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Consider a nonrelativistic particle of mass $m$ undergoing an elastic collision with a stationary free target of mass $M=Am$, where $A\ge 1$. Assume the outgoing direction is uniformly distributed over solid angle in the center of mass frame; this assumption concerns the scattering angle distribution and need not hold for every target or energy. Let $\mu$ be the cosine of that frame's scattering angle. Transforming the outgoing velocity back to the laboratory givesThe uniform distribution of $\mu$ on $[-1,1]$ makes $x$ uniform on $[\alpha ,1]$. For $A>1$, integration yieldsFor equal masses, take the limit $\alpha \to 0$: $\xi =1$. The expected value of the fractional energy after a collision is instead $\mathbb{E}[x]=(1+\alpha )/2$. These are different averages: averaging a logarithm does not give the logarithm of an average.

For independent collisions with the same distribution, the logarithms of successive energy ratios add. ThusHere $E_{\mathrm{geom}}$ is the geometric-mean energy, and $n_{\log }$ is a characteristic collision count for reaching an energy scale $E_{\ast }$. It is not an exact first-passage expected value. Individual histories fluctuate, and neither the time between collisions nor absorption is specified by this model.

In an original chosen example, set $E_0=10^6\mathrm{eV}$ and $E_{\ast }=0.025\mathrm{eV}$. Approximate hydrogen by $A=1$ and carbon by $A=12$. ThenThese are illustrative model values, not measured collision counts from 1934. The energy scale marks where neglected thermal motion matters; the calculation does not assert continued cooling to zero energy.

Momentum conservation states that total momentum stays constant when the net force exerted from outside the chosen collection is zero. Use a closed collection, meaning the same fixed-mass point masses are followed throughout, and an inertial frame. ThenTo see why, sum Newton's second law over the members. If ${\mathbf{F}}_{ij}$ is the force on member $i$ from member $j$, Newton's third law assumes ${\mathbf{F}}_{ij}=-{\mathbf{F}}_{ji}$. Each pair cancels in the total:The internal forces act on different bodies and can change their individual momenta; their contributions cancel only in the collection's sum. An isolated system has no external interaction, but balanced external forces also suffice for momentum conservation. A closed system merely excluding matter exchange need not have conserved momentum. ^[ref]

The usefulness is that the detailed internal forces during a collision need not be known. Both an elastic collision and a perfectly inelastic collision obey the same isolated momentum balance; kinetic energy conservation is an additional condition for the elastic collision. If the external impulse is small but nonzero, the total momentum change is correspondingly small. For a variable-mass system exchanging matter, the entering or leaving momentum must also be included.

In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.

Let $\overline{N}$ be the expected value of the number of radioactive product atomic nuclei, initially zero, produced at constant rate $R$. Let their nuclear decay constant be $\lambda >0$. Assume one product species, negligible target depletion and negligible capture by the product. During irradiation, production adds $Rdt$ while radioactive decay removes $\lambda \overline{N}dt$ in expectation. Multiplying the resulting ordinary differential equation by $e^{\lambda t}$ givesThe expected radioactive activity $\overline{A}=\lambda \overline{N}$ tends to the production rate $R$. This saturation is a balance of continuing production and radioactive decay, not exhaustion of target atomic nuclei.

If irradiation ends at $t_i$, let $s$ denote elapsed time afterward. Setting the production term to zero givesFor detection efficiency $\epsilon$, constant background count rate $b$, delay $d$ and counting-window duration $\Delta$, integrating the expected count rate givesThis assumes at most one recorded pulse per product transformation and negligible counter losses. Detector response, branching fractions and background must be calibrated independently. Subtracting an estimated background does not remove its counting uncertainty. Several products give a sum of these terms with distinct nuclear decay constants; a single exponential decay fit can then mislead.

For a wholly stipulated example, choose $R=120{\min }^{-1}$, half-life $2\min$ and $t_i=4\min$. Then $\lambda =\log 2/(2\min )$, $\overline{A}(t_i)=90{\min }^{-1}$ and $\overline{N}(t_i)\simeq 259.69$. Choose $\epsilon =0.25$, $b=2{\min }^{-1}$ and $d=\Delta =2\min$. The result isThe background contributes four expected counts. Multiplying the initial net count rate in that window by its duration would give $22.5$ net counts, overestimating the integrated $16.23$ because the product decays appreciably during counting. All these values and the following curves are original examples, not measured historical data or parameters assigned to a particular silver isotope.

