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For zero initial velocity and initial displacement $h(x)$ on $x\ge 0$ with $h(0)=0$, let $H$ be its odd extension to the real line. The displayed D'Alembert's formula solves the wave equation with $y(0,t)=0$, since $H(-ct)+H(ct)=0$. A pulse travelling toward the endpoint reflects with reversed displacement. For piecewise linear data the equation holds classically between moving corners and in the sense of weak derivatives globally.

On the whole line, the homogeneous one-dimensional wave equation with constant wave speed $c>0$ isHere $u$ is displacement in the string interpretation; $f$ is initial displacement and $g$ is initial velocity. Its two directions of travel giveThis is D'Alembert's formula, including both Cauchy data. The countertravelling decomposition of the one-dimensional wave equation identifies $F(x-ct)$ as the right-moving travelling wave and $G(x+ct)$ as the left-moving travelling wave. Their sum is permitted by the superposition principle. For $f\in C^2$ and $g\in C^1$, it is a classical solution. More generally, travelling waves with profiles $F$ and $G$ that are locally integrable functions solve the equation as distributions, with derivatives interpreted distributionally.^[ref]

The synthetic diagram uses the smooth bump function $b$ defined in D'Alembert's formula, with $f(x)=b(x)$ and $c=1$. Releasing this displacement from rest, $g=0$, gives $F=G=f/2$. Each emerging travelling wave has half the original peak. Launching the identical initial displacement solely to the right instead requires $g=-c{f}^{\prime }$ and gives $F=f$, $G=0$. At $t=1.5$, the released profiles have peaks $1/2$ at $x=\pm 1.5$, while the launched profile has peak $1$ at $x=1.5$. These are calculated model values. A purely left-moving launch requires $g=c{f}^{\prime }$. The initial velocity therefore selects the balance of directions; the initial shape alone does not.

An angular readout using a reflected light ray: rotating a mirror by $\theta$ rotates the reflected light ray by $2\theta$, giving spot displacement $x\simeq 2D\theta$ at distance $D$ for small angles.

By T. G. Mason and D. A. Weitz. Physical Review Letters 74, 1250–1253; published 13 February 1995. DOI: doi.org/10.1103/PhysRevLett.74.1250. The publisher's abstract reports dynamic light scattering to infer linear viscoelastic moduli from probe displacements and comparison with mechanical measurements. Publisher record and abstract. No uninspected apparatus parameters are inferred from that abstract.

A detector measures a signal, not a force. One proposed arrangement collects transmitted and scattered light onto a quadrant photodiode, using a normalized difference of opposite segments as a position-sensitive signal. A camera can supply an independent image and bead-height check. Near a chosen operating point, a voltage channel may obey $V-V_0=x/\beta$, but its linear range and sensitivity $\beta$ must be measured at that bead size, focus and collection geometry. An arbitrary scattered-light intensity is not automatically proportional to displacement.

A useful proposed control sequence is to record detector offsets with illumination blocked, calibrate image scale against known stage travel, and test the detector response with independently known bead-relative motion. Moving a free trapped bead's chamber is not by itself a known bead displacement: fluid drag and trap stiffness determine the shift. Use a separate immobilized bead or an independently imaged trajectory for the position response, then check whether the response transfers to the actual trapped bead. Repeat the calibration after changing trapping power or optics.

Measure the acquisition frequency response and choose a sample rate that resolves $f_c$, with sufficient signal bandwidth and control of aliasing. Compare thermal calibration of an optical trap with optical trap drag calibration, sweep small positive and negative flow speeds, inspect for multiple beads, and repeat at several wall distances. Check temperature and drift before interpreting a power-dependent stiffness: a changed fluid viscosity can change $f_c$ even at unchanged $k$. These are proposed diagnostics of the model, not procedures attributed to the 1970 or 1986 papers.

