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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.

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 $2\pi$.

For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.

The phase of stellar evolution in which hydrogen-shell nuclear fusion surrounds an inert helium core and the envelope expands and cools. Low-mass stars climb toward helium ignition with growing core mass and luminosity; their cores become supported by electron degeneracy pressure. A red giant is a broader observational category and need not be on this particular evolutionary branch.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

For polymer mass concentration $c$, the leading ideal chain pressure is $RTc/M_n$. Plotting $\Pi /c$ against $c$ and extrapolating removes the leading virial correction. The intercept has units of energy per mass; without the measurement temperature it determines $M_n/T$, not an absolute molar mass.

The combination of lensing shear and lensing convergence governing the local shape distortion of background sources. Galaxy ellipticities estimate it statistically under a specified ellipticity convention and intrinsic-orientation model. It is invariant under the mass-sheet degeneracy, so shape information alone does not generally fix absolute projected mass normalization.

For a semi-infinite ideal string on $x\le 0$ with tension $T$ and linear mass density $\mu$, let a free transverse endpoint mass be $m$. Use a time-harmonic wave $y=\mathrm{Re}\{A(e^{ikx}+R{e}^{-ikx})e^{-i\omega t}\}$, where $k=\omega \sqrt{\mu /T}>0$. The endpoint boundary condition from Newton's second law is $m{y}_{tt}=-T{y}_x$. It gives the title displacement reflection coefficient, with characteristic impedance of a stretched string $Z=\sqrt{T\mu }$. Since numerator and denominator have equal moduli, all mean incident power is reflected. The endpoint kinetic energy varies within a cycle, but its mean rate of change is zero. A free mass changes the reflected phase, and does not by itself provide dissipation. The limits $m\to 0$ and $m\to \infty$ give $R=+1$ and $R=-1$ respectively at fixed positive angular frequency.

In Feynman gauge, use the auxiliary photon propagator $-i{g}_{\mu \nu }/(k^2-{\lambda }^2+i0)$ and $r=\lambda /m>0$. The projected one-loop Feynman parameter integral isHere $x$ is the source's photon parameter $z$.^[ref]

To check the reduction, use $N_2=4z(1-z)m^2$ and ${\Delta }^{\prime }=(1-z{)}^2{m}^2+z{\lambda }^2$. The three-parameter expression has prefactor $e^2/(16{\pi }^2)$ multiplying $N_2/{\Delta }^{\prime }$. At fixed $z$, ${\int }_0^{1-z}du=1-z$, giving $e^2/(4{\pi }^2){\int }_0^{1}z(1-z{)}^2/[(1-z{)}^2+z{r}^2]dz$. Substituting $e^2=4\pi \alpha$ gives exactly the displayed integral. This starts from the projected QED numerator; it is not a derivation of the whole QED calculation from an arbitrarily chosen integral.

The regulator can then be removed by a short independent argument. Write its integrand as $f_r(x)$. For $0\le x<1$, $f_r(x)\to 2x$, andThe dominated convergence theorem therefore gives $I(r)\to 1$ and $F_2^{(1)}(0)=\alpha /(2\pi )$. At $x=1$ the regulated integrand is zero, whereas the limiting function can be assigned value $2$; one point has zero measure and changes no integral. For every $r>0$, $0<I(r)<1$, and decreasing $r$ increases the integral. The visible suppression near $x=1$ narrows as the regulator shrinks; it is not a uniform pointwise approximation at that endpoint.

This auxiliary mass regulates an infrared divergence, not an ultraviolet divergence; the finite-mass prescription alone is not fully gauge-invariant.^[ref]

The ultraviolet problem belongs to the full QED amplitude. Electron self-energy, wave-function renormalization and the charge vertex require consistent counterterms and renormalization conditions. The Ward identity for QED renormalization constants ties $Z_1$ to $Z_2$, rather than allowing the vertex and external electron normalization to be fitted independently. Minimal renormalizable QED has no independent Pauli term to adjust this finite one-loop $F_2(0)$: its dimension-five coefficient is predicted from the mass and electric charge inputs. Adding a free Pauli term would define a different theory with an extra parameter. Finiteness of the reduced integral alone proves neither all-order renormalizability nor finiteness of every QED observable.

