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Products of noncommuting linear operators depend on their order. For position and momentum, $xp-px=i\hslash$; substituting classical commuting expressions into a product can therefore lose an additive quantum term. For example, expansion of the oscillator product $a^{†}a$ includes a commutator contribution $-1/2$, leading to the correct zero-point energy in $H=\hslash \omega (a^{†}a+1/2)$.

A quantum time-reversal operator is an antiunitary operator reversing spatial momentum and angular momentum. For a time-reversal-invariant Hamiltonian operator, its conjugation of $i$ makes it intertwine forward and backward unitary time evolution without reversing the energy spectrum. A scalar momentum eigenstate may transform as $\hat{T}|\mathbf{p}⟩=e^{i\chi (\mathbf{p})}|-\mathbf{p}⟩$. With unit-modulus phases and a complete normalized basis, expansion of arbitrary states proves antiunitarity.

A quark-line diagram can show how existing valence quarks are regrouped into outgoing hadrons while gluons exchange momentum and colour. Alternatively, a quark-antiquark pair can annihilate and create a different flavour pair with zero net flavour. Such drawings enforce global charge, baryon number and flavour constraints; they are not controlled perturbative calculations of an exclusive low-energy hadronic amplitude because binding is nonperturbative.

The quasiparticle residue is the coherent pole weight of the single-particle Green's function. Its coherent spectral term $Z\delta (E-{\xi }_p)$ contributes $Z\Theta (-{\xi }_p)$ to the zero-temperature momentum occupation. A smooth remainder leaves a jump $Z$ at the Fermi surface; Landau quasiparticle occupations have a unit step.^[ref]

For the local illustrative distribution in the figure, put $x=(p-p_F)/p_F$ and$\Theta$ is the Heaviside step function. Its one-sided limits are $0.825$ and $0.175$, giving $Z=0.65$. The smooth term is stipulated; this local curve does not specify a normalized distribution over all momenta. In particular, $Z$ is neither the fraction of atoms above the Fermi surface nor $m/m^{\ast }$.

For $H=c(p_x{\sigma }_x+p_y{\sigma }_y)$ with $c>0$ and physical spin matrices, $⟨\mathbf{S}{⟩}_s=s(\hslash /2)\hat{\mathbf{p}}$. The positive-energy branch is parallel and the negative-energy branch antiparallel to momentum. Both are collinear; no preferred spin direction is defined at zero momentum.

For three-dimensional radial wavefunctions with measure $r^{2}dr$, this formal symmetric momentum expression obeys $[f(r),p_r]=i\hslash f^{\prime }(r)$ on appropriate test functions and has square $-{\hslash }^2({\partial }_r^2+2{\partial }_r/r)$. It is useful in separating kinetic energy as $p^2=p_r^2+L^2/r^2$. Formal symmetry and these identities do not establish a self-adjoint momentum observable on the half-line; domains and origin conditions must be specified when taking adjoints or defining products.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

Radiation reaction is the influence on a source's motion of its own electromagnetic field, including the balance of radiated energy and momentum with changes of the near field. For sufficiently weak radiation loss, an averaged power balance can give the leading damping law without solving the full self-field dynamics. Replacing a constant-frequency radiation formula by its instantaneous slowly varying value requires an additional time-scale assumption; it is not an exact treatment of arbitrarily rapid motion changes.

Depositing supernova energy into dense unresolved gas can radiate it away before a numerically resolved expansion develops. Equal injected energy therefore need not imply equal outflow momentum or heating at different resolution. A feedback model should represent the coupling and cooling phase it intends to approximate, and comparisons should include gas recycling and the resulting stellar and circumgalactic properties.

A shock wave with significant radiation losses in its downstream evolution. Often a narrow dissipative jump initially heats the gas according to adiabatic Rankine-Hugoniot conditions, followed by a much longer radiative cooling layer. The gas's specific entropy increases at the jump and may then fall as radiation carries entropy away. If both distant states have equal temperature, the combined mass and momentum balances give an isothermal shock; the gas's energy flux is not conserved across the whole radiating region.