In neutron detection by proton recoil, a neutron strikes a hydrogen atomic nucleus, namely a proton, and the moving proton produces detectable ionization. Paraffin wax provides many hydrogen targets. There are two distinct questions: how much kinetic energy can a collision transfer, and how often does it occur? Elastic collision kinematics answers the first. It does not determine a collision cross-section or a detector's efficiency.

For the following derivations the target is initially at rest in the laboratory but is free to recoil. Its mass is finite. In particular a proton must not be treated as an immovable target. All massive particles are nonrelativistic particles, their translational kinetic energies are $m{v}^2/2$, and the collision changes no internal state. Target binding and thermal motion are neglected relative to the incident kinetic energy. These assumptions suit an introductory description of fast-neutron elastic collisions; they do not cover every neutron capture or other nuclear reaction.

Neutron moderation reduces the kinetic energy of fast neutrons through repeated collisions in a neutron moderator. In a simple elastic collision model, hydrogen is especially effective because a proton and a neutron have nearly equal masses: the maximum elastic recoil energy fraction can approach one. The mean logarithmic energy decrement describes typical accumulated slowing, whereas the maximum transfer describes a single endpoint.

Slowing, absorption and escape compete. A material can remove neutrons by neutron capture before they become thermal neutrons, and some neutrons escape its boundaries. Describing a neutron moderator therefore requires more than collision kinematics. Near thermal equilibrium, target thermal motion can also give energy back to a neutron; the initially stationary target approximation must then be replaced.

For equal-mass elastic collisions, one reflection parametrization is $v^{\prime }=v-((v-v_{\ast })\cdot n)n$ and $v_{\ast }^{\prime }=v_{\ast }+((v-v_{\ast })\cdot n)n$. The fixed-$n$ transformation preserves pairwise momentum, kinetic energy and velocity-pair measure. A nonnegative kernel invariant under particle interchange and collision reversal yields a gain-loss operator. Its weak Boltzmann collision identity proves conservation and its entropy identity proves the Boltzmann H theorem. Unlike the Maxwell molecule collision operator, this general kernel may depend on relative speed.

The rubidium experiment used laser cooling, a magneto-optical trap and magnetic trapping of neutral atoms with a rotating bias field. Radio frequency transitions removed energetic atoms; elastic collisions restored thermal equilibrium. Before absorption imaging, the trap was decompressed, then nearly released with a gradient supporting against gravity. Varying the final evaporation cut revealed a narrow anisotropic core within an isotropic thermal cloud. The broad component supplied a simultaneous comparison through the same optics. Resolution and size calibration limited inference; interactions broadened the core relative to an ideal gas.^[ref]

An elastic collision with a wall moving toward the particle at speed $V$ changes the particle speed by approximately $2V$ when recoil is negligible. The collision rate is approximately $v/\ell$ for $v\gg V$, and $\dot{\ell }=-2V$, yielding $\dot{v}/v=-\dot{\ell }/\ell$. For a freely moving finite wall of mass $M$, the neglected recoil changes the increment by order $(m/M)v$, so relative accuracy also requires $(m/M)(v/V)\ll 1$.

// Target: classical-mechanics.bigb

Evaporative cooling of atoms selectively removes atoms with unusually high total energy, then elastic collisions rethermalize those remaining. Loss of particles alone is insufficient: the lost atoms must carry more than the mean energy per particle if the temperature is to fall. The process is open and dissipative, so it can increase the phase-space density of an atomic gas in the remaining sample while reducing its atom number. As the temperature falls, the cutoff must be lowered to keep replenishing a useful escaping tail; a fixed high cutoff eventually gives negligible evaporation.

