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For conservative forces and ideal constraints that do no work, mechanical energy $E=T+U$ is constant. Work-energy balance remains useful with nonconservative forces: their power gives $dE/dt$. In one dimension,locates accessible regions and turning points, and reduces the trajectory to a quadrature $dt=dq/\sqrt{2(E-U)/m}$. A local minimum of potential is a stable equilibrium; a local maximum is unstable. Near a nondegenerate minimum, $U=U_0+K(q-q_0{)}^2/2+\cdots$ gives ${\omega }^2=K/m$.

For a central force, conserved angular momentum reduces radial motion to an effective potential $U_{\mathrm{eff}}=U(r)+L^2/(2m{r}^2)$. Its extrema determine circular orbits and their radial stability. In several coordinates, small oscillations follow from the kinetic-energy mass matrix $M$ and potential Hessian $K$ through $\det (K-{\omega }^{2}M)=0$.

Energy is also central to the Lagrangian $\mathcal{L}=T-U$ and the Euler-Lagrange equations. A time-independent natural mechanical system has conserved energy; cyclic coordinates yield additional momenta. Energy alone does not normally determine every coordinate of a multi-degree-of-freedom motion. Its value is to expose allowed motion, stability and reductions before solving the full differential equations.

Capacity convention. The printed equation requires $c$ to be volumetric heat capacity, in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. If “specific heat capacity” means per unit mass instead, the storage term is ${\rho }_{m}c{\partial }_{t}T$ and both denominators in the printed equation must contain ${\rho }_{m}c$. This distinction is fixed by dimensions.

For a fixed material region $V$, conservation of thermal energy gives the heat balance with volumetric heat generation,Use the divergence theorem, $\mathbf{J}=-\kappa \nabla T$ and constant coefficients. Since the region is arbitrary,The heat source is power per volume, while $\kappa$ is thermal conductivity rather than thermal diffusivity.

For uniform electrical conductivity and imposed current, current density is $j=I/(\pi a^2)$ and Joule heating is $H=j^2/\sigma$. A long, steady, cylindrically symmetric wire therefore satisfiesIntegration gives $T^{\prime }=-Hs/(2\kappa )+C/s$. Regularity at $s=0$ makes $C=0$, and the surface condition fixes the remaining constant. Thus the Joule-heated cylindrical wire hasHere $s$ denotes the printed radial coordinate $\rho$ and avoids confusing it with mass density. The centre is hotter by $I^2/(4{\pi }^2{a}^2\kappa \sigma )$.

For the air boundary, the internal flux at $s=a$ is $-\kappa T^{\prime }(a)=Ha/2$. Equate this to $\alpha [T(a)-T_{\mathrm{air}}]$. The Newton cooling of a Joule-heated wire givesMultiplication by circumference $2\pi a$ verifies that surface loss per length equals $I^2/(\pi a^2\sigma )$, the electrical power per length. The full air-cooled profile uses this surface temperature plus the same parabolic interior excess, assuming constant material properties and negligible end losses.

With the standard liquid-water approximation $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, each portion has mass $100\mathrm{kg}$. The equal heat capacities give final temperature $T_f=310\mathrm{K}$ by energy conservation. For an insulated mixing process with no other changes,The reversible integrals $mc\int dT/T$ evaluate entropy differences between equilibrium states; the actual heat transfer is irreversible. These are two portions of the same substance, so there is no additional ideal-mixture compositional entropy.

For distributed inductance $L$ and capacitance $C$, a small segment obeys $V_z=-L{I}_t$ and $I_z=-C{V}_t$. Differentiation gives $V_{zz}=LC{V}_{tt}$ and $I_{zz}=LC{I}_{tt}$. For a forward wave,A backward voltage wave has current $I^{-}=-V^{-}/Z$. This sign matters at an impedance discontinuity.