The Rankine-Hugoniot condition expresses conservation laws across a moving discontinuity of a weak solution. Across a discontinuity $x=\sigma t$ in $u_t+f(u{)}_x=0$, with left and right states $u_{-}$ and $u_{+}$, conservation requiresWriting $[u]=u_{+}-u_{-}$ and $[f(u)]=f(u_{+})-f(u_{-})$ gives $\sigma [u]=[f(u)]$. The same relation holds componentwise for a system of conservation laws; in compressible flow, conservation of mass, momentum and energy supplies the jump conditions for a shock wave. Integrating the weak solution identity over a thin space-time region crossing the discontinuity yields this balance between the swept-up conserved quantity and the difference of fluxes.

Rapidity is a dimensionless parameter for a signed collinear velocity $u$ with $|u|<c$, defined using the inverse hyperbolic functions byHere $c$ is the speed of light, and $\tanh$ is the hyperbolic tangent. Zero velocity has zero rapidity; negative velocity has negative rapidity. Every finite real rapidity gives $|u|<c$, while $u\to \pm c$ requires $\theta \to \pm \infty$. Thus a light ray has no finite rapidity in this parametrization. The header's formula uses natural units $c=1$: its $v$ is the numerical velocity $u/c$.

For the inertial frames in the velocity-addition formula, $S^{\prime }$ moves at signed velocity $V$ along the $x$-axis of $S$. Set $\varphi =\mathrm{artanh}(V/c)$. Re-expressing the same particle in the moving inertial frame givesThe subtraction is a passive change of reference frame; the inverse adds the particle's rapidity in $S^{\prime }$ to the frame's rapidity relative to $S$. For example, $V=0.6c$ and $u^{\prime }=0.6c$ give $\varphi ={\theta }^{\prime }=\ln 2$, so $\theta =\ln 4$ and $u=c\tanh (\ln 4)=15c/17$, exactly as in the velocity-addition formula.

For a massive particle of rest mass $m>0$ moving in one spatial dimension, its energy and signed momentum areThe hyperbolic cosine is the particle's Lorentz factor, and the hyperbolic sine is $\gamma u/c$. The hyperbolic Pythagorean identity immediately gives the relativistic energy-momentum relation $E^2-c^2{p}^2=m^2{c}^4$. In the older $c=1$ notation, writing the rest mass as $M$ recovers $v=\tanh \theta$, $E=M\cosh \theta$, and $p=M\sinh \theta$. These expressions use natural units in which the factors of $c$ are numerically one; the preceding formulas restore those factors explicitly. ^[ref].

The useful simplification follows from the hyperbolic-function identity $\tanh (a+b)=(\tanh a+\tanh b)/(1+\tanh a\tanh b)$: the nonlinear velocity-addition formula becomes ordinary addition of rapidities. See iterated collinear boosts with equal rapidity for a plotted example. A common collinear Lorentz boost shifts every rapidity by the same constant, so a rapidity difference is invariant. This makes rapidity useful for relativistic factorized scattering in one spatial dimension, where two-body S-matrices depend on the difference of the incoming rapidities, with particle species fixed.

The Rashba effect, named for Emmanuel Rashba, is a momentum-dependent spin-orbit interaction associated with broken spatial inversion symmetry. In an isotropic two-dimensional model, choose $\lambda >0$ and write the Hamiltonian $H={\hslash }^2{k}^{2}I/(2m)+\lambda (\hat{\mathbf{z}}\times \mathbf{k})\cdot \mathbf{\sigma }$, where $I$ is the identity matrix and $\mathbf{\sigma }$ is the vector of Pauli matrices. The eigenvalues of the spin term give $E_{\pm }(k)={\hslash }^2{k}^2/(2m)\pm \lambda k$. Define $k_R=m\lambda /{\hslash }^2$ and $E_R={\hslash }^2{k}_R^2/(2m)$. The lower branch is $E_{-}(k)={\hslash }^2(k-k_R{)}^2/(2m)-E_R$ and has a ring of minima at $k=k_R$. A negative $\lambda$ reverses the spin directions; ordering the two energies still uses $|\lambda |$. The alternative parameter $q_R=2k_R$ must not be mistaken for the minimum radius.