An experimental sequence first loads laser-cooled atoms into a conservative trap and suppresses resonant-light heating. In a trap using magnetic trapping of neutral atoms, a radiofrequency evaporation knife drives a Zeeman splitting transition from a trapped weak-field-seeking state to an untrapped state on a surface where the splitting matches the radiofrequency photon energy. Only atoms energetic enough to reach that surface escape. Lowering the radiofrequency cutoff progressively removes less energetic atoms. Alternatively, lowering an optical dipole trap's depth permits energetic atoms to spill out. This changes confinement as well as the escape cutoff, so it can reduce the collision rate. Radiofrequency-selective evaporation and the need for favourable elastic-collision and background-loss rates are documented in the original condensate experiments.^[ref]

The elastic collision rate ${\Gamma }_{\mathrm{el}}\sim n{\sigma }_{\mathrm{el}}\bar{v}$ must be large enough for elastic rethermalization of a trapped atomic gas throughout evaporation, and much faster than unwanted loss and heating. A clean vacuum reduces background collisions; suitable trapping states suppress inelastic spin-changing loss, and density must not make three-body recombination dominant. The cutoff ramp should allow equilibration without taking so long that background losses dominate. A large evaporation truncation parameter $\eta ={ϵ}_{\mathrm{cut}}/(k_{B}T)$ selects energetically valuable atoms, but an excessively large value suppresses their escape rate. A very small value discards too many useful low-energy atoms. If elastic collisions become more frequent as evaporation proceeds, the resulting runaway evaporative cooling can improve efficiency; it is not automatic in a weakening optical trap.

The energy selectivity for evaporation to increase phase-space density can be made precise in a classical three-dimensional harmonic trap with fixed frequencies. Equipartition gives $E=3N{k}_{B}T$. If each escaping atom removes mean energy ${ϵ}_{\mathrm{ev}}$, energy balance is $dE={ϵ}_{\mathrm{ev}}dN$ with $dN<0$, soThe peak harmonic-trap phase-space density is $\mathcal{D}=N(\hslash \bar{\omega }/(k_{B}T){)}^3$. ThereforeCooling requires ${ϵ}_{\mathrm{ev}}>3k_{B}T$; increasing $\mathcal{D}$ despite atom loss requires the stronger condition ${ϵ}_{\mathrm{ev}}>4k_{B}T$. These inequalities assume rapid thermalization and no work from changing trap frequencies. They distinguish lower temperature from genuinely improved quantum degeneracy. Eventually quantum statistics and inelastic processes require a refinement of this classical estimate. Efficient evaporation combines energy-selective loss, rapid elastic redistribution, and low unwanted loss; it can bring a precooled gas to Bose-Einstein condensation.

evaporative cooling removes atoms selectively from the high-energy tail of a trapped gas. The remaining sample has less energy per atom; elastic collisions then redistribute energy and restore a thermal distribution at lower temperature. Random atom loss would generally remove average energy and would not provide the same cooling mechanism. Repeated energy selection and rethermalization, rather than one removal event, is the essential cycle.

Using the evaporation truncation parameter $\eta$, let the escape threshold be ${ϵ}_{\mathrm{cut}}=\eta k_{B}T$. As the gas cools, a fixed threshold is populated by fewer atoms, so evaporation stalls. In forced evaporative cooling the threshold is lowered during the sequence. In a magnetic trap, a radiofrequency evaporation knife flips trapped weak-field-seeking atoms into untrapped spin states where the Zeeman splitting matches the photon energy. Energetic atoms reach that resonance surface; sweeping the radiofrequency lowers the energy knife. In an optical dipole trap, lowering the trap depth offers a different implementation, although it usually also weakens confinement.

A simple evaporative cooling energy balance shows when atom loss is useful. For a classical gas in a three-dimensional harmonic trap of fixed frequencies, equipartition gives $E=3N{k}_{B}T$, counting both kinetic and potential energy. Suppose escaping atoms remove average energy ${ϵ}_{\mathrm{ev}}=(\eta +\kappa )k_{B}T$, with $\kappa$ a distribution-dependent excess above the threshold. For $dN<0$,Thus preferential removal cools when the removed energy exceeds $3k_{B}T$. More stringently, cooling must raise the phase-space density of an atomic gas, not merely lower the temperature:For this harmonic-trap phase-space density, with fixed geometric-mean trap frequency $\bar{\omega }$, $d\ln \mathcal{D}/d\ln N=1-3\alpha$. Therefore phase-space density rises during atom loss if $\alpha >1/3$. The numerical factors here belong to the stated harmonic-trap model, not every evaporative apparatus.

