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A deforming free swimmer must satisfy total linear and angular momentum balances. The unknown lateral translation and yaw are its recoil correction. If the reference translation is the true centre of mass, the prescribed deformation must first have its mass-weighted mean removed. Otherwise the reference translation differs from actual centre of mass displacement. The body mass and its hydrodynamic added mass contribute with the same sign to its effective inertial coefficient.

A recoil-free nuclear transition emits or absorbs a gamma ray while leaving the internal vibrational state of the surrounding solid unchanged. It belongs to the elastic part of the process: no phonon is created or destroyed. The photon still carries momentum $p=E_{\gamma }/c$, where $c$ is the speed of light. The emitting solid takes momentum $-\mathbf{p}$; the absorbing solid takes $+\mathbf{p}$. Conservation of momentum holds for each event.

Why does an isolated atomic nucleus pose a problem? Start with an excited state at rest and let $E_0$ be its transition energy, with nuclear mass $M$ and $E_0\ll M{c}^2$. Conservation of momentum gives a recoil momentum of magnitude $p$; its nonrelativistic kinetic energy is thereforeBy conservation of energy, emission spends part of $E_0$ on nuclear recoil, whereas absorption must supply both $E_0$ and the final kinetic energy. To leading order,Thus the relevant comparison is $2E_R$ against the overlapping spectral line widths, rather than $E_R/E_0$ alone. If $2E_R$ greatly exceeds the widths, the isolated-nucleus lines have negligible overlap. Finite Lorentzian line shape tails mean that this is suppression, rather than an absolute prohibition on absorption.

For a bound atomic nucleus, the recoil-free channel transfers momentum to the center of mass of the emitting or absorbing solid, with negligible kinetic energy $p^2/(2M_{\mathrm{solid}})$. Other channels exchange phonons and carry substantial energy into or out of the internal vibrations. A large $M_{\mathrm{solid}}$ makes the elastic channel's energy cost small; the vibrational quantum state determines its probability. Merely replacing $M$ by a large mass in the free-particle formula does not calculate that probability.

Here is a deliberately stipulated teaching system, independent of historical isotope data: $E_0=20\mathrm{keV}$, $M{c}^2=10^{11}\mathrm{eV}$, and a single-line natural linewidth ${\Gamma }_E=10^{-7}\mathrm{eV}$ measured as an energy full width at half maximum. These give $E_R=0.002\mathrm{eV}$ and $2E_R/{\Gamma }_E=40,000$. A solid of $10^{12}$ such nuclear masses has elastic recoil energy $2\times 10^{-15}\mathrm{eV}$, far below ${\Gamma }_E$. In this article ${\Gamma }_E$ is an energy width; the corresponding lifetime is $\tau =\hslash /{\Gamma }_E$, with $\hslash$ the Planck constant divided by $2\pi$.

A normalized hard rectangular spatial wavefunction has a sinc-squared momentum density. Its central lobe has finite first-zero width $h/d$, but its second momentum moment diverges. Thus that diffraction width is not a finite rms uncertainty. Smooth-edge states can have finite moments and satisfy the usual uncertainty inequality without this idealization.

A freely moving massive test particle in an FLRW metric has physical momentum $p=m\gamma v$ relative to a comoving observer, where $v$ is its peculiar velocity. For a radial geodesic, the geodesic Lagrangian gives $C=a^{2}d\chi /d\tau$ constant, and the local radial momentum is $p=mad\chi /d\tau =mC/a$. The proper-time normalization gives $dt/d\tau =\sqrt{1+C^2/(a^2{c}^2)}$, soConsequently nonrelativistic peculiar velocity scales as $a^{-1}$, while the exact relativistic law keeps $|v|<c$. The momentum magnitude of a free photon also scales as $a^{-1}$, giving cosmological redshift through $E=cp$. These statements concern force-free propagation; a bound system need not follow the cosmological expansion.

A reduced Brillouin zone is the reciprocal primitive cell for an enlarged real-space unit cell. Doubling a one-dimensional unit cell halves the momentum interval but adds a second internal sublattice degree of freedom. Band folding therefore preserves the number of normal modes.

For a two-particle collision conserving total momentum, the outgoing speeds in the centre-of-momentum frame are multiplied by the same $q$. The relative kinetic energy is multiplied by $q^2$. For a smooth oblique collision with normal coefficient of restitution $e_n$ and incident relative velocity inclined by $\chi$ to the contact normal, unchanged tangential relative motion gives $q^2=e_n^2{\cos }^2\chi +{\sin }^2\chi$. Thus $q=e_n$ is not a general oblique-impact identity: $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}$. Sticking requires all relative motion to vanish, not merely its normal component.