The source figure is the stacked-strip parallel-plate transmission line geometry, despite its use of the word coplanar. Gauss's law gives $C={ϵ}_0{ϵ}_{r}a/d$. Opposite surface currents give magnetic field $B={\mu }_{0}I/a$ between the strips by Ampère's law; the magnetic flux per unit length is $Bd$, hence $L=Bd/I={\mu }_{0}d/a$. ThereforeNo numerical $a/d$ is given; the impedance is appropriately left in terms of that ratio. The line is taken to be nonmagnetic and lossless.

Increasing the gap by $\alpha$ changes $L$ to $\alpha L$ and $C$ to $C/\alpha$, so $Z_2=\alpha Z_1$ while $v$ is unchanged. voltage and current continuity at the step giveConsequently the voltage and energy-flux transmission coefficients areHere amplitudes are peak voltage amplitudes, with mean power $|V{|}^2/(2Z)$. The reciprocal gaps have $r_V(1/\alpha )=-r_V(\alpha )$: equal reflected energy but opposite voltage phase. A voltage transmission coefficient above one does not imply energy gain.

For a finite impedance section of a transmission line, put $\ell =l_2-l_1$ and $\delta =\omega \ell /v$. Multiple reflections inside the altered section interfere. Summing their geometric series, relative to the first boundary,Transmission is unity at $\delta =n\pi$ and minimal at $\delta =(n+1/2)\pi$, with minima $4{\alpha }^2/(1+{\alpha }^2{)}^2$. Resonances are separated by $\Delta f=v/(2\ell )$; the long-wavelength limit also transmits fully. Matching the far termination removes end echoes, not the two internal reflections.

For cold, collisionless, unmagnetized electrons with stationary heavy ions, $-m_e{\omega }^2\mathbf{x}=-e\mathbf{E}$. Therefore $\mathbf{P}=-ne\mathbf{x}=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$ andThis is the cold collisionless plasma dielectric response. For a propagating transverse wave, $\mathbf{B}=\mathbf{k}\times \mathbf{E}/\omega$, so $E/B=c$ in vacuum and $E/B=c/n_p$ in plasma, where $n_p=\sqrt{{ϵ}_r}$. Equivalently $E/H=Z_0$ and $Z_0/n_p$ for $\mu ={\mu }_0$.

Converting the density gives $n=10^{18}{\mathrm{m}}^{-3}$, so ${\omega }_p=5.64\times 10^{10}{\mathrm{s}}^{-1}$ and $f_p=8.98\mathrm{GHz}$. At $10\mathrm{GHz}$,At an interface without independently imposed surface charge or sheet current, tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{B},\mathbf{D}$ are continuous, with the electron response included in the effective permittivity. Normal $E$ need not be continuous; the induced electron charge is represented by that response. In a convention treating it explicitly as a surface charge, the normal vacuum-displacement jump equals that charge instead.

For the downward transmitted wave, transversality gives $k_x{E}_x+k_z{E}_z=0$. The measured field therefore has $k_x/(-k_z)=3$, so $\tan {\theta }_t=3$ and $\sin {\theta }_t=3/\sqrt{10}$. Tangential phase matching givesFor inferring incident power from a transmitted plasma field, choose ${\mathbf{E}}_i=E_i(\cos {\theta }_i,0,\sin {\theta }_i)$ and ${\mathbf{E}}_r=E_r(\cos {\theta }_i,0,-\sin {\theta }_i)$. Their magnetic fields have opposite $y$ directions. The two tangential conditions areHence $E_i\simeq 1.2464\mathrm{V}{\mathrm{m}}^{-1}$ and $E_r\simeq -0.1458\mathrm{V}{\mathrm{m}}^{-1}$. Since these are peak phasor amplitudes,This is incident power per area perpendicular to its propagation. Per area of the interface the incoming normal flux is $⟨S_i⟩\cos {\theta }_i\simeq 1.87\times 10^{-3}\mathrm{W}{\mathrm{m}}^{-2}$. The reflected sign is a phase reversal, not negative reflected energy.