There must be enough elastic collisions to repopulate the tail: ${\Gamma }_{\mathrm{el}}\sim n\sigma \bar{v}$ should exceed damaging loss and heating rates, and the ramp should allow rethermalization. Increasing density can improve elastic collisions but also enhances inelastic and three-body losses. Very rapid removal can leave a nonequilibrium cloud; very slow removal exposes it to background-gas losses. Lowering an optical trap can reduce density and collision rate, so a falling temperature is not by itself evidence of efficient evaporation.

The dilute-gas condensation experiments made this tradeoff concrete. Eric A. Cornell and Carl Wieman's account explains how selective escape and contraction of a trapped cloud raise phase-space density, while the need for rapid two-body equilibration and slow three-body loss restricts usable density. It also describes expansion imaging as an experimental diagnostic.^[ref] Useful controls compare different knife ramps and hold times, measure both atom number and expansion temperature, and check trap-frequency calibration. Pure adiabatic expansion can cool without increasing phase-space density. Successful evaporation pays an atom-number cost to increase the remaining gas's phase-space occupancy.

To obtain the Boltzmann equation, take a dilute gas with a short-range pair potential. Its interaction range is small compared with the mean free path and the spatial scale on which the phase-space distribution function changes. The binary collision approximation treats encounters as isolated elastic collisions; simultaneous encounters involving three or more particles can be neglected. The encounter duration is short compared with the collision relaxation time and macroscopic evolution time, and the external force changes negligibly during an encounter. This permits local, effectively instantaneous scattering at one position $\mathbf{x}$.

The essential statistical closure is molecular chaos for the Boltzmann equation: the distribution of incoming pairs factorizes as $f{f}_{\star }$. Outgoing pairs need not be independent. Follow each incoming pair through its two-body scattering, and use conservation of momentum and kinetic energy, together with time reversal and the invariant scattering measure, to express both gain and loss in the same variables. This givesHere $d\Omega =\sin \theta d\theta d\phi$; when the measure is written as $d\theta d\phi$, its Jacobian is included in the Boltzmann scattering kernel. In a scattering-angle convention, that kernel contains relative speed times differential scattering cross-section. The hemisphere prevents double counting equivalent outcomes for indistinguishable particles; the precise factor is included in $B$. The primed velocities are connected to the unprimed ones by the reversible elastic collision map. This closure is appropriate for a dilute short-range gas, rather than a generic long-range interacting system with an important collective mean-field theory contribution.

The further replacement by the BGK model is a phenomenological relaxation closure. Replace the full nonlinear Boltzmann collision operator by relaxation towards a local Maxwellian, with a single, velocity-independent collision relaxation time $\tau >0$:It models relaxation towards a local Maxwellian without resolving the detailed scattering. Treating $\tau$ as constant is an additional simplification; it is of the order of a mean time between effective collisions. Near a local Maxwellian, departures in the nonconserved velocity moments decay on this timescale.

Choose $f^{(0)}$ to match the collision invariants $1$, $\mathbf{v}$ and $|\mathbf{v}{|}^2$ of $f$. For unit particle mass, this meansHere $\rho$ is the mass density, $\mathbf{u}$ is the mean velocity, and $\Theta$ is the temperature in squared-velocity units, $\Theta =k_{B}T/m$ before setting $m=1$. In particular,Thus relaxation preserves local mass, momentum and kinetic energy rather than relaxing them to externally prescribed values.

