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Twice differentiating the proposed normal mode givesIt therefore solves the two-dimensional wave equation precisely whenThe fixed-edge boundary conditions require $z(x,0,t)=z(x,b,t)=0$. The first is automatic, and the second requires $\sin (k_{y}b)=0$. The nonzero normal modes are thereforeThe apparent $n=0$ choice gives identically zero displacement for this sine solution. There are no imposed end conditions in $x$, so $k_x$ is continuous rather than quantized by a second length. For a propagating mode it is real; the plus sign in the stated $\cos (\omega t-k_{x}x)$ convention describes propagation in $+x$. Each positive integer $n$ labels a transverse standing wave across the strip.

A mass $M$ on a spring of stiffness $K$, subject to linear viscous damping and a sinusoidal drive, obeys the forced harmonic oscillator equationHere $x$ is displacement, $2M\gamma$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, and $F_0$ is a force in newtons. A series RLC circuit is an electrical analogue, with $M\leftrightarrow L$, $2M\gamma \leftrightarrow R$, $K\leftrightarrow 1/C$ and $x\leftrightarrow q$.

For free motion set $F_0=0$ and solve the characteristic equation $s^2+2\gamma s+{\omega }_0^2=0$. The roots $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$ distinguish three cases:

The constants are set by initial position and velocity. The mechanical energy and its rate of change in free motion areFor weak damping, the cycle-averaged energy decays approximately as $e^{-2\gamma t}$. The quality factor is $Q={\omega }_0/(2\gamma )$; for $Q\gg 1$ it equals $2\pi$ times stored energy divided by the loss per cycle. Logarithmic decrement of successive free-decay peaks determines $\gamma$ experimentally.

For driven motion the complete answer is the decaying free transient plus a steady response at the driving angular frequency:At low frequency, $X\to F_0/K$ and the displacement is nearly in phase with the drive. At $\Omega ={\omega }_0$, the lag is $\pi /2$ and $X=F_0/(2M\gamma {\omega }_0)$. At high frequency, $X\simeq F_0/(M{\Omega }^2)$ and the lag tends to $\pi$. The displacement resonance maximum lies atonly when $\gamma <{\omega }_0/\sqrt{2}$. Thus an underdamped oscillator need not have a nonzero-frequency displacement resonance peak.

In the steady state the mean input power equals the mean dissipated power:This absorption peaks exactly at $\Omega ={\omega }_0$, rather than at ${\Omega }_r$. Its half-power points have bandwidth $\Delta \Omega =2\gamma$, so $Q={\omega }_0/\Delta \Omega$ for this linear model. For weak damping, a high quality factor therefore means a narrow strong resonance and a long free-decay time. Exactly zero damping at resonant driving has no finite steady response: a secular amplitude grows with time.

displacement current completes Ampère's law:In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$. Taking the divergence and using $\nabla \cdot \mathbf{D}={\rho }_{\mathrm{free}}$ gives $\nabla \cdot {\mathbf{J}}_{\mathrm{free}}+{\partial }_t{\rho }_{\mathrm{free}}=0$, exactly the continuity equation. Without the added term a time-dependent charge distribution would contradict the curl equation.

A charging parallel-plate capacitor illustrates the physics. For plate area $A$, $Q={ϵ}_{0}AE$, so the gap displacement current is $I_d={ϵ}_{0}A\dot{E}=\dot{Q}$, equal to the conduction current in its leads. An Ampère contour therefore gives the same magnetic circulation whether its spanning surface crosses the wire or the gap. No electrons have to cross an ideal vacuum gap. In matter, ${\partial }_t\mathbf{D}$ includes the polarization response as well as the vacuum term. Combined with Faraday induction, the displacement term permits source-free electromagnetic waves.