At $1\mathrm{GHz}<f_p$, the relative permittivity is negative and there is no propagating transverse bulk wave. The field decays evanescently into the plasma and the ideal collisionless half-space reflects all incoming normal power. Collisions would allow absorption; a complex refractive index here does not by itself imply dissipative heating.

A normal mode is a pattern in which all coordinates share one frequency and fixed relative amplitudes. For a conservative linear system, it solves a generalized eigenvalue equation with stiffness matrix $K$ and mass matrix $M$. For this free three-mass spring chain,Solving $K\mathbf{u}={\omega }^{2}M\mathbf{u}$, for $\alpha >0$, givesTheir overall normalization is arbitrary; distinct modes are orthogonal in the mass inner product. The zero mode is rigid translation because neither end is fixed. In the antisymmetric mode the middle mass stays still, so changing its mass cannot alter the frequency. The symmetric internal mode keeps the centre of mass fixed through its middle displacement $-2/\alpha$. Its frequency rises without bound as $\alpha \to 0$ and approaches $\sqrt{k/m}$ as $\alpha \to \infty$, becoming degenerate with the antisymmetric frequency in that limit.

For equal masses put ${\omega }_1=\sqrt{k/m}$ and ${\omega }_2=\sqrt{3k/m}$. Positions are equilibrium positions plus displacementsThe translational mode needs the linear-in-time term; a cosine at zero frequency alone would miss arbitrary centre-of-mass velocity.

Taking the springs initially at equilibrium, the impulse gives $\dot{\mathbf{q}}(0)=(v,0,0)=(v/3){\mathbf{u}}_0+(v/2){\mathbf{u}}_1+(v/6){\mathbf{u}}_2$. Their squared Euclidean norms are $3,2,6$, so their initial kinetic energies are $m{v}^2/6,m{v}^2/4,m{v}^2/12$. Relative to total $m{v}^2/2$,Subsequent motion conserves each mode's total energy, exchanging its kinetic and spring contributions. The zero-mode share remains centre-of-mass kinetic energy.

Use the rotating reference frame equilibrium model, in which the centrifugal potential energy of a particle of mass $m_i$ is $U_i(r)=-m_i{\omega }^2{r}^2/2$. The Boltzmann distribution then givesDefine the concentration ratio as $c=n_1/n_2$. At the axis,Thus the axis favours the lighter species. The expression assumes a dilute classical mixture at one temperature, equilibrium in the rotating reference frame, and sampling sufficiently close to $r=0$.

An applied electric field displaces bound charges or reorients permanent dipoles, producing electric polarization $\mathbf{P}$, the dipole moment per unit volume. It gives bound charges ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \mathbf{n}$. Definefor a linear isotropic dielectric. Then $\nabla \cdot \mathbf{D}={\rho }_f$ separates free charge from the bound-charge response. Electronic, ionic and orientational electric polarization respond on different timescales; real permittivity is frequency dependent, and its dissipative part describes dielectric loss.

A uniform dielectric filling a parallel-plate capacitor raises its capacitance to $ϵA/d$. At fixed charge this reduces field energy $Q^2/(2C)$; at fixed voltage the stored energy $C{V}^2/2$ increases because the voltage source supplies charge and energy. Insulation and energy storage are useful applications, limited by loss and dielectric breakdown. Static interface conditions are continuous tangential $\mathbf{E}$ and a jump in normal $\mathbf{D}$ equal to free surface charge.

A current in one circuit links magnetic flux through another. It produces magnetic flux linkage through another. With fixed geometry and a linear reciprocal medium, choose consistent winding orientations and writeThe shared coefficient is mutual inductance; reversal of one winding reverses its sign. Faraday's law of induction gives the mutual induced emf ${\mathcal{E}}_2=-M{\dot{I}}_1$ for fixed $M$. The minus sign expresses Lenz's law.