Choose the positive direction along the incoming car's velocity. Treat each car as a translating body, take $m_1,m_2>0$, and neglect external impulse, rotation and energy losses during the short elastic collision. If $v_1,v_2$ are the final signed velocities, conservation of momentum and conservation of energy giveFactoring the second relation and using the first gives $U_1+v_1=v_2$ on the collision branch. The other algebraic branch, $v_1=U_1,v_2=0$, describes no impulse and cannot describe separating bodies after contact. Solving the collision branch yieldsFor $0<R<1$ the first car rebounds and the second moves forwards. At $R=1$ the first stops and the second takes its original velocity. For $R>1$ both move forwards, with $v_2-v_1=U_1>0$, so they separate. In the limits $R\to 0$ and $R\to \infty$, the first car's velocity approaches $-U_1$ and $U_1$, respectively.

For the deformable car, take $y=0$ at first contact. The resisting force does negative work. The work-energy theorem therefore givesAt the first stop $v=0$, whence $\sqrt{L-y}=\sqrt{L}-M{V}^2/4$. The speed bound ensures that this right-hand side is positive, so squaring introduces no extraneous solution. The stopping compression isFor $0<V<2L^{1/4}/M^{1/2}$ this lies strictly between $0$ and $L$, as required before complete crushing. The coefficient in the stipulated force law is taken in its stated normalization.

Take rightward velocities as positive just before the right-wall collision. The incoming particle velocity is $v$ and the wall velocity is $-V$. Their total momentum divided by total mass gives the centre of mass velocityIn an elastic collision, relative velocity reverses, so the particle's signed outgoing velocity is $v_{\mathrm{out}}=2v_{\mathrm{cm}}-v$. For $M>m$ it is negative, and its speed isConsequently the exact change in speed isFor $m/M\to 0$ with $v,V$ fixed, this proves the requested leading increment $v^{\prime }-v\simeq 2V$. Reflection gives the same increment at the left wall.

This approximation is not uniform when $v/V$ is allowed to become arbitrarily large. To neglect recoil relative to the increment $2V$ requiresThe inequalities $m\ll M$ and $v\gg V$ alone do not imply it: for example, $m/M=10^{-3}$ and $v/V=10^3$ give an exact zero speed increment. The subsequent slowly moving-wall approximation therefore uses the consistent joint range $1\ll v/V\ll M/m$, or treats the walls as effectively prescribed massive reflectors. The assumed slow change of wall speed over many impacts embodies the same negligible-recoil requirement.

Within this approximation, the separation derivative is the right-wall velocity minus the left-wall velocity:For a time window containing many crossings but little separation change, $\ell /v\ll \Delta t\ll \ell /V$, the flight time between successive wall collisions is approximately $\ell /v$. Thus their number and the cumulative speed change arePassing to the averaged differential equation givesHence the particle-speed invariant between slowly approaching walls isIt is an adiabatic invariant for the rapid elastic bouncing motion while the separation and wall velocities vary slowly on the flight-time scale.

Label the initially left-hand rod $A$ and the right-hand rod $B$. At impact their centres are $(0,\ell )$ and $(0,-\ell )$, and their contact points are both the origin. Each rod has total mass $2m$ and moment of inertiaLet the impulse on $B$ be $+J{\mathbf{e}}_x$, so the impulse on $A$ is $-J{\mathbf{e}}_x$. The impulsive form of Newton's second law givesThe contact lever arms from the centres are $(0,-\ell )$ for $A$ and $(0,+\ell )$ for $B$. Both angular impulses are $-\ell J{\mathbf{e}}_z$. Thus, taking counterclockwise rotation as positive,Equal and opposite linear impulses therefore produce equal clockwise rotations, not opposite rotations.

The restitution law uses the velocities of the contacting endpoints, including the rotational terms. For $A$ that velocity isand for $B$ it is $u_B=-V+J/m$. The approach speed before impact is $2V$. The coefficient of restitution $e$ therefore givesSubstituting into the centre and angular velocities yields the endpoint collision of two freely rotating equal rods:For the first, perfectly elastic case, $e=1$, soAll initial translational energy has become rotational energy. Indeed the total initial kinetic energy is $2m{V}^2$, while the final energy for general $e$ isThe loss is $m{V}^2(1-e^2)$, zero for an elastic collision. Conservation of total angular momentum about the origin provides another check: initially it is $-4m\ell V$; after collision the centre-motion contribution is $-2m\ell V(1-e)$ and the spins contribute $-2m\ell V(1+e)$, with the same total. The initial distance $d$ determines when the collision occurs but does not enter the impulsive velocities.

