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For one daughter of mass $m$ and two massless daughters, set $r=m^2/M^2$, where $M$ is the parent mass. In natural units, integrating over all directions and all kinematically allowed final momenta givesThis follows by splitting the final state into a massless spectator and a two-particle cluster of invariant squared mass $s$, and using $d{\Phi }_3=d{\Phi }_2(M;\sqrt{s},0)d{\Phi }_2(\sqrt{s};m,0)ds/(2\pi )$. The expression counts phase space alone; a physical decay rate must weight it with the squared decay amplitude. Near threshold the dimensionless factor vanishes as $(1-r{)}^3/3$.

The gravitational motion of three interacting bodies under their coupled Newtonian mechanics equations. When one body's mass is negligible, the restricted three-body problem neglects its influence on the other two while retaining both forces on its trajectory.

For three massive daughters with a smooth nonzero decay amplitude at the decay threshold, the integrated Lorentz-invariant phase space vanishes as $Q^2$, where $Q$ is the parent's mass minus the sum of daughter masses. The spatial Dirac delta function leaves six independent momentum coordinates. Rescaling each by $Q^{1/2}$ contributes $Q^3$, while the conservation of energy Dirac delta function contributes $Q^{-1}$. The factors $1/(2E)$ are smooth nonzero constants at the decay threshold. Thus the decay width also scales as $Q^2$ when the decay amplitude has a nonzero constant term. A vanishing decay amplitude, a massless daughter, or a threshold singularity can change this law. In three spatial dimensions the same argument gives $Q^{(3N-5)/2}$ for $N$ massive daughters.

For a tree-level left-handed three-body decay with one massive final fermion and two massless ones, the integrated width relative to the all-massless result includes $f(r)=1-8r+8r^3-r^4-12r^2\ln r$, where $r$ is the squared final-to-parent mass ratio. It satisfies $f(0)=1$ and $f(1)=0$. It is not a universal factor for channels with several massive daughters.

For the unit-modulus hyperbolic scattering block with elementary mass $m$, a pair bound at rapidity difference $iu$ has mass $m_2=2m\cos (u/2)$. Bootstrap fusion with another elementary particle gives a further direct pole at $3iu/2$. It is inside the physical rapidity strip with positive imaginary residue only for $0<u<2\pi /3$. In that range the new bound state has rest-frame constituent rapidities $iu,0,-iu$, so its energy is $m(1+2\cos u)=m\sin (3u/2)/\sin (u/2)$. Charges add at the bound-state vertex. At $u=2\pi /3$ the apparent pole cancels; beyond that point it lies outside the physical strip. The formula is conditional on this particular scattering block and its direct-pole interpretation, not a universal three-body binding law.

For an original classical calculation of the return step, neglect the residual ion's Coulomb force after release. Take a spatially uniform driving electric field $E(t)=E_0\cos \omega t$, an electron of electric charge $-e$ and mass $m$, and zero initial position and velocity at $t_i$. With $\alpha =\omega t_i$ and $\beta =\omega t$, integrating Newton's second law givesA return requires $\beta >\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver kinetic energySearching the first-return solutions for release phases $0<\alpha <\pi /2$ gives a maximum near $\alpha =0.3134$ and $\beta =4.3990$, with $K_{\mathrm{return}}\simeq 3.1731U_p$. For a chosen ionization energy $I_p$, recombination into the initial state can therefore emit a photon with energy up to approximately $I_p+3.17U_p$ in this model. This is an upper return-energy scale, not a prediction of the emitted yield. It omits release probabilities, quantum interference, propagation, phase matching and depletion of the neutral gas.

The calculation explains why a long driving wavelength can increase the available return energy: at fixed $E_0$, $U_p$ grows as ${\omega }^{-2}$. It does not say that conversion efficiency improves. The electron must return with a probability amplitude for emission that survives the microscopic dynamics and adds coherently across the gas.

An upper mass $3m$ and lower mass $m$ joined by equal light strings of length $\ell$ have small-angle mass and stiffness matrices $m{\ell }^2\left(\begin{array}{cc}4& 1\\ 1& 1\end{array}\right)$ and $mg\ell \left(\begin{array}{cc}4& 0\\ 0& 1\end{array}\right)$. The squared frequencies are $2g/(3\ell )$ and $2g/\ell$, with angular eigenvectors $(1,2)$ and $(1,-2)$ respectively. Equal-angle release decomposes with coefficients $3\alpha /4$ and $\alpha /4$. The angles are measured from vertical, not between strings.