An astronomer studying the Milky Way centre.^[ref]

For a chemical element in a specified population, its relative atomic mass or atomic weight is the dimensionless ratio $A_{\mathrm{r}}=\bar{m}/\mathrm{u}$, with $\bar{m}$ the mean neutral atomic mass weighted by isotopic abundance. For the illustrative neon population, $A_{\mathrm{r}}\approx 20.1800$, whereas $\bar{m}\approx 20.1800\mathrm{u}$ has mass units.^[ref]

Relative centrifugal force is the dimensionless ratio of the local centrifugal acceleration magnitude $a$ to the conventional gravitational reference $g_0=9.80665\mathrm{m}{\mathrm{s}}^{-2}$. Despite its name, it is an acceleration ratio: the actual force on a specified mass $m$ is $ma$. With angular speed $\omega$ and rotor radius $r$, $a={\omega }^{2}r$. If $n$ is the numerical rotational speed in revolutions per minute and $r_{\mathrm{cm}}$ the numerical rotor radius in centimetres, unit conversion givesThus rpm alone does not specify the centrifugal acceleration.^[ref]

For a stipulated $3000\mathrm{rpm}$, rotor radii of $5$ and $10\mathrm{cm}$ give approximately $503$ and $1006$ times $g_0$. Doubling the rotor radius doubles relative centrifugal force; doubling rpm quadruples it. Even equal relative centrifugal forces do not guarantee equal separation: travel distance, elapsed time, viscosity, sample density and particle shape still matter. These are comparisons of a mathematical model, not instructions for running an instrument.

For sufficiently regular positive probability densities, this compares their spatial logarithmic derivatives. It is the dissipation functional for relative entropy under suitable diffusion equations. For standard Gaussian $g=\gamma$, it becomes $\int |\nabla \log f+v{|}^{2}f$, and in one dimension equals $\int |f^{\prime }{|}^2/f+\int v^{2}f-2$ at unit mass.

The relative molecular mass is the dimensionless ratio of a specified mean molecule mass to the unified atomic mass unit, written $m_{\mathrm{u}}$ when treated as a physical quantity:Thus a relative molecular mass of approximately $18.015$ and a molar mass of approximately $18.015\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ describe related quantities with different units. The conversion uses the molar mass constant, rather than identifying a dimensionless number with a mass per mole.

At relative rapidity $iu$, the analytically continued sum of the constituent four-momentum vectors is on the bound-state mass shell with the displayed mass. For equal constituents, rapidities $\chi \pm iu/2$ sum to $2m_a\cos (u/2)(\cosh \chi ,\sinh \chi )$. A pole at $u=0$ is at threshold rather than a strictly bound state.

For metric signature $(-,+,+,+)$, a particle of mass $m$ and charge $q$ coupled to a four-potential has the reparametrization-invariant action

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.

For the real scalar field with metric $(+,-)$ and potential $U={\kappa }^2({\varphi }^2-1{)}^2/2$, the static phi-four kink is ${\varphi }_0(x)=\tanh (\kappa x)$ and satisfies ${\varphi }_0^{\prime 2}=2U({\varphi }_0)$. Its mass is $M=\int {\varphi }_0^{\prime 2}dx=4\kappa /3$. A Lorentz boost gives $\varphi (t,x)={\varphi }_0(\gamma (x-vt))$. The stress-energy tensor gives $P=-\int {\varphi }_t{\varphi }_{x}dx=\gamma v\int {\varphi }_0^{\prime 2}dx=\gamma Mv$. Similarly, $E=\frac{1}{2}[\gamma (1+v^2)+{\gamma }^{-1}]M=\gamma M$, proving the energy–momentum relation $E^2-P^2=M^2$.