For $H_R=\lambda (k_x{\sigma }_y-k_y{\sigma }_x)$ with $\lambda >0$, the lower- and upper-band spinors of unit norm can be chosen asTheir spin expectation values are $⟨\mathbf{S}{⟩}_{\pm }=\pm (\hslash /2)(-k_y,k_x,0)/k$. They are tangential to circles of constant momentum magnitude, opposite between branches, and reversed at opposite momenta as required by time-reversal symmetry. When $E_F>0$, the outer contour belongs to the lower branch and the inner one to the upper branch. For $E_F<0$ the two Fermi contours both belong to the lower branch and have the same tangential spin orientation at a given angle in polar coordinates. At $k=0$ the Hamiltonian has quantum degeneracy in spin, and no unique spin texture direction is selected.

The geometrical optics account of an optical trap adds the momentum flux changes of refracted and reflected light rays. It applies when the bead is large compared with the surrounding medium's wavelength. When the bead is much smaller, an induced electric dipole gives a different approximation. For a homogeneous dielectric sphere of radius $a$, define $\mathbf{p}(t)=\mathrm{Re}[\alpha {\mathbf{E}}_0{e}^{-i\omega t}]$, with ${\mathbf{E}}_0$ the peak complex electric field and $\alpha$ its SI polarizability. In a transparent, nonmagnetic, nondispersive host,where $n_p,n_m$ are particle and host refractive indices, $I$ is cycle-averaged light intensity, ${ϵ}_0$ is the vacuum permittivity and $c$ is the speed of light. The quasistatic electric dipole approximation requires $2\pi n_{m}a/{\lambda }_0\ll 1$, with ${\lambda }_0$ the vacuum wavelength, and a field nearly uniform across the bead.

For real quasistatic $\alpha$, building an induced electric dipole gives instantaneous potential energy $-\alpha E^2/2$. Averaging over one optical cycle givesThus $n_p>n_m$ gives positive polarizability and attraction toward larger light intensity; $n_p<n_m$ reverses this contribution. The factor $1/4$ uses a peak field: substituting a root mean square field without changing that factor would halve the force incorrectly. Dimensionally $\alpha |E_0{|}^2$ is energy, so its gradient is force.

This leading conservative term does not remove the optical scattering force. Even a nonabsorbing induced electric dipole reradiates light: at fixed wavelength and index contrast, its Rayleigh scattering cross-section scales as $a^6$, while $\alpha$ scales as $a^3$. A smaller bead can therefore improve the ratio of gradient to scattering forces while weakening the well relative to $k_{B}T$, where $k_B$ is the Boltzmann constant and $T$ the temperature. Optical force balance and resistance to thermal escape are separate questions. Intermediate bead sizes require a wave-scattering calculation; neither a ray calculation nor this small-dipole expression is uniformly valid there.

For tightly wound perturbations in a locally uniform, razor-thin gas disk, continuity equation and rotating-fluid momentum equations couple the surface-density perturbation to radial and azimuthal velocities. The sheet potential is $\delta \Phi =-2\pi G\delta \Sigma /|k|$. Eliminating these perturbations gives the displayed dispersion relation. radial epicyclic frequency supplies the rotation term, the sound-speed term resists short waves, and self-gravity supplies the negative term. For an axisymmetric mode $n=0$, use ${\omega }^2$ directly; the pattern speed $\omega /n$ is undefined.

The kinematic minimum energy needed to produce an allowed final state. In the centre-of-momentum frame, the threshold for freely separated massive final particles is the sum of their rest energies, with no relative kinetic energy. Conservation of total energy and momentum then determines the corresponding laboratory threshold. Charges and other selection rules must also permit the reaction; the mass bound alone does not guarantee an observable transition.

Recoil is the motion acquired by a body when it exchanges momentum with an emitted, absorbed or scattered particle. For isolated emission from a body initially at rest, momentum conservation makes the remaining body acquire the opposite momentum. If its final mass is $M$ and the recoil is nonrelativistic, its kinetic energy is $E_R\simeq p^2/(2M)$. A photon of energy $E_{\gamma }$ has momentum magnitude $E_{\gamma }/c$, so $E_R\simeq E_{\gamma }^2/(2M{c}^2)$. The transition energy is shared between the photon and the recoiling body; nuclear recoil is an important example.