The magnetic energy isNonnegative energy for every current pair requires $|M|\le \sqrt{L_1{L}_2}$, motivating the coupling coefficient $M/\sqrt{L_1{L}_2}$. Transformers use strong common flux to exchange energy between circuits; leakage flux reduces coupling. A shared core increases linkage, but saturation, hysteresis and eddy-current losses limit the linear description. A moving geometry also contributes $-I_1\dot{M}$ to the mutual emf.

Magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI units,in a linear isotropic response. The internal field matters: for a uniformly magnetized ellipsoid it is ${\mathbf{H}}_{\mathrm{int}}={\mathbf{H}}_{\mathrm{applied}}-{\mathbf{N}}_d\mathbf{M}$. Sample shape therefore affects an apparent magnetic susceptibility.

Diamagnetism opposes the applied field, giving small negative magnetic susceptibility from the induced orbital response. It is often only weakly temperature dependent. Paramagnetism aligns pre-existing moments and gives positive magnetic susceptibility; independent localized moments obey approximately $\chi =C/T$ in weak fields before saturating. Pauli paramagnetism of conduction electrons is a different, weakly temperature-dependent mechanism.

Ferromagnetism involves collective exchange-driven alignment and spontaneous magnetization below the Curie temperature. A specimen can nevertheless have nearly zero total moment because magnetic domains point in different directions. Field-driven domain-wall motion and rotation change the net magnetization. Antiferromagnets have compensating ordered sublattices below a Néel temperature; ferrimagnets have oppositely ordered but unequal sublattice moments and retain a net moment. Ferrites are useful examples of electrically resistive ferrimagnetic core materials.

Ferromagnetic response is nonlinear and history dependent. A hysteresis loop displays remanent magnetization, magnetic coercivity and saturation. A single constant magnetic permeability cannot describe the entire loop: differential magnetic permeability describes its local slope. The quasistatic energy loss per unit volume and cycle is the loop area $\oint HdB$. Soft magnetic materials have low magnetic coercivity and usually high useful magnetic permeability; hard magnetic materials retain magnetization and suit permanent magnets.

An electromagnet normally uses a soft, high-permeability core to guide flux generated by a coil. For a uniform-section core with length ${\ell }_c$ and air gap $g$, neglecting leakage and fringing,An air gap can dominate the magnetic reluctance even when the core magnetic permeability is large. Core saturation limits the improvement with current, while winding resistance causes heating. For alternating operation, laminations suppress eddy-current loops in metallic cores; high-resistivity ferrites also reduce those losses. Low hysteresis loss is important in repeatedly cycled cores, whereas high magnetic coercivity is valuable in a permanent magnet. Useful design therefore depends on the full magnetic response, electrical losses and operating frequency, rather than the largest quoted magnetic permeability alone.

For an attractive inverse-square potential, let the reduced mass be $\mu$, potential $-k/r$, and nonzero orbital angular momentum $\ell$. Conservation of angular momentum makes the orbit planar. The orbit equation isPositive relative energy gives $e>1$ and an unbound hyperbolic Kepler orbit, arriving from and returning to infinity. Its asymptotes satisfy $\cos {\phi }_{\infty }=-1/e$. At infinity $E=\mu v_{\infty }^2/2$ and $\ell =\mu b{v}_{\infty }$ for impact parameter $b$, so the deflection angle isThis describes gravitational encounters in the Newtonian two-body approximation. The body speeds up near periapsis while total energy remains positive. It is distinct from an elliptic bound orbit and from the zero-energy parabolic limit; repulsive Coulomb scattering also gives hyperbolae but uses a different orientation of the force and orbit.

For torque-free rotation, angular momentum is fixed in inertial space, but angular velocity need not point along it. In principal body axes,and cyclic permutations. A symmetric top with transverse moment $A$ and axial moment $C$ has constant ${\omega }_3$, while ${\omega }_1+i{\omega }_2$ rotates in body coordinates at signed frequency $(C-A){\omega }_3/A$. The symmetry axis precesses around the fixed angular momentum in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.