For a bead in a stationary viscous liquid, linear drag often dominates inertia on the observed timescale. Neglecting both inertia and the random thermal force, the deterministic force balance givesHere $\zeta >0$ is the drag coefficient and $\tau$ the optical trap relaxation time: the time to reduce the deterministic displacement by a factor $e$. The bead approaches equilibrium without oscillation in this overdamped model. The approximation requires times long compared with the inertial time $m/\zeta$ and, for separated timescales, $m/\zeta \ll \zeta /k$, where $m$ is the bead's mass. For a no-slip spherical bead of radius $a$ in an unbounded liquid of dynamic viscosity $\eta$, Stokes drag law gives $\zeta =6\pi \eta a$. Nearby walls change that drag coefficient.

For example, $a=0.5\mu \mathrm{m}$, $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ and $k=0.01pN/nm=10^{-5}N/m$ give $\zeta \approx 9.42\times 10^{-9}\mathrm{N}s/m$ and $\tau \approx 0.942\mathrm{ms}$. A displacement of $100\mathrm{nm}$ gives a restoring force of magnitude $1\mathrm{pN}$, and its deterministic displacement drops to $100/e\approx 36.8\mathrm{nm}$ after one optical trap relaxation time. These are predictions for the stated model, not measured trajectories.

The figure uses $q=x/\ell$, $\kappa =k/k_0$ and $s=t/(\zeta /k_0)$, with arbitrary positive reference length $\ell$ and stiffness $k_0$. Thus $U/(k_0{\ell }^2)=\kappa q^2/2$, $F/(k_0\ell )=-\kappa q$ and, from $q(0)=1$, $q(s)=e^{-\kappa s}$. At fixed drag, greater stiffness gives both a steeper potential energy well and faster relaxation. Thermal Brownian motion is omitted from these deterministic curves.

Optical tweezers hold and move a small particle using the force exerted by a tightly focused beam of light, usually from a laser. They can position transparent beads in a liquid and manipulate suitable biological cells. Attaching a molecule to a trapped bead turns the bead into a handle: a measured bead displacement, together with a calibrated spring constant, reveals the force exerted on the attachment. This makes microscopic manipulation and measurement possible without a mechanical tool touching the bead. Living biological cells were manipulated with infrared beams by Ashkin, Dziedzic and Yamane in 1987; Svoboda and Block used trapped beads to measure forces from individual molecules in 1994. ^{[ref] [ref]}.

Why can light pull a bead towards a focus? Light carries momentum. A transparent bead with a higher refractive index than the surrounding liquid bends transmitted light rays by refraction, changing their momentum; conservation of momentum requires an opposite transfer to the bead. In the geometrical optics picture, unequal illumination of a displaced bead produces a net restoring force towards the brighter region. For a particle much smaller than the wavelength, a physical optics description replaces that ray picture. The detailed force depends on particle size, refractive index, beam shape and wavelength; the qualitative attraction does not supply a universal formula for every particle.

The optical gradient force draws a suitable transparent bead towards high illumination, while the optical scattering force generally pushes along beam propagation. A stable three-dimensional optical trap needs a balance with a restoring force in each direction: transverse confinement alone does not ensure axial confinement. Strong focusing can make the backward axial gradient contribution large enough to balance the forward scattering contribution, with the equilibrium displaced from the focus. Higher refractive index alone is insufficient; absorbing particles, different index contrasts and different beam geometries can behave differently. ^[ref].

Historically, Ashkin's 1970 experiments accelerated and trapped particles using radiation pressure. The 1986 paper by Ashkin, Dziedzic, Bjorkholm and Chu demonstrated the single-beam gradient optical trap for transparent particles. These are distinct steps: the earlier work established optical manipulation, and the later work supplied the tightly focused single-beam arrangement. ^{[ref] [ref]}.

Near a stable equilibrium, a harmonic optical trap behaves like a spring; thermal calibration of an optical trap uses its Brownian motion to determine the effective spring constant. The later passage through a translating optical trap sections instead consider a moving localized potential energy, which need not remain harmonic throughout a passage.