An Euler strut is a slender initially straight elastic column under axial compression. For a uniform bending stiffness $EI$ and a small transverse displacement $y(x)$, the second-order potential energy under a conservative compressive load $P$ isThe bending term is positive; shortening releases load potential. Stationarity yields $EI{y}^{(4)}+P{y}^{\prime \prime }=0$. Pinned ends impose $y(0)=y(L)=0$ and zero end bending moments, $y^{\prime \prime }(0)=y^{\prime \prime }(L)=0$. Substitution of $y=A\sin (n\pi x/L)$ givesThe straight configuration loses stability first in the longest admissible bending mode. The second moment of area $I=\int z^{2}dA$ is geometric, not a mass moment of inertia. Boundary conditions matter: a fixed-free column has effective length $2L$, while fixed-fixed ends have effective length $L/2$, changing the threshold by factors $1/4$ and $4$. Slenderness, elastic material response, alignment and the weak bending direction must be checked. Imperfections produce bending before the ideal threshold, and a short column can yield or crush first. Linear buckling theory does not predict an unlimited deflection or a complete postbuckling curve.

stress describes internal force per area. A cut with unit normal $\hat{\mathbf{n}}$ has traction $\mathbf{t}=\mathbf{\sigma }\hat{\mathbf{n}}$; its normal component is $\hat{\mathbf{n}}\cdot \mathbf{\sigma }\hat{\mathbf{n}}$ and its remainder is shear traction. force balance constrains the stress divergence, while ordinary local angular-momentum balance makes the Cauchy stress tensor symmetric in a medium without body couples.

strain describes deformation rather than force. For small displacement gradients,The antisymmetric displacement-gradient part is a rigid rotation to first order and contributes no infinitesimal strain. Engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$, so it must not be confused with the tensor component. stress has units of pressure; strain is dimensionless.

An isotropic linear elastic solid obeysTwo independent moduli are required. The shear modulus $G$ and bulk modulus $K=\lambda +2G/3$ must be positive for stability. Young's modulus and Poisson ratio satisfyFor a tensile specimen with free side faces, ${\sigma }_{xx}=F/A$ gives ${ϵ}_{xx}={\sigma }_{xx}/E$ and ${ϵ}_{yy}={ϵ}_{zz}=-\nu {\sigma }_{xx}/E$. Pure shear gives $\tau =G\gamma$; hydrostatic loading gives $\Delta V/V=-p/K$. The incompressible limit has $\nu \to 1/2$ at finite shear stiffness, not vanishing shear resistance. Crystals can be anisotropic and need a fourth-rank elastic stiffness tensor instead of the isotropic two-constant law.

elastic deformation is recoverable on unloading; it need not remain linear at large strain. Beyond yield, irreversible rearrangements such as dislocation motion cause plastic strain, often followed by hardening. Ductile samples can neck before fracture; brittle failure has much less plastic deformation. Engineering stress uses initial area, while true stress uses the changing instantaneous area. A decreasing engineering stress during necking therefore need not mean that local true stress decreases. stress concentrations, flaws and loading geometry can determine failure before a uniform-material strength is reached. Viscoelastic materials also show rate dependence, creep and stress relaxation; their losses cannot be described by a static Hooke law alone.

A tensile test should measure force and gauge-length extension independently, correct machine compliance, align the grips, and control strain rate and temperature. Unloading distinguishes recoverable strain from a residual plastic extension. Separate shear or hydrostatic tests determine moduli that one axial measurement does not fix. Compression specimens need friction and buckling checks. Work per volume is $\int {\sigma }_{ij}d{ϵ}_{ij}$; the recoverable part is elastic energy, while a dissipative cycle produces a loss rather than creating energy. The original loading/unloading curve below is a chosen constitutive sketch, not measured material data.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

Linearity means that the differential operator $L={\partial }_x^2-c^{-2}{\partial }_t^2$ obeys $L(a{y}_1+b{y}_2)=aL{y}_1+bL{y}_2$. Therefore any linear combination of homogeneous solutions is again a solution; associated initial or boundary data must be combined by the same rule.