A kinetic model of a dilute gas. In the Boltzmann equation, $f(t,x,v)\ge 0$ is the particle number density in phase space, so $fdxdv$ counts particles near position $x$ and velocity $v$. For unit particle mass in three dimensions, without external forces, the nonlinear equation isThe left side is the free transport equation. The quadratic right side describes binary elastic collisions occurring locally in space. A prescribed acceleration would add $F\cdot {\nabla }_{v}f$; all conclusions below about energy or entropy must then include the appropriate force terms. Assume enough smoothness, velocity decay and finite moments to justify the calculations, and either periodic space or decay that removes spatial boundary fluxes. These are assumptions on a classical solution, not a general existence theorem.

The dilute-gas approximation neglects simultaneous many-body collisions and treats the interaction region as small relative to macroscopic spatial scales. Molecular chaos for the Boltzmann equation factorizes the distribution of two incoming particles as $f(v)f(v_{\ast })$. It applies to incoming collision pairs; it does not assert that an interacting many-particle system stays exactly independent. This approximation closes the equation at the one-particle level and explains its nonlinearity.

Collision geometry and the gain-loss operator. Let $g=v-v_{\ast }$ and $n\in S^2$. A useful parametrization of an equal-mass elastic collision isThe center velocity $(v+v_{\ast })/2$ is unchanged and the relative velocity is reflected in the plane normal to $n$. ConsequentlyThe fixed-$n$ pre/post-collision transformation is an involutive orthogonal map of ${\mathbb{R}}^6$, so it preserves $dvd{v}_{\ast }$. Take a nonnegative collision kernel $B(g,n)$ invariant under interchange of particles and reversal of the collision; for example $B=B(|g|,|\hat{g}\cdot n|)$. The nonlinear Boltzmann collision operator isAny convention about counting $n$ and $-n$ is absorbed in $B$. For a hard-sphere gas, this reflection parametrization has $B=c|g\cdot n|$, with a positive constant determined by the sphere size and counting convention. Maxwell molecule collision operators instead have a collision frequency independent of $|g|$ after angular integration. A Grad angular cutoff requires a finite angular integral; singular kernels need additional cancellation analysis and cannot be handled by every elementary gain-loss argument below.

Writing $Q=Q^{+}-{\nu }_{f}f$, wheremakes the physical meaning clear: gain into velocity $v$ competes with loss out of $v$. Along a free characteristic curve, put $F(t)=f(t,x+tv,v)$. Then $F^{\prime }+{\nu }_f(t,x+tv,v)F=Q^{+}(t,x+tv,v)$. The integrating factor givesFor finite collision integrals, this representation shows preservation of nonnegativity. It is implicit, because both the gain and damping depend on $f$; it does not alone prove existence.

The weak collision identity and conserved quantities. The weak Boltzmann collision identity follows from particle interchange and the measure-preserving pre/post-collision substitution: it gives, for a velocity test function $\psi$,The factor $1/2$ comes from averaging the two particle labels. Equivalently the right side isA collision invariant has zero square-bracketed change in every collision. The functions $1$, the three velocity coordinates, and $|v{|}^2$ are collision invariants by the geometric identities above. Hence the collision term conserves particle number, momentum and kinetic energy locally in position. Multiplying the full Boltzmann equation by these functions and integrating in velocity gives local conservation laws; integrating in space yields constant total mass, momentum and energy when boundary fluxes vanish.

The smooth collision invariants are exactly their linear span, provided the kernel allows all scattering directions. Here is a direct proof. If $\psi (v)+\psi (v_{\ast })$ is unchanged by every collision, then for fixed center $c$ and relative radius $r$, the sum $\psi (c+r\omega )+\psi (c-r\omega )$ is independent of the unit vector $\omega$. Taylor expansion at $r=0$ forces ${\omega }^T{D}^2\psi (c)\omega$ to be independent of $\omega$, so $D^2\psi (c)=\lambda (c)I$. Mixed derivatives vanish. Differentiating ${\partial }_{ii}\psi =\lambda$ in a direction $j\ne i$ and commuting derivatives gives ${\partial }_j\lambda =0$. Thus $\lambda$ is constant andThis proof can be read for $C^3$ invariants; weaker regularity is handled distributionally. The multidimensional scattering hypothesis matters: in one dimension equal-mass elastic collisions only exchange the two velocities, and every function is then a collision invariant.