For the damped harmonic oscillator, the mechanical energy is the sum of kinetic energy and spring potential energy:Thus $b{\dot{x}}^2$ is the power transferred out of the oscillator's mechanical energy. For $b>0$, the rate vanishes instantaneously at a turning point, despite nonzero stored potential energy; it is not a constant loss per unit time. In eddy-current damping the loss becomes Joule heating. ^[ref]

For an underdamped response, advancing one oscillation period multiplies both displacement and velocity by $e^{-\zeta {\omega }_0{T}_{\mathrm{d}}}$, so $E(t+T_{\mathrm{d}})=e^{-2\zeta {\omega }_0{T}_{\mathrm{d}}}E(t)$. This exact cycle relation does not imply that instantaneous energy is a pure exponential function throughout each cycle.

Take incidence from A into B and first specify pressure amplitude. For a plane acoustic wave traveling normally to the interface, the acoustic impedance is $Z=\rho c_s$. A right-moving wave has pressure amplitude $P=ZU$, where $U$ is its fluid-velocity amplitude; a left-moving reflected wave has $P=-ZU$.

Let $P_i,P_r,P_t$ denote incident, reflected, and transmitted pressure amplitudes. Continuity of pressure and normal velocity at the interface givesSolving the two linear equations gives the acoustic reflection and transmission at an interface coefficientsFor the same perfect gas, take a common adiabatic index $\gamma$. The adiabatic sound speed satisfies $c_s^2=\gamma p/\rho$. The equal equilibrium pressure therefore implies $Z_B/Z_A=\sqrt{{\rho }_B/{\rho }_A}=\sqrt{\delta }$, givingThe word amplitude is otherwise ambiguous. For fluid velocity, or displacement at the same frequency, the corresponding answer isThese are consistent because pressure and velocity amplitudes differ by the impedance. As an energy check, the time-averaged acoustic intensity is $|P{|}^2/(2Z)$, so its transmission coefficient is $\mathcal{T}=4\sqrt{\delta }/(1+\sqrt{\delta }{)}^2\le 1$, even when the transmitted pressure amplitude exceeds the incident one.

Equal mean molecular weight alone does not logically guarantee equal $\gamma$. If the gases have different values of the adiabatic index, replace $\sqrt{\delta }$ in these formulas by $\sqrt{{\gamma }_B\delta /{\gamma }_A}$; a density-only answer requires the common-$\gamma$ assumption.

For the work of a spatially gated myosin head, the mechanical force exerted by an attached head in the increasing-$x$ direction is $f(x)=-dU/dx=\kappa (x_0-x)$. It assists motion before $x_0$ and resists it afterwards. For one site encounter the expected work isIn the narrow binding-window limit, retain the capture probability $p_b=1-e^{-k_{on}\Delta x/v}$ but neglect the attachment displacement compared with $x_0$. The positive work is $p_b\kappa x_0^2/2$. After $x_0$, use $s=x-x_0$ to obtain the negative contribution $-p_b\kappa {\int }_0^{\infty }s{e}^{-k_{off}s/v}ds=-p_b\kappa (v/k_{off}{)}^2$. HenceAn independent energy calculation gives the same factor: binding near zero loads energy $\kappa x_0^2/2$, while an exponentially distributed post-threshold attachment time has $⟨t^2⟩=2/k_{off}^2$, leaving mean elastic energy $\kappa v^2/k_{off}^2$ to be lost upon detachment. The factor of two in that second moment is essential.

One site passes every distance $D$, so average work per distance, or power divided by $v$, givesThe force tends to $\kappa x_0^2/(2D)$ as $v\downarrow 0$ and changes sign at $v=k_{off}{x}_0/\sqrt{2}$ in this idealized model. Above that speed forced sliding is resisted on average. The formula uses the reset condition in b, as well as $\Delta x\ll x_0$; if reset fails, replace $p_b$ by the periodic capture probability to leading order in $\Delta x/x_0$. For arbitrary window width, the isolated-site work is exactly ${\int }_0^{\Delta x}(k_{on}/v)e^{-k_{on}x/v}[\kappa (x_0-x{)}^2/2-\kappa v^2/k_{off}^2]dx$. The loaded spring energy must be supplied by the binding/power-stroke chemistry; this model does not specify an ATP efficiency. Real force-dependent transition rates and energy budgets limit extrapolation of the toy force-velocity law.

