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A single recoil endpoint leaves a tradeoff between incident speed and projectile mass. Two targets can remove the unknown incident speed. Assume both runs sample the same maximum projectile speed $v$ and that single elastic collisions reach the endpoint. For initially stationary target masses $M_1<M_2$, let their maximum recoil speeds be $u_1,u_2$. The maximum elastic recoil energy fraction calculation givesIf energies rather than speeds are inferred, use $R=\sqrt{T_{1,\max }{M}_2/(T_{2,\max }{M}_1)}$. The energy ratio itself is not the speed ratio.

For the stipulated $4\mathrm{MeV}$ example, hydrogen and nitrogen giveThe inferred mass then determines $v=u_1(m+M_1)/(2m)$. Changing the assumed mass to improve one recoil endpoint changes the prediction for the other: this is the constraining power of the comparison.

James Chadwick's June paper instead used approximate endpoints $u_p=3.3\times 10^{7}m/s$ and $u_N=4.7\times 10^{6}m/s$, obtaining $m/m_p\simeq 1.15$; it allowed roughly ten percent uncertainty in $u_N$. This was a rough mass comparison.^[ref]

The inverse calculation has useful built-in checks. A positive finite mass requires $1<R<M_2/M_1$. As $R$ approaches $1$, distinguishing large projectile masses becomes difficult. Differentiation givesUncertain recoil endpoints therefore produce an uncertain inferred mass; the collision model cannot turn imprecise range measurements into exact spectroscopy. If different target runs select different incident energy groups, the common-$v$ cancellation also fails.

For comparison, Compton scattering of a photon of energy $E_{\gamma }$ from a stationary free target of mass $M$ has the exact relativistic endpointThis follows by combining momentum conservation with the target's relativistic energy, maximizing transfer when the photon reverses direction. Unlike the massive-projectile calculation, this equation does not approximate the incident photon as a nonrelativistic particle.

As a separate, explicitly hypothetical test, take $m_p{c}^2=938\mathrm{MeV}$ and target rest energies $938\mathrm{MeV}$ and $14\times 938\mathrm{MeV}$. To reproduce the stipulated hydrogen recoil endpoint $T=4\mathrm{MeV}$, the photon would need $E_{\gamma }\simeq 45.36\mathrm{MeV}$. The same photon would give the heavier target at most $0.311\mathrm{MeV}$, whereas the stipulated massive projectile gave $0.996\mathrm{MeV}$. This is the structure of a two-target discrimination, not a transcription of a 1932 measurement. Actual radiation can contain more than one energy group or both photons and neutrons; a useful model must account for the observed endpoints and their origin together.

Orbital migration of a low-mass embedded planet driven by its interaction with disk gas, without a deep gap. Spiral wave torques and corotation-region torques exchange orbital angular momentum. The sign depends on the disk gradients and thermodynamics; this regime is not restricted to inward migration.

For a circular fixed-mass planet, $L=M_p{a}^2\Omega$ and $\dot{a}=2\Gamma /(M_{p}a\Omega )$. With $\Gamma =C(q/h{)}^2\Sigma a^4{\Omega }^2$, the radial migration timescale is $a/|\dot{a}|=(M_{\star }/M_p)[M_{\star }/(\Sigma a^2)]h^2/(2|C|\Omega )$. The factor of two distinguishes radial evolution from $L/|\Gamma |$. The dimensionless coefficient and its sign depend on the disk model.

A neutrino with energy much larger than its mass. At fixed momentum, expanding the relativistic energy-momentum relation gives $E_i\simeq p+m_i^2/(2p)$, so differences of propagation phases are governed by $\Delta m^2/(2E)$.

For one unit-charge Dirac field in $d=4-2ϵ$, define the insertion as $i{\Pi }_{\mu \nu }=i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })\Pi$. The one-loop ultraviolet pole is ${\Pi }_{\mathrm{UV}}=-e^2/(12{\pi }^2ϵ)$. The Ward identity cancels any apparent divergent vector mass term. With photon counterterm insertion $-i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu }){\delta }_3$, this gives ${\delta }_3=-e^2/(12{\pi }^2ϵ)$ in the minimal subtraction scheme.

A uniform circular disc of radius $a$ and mass $m$, pivoted on a perpendicular axis through its rim, has moment of inertia $I_A=3m{a}^2/2$ by the parallel axis theorem. Its hanging small oscillations have period $2\pi \sqrt{3a/(2g)}$. Measured from the upright position, the falling equation is $\ddot{\theta }=2g\sin \theta /(3a)$. Exact upright rest remains an unstable equilibrium; the energy level of that state describes a limiting separatrix after perturbation, with ${\dot{\theta }}^2=4g(1-\cos \theta )/(3a)$. The center acceleration and pivot reaction follow by resolving radial and tangential components.