At larger kinetic energy, special relativity changes the relation between momentum and velocity. With $m$ the invariant mass, the Lorentz factor is $\gamma =1+K/(m{c}^2)$, and a relativistic cyclotron orbit in a fixed uniform magnetic field satisfiesThe cyclotron frequency decreases as the kinetic energy increases. If the drive stays at the classical value ${\omega }_0=|q|B/m$, the half-turn time is $\pi \gamma /{\omega }_0$. During it the drive advances through $\pi \gamma$ rather than $\pi$ radians. Its excess phase per half-turn is thereforeEven a small fractional frequency mismatch accumulates across many turns. A particle eventually reaches a weak or decelerating gap phase, so the classical fixed-field, fixed-frequency design has a practical kinetic energy limit. The limit depends on the kick size and acceptable accumulated phase error, rather than one universal cutoff energy. Changing drive frequency or shaping the magnetic field can restore synchronism; the argument does not forbid relativistic circular accelerators.

For the illustrative proton at $1.00\mathrm{MeV}$, $\gamma -1\simeq 0.00106579$: the frequency is lower by about $0.1065\%$, and the next half-turn adds about $0.19184^{\circ }$ of drive phase excess. To see cumulative sensitivity, prescribe an ideal ramp $K_j=jϵ$ with $ϵ=10\mathrm{keV}$ and sum the errors of the first $N$ coasts after kicks:This gives ${\Phi }_{100}\simeq 9.688^{\circ }$ after the coast following kick $100$; the arrival at kick $100$ instead has ${\Phi }_{99}\simeq 9.496^{\circ }$. The prescribed ramp diagnoses detuning but does not feed the resulting phase error back into the kicks. A self-consistent thin-gap model uses the cyclotron gap energy gain with $\cos {\varphi }_n$, then advances the phase with the new kinetic energy during the following coast.

In units with $c=1$, a particle of rest mass $M$ and total energy $E$ strikes a stationary target of rest mass $m$. For the nontrivial one-dimensional elastic collision, reverse the incident momentum in the centre-of-mass frame and transform back to obtain the displayed outgoing projectile energy. The invariant squared total four-momentum is $s=M^2+m^2+2mE$. If $M\gg m$ and $Em\gg M^2$, then $E^{\prime }\sim M^2/(2m)$. If $m\gg M$, retaining arbitrary $E/m$ gives $E^{\prime }\simeq Em/(m+2E)$; only the further fixed-$E$, infinite-target-mass limit gives $E^{\prime }\to E$.

For a one-dimensional free particle with $m>0$ and $p>0$, take the positive total energy from the relativistic energy-momentum relation:A fixed plane wave quantum phase obeys $kx-\omega t=\text{constant}$. Its phase velocity is thereforeFor a narrow wave packet centered at $p_0$, expand its dispersion relation as $E(p)\simeq E(p_0)+E^{\prime }(p_0)(p-p_0)$. In the Fourier inversion integral building the wave packet, the envelope then depends on $x-E^{\prime }(p_0)t$. Thus its group velocity isThe superluminal phase velocity follows a repeating phase marker; it does not describe a localized signal or superluminal transport of energy. Higher derivatives of the dispersion relation produce wave packet spreading, so the translated-envelope argument is a narrow-band approximation. For $v=0.6c$, the Lorentz factor is $1.25$, $p=0.75mc$ and $E=1.25m{c}^2$: the group velocity is $0.6c$ while the phase velocity is $5c/3$.

The total-energy convention matters. Define $\psi =e^{im{c}^{2}t/\hslash }\Psi$ to remove the common rest energy from a wavefunction $\Psi$ with total-energy time dependence. ThenThis is exactly the free-particle momentum convention used by the Schrödinger equation here. Removing a common time-dependent global phase changes the reported phase velocity, while leaving the de Broglie wavelength, relative spatial quantum phases and group velocity unchanged. Adding $m{c}^2$ again to a total energy would count the rest energy twice.

For a massive particle, its spatial momentum in an inertial frame is $\gamma m\mathbf{v}$, with m its invariant rest mass and $\gamma$ the Lorentz factor. Together with its relativistic energy it forms the four-momentum. In a magnetic quadrupole's paraxial approximation, replacing $mv$ by the longitudinal momentum extends the transverse focusing scale to relativistic beams.

Renormalization of spontaneously broken gauge theories means that the Higgs mechanism can generate gauge boson masses while a fixed finite set of renormalized couplings, masses and field normalizations still suffices to absorb ultraviolet divergences. The symmetry constrains the counterterms even when the chosen vacuum expectation value hides it. This makes radiative corrections to the electroweak interaction calculable after inputs have been fixed by renormalization conditions.

A useful distinction is between choosing variables that display the physical particles and choosing variables that display good power counting. Unitary gauge removes the would-be Goldstone bosons from the explicit fields. R-xi gauge keeps them and the required Faddeev-Popov ghosts in intermediate Feynman diagrams, exposing simpler large-momentum behavior. The physical scattering amplitude must agree after all contributions and counterterms have been included consistently.

