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Translating a Fermi circle of radius $p_F$ by $p_0\hat{\mathbf{x}}$ preserves its area. For $p_0<p_F$, its radial displacement from the original centre is the displayed expression. Its Taylor expansion is $\nu =p_0\cos \theta -p_0^2{\sin }^2\theta /(2p_F)+O(p_0^4/p_F^3)$. The second-order term cancels the apparent area increase from the first-order shape: ${\int }_0^{2\pi }(p_F\nu +{\nu }^2/2)d\theta =0$. This distinction matters when converting a grand-potential variation into a fixed-particle number energy change.

Particle decay is a spontaneous transformation of an unstable particle into allowed final particles. Conservation of energy, momentum, electric charge and the relevant interaction's other conserved quantities constrain the possibilities. A decay can be energetically allowed yet absent or weak because its transition amplitude is suppressed.

For an isolated state with an approximately constant rest-frame decay rate, the surviving population follows an exponential in proper time. The mean lifetime $\tau$ and energy decay width $\Gamma$ obey $\Gamma =\hslash /\tau$. A moving particle's laboratory flight-time distribution also includes time dilation.

Experimentally, detectors reconstruct decay products and sometimes a separated decay vertex. Correcting for acceptance and reconstruction efficiency is necessary before interpreting their counts as a lifetime or branching fraction. Different decay channels compete to make the total width.

The particle-hole continuum is the range of energy and momentum transfers that can excite an occupied fermion state into an unoccupied state, leaving a hole behind. For a free gas at zero temperature, choose $|\mathbf{k}|<k_F$ and $|\mathbf{k}+\mathbf{q}|>k_F$; the excitation energy is ${ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$.

Varying the initial momentum produces a continuum at fixed transferred wavevector. For small $q$ in an isotropic gas, its positive frequencies extend approximately up to $v_{F}q$. A collective mode overlapping these excitations can decay by creating particle–hole pairs; a zero sound branch outside that region can avoid this particular damping channel. Interactions change the response and effective parameters, so the free-gas boundary is a useful starting model rather than a universal exact spectrum.

Particle-hole symmetry relates $H_{\mathrm{BdG}}(k)$ to $-H_{\mathrm{BdG}}^{\ast }(-k)$. Consequently every energy $E$ at momentum $k$ is paired with energy $-E$ at momentum $-k$.

A nearest-neighbour tight-binding model on a bipartite lattice, with identical on-site energy ${ϵ}_0$, has paired single-particle energies ${ϵ}_0\pm x$. At half filling, $f({ϵ}_0+x)+f({ϵ}_0-x)=1$ for the Fermi-Dirac distribution with $\mu ={ϵ}_0$. Thus its fixed-density chemical potential equals ${ϵ}_0$ at all positive temperatures within the isolated-band model. An anisotropic rectangular nearest-neighbour band retains this property; additional same-sublattice hopping or unequal on-site energies can remove it.

A particle of mass $M$ confined by infinite walls to $0<x<a$, $0<y<b$, $0<z<c$ has Dirichlet boundary conditions. Separation of variables gives the normalized product energy eigenstatesEach factor comes from the infinite square well, and its second derivative contributes one term to the kinetic energy. The minimum uses $(1,1,1)$; zero quantum numbers would give the zero wavefunction, not a lower energy state.

For a cube, permuting unequal quantum numbers can produce degenerate energy levels. The first excited energy has the three basis states $(2,1,1)$, $(1,2,1)$ and $(1,1,2)$, each with $E=6{\pi }^2{\hslash }^2/(2M{a}^2)$. A general state in that eigenspace is a linear combination of those products, so degeneracy permits stationary states that are not a single separated product.

A collider isolation requirement limits additional tracks or deposited energy near a particle candidate. It suppresses candidates embedded in jets while preserving suitably separated prompt leptons or photons. The cone geometry, energy threshold and collision-pileup correction form part of the observable's experimental definition.