Poinsot construction expresses the same torque-free motion geometrically. The energy ellipsoid $\sum I_i{\omega }_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf{L}$, and $\mathbf{L}\cdot \mathbf{\omega }=2T$ places that point in a fixed plane perpendicular to $\mathbf{L}$. The ellipsoid rolls on this invariable plane without slipping: the instantaneous velocity of its contact point is proportional to $\mathbf{\omega }\times \mathbf{\omega }=0$. The contact curve is the polhode on the body ellipsoid and the herpolhode on the plane.^[ref]

For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis givesThe extreme-moment axes are stable; the intermediate one is unstable. The major-axis relaxation theorem adds weak internal dissipation while retaining negligible external torque. At fixed $L$,Energy loss favours rotation about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. A perfectly rigid torque-free body does not dissipatively settle; its energy is conserved. Distinguishing stability from relaxation also explains why minimum-inertia rotation can be stable in the ideal rigid model.

Forced precession occurs when an external torque changes $\mathbf{L}$, with $\dot{\mathbf{L}}=\mathbf{\tau }$. For steady precession, $\mathbf{\tau }={\mathbf{\Omega }}_p\times \mathbf{L}$. A rapidly spinning, pivoted heavy gyroscope has $L\simeq C{\omega }_3$ along its symmetry axis, givingThe torque is perpendicular to the nearly axial angular momentum, so it mainly changes its direction. For a steady inclined symmetric top, the more complete condition isIt exhibits slow and fast precession branches when real solutions exist. A generic release excites nutation as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which angular momentum and pivot are being used.

For speed distribution $f(v)$ normalized to one and isotropic directions, the incident number flux isThe positive normal velocity weights the population reaching the wall. Consequently its mean translational energy isThe bulk value $3k_{B}T/2$ would omit the arrival weighting.

For the two vessels, let $r=\sqrt{T_1/T_0}$, with subscripts h and c for hot and cold. Equal opposing effusion rates give $p_h/\sqrt{T_1}=p_c/\sqrt{T_0}$. Number conservation in equal, fixed volumes adds $p_h/T_1+p_c/T_0=2p_0/T_0$. HenceBoth final pressures must be determined; holding the cold temperature fixed does not hold its particle number fixed.

For aperture area $A_h=\pi d^2/4$, the common one-way flux is $J=p_c/\sqrt{2\pi m{k}_B{T}_0}$. The translational heat transfer from hot to cold is thereforeThis is the total heat transfer for a monatomic gas with no active internal modes, neglecting radiation and conduction through vessel walls. For a molecular gas, add $A_{h}J[u_{\mathrm{int}}(T_1)-u_{\mathrm{int}}(T_0)]$ for its internal energy per molecule. The maintained unequal temperatures describe a stationary nonequilibrium state even though the net particle current is zero.

A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=TdS-pdV+\mu dN$. Legendre transforms give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with natural variables of a thermodynamic potential $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T,V$; Gibbs free energy at fixed $T,p$.

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work $pdV$ in addition to internal energy. The steady first-law balance isWith no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking $H$ as the printed specific enthalpy, the steady-flow energy balance givesFor fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value givesThis uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq 8.651\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and $\alpha \simeq 0.193$. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_{\ell }$ is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

A heat pump delivers $Q_H=Q_S+W$: electrical work moves heat from a source as well as contributing energy itself. Its heating coefficient of performance can therefore exceed one, whereas direct resistance heating supplies at most its electrical input as useful room heat. For an ideal reversible pump,with absolute temperatures and $T_S<T_H$. This is an ideal limit; actual costs also depend on compressor performance and tariffs.

For the annual ground-temperature wave, model the soil as a homogeneous semi-infinite conductor, with depth positive downward and a prescribed surface temperature. Here $\kappa$ is thermal conductivity and $C$ is heat capacity per unit volume, so the thermal diffusivity is $\kappa /C$. Conservation of energy gives $C{\partial }_{t}T=\kappa {\partial }_z^{2}T$. After transients decay, put $T-T_0=\mathrm{Re}[A{e}^{i\omega t-qz}]$. Substitution gives $q^2=i\omega C/\kappa$; the solution bounded at depth uses $q=(1+i)/\delta$. Matching the surface then givesThis approximation identifies the ground surface with air temperature and neglects the pump's perturbation of the ground, as required by the model. The attenuation and phase delay arise from heat diffusion, not propagation at one fixed wave speed.