Inside a homogeneous spherical mass, gravity is proportional to position. Projection onto a fixed straight chord gives $\ddot{y}=-{\omega }_0^{2}y$, independently of its perpendicular offset from the sphere centre. Release from one surface endpoint at rest reaches the opposite endpoint after $\pi /{\omega }_0$. The transverse force is supplied by the tunnel constraint. This idealization neglects removed tunnel mass, rotation, density variation and dissipative forces.

For $N$ Dirac fields with common mass, $\mathcal{L}=\bar{\Psi }(i/\partial -m{1}_N)\Psi$ is invariant under constant complex-linear rotations $\Psi \mapsto U\Psi$ with $U\in U(N)$. A Hermitian flavour generator $T^A$ has Noether current $j_A^{\mu }=\bar{\Psi }{\gamma }^{\mu }{T}^A\Psi$. A nondegenerate fermion mass matrix restricts the symmetry to unitary matrices commuting with it. If the theory is entirely free and rotations into charge-conjugate fields are allowed, its enhanced internal symmetry of free Dirac fields can be larger than this subgroup.

In a uniform U-tube containing incompressible liquid of density $\rho$, cross section $A$ and total moving length $l$, let one surface fall by $y$ while the other rises by $y$. The height difference is $2y$, and potential energy relative to equilibrium is $U=\rho gA{y}^2$. The whole moving mass is $\rho Al$, so $K=\rho Al{\dot{y}}^2/2$. Newton's second law or energy conservation gives $\ddot{y}+(2g/l)y=0$ and ${\omega }_0=\sqrt{2g/l}$. This assumes small excursions that keep the free surfaces in the uniform vertical limbs.

When a vacuum vessel's window shatters, a nearly atmospheric initial pressure difference gives inward pressure force $\Delta pA$ on frontal area $A$. Dividing by the glass mass $M$ estimates its initial collective acceleration. An ideal atmospheric pressure-work budget $p_{0}V$ for vessel volume $V$ would give a collective kinetic energy estimate $M{v}_{\mathrm{rms}}^2/2\lesssim p_{0}V$ if assigned entirely to the fragments. This is a model-dependent energy budget, not a universal bound on every fragment's speed: air motion, fragmentation, pressure equalization and wall impacts alter the transfer, and unequal sharing can accelerate a small fragment faster than the collective estimate.

The valley of stability is the region of nuclear masses near the minima with respect to proton-neutron imbalance. Its beta-stable band bends toward neutron excess for heavier nuclei, as nuclear Coulomb energy competes with the cost of neutron-proton asymmetry. The band is distinct from the particle-emission boundaries set by nuclear separation energies.

An ideal gas law measurement obtains amount of substance without knowing the chemical formula in advance:Here $p$ is vapor pressure, $V$ is its occupied volume, $T$ is absolute temperature, $R$ is the molar gas constant, and $m$ is the mass of vapor occupying that volume at those conditions.

Santa Monica College's teaching experiment uses a $125\mathrm{mL}$ Erlenmeyer flask with a pinhole foil cap, a $600\mathrm{mL}$ water bath on a hot plate, a thermometer, and a barometer. Excess volatile liquid displaces air. A dissolved dye loses its colour and deposits when the liquid evaporates; three more minutes of heating follow. Cooling condenses retained vapor. Subtracting the dry assembly mass from the cooled assembly mass supplies $m$. The flask volume comes from brim-full water mass and mass density at its measured temperature. Controls include a dry exterior, excluding bath water, and three trials.^[ref]

The inference assumes vapor pressure equals atmospheric pressure, vapor and bath reach thermal equilibrium, the retained vapor mass is recovered on cooling, and the vapor obeys the ideal gas law. Residual air, thermal expansion of the flask, and vapor temperature below bath temperature are explicit error questions in the teaching experiment. Repeating trials assesses repeatability; it does not remove a shared bias.^[ref]

A body exchanging mass with material of inertial velocity $\mathbf{u}$ satisfies the displayed momentum conservation law for one stream, with signed $\dot{m}$ positive for entry and negative for exit. The force $\mathbf{F}$ excludes the incoming or outgoing momentum flux. A short time interval gives $d(m\mathbf{v})/dt=\mathbf{F}+\mathbf{u}\dot{m}$, proving the formula. For stationary incoming matter, $d(m\mathbf{v})/dt=\mathbf{F}$; for matter leaving at the body's own velocity, $m\dot{\mathbf{v}}=\mathbf{F}$. The rocket equation and accreting-drop motion are applications.

An interaction making particle velocities approach those of their neighbors. Symmetric nonnegative weights $K_{ij}$ conserve total momentum for equal particle masses and dissipate relative kinetic energy. Connectivity and quantitative lower bounds on the interaction matter for eventual consensus; nonnegativity alone permits noninteracting groups.

Directional inequality of velocity dispersions. Inferring a three-dimensional dynamical mass from one line-of-sight dispersion requires an assumption about this anisotropy.

