\ast }$, so the positive constant multiplying $u_1^2$ forces that speed to remain unchanged. The same momentum relation fixes the second speed:Only their directions can change; the two velocities remain antiparallel in this frame. This derivation does not require a head-on collision.

In the parent's rest frame, the identical daughters have equal and opposite momenta, and each has total energy $110\mathrm{GeV}$. With daughter rest energy $Q=90\mathrm{GeV}$, each speed in this frame isTransform to the rest frame of one daughter using the collinear relativistic velocity addition law. The other daughter then has speedAn independent invariant mass check is useful. In the rest frame of one daughter the other's total energy satisfies $W^2=2Q^2+2Q{E}^{\prime }$, so $E^{\prime }=178.8889\mathrm{GeV}$ and the relative Lorentz factor is ${\gamma }_{\mathrm{rel}}=E^{\prime }/Q=161/81$. Hence $w/c=\sqrt{1-{\gamma }_{\mathrm{rel}}^{-2}}$, the same result.

The answer is unchanged if the parent is produced by a fixed-target collision. Its laboratory velocity then differs, but its rest mass and two-body decay kinematics do not. The scalar product of the daughters' four-momenta, equivalently their relative Lorentz factor, depends only on $W$ and $Q$. A common boost cannot change the speed of one measured in the rest frame of the other.^[ref]

Let $P=m_p{c}^2$ denote the proton rest energy and $W=220\mathrm{GeV}$ the desired particle's rest energy. For an antiproton of total energy $E_{\bar{p}}$ striking a stationary proton, the incoming squared centre-of-mass energy isFor production of the stated single particle, four-momentum conservation requires $s=W^2$. ThusThe numerical value uses $P=0.938272\mathrm{GeV}$. It is a total incident energy; the corresponding kinetic energy is $E_{\bar{p}}^{\mathrm{fixed}}-P\simeq 25.7902\mathrm{TeV}$. Much of the laboratory energy supplies motion of the outgoing system rather than its invariant mass.

For equal and opposite beam momenta, the laboratory is the centre-of-mass frame. Each beam has the same total energy and their sum is $W$, givingThese are the fixed-target production threshold and collider kinematic comparison for the specified final state. Extra final particles or restrictions from conserved quantum numbers would require a separately specified reaction.

Allowing azimuthal motion makes the endpoint speed satisfy $v^2=l^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\phi }}^2)$. The light-rod spherical pendulum therefore hasThe azimuth $\phi$ is a cyclic coordinate, since the Lagrangian is independent of it. Its conjugate canonical momentum is the conserved vertical angular momentum:For a finite uniform rod, replace $m{l}^2$ by $I$ and $mgl$ by $A$ from i. The same azimuthal symmetry then gives $p_{\phi }=I{\sin }^2\theta \dot{\phi }$.

The Hamiltonian is the Legendre transform of this Lagrangian. With $p_{\theta }=I\dot{\theta }$,There is no explicit time dependence, so $dH/dt=\partial H/\partial t=0$. Here $H$ also equals the total mechanical energy: the constraint is stationary, the kinetic energy is quadratic in generalized velocities, and gravity is conservative. The apparent singularity of the momentum expression at $\sin \theta =0$ is a spherical-coordinate singularity; trajectories reaching a pole must be described consistently, and nonzero $J_z$ prevents reaching it at finite energy.

Four-momentum conservation fixes the outgoing system's invariant mass to the same $2E/c^2$. In the rest frame of particle 3, let particle 4 have energy $E_4^{\prime }$ and momentum ${\mathbf{p}}_4^{\prime }$. Each outgoing particle has rest mass $M_2$, soConsequentlyHere $E_4^{\prime }$ is the total energy, including rest energy. At the production threshold $E=M_2{c}^2$, it reduces to $M_2{c}^2$, as it should. Production is kinematically possible only for $E\ge M_2{c}^2$. The initial rest mass $M_1$ drops out because the conserved total centre-of-mass energy, rather than the identity of the incoming particles, determines this outgoing relative motion.