The graph below covers a full annual cycle. At $z/\delta =\pi /2$, the amplitude is $20e^{-\pi /2}=4.16$ K and the minimum lags January by a quarter-year. At $z/\delta =\pi$, the amplitude is only $0.864$ K and the minimum lags by half a year: the ground is slightly warmer than its annual mean in January.

The seasonal source depth for a reversible heat pump has a subtle limit: a deep source approaches $T_0$ and avoids severe winter cooling, but infinite depth is not the thermal optimum of this single-frequency model. In January, with $x=z/\delta$, maximize $T_S=T_0-T_A{e}^{-x}\cos x$. The first and largest maximum is at $x=3\pi /4$, where $\cos x+\sin x=0$, giving $z=3\pi \delta /4$ and $T_S\simeq 16.34^{\circ }\mathrm{C}$. It is also the optimum of the ideal annually integrated heating work: the heating load is even about January, so its weighted sine term integrates to zero and the depth dependence is again proportional to $e^{-x}\cos x$. Installation cost, finite tube spacing, heat extraction, weather and real pump performance alter this result. The supplied model therefore supports a finite seasonal thermal optimum, while not fixing a practical burial depth in metres.

For rolling without slipping through height $h$, conservation of energy givesA uniform solid sphere has $I/(m{R}^2)=2/5$, a solid cylinder $1/2$, and a pipe $(1+r_{\mathrm{in}}^2/R^2)/2>1/2$. Thus the ball is fastest, then the solid cylinder, then the pipe. A smaller moment of inertia leaves more energy in translation. This assumes release from rest through the same height, adequate static friction and negligible rolling resistance; a thin pipe has $v^2=gh$.

For isotropic planar motion, the planar effusion flux gives an escape rate through a gap of width $w$ is $\Gamma =w(N/\mathcal{A})⟨v⟩/\pi$, where $\mathcal{A}$ is tray area. At fixed equilibrium kinetic energy, or fixed shape of the scaled speed distribution, $v\propto m^{-1/2}$. Hence halving the mass increases the rate by $\sqrt{2}$. This planar effusion result assumes the same spatial density and an unrestricted small opening. Strictly, average energy alone fixes $⟨v^2⟩$, not $⟨v⟩$; an arbitrary change of distribution shape would need more information.

$Z$ is the terminating load impedance, $Z_0$ the line's characteristic impedance, and $k$ the propagation wave number; $l$ is its length. In the lossless convention printed, an open circuit gives $Z_{\mathrm{in}}=-i{Z}_0\cot kl$, a short circuit gives $i{Z}_0\tan kl$, and a matched load $Z=Z_0$ gives $Z_{\mathrm{in}}=Z_0$ for every length. The divergent or zero limits at isolated electrical lengths are understood continuously.

Put $\theta =kl$, $c_{\theta }=\cos \theta$, $s_{\theta }=\sin \theta$. The voltage/current matrix of one section isFor a pair beginning with $Z_1$, write $M_1{M}_2=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Its entries are $A=c_{\theta }^2-(Z_1/Z_2)s_{\theta }^2$, $D=c_{\theta }^2-(Z_2/Z_1)s_{\theta }^2$, $B=i(Z_1+Z_2)s_{\theta }{c}_{\theta }$, and $C=i(1/Z_1+1/Z_2)s_{\theta }{c}_{\theta }$. The alternating-impedance transmission line has unchanged input $X$, its Bloch impedance, obeying $X=(AX+B)/(CX+D)$. Hence, away from the removable dividing zeros,For real positive $Z_1,Z_2$, select the branch carrying energy into the line or decaying into it. A positive real part selects the pass-band root; infinitesimal passive loss resolves the stop-band branch. At $\theta =m\pi$ the cell matrix is the identity, so the exact fixed-point condition alone does not determine $X$; at odd quarter-wave lengths the appropriate limiting impedance can be zero or infinite.