Entropy dissipation and the H theorem. Define the Boltzmann H functional $H(f)=\int f\log fdxdv$, with $0\log 0=0$. The physical kinetic entropy has the opposite sign, up to constants and the particle-number normalization. Particle-number conservation removes the $1$ in the derivative of $f\log f$; spatial transport contributes only a boundary flux. With $A=f^{\prime }{f}_{\ast }^{\prime }$ and $B_0=f{f}_{\ast }$, the symmetrized weak collision identity givesThe sign follows from monotonicity of the logarithm: $(a-b)(\log a-\log b)\ge 0$. Zeros are treated by the usual extended nonnegative limit. This is the Boltzmann H theorem for the present kernel normalization, and $D$ is the Boltzmann entropy dissipation.

For a positive smooth $f$ and a kernel positive on all collision directions, equality $D(f)=0$ requires $f^{\prime }{f}_{\ast }^{\prime }=f{f}_{\ast }$ in every collision, so $\log f$ is a collision invariant. Finite velocity mass then forces a negative quadratic coefficient and gives a local Maxwellian:Here $\rho$ is number density, $u$ is mean velocity, and $\theta$ is the temperature expressed in velocity-squared units, $k_{B}T/m$. Completing the square proves the formula; conservation of momentum and kinetic energy makes $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$, so $Q(M,M)=0$. Vacuum is another collision equilibrium.

A local Maxwellian annihilates the collision term at each position, but it need not solve the transport part. A spatially uniform time-independent Maxwellian is a full equilibrium on periodic space. On all of ${\mathbb{R}}_x^3$ it has infinite total mass unless it is vacuum. Spatially varying local Maxwellians must satisfy their additional transport equations; collision equilibrium alone is insufficient.

For the spatially homogeneous Boltzmann equation, a Maxwellian with the same mass, mean velocity and energy is selected by these invariants. The relative entropy $\int f\log (f/M)dv$ is nonnegative. Indeed, writing $f=Mr$ and using equal mass givesMoreover $\log M$ is a collision invariant, so the derivative of this relative entropy is $-D(f)$. Thus $M$ minimizes $H$ at fixed conserved moments. Entropy decrease and the characterization of its equality case are structural information; convergence to equilibrium or a decay rate additionally needs compactness or a quantitative coercive estimate.

Moment equations and fluid closure. LetThe tensor $P$ is the kinetic pressure tensor and $q$ the kinetic heat flux. Put $e=(2\rho {)}^{-1}\int |v-u{|}^{2}fdv$. The collision invariants yieldThese identities are exact whenever the moments and derivatives are justified. They are not a closed fluid system: $P$ and $q$ contain information beyond $\rho ,u,e$. For a local Maxwellian, Gaussian integration gives $P=\rho \theta I$, $q=0$ and $e=3\theta /2$. Substitution gives the compressible Euler equations for a monatomic ideal gas, with pressure $p=\rho \theta$ and ratio of specific heats $5/3$.

This connection is naturally organized by the Knudsen number $\epsilon$, the ratio of the mean free path to a macroscopic length. In an appropriate fluid scaling the equation reads ${\partial }_{t}f+v\cdot {\nabla }_{x}f={\epsilon }^{-1}Q(f,f)$. Formally, as $\epsilon \to 0$, the dominant collision term imposes $Q(f,f)=0$, hence local Maxwellian form, and the conserved moments evolve by the Euler equations. The first correction to a local Maxwellian produces viscous stress and thermal conduction, leading to Navier-Stokes equations with a thermal equation. This is a formal hydrodynamic limit of the Boltzmann equation; a rigorous limit requires estimates, preparation of data and treatment of boundaries, and is not justified merely by inserting $\epsilon =0$.

