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For a partially condensed equilibrium ideal Bose gas in a fixed three-dimensional box, the energy scales as $T^{5/2}$. Unbiased random particle removal keeps mean energy per particle fixed, so re-equilibration gives $T\propto N^{2/5}$ until condensation disappears. Removal of an excited particle can be followed by partial replenishment from the zero-energy condensate. This differs from selective evaporative cooling of atoms, which preferentially removes high-energy atoms and reduces mean energy per particle.

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.

In a dilute nonmagnetic medium of isotropic independent polarizable particles, $n-1\simeq N\alpha /(2{ϵ}_0)$ and the Rayleigh scattering cross-section is $8{\pi }^3{\alpha }^2/(3{ϵ}_0^2{\lambda }^4)$. Eliminating polarizability and setting direct-beam attenuation coefficient to $N\sigma$ gives the title relation. It describes scattering out of a narrow beam, not absorption of all radiant energy. It assumes fixed number density, negligible correlations/local-field corrections and a scalar polarizability; molecular anisotropy, aerosols and stratification require additional information.

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.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

A reaction energy barrier is the height of a specified peak relative to the starting state in an energy profile. State which energy is used and which direction is being considered. A schematic barrier is not automatically the empirical activation energy obtained from the temperature dependence of a rate constant.

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.

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 $2\pi$.

RABBITT measures adjacent harmonic spectral phases through two-photon photoionization. A weak delayed infrared reference opens two interfering paths to a shared photoelectron energy; the two-path attosecond phase reconstruction spells out the inference.

For rectangular real-space periods $a>b$, the free first-zone corner has energy ${\hslash }^2{\pi }^2(a^{-2}+b^{-2})/(2m)$, whereas the free second-zone minimum is ${\hslash }^2{\pi }^2/(2m{a}^2)$. Their difference is the displayed overlap. A divalent spin-degenerate system is a band insulator only when the actual first-band maximum lies below the second-band minimum. A rigid symmetric splitting $\Delta$ removes the overlap if $\Delta >{\Delta }_{\mathrm{overlap}}$, but an arbitrary periodic potential does not shift every corner and face by the same amount. The exact criterion uses the indirect band gap, not a geometry-only local direct-gap width.

For a massless join between two ideal strings with a common tension $T$, require continuous transverse displacement and continuous transverse force $T{\partial }_x\Psi$. With $Z_j=\sqrt{T{\rho }_j}$ for linear mass densities ${\rho }_j$, the displacement reflection coefficient and transmission amplitude areThe reflected and transmitted mean-power fractions are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(Z_2/Z_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. Reflection from the heavier side reverses phase; reflection from the lighter side does not. Total reflection occurs in the limits $Z_2/Z_1\to \infty$ or zero, while a finite positive impedance mismatch transmits some energy. Equal mechanical impedances remove reflection.

A refrigerator removes heat from the region we want to keep cold and releases heat to warmer surroundings. Here we consider a work-driven cyclic device: its useful output is the cold-side extraction $Q_c$, and its input is net work $W$. A food compartment is a familiar example; the same cooling objective applies when removing heat from a room. The coefficient of performance measures extraction per unit work, rather than the fraction of input energy converted into output. ^[ref]

Choose the complete machine as the thermodynamic system. Let $Q_c>0$ be heat entering from the colder side, $Q_h>0$ be heat delivered to the hotter side, and $W>0$ be net work supplied, all over the same complete cycle. Assume these are its only energy transfers, no matter crosses the outer boundary, and there is no net change in stored energy. Returning to the initial state gives $\Delta U=0$, so the first law of thermodynamics readsThe heat leaving the warmer side includes both the extracted heat and the supplied work. For instance, stipulated transfers $Q_c=3\mathrm{kJ}$ and $W=1\mathrm{kJ}$ require $Q_h=4\mathrm{kJ}$; the cooling coefficient of performance is $3$. This is an illustrative budget, not measured performance. At steady operation the same balance holds for average transfer rates, ${\dot{Q}}_h={\dot{Q}}_c+P_{\mathrm{in}}$, where $P_{\mathrm{in}}$ is input power.

This budget explains a useful counterexample: operating a refrigerator with its door open inside an otherwise isolated room cannot cool the room as a whole. If both heat transfers occur within the room and the electrical input comes from outside, the room gains $Q_h-Q_c=W$ each cycle. Locally cold air does not imply net cooling. The Second law of thermodynamics permits cold-to-hot transfer because work is supplied; it forbids a cyclic device whose sole effect is that transfer. See the Clausius statement of the second law.

At fixed particle number and volume, a classical monatomic ideal gas has $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$. The canonical energy fluctuation identity $\mathrm{Var}(E)=k_B{T}^2{C}_V$ therefore gives ${\sigma }_U/U=\sqrt{2/(3N)}$. Absolute fluctuations grow as $\sqrt{N}$, while relative fluctuations decrease as $N^{-1/2}$. A particle reservoir changes the fluctuation problem because energy and particle-number fluctuations are then correlated.

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 a massive free particle with invariant rest mass $m$, speed $v<c$, and Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, its total energy is $E=\gamma m{c}^2$. It includes the rest energy $m{c}^2$ and the relativistic kinetic energy $(\gamma -1)m{c}^2$. Here $c$ is the speed of light. At fixed total energy, a larger speed requires a smaller rest mass.

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 $E^{\prime }\to E$.

An irradiated accretion disk can produce iron fluorescence with a known local energy scale. relativistic Doppler effect, gravitational redshift, beaming and light bending redistribute those photons into a broadened asymmetric line. The red wing is sensitive to the deepest emitting radii, while inclination and orbital motion influence the profile. Coronal illumination, ionization, absorption and the continuum model must be controlled before inferring an inner radius or testing a spacetime geometry.