A point mass constrained to the fixed graph $y=y(x)$ has velocity $(\dot{x},y^{\prime }\dot{x})$. Under uniform downward gravity, its Lagrangian is $L=m(1+y^{\prime 2}){\dot{x}}^2/2-mgy$, so its canonical momentum is $p_x=m(1+y^{\prime 2})\dot{x}$, not the horizontal mechanical momentum. The Euler-Lagrange equation gives the equation in the title, and the conserved energy is $m(1+y^{\prime 2}){\dot{x}}^2/2+mgy$. A straight segment with constant slope $s$ has constant horizontal acceleration $-gs/(1+s^2)$; zero curvature does not imply zero acceleration. A cart held to a track satisfies this mechanical constraint only while the assumed contact or retaining mechanism remains effective.

For a smooth plane at inclination $\theta$ rotating about a vertical axis at constant angular velocity $\omega$, choose rotating coordinates $\xi$ horizontally and $\eta$ uphill, with their cross product the upward normal vector. A sliding particle under gravity satisfiesThe normal reaction is $N=m(g\cos \theta -2\omega \dot{\xi }\sin \theta +{\omega }^2\eta \sin \theta \cos \theta )$ while contact is maintained. The Coriolis force does no work in the rotating frame, giving the conserved specific energyThe last term is the centrifugal potential; this quantity need not equal the inertial mechanical energy.

Particle physics studies the elementary constituents of matter and their interactions. It connects quantum fields and symmetry principles with measurements of scattering, decays and particle production.

An accelerator experiment does not photograph an elementary particle's internal form directly. Detectors record tracks, energy deposits and timing signals; a model reconstructs the event and compares distributions with predicted processes. Collider studies, neutrino measurements and nonaccelerator experiments provide complementary tests of the Standard Model and searches for additional phenomena.

The lowest initial centre-of-mass frame energy at which a specified final state can be made. For free massive final particles in natural units, the minimum is the sum of their masses, attained when their relative momenta vanish. The cross section at the exact threshold can vanish even though the channel becomes kinematically accessible there.

At fixed temperature, the continuum excited population of a three-dimensional ideal Bose gas reaches a finite maximum as the chemical potential approaches the ground energy. Adding particles beyond that maximum creates macroscopic ground occupation. Thus $N^{\prime }=\min (N,N_c)$ and $N_0=\max (0,N-N_c)$ to leading thermodynamic-limit order. Finite systems retain nonmacroscopic ground occupation below the nominal threshold.

For a one-electron P-to-S transition with nuclear spin zero, ten electric-dipole sublevel channels have finite strength at weak nonzero field. Six retain finite strength in the complete orbital-spin decoupling limit. Channel count is distinct from spectral frequency count: retaining first-order residual spin-orbit interaction energy gives five distinct frequencies, with two central channels degenerate.

With pressure amplitude $P=ZV$ and velocity $V$ directed into a boundary, the mean absorbed acoustic energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. A passive boundary cannot supply net energy, so $\mathrm{Re}Z\ge 0$ at a real frequency. Its reflection from a propagating acoustic plane wave has magnitude at most one. A pole continued to a complex incidence angle or frequency must be interpreted separately from real-frequency passive scattering.

A passive mechanical network can store and dissipate energy, but cannot supply more net work than was previously stored or delivered through its terminals. For a two-terminal mechanical element, choose signed force $F$ and relative velocity $v$ so that $p=Fv$ is power entering the network. With a nonnegative mechanical energy storage function $E$, passivity requiresIn particular, the net extracted energy over that interval cannot exceed $E(t_0)$. This permits temporary outward power: a stretched spring can release stored elastic potential energy. A requirement that $Fv$ be nonnegative at every instant would incorrectly exclude such a spring.^[ref]

For a spring, dashpot and inerter connected between the same two terminals, let $x$ be relative displacement and $v=\dot{x}$. Their constitutive equations giveFor $k,b,c\ge 0$, integration proves the passivity inequality: the extra term $\int c{v}^{2}dt$ is dissipated energy. The inerter stores kinetic energy associated with relative motion, so a passive mechanical network need not be a device that merely resists displacement or converts all input work into heat. This is central to mechanical network synthesis using springs, dashpots and inerters.^[ref]

For example, a spring with $k=1000N/m$ initially extended by $0.10\mathrm{m}$ stores $5\mathrm{J}$. It can deliver up to $5\mathrm{J}$ while relaxing, despite being passive. A purely viscous element with $c=-10\mathrm{N}s/m$ would instead have input power $-10v^2$: maintaining $v=1m/s$ for one second would extract $10\mathrm{J}$ even from an initially unenergized model. An apparatus producing that sustained behavior needs another energy source.