Linearization and its dissipative structure. Around a fixed positive Maxwellian $M$, write $f=M(1+h)$. The linearized collision operator on the relative perturbation isThis follows by expanding the two quadratic products to first order and using $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$. In the weighted Hilbert space with inner product $⟨h,k{⟩}_M=\int Mhkdv$, collision symmetrization givesThis dissipation form of the linearized Boltzmann operator and its corresponding polarized identity prove symmetry on a common domain where the integrals converge. Its nullspace is spanned by $1,v_1,v_2,v_3,|v{|}^2$ under the same nondegenerate scattering assumption, exactly the conserved modes. For unbounded collision frequencies, the domain and closed realization must be specified before calling the operator self-adjoint. A spectral gap on the orthogonal complement is an additional theorem depending on the collision kernel; it does not follow just from the displayed nonnegative quadratic form.

Analytic scope and irreversibility. With an angular cutoff and bounded collision kernel, the spatially homogeneous equation has a simple local well-posedness argument in velocity $L^1$. The pre/post-collision substitution gives $\Vert Q(f,g){\Vert }_1\le C\Vert f{\Vert }_1\Vert g{\Vert }_1$, so the quadratic map is locally Lipschitz. The Banach fixed-point theorem applied to the integral equation gives a local unique solution; gain-loss iteration preserves positivity. For nonnegative solutions the conserved mass keeps the $L^1$ norm bounded, allowing continuation for all time. This bounded-kernel homogeneous argument is a precise special case, not a proof of global classical well-posedness for arbitrary inhomogeneous hard-sphere data.

More difficult collision kernels and spatial dependence require different solution frameworks. One should distinguish classical solutions, weak solutions and renormalized solutions rather than infer smoothness from conservation and entropy alone. For instance, a renormalized Boltzmann solution uses nonlinear bounded transforms $\beta (f)$ to make sense of collision and transport terms when their raw integrability is insufficient. No general smooth global existence conclusion for unrestricted large inhomogeneous data follows from the elementary calculations above.

Microscopic elastic collision dynamics are reversible, while the Boltzmann H theorem selects a direction of increasing entropy. The step that introduces this direction is the incoming-pair factorization: time reversal creates correlations in incoming pairs that the same factorization discards. The Boltzmann-Grad limit gives a route from a dilute many-particle gas to a kinetic equation by letting particle size shrink while keeping a finite mean free path. Propagation of an appropriate chaos hypothesis and control of recollisions are substantive parts of that limit; the collision geometry alone is not a derivation.

Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit vector from A to B is $n=(-c,s)$. The impulse is normal, so tangential velocity is unchanged, while the normal components undergo a one-dimensional elastic collision. Therefore

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 $H$ be the common drop height in the small-gap, small-ball approximation and put $u=\sqrt{2gH}$. Use upward positive velocity. Just before ground impact both balls have velocity $-u$. The massive fixed ground reverses the lower ball's velocity elastically, so immediately before the ball-ball encounterOne-dimensional conservation of momentum and reversal of relative velocity in an elastic collision giveThe sequence is shown for the optimal ratio in b. At that ratio the lower ball stops and the upper ball departs upward, so no immediate third encounter needs to be included. Finite gaps introduce unequal contact times and gravitational corrections; no gap or ball radii are supplied to calculate those corrections.

Write $r=m_1/m_2>0$. The upper ball's fraction of the available kinetic energy after the separated elastic collisions isThus the maximal possible fraction is unity, attained atThe identity proves the global bound without selecting a spurious stationary point. At the optimum the upper ball receives all the initial translational kinetic energy. Its subsequent rise is $4H$ above the mutual-collision level. For arbitrary large mass ratios further ground encounters may occur; the unity bound is already attained in the stated two-impact sequence.

Let $\mathbf{a},\mathbf{b}$ be the outgoing velocities. In the nonrelativistic elastic collision, equal masses giveby conservation of momentum and conservation of energy. Squaring the first relation and comparing with the second yieldsIf both particles move, their paths are perpendicular. If either velocity vanishes, its direction is undefined and the other particle carries all the incident momentum and kinetic energy. This conclusion depends on the stated $v\ll c$ approximation.