In spectrin unfolding by force spectroscopy, the cantilever measures tension through its calibrated deflection, while displacement of the pulling point increases the separation of the protein attachments. The protein extension is obtained after allowing for cantilever compliance and any linker extension; it is not simply the commanded displacement.

A folded spectrin repeat stores much of its backbone in a compact domain. Between unfolding events, the already released chain is stretched: increasingly straight conformations have fewer accessible configurations, and the resisting entropic force rises nonlinearly. A peak is reached when a still-folded domain crosses an unfolding barrier on the timescale of the imposed loading. Its released backbone increases the available contour length abruptly. At nearly the same end-to-end distance the fractional extension $x/L$ is then smaller, so tension drops. Pulling resumes on the next, longer-chain elastic branch. Repeated release of domains produces the saw-tooth sequence.

The record is therefore evidence for sequential mechanical unfolding of modular structure, not repeated stretching of a fixed-length rod. For the nominal eight-domain construct, successive branches correspond to increasing numbers of unfolded repeats. The order and rupture forces can fluctuate because barrier crossing is thermally activated and depends on loading rate, structural heterogeneity and pulling geometry; equivalent repeats need not rupture at identical force.

The horizontal separation between worm-like-chain fits at the same force estimates the added contour length, whereas the vertical force drop at one extension reflects redistribution of tension. These are distinct measurements. A repeat increment near the specified $31.7\mathrm{nm}$ is consistent with repeated similar structural units. The gain is a net release relative to the folded span and attachment geometry; $31.7/106\mathrm{nm}$ is not automatically the universal contour length of a peptide residue.

During unloading, relaxation and possible refolding can restore compact domains. Hysteresis would indicate dissipative or kinetically delayed processes. Neither the peak force alone nor an arbitrary saw-tooth area directly equals an equilibrium unfolding free energy: one must separate chain elasticity, cantilever storage, irreversible loss and pulling-rate effects. The worm-like-chain branches describe released-chain elasticity, not the molecular transition state responsible for a rupture peak.

A conceptual single-molecule measurement follows a labelled bound molecule along an independently mapped DNA contour. Project fitted positions onto contour distance and estimate the mean-squared displacement at several lag times. For unbiased one-dimensional diffusion,Independent localization errors of variance ${\sigma }_{\mathrm{loc}}^2$ add an intercept $2{\sigma }_{\mathrm{loc}}^2$. Finite exposure adds motion blur, and drift, binding transitions and the finite DNA contour must be distinguished from free sliding. A slope fit with those limitations can estimate $D_1$; simply fitting all observed displacements to a noiseless line can bias it.

Retain the printed coefficient. The original PDF really prints $D_1=5\times 10^{10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, not a negative exponent. In sliding diffusivity with per-step dissociation, for a basepair step $a\simeq 0.34\mathrm{nm}$, an unbiased nearest-neighbour walk has step interval $a^2/(2D_1)$. If the dissociation probability per step is $P$, the mean number of steps is approximately $1/P$, givingFor the printed values, ${\tau }_{\mathrm{bound}}\simeq 1.16\times 10^{-24}\mathrm{s}$ and ${\ell }_{\mathrm{rms}}\simeq 0.34\mu \mathrm{m}$. The latter is a random-walk displacement, not the contour distance travelled in a million steps. The step interval is about $1.16\times 10^{-30}\mathrm{s}$, even shorter than the light-travel time across a basepair by many orders of magnitude. This is an unphysical extrapolation for a protein, not a meaningful experimentally measurable diffusivity.