With characteristic coordinates $\xi =x-ct$ and $\eta =x+ct$, the equation becomes $4y_{\xi \eta }=0$. Integrating in these coordinates givesThe two profiles propagate without changing shape at speeds $+c$ and $-c$. Equivalently write $y=P(t-x/c)+Q(t+x/c)$. At $x=0$, the data imply $P+Q=f$ and $-P^{\prime }+Q^{\prime }=0$. Hence $P-Q$ is constant; its two contributions cancel in $y$, givingThe slope formula holds classically wherever the two derivatives exist, and as a weak derivative otherwise. Since the string tension is $\rho c^2$, $\rho$ here must be mass per unit length. These are two-sided string data with zero slope. The stipulated boundary data specify incoming as well as outgoing waves; this is not a pulse generated on an everywhere initially resting infinite string. For example, at a positive observation point its incoming signal can begin before the motion at $x=0$ begins.

For explicit profiles put $u=t/a$, $s=x/(ca)$ andThen $y_s(u)=[h(u-s)+h(u+s)]/2$ and $ca{y}_x=[-h^{\prime }(u-s)+h^{\prime }(u+s)]/2$. At ordinary points, $h^{\prime }(v)=\pi \cos (\pi v)$ for $|v|<1$ and zero for $|v|>1$. Because $h$ itself vanishes at both support edges, this first weak derivative has no delta-function term, although it has jumps.

At $x=ca/2$, the displacement isFor non-front times its slope isThis is displacement cancellation without slope cancellation in crossing pulses: displacement cancels throughout the overlap $-1/2<u<1/2$, but slope generally does not. Cancellation at one spatial position for an interval of time does not imply cancellation of the spatial derivative there. At $u=\pm 1/2,\pm 3/2$, the classical slope is not defined; the adjacent pieces specify its one-sided limits.

At $x=ca$ the two windows meet without an open overlap. Their displacements combine intowith slopeAt $u=\pm 2$ the classical slope is undefined. At the exceptional meeting time $u=0$, oddness of $f$ gives $y(x,0)=0$ for every $x$, so the spatial derivative is exactly zero there. Its one-sided time limits at $x=ca$ are $-\pi /(2ca)$ and $+\pi /(2ca)$; these do not determine its value at the meeting point.

The original graphs distinguish continuous displacements, open-ended one-sided slope traces, and the zero slope at the isolated meeting point. The sharp temporal window is not differentiable at its endpoints, so the overall solution is piecewise classical and globally weak. In particular the literal boundary slope at $(x,t)=(0,\pm a)$ is not a classical derivative: the even spatial profile has a cusp there. The zero-slope boundary prescription is satisfied at regular times and in its almost-everywhere trace sense; it cannot be read as a globally smooth classical condition for this nonsmooth $f$.

Separation of variables replaces a compatible partial differential equation by ordinary differential eigenvalue equations. For example, with a thermally diffusive rod, $u_t=D{u}_{xx}$, try $u=X(x)T(t)$. Wherever division is valid,The two sides depend on independent variables, so each must be constant. Homogeneous end temperatures $u(0,t)=u(L,t)=0$ give $X_n=\sin (n\pi x/L)$ and ${\lambda }_n=(n\pi /L{)}^2$. The general initial-temperature profile is represented by superposing these separated modes:Each spatial mode decays at its own rate. For a fixed-end wave equation the same spatial eigenfunctions instead have oscillatory time factors $A_n\cos (cn\pi t/L)+B_n\sin (cn\pi t/L)$, with coefficients fixed by initial displacement and velocity. Similar constructions separate Laplace's equation in suitable coordinate systems.

A single product is usually only one mode, not the most general solution. Linear superposition and completeness of the spatial eigenfunctions are what extend the method to general allowed data. Nonhomogeneous boundary values can first be removed by a known stationary or particular solution. Separation is effective when the geometry, coefficients and boundary conditions admit a common separated eigenproblem; an arbitrary nonlinear equation or inseparable geometry need not do so. Self-adjoint Sturm-Liouville problems explain the weighted orthogonality and completeness in many physical applications.