In a nonrelativistic elastic collision of incident mass $M$ at speed $u$ with initially stationary mass $m$, the centre-of-momentum frame moves at $V=Mu/(M+m)$. The projectile's outgoing velocity in that frame has fixed magnitude $w=mu/(M+m)$. Its laboratory velocity tips lie on a circle of radius $w$ centred at $V\hat{\mathbf{x}}$. For $M>m$, a tangent from the origin gives the largest laboratory scattering angle, $\sin {\theta }_{\max }=w/V=m/M$. This geometrical result follows from Galilean velocity addition and conservation of energy.

Integrating a phase-space distribution function against products of particle velocity components produces its velocity moments. The zeroth gives number density, the first gives number flux, and, after multiplying by particle mass, the second gives momentum flux. Replacing $\mathbf{v}$ by $\mathbf{v}-\mathbf{u}$, where $\mathbf{u}$ is the mean velocity, defines central velocity moments; the second is the pressure tensor divided by particle mass.

For a particle of mass $m$ with linear drag $-\zeta v$, the velocity relaxation time is the exponential relaxation time ${\tau }_v=m/\zeta$. Under a constant applied force $F$ and zero-mean random forcing, the averaged Langevin equation is $md⟨v⟩/dt=-\zeta ⟨v⟩+F$. Solving givesThus the initial mean-velocity deviation decays by a factor $e^{-1}$ after time ${\tau }_v$. Since the force-per-velocity coefficient $\zeta$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $m/\zeta$ has units of time. If the damping parameter is instead defined as a rate $\gamma =\zeta /m$, the same time is $1/\gamma$. For a sphere obeying Stokes's law, ${\tau }_v=m/(6\pi \eta R)$. The fluctuation-dissipation relation for a Langevin particle gives the same decay time for its equilibrium velocity covariance.

Reuse the velocity selector analysis in the Thomson charge-to-mass experiment, but now send positive ions with $q=ze$ into a separate magnetic analyzer. The selector fields $E_s$ and $B_s$ give $v_0=E_s/B_s$. If the analyzer field $B_a$ is perpendicular to the selected velocity, its gyroradius givesfor a semicircular return to the entrance plane. The displayed impact separation $D$ is a diameter. This formula is linear in $r$ at fixed selected speed; the fixed-electric potential difference model above instead gives $m/q\propto r^2$. They impose different experimental conditions.

For a chosen example, take $E_s=2.00\times 10^{4}V/m$, $B_s=0.200\mathrm{T}$, and $B_a=0.500\mathrm{T}$. Thus $v_0=1.00\times 10^{5}m/s$. Approximating an ion mass by $24\mathrm{u}$ and setting $q=e$ gives $r\approx 49.75\mathrm{mm}$ and $D\approx 99.50\mathrm{mm}$. Here $\mathrm{u}$ denotes the unified atomic mass unit. These are stipulated teaching inputs, not a reported Aston measurement or an exact magnesium atomic mass.

A finite accepted speed range broadens the radius distribution. At fixed $m/q$ and $B_a$, $\delta r/r=\delta v/v$. At fixed electric charge and mass, $K=m{v}^2/2$ gives $\delta v/v\simeq \delta K/(2K)$ for a small kinetic energy spread. A one-percent kinetic energy spread therefore produces roughly a half-percent radius spread before other broadening is included. Its width convention must be specified before converting that statement into a mass resolving power.

A velocity selector uses perpendicular electric field and magnetic field to transmit a narrow range of speeds along a chosen beam direction. Assume uniform static electric field $\mathbf{E}=E\hat{\mathbf{y}}$ and magnetic field ${\mathbf{B}}_s=B_s\hat{\mathbf{z}}$, with $E,B_s>0$, and an incident velocity $\mathbf{v}=v\hat{\mathbf{x}}$. The cross product gives $\mathbf{v}\times {\mathbf{B}}_s=-v{B}_s\hat{\mathbf{y}}$, so the transverse Lorentz force isFor an electron, the electric force points downward and the magnetic force upward. A faster electron initially bends upward; a slower one initially bends downward. At the selected speed, both forces cancel along the whole straight trajectory. The cancellation is independent of mass and nonzero electric charge; it selects velocity, not a particular charge-to-mass ratio. A component of velocity parallel to the magnetic field is not constrained by the cancellation, so the incident beam must also have its direction restricted.

A real exit slit accepts a finite range of speeds and incident directions. A zero displacement at one screen alone does not establish zero force throughout the path: a curved trajectory could return to that screen position. The ideal velocity selector interpretation requires a straight aligned beam, known overlapping electric field and magnetic field, and sufficiently narrow entrance and exit slits. Its useful role in the Thomson charge-to-mass experiment teaching model is to supply speed independently of the subsequent magnetic bending.