Consequently no fluorescence camera could temporally resolve trajectories under the stated formal values, regardless of the spatial excursion's length. An actual corrected exponent cannot be inferred uniquely from the paper. If a physically plausible $D_1$ were supplied instead, resolving sliding would require exposures substantially shorter than ${\tau }_{\mathrm{bound}}$ and diffusion displacements larger than localization error, with enough detected photons. The dependence on $D_1$ is explicit above, so no guessed source correction is needed.

The Euler-Lagrange equations of the quadratic Lagrangian areThe mechanical mass matrix is the identity. A normal mode $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\varphi )$ therefore satisfies the eigenvalue problem $\mathsf{K}\mathbf{a}={\omega }^2\mathbf{a}$. Direct multiplication givesThese are normal modes of two spring-coupled equal pendulums. The corresponding normal mode coordinates $u={\theta }_1+{\theta }_2$ and $v={\theta }_1-{\theta }_2$ obey $\ddot{u}+u=0$ and $\ddot{v}+(1+2k)v=0$.

For the in-phase normal mode, the bobs undergo identical displacements, leaving their separation exactly $r$. The spring stays at its natural length and exerts no force, so this angular frequency is independent of $k$. The value $1$ is the individual simple pendulum angular frequency $\sqrt{g/l}$ in the chosen units.

Inelastic neutron scattering measures both momentum and energy exchanged with the crystal. Comparing incident and outgoing neutron wavevectors gives $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$, while their energy difference gives a created or annihilated phonon energy $\hslash \omega$. Crystal momentum conservation identifies the reduced phonon wavevector modulo a reciprocal-lattice vector. Varying scattering geometry maps dispersion curves and polarization-dependent intensities across a substantial part of the Brillouin zone.

Inelastic X-ray scattering provides a complementary energy- and momentum-resolved probe, particularly when small samples or high-pressure environments favor photon beams. Raman and infrared spectroscopy measure optical-frequency resonances subject to different symmetry selection rules. Photon momentum at these frequencies is usually small on the Brillouin-zone scale, so these methods mostly sample near the zone center rather than an arbitrary phonon wavevector. Brillouin scattering and ultrasonic measurements probe low-frequency acoustic motion and infer sound velocities or elastic constants.

Heat-capacity measurements constrain the thermally weighted vibrational density of states and its low-frequency behavior, but do not by themselves reconstruct a unique wavevector-resolved dispersion. Elastic diffraction and temperature-dependent displacement factors also constrain mean-square motion, while energy-resolved inelastic measurements distinguish individual excitations directly. A monatomic scalar chain has only the acoustic branch derived below; it should not be assigned an optical branch merely because Raman spectroscopy exists for more complicated crystals.

The right spring changes length by $u_{n+1}-u_n$ and pulls atom $n$ to the right with force $C(u_{n+1}-u_n)$. The left spring contributes $-C(u_n-u_{n-1})$. Adding these restoring forces and applying Newton's second law givesUniform displacement makes both spring extensions vanish, as translational invariance requires. Search for a normal mode $u_n=u_0{e}^{i(kna-\omega t)}$. Substitution yieldsOne representative first Brillouin zone is $-\pi /a\le k\le \pi /a$. This is one positive-frequency acoustic branch: near zero wavevector, $\omega \simeq a\sqrt{C/m}|k|$, so the sound speed is $v_s=a\sqrt{C/m}$. Opposite signs of $k$ label opposite propagation directions, not a second polarization branch.

The model assumes small displacements, harmonic nearest-neighbor forces, one scalar displacement per atom and uniform equilibrium spacing. It excludes damping, anharmonic expansion and transverse polarizations. In a three-dimensional monatomic crystal there would normally be three acoustic polarizations; that counting cannot be imported into this explicitly one-dimensional scalar model.