Use a local horizontal plane, constant gravity and first order in Earth's angular speed. Let $x$ point along the initial horizontal shot, $y$ to the gunner's left and $z$ upward. At the North pole, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. Maximum unperturbed range at equal launch and landing heights requires a $45^{\circ }$ elevation, givingThe leading Coriolis acceleration is $\ddot{y}=-2\Omega v_x$, while vertical motion has no first-order Coriolis correction. With zero initial lateral displacement and velocity,Here $u=500\mathrm{m}{\mathrm{s}}^{-1}$, $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\Omega \simeq 7.2921\times 10^{-5}{\mathrm{s}}^{-1}$. The miss is to the right of the aiming direction, independent of its azimuth at the pole. The flight time is about $72.1\mathrm{s}$, so $\Omega t_f\simeq 0.0053\ll 1$. This polar projectile Coriolis deflection uses the usual flat, uniform-gravity approximation; planetary curvature and variation of gravity are beyond its accuracy. Using the polar gravity rather than $9.81$ changes the estimate by less than one percent. Earth's sidereal rotation period, rather than the solar day, sets $\Omega$.^[ref]

A tidal force measures the difference of gravitational acceleration across an extended object. A freely falling reference frame removes a spatially uniform acceleration, but not its gradient. For a distant point mass $M$ at distance $R$ and direction $\hat{\mathbf{n}}$ from the reference centre, a small internal displacement $\mathbf{x}$ givesThe tidal tensor has eigenvalues $2GM/R^3$ radially and $-GM/R^3$ in each transverse direction: radial stretching and transverse squeezing. Its trace vanishes in the source-free region, consistent with the gravitational Gauss law. An associated relative potential is ${\Phi }_{\mathrm{tide}}=-GM[3(\hat{\mathbf{n}}\cdot \mathbf{x}{)}^2-x^2]/(2R^3)$.

The scale is $GMl/R^3$ across size $l$, not the centre's absolute gravity $GM/R^2$. This distinction explains how a more distant massive object can exert a large force but a modest tide. Opposite-side stretching supports the elementary two-bulge picture; real ocean tides also involve rotation, basin geometry and fluid dynamics, so the static gradient alone does not predict coastal heights or timing. The same gradient stresses satellites and can disrupt a self-gravitating body when it competes with the body's own gravitational restoring acceleration.

A normal mode is a motion in which all displacement coordinates oscillate at one angular frequency with fixed amplitude ratios. For small conservative oscillations, the symmetric stiffness and mass matrices give the generalized eigenvalue problem for small oscillations, $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. For positive mass and a stable equilibrium, its mass-orthogonal eigenvectors span the motion. A superposition of these modes, with phases and amplitudes fixed by initial displacements and velocities, gives the general solution; different modes need not share a phase.

Use upward end displacements. The spring compressions are $x_0-x_1$ and $x_0-x_2$, while the beam's centre rises by $q=(x_1+x_2)/2$. The potential energy isStatic force balance gives $2k{x}_0=mg$, cancelling the linear terms and leavingFor a uniform beam, its central moment of inertia in the vertical plane is $m{l}^2/12$. To first order in the tilt, $\theta =(x_2-x_1)/l$. Its translational and rotational kinetic energies areThus $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. The normal modes of a rigid beam on two end springs have eigenvectorsThe first translates without tilt; the second rocks about a stationary centre. Ordinary frequencies are ${\omega }_s/(2\pi )$ and ${\omega }_a/(2\pi )$.

Assume the system had settled before the end load leaves, and put that load at end 1. Its extra static spring load is $Mg$ at that end and zero at the other. Therefore $x_1(0)=-Mg/k$, $x_2(0)=0$, and both initial velocities vanish because no impulse is imparted. Decomposing into the two normal modes givesOnly the symmetric mode contributes to the centre position:The datum $z_{\mathrm{eq}}$ is the unloaded beam's equilibrium height; no absolute height is supplied. The rocking mode is present even though it cancels out of this coordinate. These results use small tilt, $Mg/(kl)\ll 1$, ideal light vertical springs and negligible damping.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

Neglect ends and take the exterior as vacuum. With polar angle measured from the applied field, regularity inside and the far-field condition allowContinuity of potential gives $B=1-A$; continuity of normal electric displacement, with no free surface charge, gives ${ϵ}_{r}B=1+A$. Hence the transverse electrostatic field inside a dielectric cylinder is uniform, parallel to the applied field, with magnitudeThis cylindrical result differs from the spherical depolarization factor. It returns $E_0$ for ${ϵ}_r=1$ and tends to zero for large positive ${ϵ}_r$.