The instantaneous acceleration contribution from rocket thrust $T$ acting on remaining rocket mass $M$ is $a_T=T/M$. Gravity and other external forces contribute separately. A constant rocket thrust therefore produces an increasing thrust acceleration as rocket propellant is consumed.

Tidal capture converts sufficient relative orbital energy into internal tidal heating to leave two initially unbound galaxies on a bound orbit. For identical spherical galaxies of mass $m_g$, a distant rapid flyby of impact parameter $p$ and relative speed $v$ gives the approximate capture condition $pv<[(32/3)G^2{m}_g^2⟨r^2⟩{]}^{1/4}$. This requires the impulse approximation and weak gravitational focusing; it is not a general criterion for subsequent coalescence.

Disruption of a self-gravitating body when differential gravitational acceleration across it exceeds its binding acceleration. For a body of radius $R$, mass $m$ at distance $r$ from a much more massive body $M$, the competing scales are $GMR/r^3$ and $Gm/R^2$; their equality gives $r\sim R(M/m{)}^{1/3}$, up to geometry and internal-structure factors. The tidal tensor gives the directional differential acceleration. A tidal disruption event near a black hole is a specific astronomical application.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

A tidally enhanced stellar wind is a model in which a companion increases a giant's wind mass loss as the giant approaches its Roche lobe. Removing envelope mass before contact can reduce or reverse the mass ratio and make subsequent Roche-lobe overflow more stable. This is one proposed route to wide systems with Algol-like reversed evolutionary ordering. Van der Sluys's compact-binary study, section 5.7.1 discusses the enhanced-wind scenario and its effect on mass-transfer stability.

A mass above a sufficiently wide estuary experiences additional downward gravity when extra water is present. Approximating the tide as a plane slab of density $\rho$ and added thickness $h$ gives $\Delta g=2\pi G\rho h$. A linear spring with initial extension $x_0=mg/k$ changes by $\Delta x=(m/k)\Delta g=x_0\Delta g/g$. This isolates direct water attraction; it omits finite shoreline geometry, buoyancy, structural motion and terrestrial tides.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

Titrant standardization determines the actual titrant concentration by reaction with a known inventory, often a primary standard. It is distinct from calibrating the delivered volume or a glass electrode's response. For a known monoprotic acid inventory $n_s$ and a suitable blank-corrected endpoint,An original illustrative standard of $0.980\mathrm{mmol}$ requiring $9.85\mathrm{mL}$, with matched blank $0.05\mathrm{mL}$, gives $c_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. The measured quantities here are stipulated examples. Independent solution preparation and volumetric calibration tests the pipette, burette, and volumetric flask; titrant standardization tests the reacting inventory per delivered volume.

The useful chain is known mass, known composition, known amount of substance, and then a measured titre through a specified stoichiometry. If each standard unit consumes $r$ units of titrant, the general inventory equation is $c_{B}V=r{n}_s$, where $V$ is the corrected titre. The one-to-one formula above has $r=1$. A titration endpoint supplies an observable stopping rule; whether it represents the equivalence point accurately remains a separate question. Standardization endpoint and blank bias makes the direction of the resulting error explicit.

A top-hat filter averages a field with uniform weight inside a ball and zero weight outside it.

The top quark is the up-type member of the third quark generation. Its mass exceeds the W boson mass, so a real W boson cannot decay into an on-shell top quark. A massless approximation for accessible decay products must not silently remove this channel threshold.

Here $x=m_W^2/m_t^2$, and the tree-level decay width neglects the bottom quark mass. The open two-body W channel gives a top quark lifetime of a few $10^{-25}\mathrm{s}$, shorter than the hadronization time. A fifth-power Sargent's rule extrapolation misses this physical regime.^[ref]

The top squark, or stop, is the scalar superpartner of the top quark. Its mass controls the leading residual top-sector correction to the Higgs mass after soft supersymmetry breaking.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.