The in-plane equilibrium equations are ${\tau }_{xx,x}+{\tau }_{xy,y}=0$ and ${\tau }_{xy,x}+{\tau }_{yy,y}=0$. On a simply connected patch, each divergence-free row has a streamfunction: write ${\tau }_{xx}={\varphi }_y$, ${\tau }_{xy}=-{\varphi }_x=-{\psi }_y$, ${\tau }_{yy}={\psi }_x$. Symmetry gives ${\varphi }_x={\psi }_y$, so another potential exists with $A_y=\varphi$, $A_x=\psi$. HenceIf two such Airy stress functions give the same stress, their difference has zero Hessian and is therefore an arbitrary affine function $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed stress. The existence argument is local, or global on a simply connected domain; domains that are multiply connected regions require the associated force and moment period conditions for a single-valued global potential. Equilibrium alone should not silently erase that topological qualification.

For plane strain put $s=e_{xx}+e_{yy}$. The constitutive equation gives ${\tau }_{xx}+{\tau }_{yy}=2(\lambda +\mu )s$. ConsequentlyThe displacement Saint-Venant compatibility equations $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomesFor stable elastic moduli the coefficient is nonzero, so $\boxed{{\Delta }^{2}A=0}$. If $D$ is bounded and smooth and $A={\partial }_{n}A=0$ on its boundary, integrating twice by parts gives $0={\int }_{D}A{\Delta }^{2}AdV={\int }_D(\Delta A{)}^{2}dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the Dirichlet problem. The normal-derivative condition is what removes the extra fourth-order boundary terms.

For the pressurized annulus, substituting the radial constitutive equations into equilibrium gives $u^{\prime \prime }+u^{\prime }/r-u/r^2=0$, so $u=Cr+D/r$. The stresses areUsing gauge stresses relative to ambient, the boundaries are ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$. ThereforeandThe maximum occurs at the bore. Once the chosen yield criterion is reached, Hooke's law cannot describe the total strain alone: write total strain as elastic plus plastic strain and supplement the elastic stress law with a yield condition, a plastic flow rule and any required work hardening law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-stress difference and the quoted yield stress depends on whether the material criterion is Tresca or von Mises. This is the elastic stress concentration in a pressurized thick cylinder.

The tension term $T{w}_{xx}$ is the transverse restoring force from a nearly straight, tensioned rod. The bending term $-EI{w}_{xxxx}$ is the force from Euler-Bernoulli flexure, where $E$ is Young's modulus and $I$ the second moment of area. The right side is transverse inertia. In the displayed one-dimensional equation $\rho$ denotes linear mass density; if it denotes volumetric density instead, the inertial coefficient must include cross-sectional area.

For a Fourier mode, $w_{xx}=-k^{2}w$, $w_{xxxx}=k^{4}w$ and $w_{tt}=-{\omega }^{2}w$, givingFor $k>0$ the phase speed is $c_p=\omega /k=\sqrt{T/\rho +EI{k}^2/\rho }$, strictly increasing with $k$ when $EI>0$ and $T\ge 0$. Thus shorter wavelengths propagate faster in phase. The group velocity $c_g=(T+2EI{k}^2)/\sqrt{\rho (T+EI{k}^2)}$ is also increasing, so the statement is consistent with wave-packet transport.

At a simple support the displacement and bending moment vanish: $w=0$ and $EI{w}_{xx}=0$ at $x=0,L$. The eigenmodes are $\sin (n\pi x/L)$ for $n=1,2,\dots$. Setting $T=0$ in the dispersion relation therefore yieldsThe integer is positive for a nonzero mode; $n=0$ gives the identically zero sine function, and negative integers repeat the same modes. Bending stiffness makes the frequencies quadratic in mode number, unlike a purely tensioned string.

Decomposing a plane displacement into components parallel and perpendicular to its wavevector gives compressional and shear speedsThe incident polarization is longitudinal. Its reflected longitudinal wave reverses the vertical wavevector component while retaining its horizontal component. The converted shear wave has wavevector $k_s(\sin \beta ,-\cos \beta )$ perpendicular to polarization $(\cos \beta ,\sin \beta )$.