In Renormalizable Lagrangians for massive Yang-Mills fields (1971), Gerard 't Hooft constructed spontaneously broken gauge theories with massive gauge bosons. His worked SU(2) gauge theory with an adjoint Higgs field leaves a massless photon; it is a demonstration model, not the full chiral Standard Model. The gauge fixing, Faddeev-Popov ghosts and Ward identities connect renormalizability with physical unitarity.^[ref]

Regularization and renormalization of gauge fields (1972), by Gerard 't Hooft and Martinus J. G. Veltman, developed dimensional regularization with recursive subtraction of ultraviolet subdivergences. It continued Feynman integrals in the dimension, keeping the relevant Ward identities in non-anomalous cases; its final section examined quantum anomalies. The procedure does not make an inconsistent chiral gauge theory consistent: gauge anomaly cancellation remains essential.^[ref]

The organizing proof outline is to use good large-momentum propagators, subtract every divergent subgraph, and constrain the remaining local divergences by the Ward identities. A finite family of allowed counterterms must close under this operation at every loop order. The same identities must remove unphysical intermediate contributions from physical amplitudes. Power counting alone establishes neither closure nor these cancellations. Modern BRST symmetry packages these restrictions; the scalar subtraction below illustrates only locality.

For a condensate with uniform bulk complex field one, its convergent localized momentum isThe subtraction removes the uniform-background phase-gradient contribution. If $\psi -1=O(r^{-1})$ and $\nabla \psi =O(r^{-2})$, its integrand is $O(r^{-3})$, integrable in two dimensions. Under decaying variations its functional derivative is $\delta p_j/\delta {\psi }^{\ast }=-i{\partial }_j\psi$; the subtraction is a boundary correction and does not change the bulk equation.

The residual resistivity is the low-temperature electrical resistivity remaining in a normal metal when thermally excited scattering becomes negligible. Static disorder, impurities and boundaries can relax carrier momentum without requiring a thermal population of phonons. Its value depends on sample purity and structure; a superconducting transition is a different phenomenon.

A complete orthonormal basis in a Hilbert space resolves the identity as ${\sum }_j|j⟩⟨j|=I$. In infinite dimension the partial sums converge strongly: their action on each fixed vector converges in norm. Generalized position and momentum eigenstates instead give distributional integral resolutions, such as $\int dx|x⟩⟨x|=I$, with their normalization interpreted through the Dirac delta function.

The rest frame of a massive particle is an inertial frame in which its spatial momentum is zero. Its four-momentum is then $(mc,0)$ and its energy is $m{c}^2$. Such a frame exists for a future-directed timelike four-momentum: the Lorentz transformation with boost velocity $\mathbf{v}=c^2\mathbf{p}/E$ makes the spatial momentum zero, and the mass shell relation implies $|\mathbf{v}|<c$. A massless particle has no rest frame, because $E=c|\mathbf{p}|$ would require a boost with speed $c$.

For a photon and a massive particle with zero total momentum, their momenta are opposite and equal. Conservation of total energy fixes the same momentum magnitude before and after elastic scattering, independently of direction. A right-angle photon turn therefore makes the recoiling particle move perpendicular to the original photon direction in that frame. Transform both velocity components, or the complete four-momentum, to obtain the laboratory recoil; a transverse component is reduced by the boost Lorentz factor when the particle has zero longitudinal velocity in the zero-momentum frame.

The Robertson-Schrödinger uncertainty relation keeps the real part discarded in the proof of the Robertson uncertainty principle. Write $A^{\prime }=A-⟨A⟩I$ and $B^{\prime }=B-⟨B⟩I$. Their quantum covariance is $C_{AB}=\mathrm{Re}⟨A^{\prime }\psi |B^{\prime }\psi ⟩$. On the same finite-moment and product operator domains used above, the Cauchy-Schwarz inequality givesFor position and momentum, form the real, symmetrized covariance matrixA larger coordinatewise product can therefore encode a tilted covariance ellipse, rather than a larger determinant. This explains why the equality case of the Heisenberg uncertainty relation is more restrictive than equality in the Robertson-Schrödinger uncertainty relation.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

A roton is an excitation near a local minimum of a superfluid's dispersion relation at nonzero momentum $p_0$. The displayed local approximation defines a positive energy gap $\Delta$ and positive curvature effective mass $m_r$; it is not a free helium atom or a macroscopic quantum vortex.^[ref] In the stipulated plot, these correspond to $p_0=2p_{\ast }$, $\Delta =0.6E_{\ast }$ and $m_r=p_{\ast }^2/(4E_{\ast })$. The plot illustrates how a low energy at appreciable momentum can lower the Landau criterion below the speed of sound, without assigning any measured parameter to real helium.

Landau's 1947 revision joined the low-momentum linear part to a spectrum with a finite-momentum minimum, retaining the macroscopic superfluid theory while changing its excitation model.^[ref]

The Saffman integral is $L=\int ⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩d^{3}r$. With integrable trace correlation it equals the large-volume limit of $⟨|{\int }_V\mathbf{u}dV{|}^2⟩/|V|$. It is nonnegative and measures momentum-fluctuation density. Under appropriate unforced far-field flux conditions it is invariant.