In a transmission-line stop band, a lossless semi-infinite ladder totally reflects because its physical input is reactive. Equivalently its cell hasThus, for nonzero contrast, the stop-band interiors satisfyWith ${\theta }_c=\arcsin [2\sqrt{Z_1{Z}_2}/(Z_1+Z_2)]$, these are $m\pi +{\theta }_c<kl<(m+1)\pi -{\theta }_c$; band-edge reflection follows by a limit. If $k=\omega /u$, multiply the endpoints by $u/l$ for angular frequencies. Identical impedances have no stop band. A finite ladder can transmit by tunnelling, unlike the semi-infinite idealization.

A normal mode is a small oscillation with fixed relative displacement amplitudes, all coordinates oscillating at one frequency. linearization gives $M\ddot{\mathbf{y}}=-K\mathbf{y}$, so mode shapes solve $K\mathbf{u}={\omega }_n^{2}M\mathbf{u}$.

Write $C=q^2/(4\pi {ϵ}_0)$. force balance on an outer ion gives $m{\omega }^2{x}_0=C/x_0^2+C/(2x_0{)}^2$, henceA pair separated by $d$ has quadratic energy $C(y_j-y_i{)}^2/d^3$, equivalent to spring constant $2C/d^3$. The adjacent-pair constants are $8m{\omega }^2/5$, and the outer-pair constant is $m{\omega }^2/5$. Adding the trap stiffness givesIn the three-ion axial normal modes, the translation mode leaves every separation unchanged. The antisymmetric breathing mode leaves the central ion fixed; the remaining vector is orthogonal to both. Direct multiplication yieldsFor breathing excitation after a trap-frequency quench, put ${\omega }^{\prime }=\omega +\delta \omega$ and $x_0^{\prime }=x_0(\omega /{\omega }^{\prime }{)}^{2/3}$. The displacement from the new equilibrium is solely along the breathing vector and the initial velocity is zero. To first order in the fractional quench,The frequency shown is the linearized frequency about the new equilibrium. Nonlinear amplitude corrections are beyond this small-quench result.

Electromagnetic induction produces an electromotive force when magnetic flux linkage changes. For an oriented moving loop,For a stationary loop this reduces to Faraday's law of induction, $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. Motion in a static field can instead supply the motional term. Lenz's law gives the minus sign: the induced electric current opposes the flux change, with the required electrical energy supplied by mechanical work or the field source.

For linear fixed-geometry circuits, flux linkages satisfy ${\Lambda }_1=L_1{I}_1+M{I}_2$ and ${\Lambda }_2=M{I}_1+L_2{I}_2$. Self-inductance gives ${\mathcal{E}}_1=-L_1{\dot{I}}_1$; mutual inductance gives the additional term $-M{\dot{I}}_2$. Reciprocity gives the same mutual coefficient for both circuits, with its sign determined by winding orientation. The stored field energy isThe bound follows from nonnegative energy for arbitrary electric currents. In a linear medium the equivalent field expression is $W_B=\frac{1}{2}\int \mathbf{B}\cdot \mathbf{H}dV$.

An ideal electrical transformer links both windings with one changing core flux, giving $V_1/V_2=N_1/N_2$ for corresponding voltage magnitudes. A loaded secondary drives an opposing magnetomotive force and increases primary electric current; ideally $I_1/I_2=N_2/N_1$ and input and output powers agree. A electrical transformer exchanges voltage for electric current, not for extra power. Real devices have winding resistance, leakage flux, eddy-current and hysteresis losses, and finite magnetizing electric current. A steady DC input gives no continuing secondary voltage after the switching transient and may cause excessive primary electric current or core saturation.