Polarization density $\mathbf{P}$ is electric dipole moment per unit volume. Displacing positive and negative bound charges leaves the bulk nearly neutral locally but can produce uncompensated charge where polarization varies or terminates. For a dielectric occupying a volume with outward unit normal $\mathbf{n}$,The divergence theorem makes their total sum zero for a body described entirely by bound dipoles. Uniform polarization gives no interior volume charge but generally gives opposite surface charges on different faces.

In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}=ϵ\mathbf{E}$. Gauss's law becomes $\nabla \cdot \mathbf{D}={\rho }_f$, separating free from bound charge. Across a static interface, tangential electric field is continuous, while the normal displacement jumps by the free surface charge. Polarization charges alter the internal field, as in the dielectric cylinder calculation. Molecular distortion and orientation contribute differently, and response can depend on temperature and frequency. Bound charge is a redistribution of material charge, not an additional free-charge reservoir.

Use a cold, unmagnetized, collisionless electron plasma with immobile ions. For an oscillating field proportional to $e^{-i\omega t}$, the electron displacement follows $-m_e{\omega }^2\mathbf{x}=-e\mathbf{E}$. Hence $\mathbf{P}=-n_{e}e\mathbf{x}=-n_e{e}^2\mathbf{E}/(m_e{\omega }^2)$ andThe induced polarization opposes the field because electron inertia changes the phase of its displacement. Maxwell's equations give transverse electromagnetic dispersionThus transverse waves propagate above the plasma frequency, while below it the wavenumber is imaginary and a semi-infinite lossless plasma reflects an incident wave with an evanescent interior field. In the cold electrostatic model, a longitudinal density oscillation instead has frequency ${\omega }_p$.

Collisions at rate $\nu$ give ${ϵ}_r=1-{\omega }_p^2/[\omega (\omega +i\nu )]$ for this time convention, introducing absorption. A background magnetic field makes the response anisotropic, and thermal motion supplies additional dispersion and static screening. The divergent zero-frequency cold collisionless formula should not be mistaken for a complete description of static Debye screening. The simple dielectric function is a dynamical electron-inertia response under stated assumptions.

A small displacement $\mathbf{u}$ in a homogeneous isotropic elastic solid obeys linear elasticity. With Lamé constants $\lambda ,\mu$ and mass density $\rho$, momentum balance givesA longitudinal plane wave has displacement parallel to its propagation direction and changes local volume. A transverse wave has perpendicular displacement and produces shear. Substitution givesAn ideal fluid has no static shear modulus and cannot support this bulk transverse elastic wave. A thin rod with free lateral surfaces instead supports an extensional long-wavelength mode with speed $\sqrt{E/\rho }$; its transverse contraction changes the effective stiffness, so it should not be identified with the unbounded-solid longitudinal mode.

Stress and displacement boundary conditions produce reflection, transmission, mode conversion and guided waves. Viscoelasticity and scattering can attenuate waves; finite geometry can make guided modes dispersive even when the bulk elastic constants are frequency independent. Wave speed depends on both material stiffness and the allowed deformation.

In electrostatics, the Maxwell equations in matter areHere ${\rho }_f$ is free charge density and $\mathbf{P}$ is polarization density. For a homogeneous isotropic linear dielectric, the constitutive equation is $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf{n}$ from medium 1 to medium 2, gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$. These are the electrostatic interface boundary conditions. When there is no free surface electric charge, the normal electric displacement field and tangential electric field are continuous; the normal electric field generally is not.

Neglect end fringing field of a capacitor. In either axial segment of the coaxial cylindrical capacitor, a cylindrical Gaussian surface gives $D_r={\lambda }_f/(2\pi r)$, where ${\lambda }_f$ is free electric charge per length. Integrating the radial electric field between the conductors givesThe liquid and air segments share the same conductor electric potential difference, so their capacitances add in parallel. Thus the partially filled coaxial capacitor hasHere the question's $ϵ$ is relative permittivity, and air's relative permittivity is approximated by one.

Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied voltage, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C{V}^2/2$. The upward dielectric insertion force at fixed voltage is consequentlyThe excess hydrostatic pressure $\rho gx$ acts over the annular area $A=\pi (b^2-a^2)$. Alternatively its gravitational potential energy is $\rho gA{x}^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives givesThis assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long capacitor that end effects are small. Differentiating gives the dielectric-rise voltmeter's sensitivity at a bias $V$:To obtain its numerical value in $\mu \mathrm{m}{\mathrm{V}}^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta =b/a$, isFor a concrete estimate, choose $a=1\mathrm{mm}$, $b=2\mathrm{mm}$, mass density $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, relative permittivity $ϵ=2$ and operating voltage $V=100\mathrm{V}$. These are illustrative choices, not supplied apparatus data. They giveThus this millimetre-scale example requires fine optical displacement measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.

The displacement is quadratic in voltage, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger relative permittivity increase sensitivity, but capillarity, leakage, dielectric breakdown, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws electric charge and leakage is possible. A high voltage can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.

Maxwell's displacement current is the electric-field contribution needed to make Ampère's law consistent with charge conservation. In macroscopic form,The displacement current density ${\mathbf{J}}_d={\partial }_t\mathbf{D}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Since $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$, it contains the vacuum field term and, in matter, the polarization-current contribution. It need not represent free charge moving through a material.

Taking the divergence gives $0=\nabla \cdot {\mathbf{J}}_f+{\partial }_t(\nabla \cdot \mathbf{D})$. Gauss's law, $\nabla \cdot \mathbf{D}={\rho }_f$, then yields the continuity equation ${\partial }_t{\rho }_f+\nabla \cdot {\mathbf{J}}_f=0$. Without the displacement term, Ampère's original law would incorrectly require every current distribution to be divergence-free even while charge accumulates.

A charging parallel-plate capacitor makes this concrete. Ignoring fringing, $D=Q/A$, so the displacement current across its dielectric gap is $I_d=A{\partial }_{t}D=dQ/dt$, equal to the conduction electric current in its wire. The integral lawtherefore gives the same answer whether the spanning surface cuts the wire or passes through the gap. Together with Faraday's law, this field term also permits self-consistent electromagnetic waves in vacuum, with speed $c=1/\sqrt{{\mu }_0{ϵ}_0}$.

For a cold, unmagnetized, collisionless plasma, take an equilibrium electron density $n_e$, electrons of mass $m_e$ and charge $-e$, and approximately stationary ions. Use fields proportional to $e^{-i\omega t}$. The electron displacement satisfiesIts electric polarization is $\mathbf{P}=-n_{e}e\mathbf{x}$, hence the cold collisionless plasma dielectric responseThe negative susceptibility arises from the inertial electron response, rather than a restoring force binding an electron to a particular atom.

A transverse electromagnetic wave then obeysFor $\omega >{\omega }_p$, the plasma supports propagation, with refractive index $n_r=\sqrt{1-{\omega }_p^2/{\omega }^2}<1$. The phase velocity is greater than $c$, while the group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$; their product is $c^2$, and no superluminal signal is implied. Below the plasma frequency, the bulk wavenumber is imaginary and the field is evanescent. A sufficiently thick plasma largely reflects such radiation in the lossless model.

Including an electron momentum-relaxation rate ${\nu }_c$ replaces the denominator by $\omega (\omega +i{\nu }_c)$:Its imaginary part describes absorption with the stated time convention. Finite electron temperature also produces plasma screening of slowly varying electrostatic fields, with ${\lambda }_D=\sqrt{{ϵ}_0{k}_B{T}_e/(n_e{e}^2)}$. That spatial, longitudinal screening problem is distinct from simply taking $\omega =0$ in the cold transverse-wave formula. Magnetic bias, ion motion and kinetic resonances require more general dielectric responses.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.