Every term must have the same phase along the rigid boundary. ThusChoose $\cos \beta >0$ so the reflected shear wave propagates downward into $y<0$. At $y=0$, both displacement components must vanish for every boundary phase:Solving this two-by-two system yieldsFor a stable isotropic material $0<c_s/c_p<1$. An incident wave directed toward the boundary has $0<\alpha <\pi$; choosing $\beta \in (-\pi /2,\pi /2)$ by the sine relation gives real reflected waves throughout this range and no zero denominator. In the usual convention restricting incident horizontal propagation to positive $x$, this is $0<\alpha \le \pi /2$. Grazing endpoints have no incoming normal energy flux and are limiting cases. There is no critical angle for incident compressional-to-shear conversion here, because the converted wave is slower; the opposite incident-wave type has different critical-angle behaviour. At normal incidence $\alpha =\pi /2$, $\beta =0$, $R_1=1$ and $R_2=0$, cancelling the incident displacement with the reversed longitudinal polarization. This is compressional-wave reflection from a rigid elastic boundary.

Write $e_{xx}=u_{x,x}$, $e_{yy}=u_{y,y}$ and $2e_{xy}=u_{x,y}+u_{y,x}$. Differentiating twice and cancelling the mixed partial derivatives givesA convenient set of six three-dimensional Saint-Venant compatibility equations isThey follow by inserting $e_{ij}=(u_{i,j}+u_{j,i})/2$ and commuting derivatives. They form the six symmetric components of the incompatibility tensor $C_{ij}={ϵ}_{ik\ell }{ϵ}_{jmn}{e}_{\ell n,km}$, up to the indicated signs and permutation of components.

They obey the three differential identities ${\partial }_i{C}_{ij}=0$, because a symmetric pair of derivatives contracts with an antisymmetric Levi-Civita symbol. Thus there are three independent differential constraints, represented symmetrically by six component equations; they are not six unrelated constraints on the six strain functions. For example, at a nonzero wavevector directed along the third axis, compatibility imposes $e_{11}=e_{22}=e_{12}=0$ while $e_{13},e_{23},e_{33}$ can arise from the three displacement components. This confirms rank three. One cannot simply discard the same arbitrary three component equations in every direction; their independence is differential, with the stated identities. On a simply connected domain these are the local Saint-Venant compatibility equations for strain to be a symmetric displacement gradient, while domains that are multiply connected regions require global closure conditions as in the screw example. This is Saint-Venant strain compatibility and differential identities.

For antiplane displacement $u=(0,0,w)$ the only nonzero strain components are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$. Commutation of mixed partial derivatives gives the necessary local compatibility conditionThe supplied stress is symmetric, independent of $z$ and trace-free. Its first two equilibrium equations are automatic; its third givesaway from the origin. Antiplane shear has zero trace, so isotropic Hooke's law reduces to $\boxed{e=\tau /(2\mu )}$. HenceLocally this integrates to $w=\arg (x+iy)+C$, represented on a suitable branch by $\arctan (y/x)+C$. The strain condition is satisfied at every point of the punctured plane, but the displacement cannot be made globally single-valued there:A gradient of a globally single-valued smooth displacement would instead integrate to zero around any closed curve. Local compatibility is therefore not sufficient on this multiply connected region.

A physical construction is a Volterra screw dislocation: cut the cylinder along a radial half-plane ending on its axis, shift one cut face relative to the other parallel to the axis, and rejoin. The axial mismatch is the Burgers vector. In the given normalization its magnitude is $2\pi$; with a dimensional mismatch $b$ one uses $w=b\arg (x+iy)/(2\pi )$. An intact pristine solid cannot reach this state by a smooth single-valued elastic deformation alone. The axis is a dislocation core where the continuum elastic field is singular and must be regularized by microscopic structure or plastic deformation. On the punctured domain the differential condition holds locally; globally its nonzero circulation detects the defect. In a distributional description, that incompatibility is concentrated on the axis. This is local strain compatibility versus screw-dislocation circulation.