An experimental check records terminal force and relative velocity simultaneously and integrates their product, accounting for initial energy and every power-carrying connection. A brief interval of negative measured power is compatible with stored-energy release. For a linear zero-initial-state network, the mechanical driving-point impedance provides a complementary frequency description; a nonnegative real part at measured sinusoidal frequencies checks mean absorbed power there, but a finite frequency sweep alone does not establish the full time-domain passivity inequality.

The passive sign convention defines terminal electric potential difference $V=\varphi (A)-\varphi (B)$ and positive electric current entering at $A$. Then $VI$ is absorbed electrical power: positive means the component receives energy, negative means it supplies energy. Reversing both reference directions leaves the product unchanged. For a positive electrical resistance obeying Ohm's law, both $V$ and $I$ reverse together and Joule heating remains $I^{2}R\ge 0$.

A passive viscoelasticity law cannot deliver net work in a periodic steady deformation without an external source of energy. With strain $ϵ={ϵ}_0\cos \omega t$ and stress $\sigma ={ϵ}_0(G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t)$, the cycle-averaged supplied power is $⟨\sigma \dot{ϵ}⟩=\omega {ϵ}_0^2{G}^{\prime \prime }/2$. Thus a necessary passivity condition, using the convention $e^{i\omega t}$, is a nonnegative loss modulus at every positive frequency. Equivalently the work per cycle is $\pi {ϵ}_0^2{G}^{\prime \prime }$. A stable differential equation for stress does not by itself ensure this energetic condition.

The group velocity, which gives wave-packet and energy propagation, is the gradient of the dispersion relation:For $\mathbf{k}=(k_{\perp },0,k_{\parallel })$ this becomesThus for $ϵ=|k_{\parallel }|/k_{\perp }\ll 1$, energy travels almost along the rotation axis, with a small horizontal component. By contrast, the normal-to-phase-front phase velocity ${\mathbf{v}}_{\mathrm{ph}}=\omega \mathbf{k}/k^2$ is almost horizontal. Indeed $\mathbf{k}\cdot {\mathbf{v}}_g=0$, so these phase and group directions are perpendicular. Specifying which meaning of propagation is intended matters for an inertial wave.

For the wave polarization, incompressibility and the $y$ equation in part a giveThe horizontal component parallel to ${\mathbf{k}}_{\perp }$ is small, but $|{\hat{u}}_y|\simeq |{\hat{u}}_z|$: the other horizontal and vertical components are in quadrature. More exactly, introduce the orthonormal transverse basisThen $\hat{\mathbf{u}}=A({\mathbf{e}}_1-\sigma i{\mathbf{e}}_2)$. The physical velocity at a fixed point traces a circle in the plane perpendicular to $\mathbf{k}$; its handedness reverses between branches. In the anisotropic limit this plane is nearly vertical. The group velocity and polarization of an inertial wave therefore have quite different geometric roles.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

In ideal magnetohydrodynamics, the first-order electric field is ${\mathbf{E}}_1=-\mathbf{v}\times B_0{\mathbf{e}}_z$. Retaining its product with both the background and first-order magnetic field gives the Poynting vectorThe first term is transverse and oscillatory, with zero period average. The second is quadratic and transports wave energy. A potential quadratic ideal electric-field term $-\mathbf{v}\times \mathbf{b}$ vanishes here because the real traveling-wave fields are parallel or antiparallel.

For the $+z$ branch, $\mathbf{b}=-\sqrt{{\mu }_0{\rho }_0}\mathbf{v}$, soFor the opposite branch, its sign reverses. The averaged Poynting flux is the total wave energy density, including equal kinetic and magnetic shares, multiplied by the group velocity. This agrees with Alfvén wave energy transport. Although the flux in this expression is electromagnetic, it supplies the transport of both components of the coupled fluid-field wave.