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For small displacement gradients write $F=I+H$. Taylor expansion around the stress-free reference state givesThe linear term vanishes because $DW(I)=0$. Drop the irrelevant constant and retain the quadratic term:The major symmetry $\boxed{C_{i\alpha j\beta }=C_{j\beta i\alpha }}$ follows from equality of mixed second derivatives. For a skew matrix $K$, stress covariance along $e^{tK}$ givesDifferentiation at zero yields $D\hat{T}(I)[K]=C[K]=0$. Thus $C_{i\alpha j\beta }{K}_{j\beta }=0$ for every skew matrix, which forces the input minor symmetry $C_{i\alpha j\beta }=C_{i\alpha \beta j}$. Combining it with the major symmetry gives the output minor symmetry as well:These identify reference and spatial indices at $F=I$. Decompose $H=e+K$ into symmetric and skew parts. Since $C[K]=0$ and the major symmetry also eliminates the other cross term,Thus infinitesimal rotations cost no energy in the quadratic approximation.

An antiplane displacement has the form $v(x_1,x_2,x_3)=(0,0,u(x_1,x_2))$, independent of the out-of-plane coordinate. Only $e_{13}=e_{31}=u_{x_1}/2$ and $e_{23}=e_{32}=u_{x_2}/2$ are nonzero, so its strain energy per unit out-of-plane thickness is $\mu |\nabla u{|}^2/2$.

The equilibrium minimizes total potential energyHere $f$ is the out-of-plane body-force component; the two crack faces have independent traces. For a variation $\eta$ with zero outer-boundary trace, stationarity givesIntegrating by parts first for interior test functions and then for variations free on either crack face yieldsThese are the weak equilibrium equation, prescribed displacement, and zero traction. The tip behavior is selected by finite energy; it is not an additional prescribed tip value.

For a differentiable crack path, the local quasistatic Griffith laws with constant positive toughness areThey express irreversibility, local stability, and energy balance during growth. A strictly subcritical release rate forces the crack to stop.

Fix a crack parameter $s$ and let $u(s,t)$ denote the equilibrium displacement with the prescribed time-dependent loads. The scalar equilibrium problem is linear and has a unique solution for its fixed Dirichlet data. Both body force and boundary displacement scale by $t$, soThe minimized potential energy is quadratic under this scaling: the strain term scales by $t^2$ and so does the load term $-f(t)u(s,t)$. Hence $E(s,t)=t^{2}E(s,1)$ andEquivalently the crack-tip coefficient scales by $t$ and its squared energy-release formula gives the same result.

Put $s_0=\ell (0)$. Finiteness and continuity of $G(s,1)$ make it bounded by some finite $M\ge 0$ on a neighborhood of $s_0$. Continuity of $\ell$ supplies a positive interval on which $\ell (t)$ stays in that neighborhood. Shorten it to $[0,t_0]$ so that $t_0^{2}M<G_c$. Then $G(\ell (t),t)\le t_0^{2}M<G_c$, and complementarity forces $\dot{\ell }(t)=0$ throughout. ThereforeNo differentiating an unknown crack path or assuming its length already constant is needed; its continuity first gives the local release-rate bound.

Use metric signature $(+,-,-,-)$ and units with propagation speed one. The scalar Euler-Lagrange equation givesWith the stated spatial Fourier convention, $-{\nabla }^2$ becomes $k^2$. Each transformed mode obeysHere $F$ and $V$ are the initial transformed field and velocity. Reality requires $F(-\mathbf{k})=F(\mathbf{k}{)}^{\ast }$ and the same for $V$; the Fourier convention places no extra $(2\pi {)}^3$ in these evolution formulas.

Including the interaction gives the two coupled real Klein-Gordon fields,Adding and subtracting diagonalizes them:Both squared masses are positive for $0<g<m^2$. For arbitrary initial data $F_i={\tilde{\varphi }}_i(0)$ and $V_i={\dot{\tilde{\varphi }}}_i(0)$, defineThen the general solution is ${\tilde{\varphi }}_1=(U_{+}+U_{-})/2$, ${\tilde{\varphi }}_2=(U_{+}-U_{-})/2$, with the corresponding reality conditions on initial data.

For the specified sinusoidal initial displacement, put ${\Omega }_{\pm }=\sqrt{q^2+m^2\mp g}$. Both normal fields initially have amplitude $A\sin (\mathbf{q}\cdot \mathbf{x})$ and zero time derivative. ConsequentlyWriting $\bar{\Omega }=({\Omega }_{-}+{\Omega }_{+})/2$ and $\delta \Omega =({\Omega }_{-}-{\Omega }_{+})/2$ gives ${\varphi }_1=A\sin (\mathbf{q}\cdot \mathbf{x})\cos \bar{\Omega }t\cos \delta \Omega t$ and ${\varphi }_2=A\sin (\mathbf{q}\cdot \mathbf{x})\sin \bar{\Omega }t\sin \delta \Omega t$. This is field-amplitude beating of coupled Klein-Gordon modes. The conserved energy includes the interaction term $-g{\varphi }_1{\varphi }_2$; separate uncoupled-field energies need not be conserved, and an envelope zero is not by itself exact complete energy transfer.

Put $h=\Delta x$ and suppose $L=(n+1)h$, so there are $n$ interior grid points $x_j=jh$, $j=1,\dots ,n$. Write $u_j(t)\simeq u(x_j,t)$. The fixed ends impose homogeneous Dirichlet boundary conditions,For a sufficiently smooth displacement field, the two Taylor series at fixed time givewhere all displayed derivatives on the right are evaluated at $(x_j,t)$. Adding cancels the odd terms, leaving the second-order central differenceSubstitute this finite difference into the wave equation and retain continuous time. The resulting method of lines gives coupled second-order ordinary differential equations,In particular, the first and last equations are ${\ddot{u}}_1=(c^2/h^2)(-2u_1+u_2)$ and ${\ddot{u}}_n=(c^2/h^2)(u_{n-1}-2u_n)$ for $n\ge 2$. With the interior displacement vector only,Equivalently, $A_{jj}=-2c^2/h^2$, $A_{j,j+1}=A_{j+1,j}=c^2/h^2$, and all other entries vanish. This entrywise definition also covers small $n$, including $A=(-2c^2/h^2)$ when $n=1$. The matrix is a symmetric matrix and a tridiagonal matrix. The spatial approximation has error $O(h^2)$ in Big O notation; time has not yet been discretized.

For a nonzero, time-independent vector $\hat{\mathbf{u}}$, substitute $\mathbf{u}(t)=\hat{\mathbf{u}}{e}^{i\omega t}$. Taking two time derivatives givesCancel the nonzero exponential to obtain the eigenvalue problemThus an eigenvector specifies a normal mode shape and its negative eigenvalue specifies the square of its angular frequency, $\omega =\sqrt{-\lambda }$. Real displacements are obtained by taking the real part, or by combining the two temporal solutions as $\hat{\mathbf{u}}(a\cos \omega t+b\sin \omega t)$.

The interior grid is $x=1,2,3$, with fixed-end displacements $u_0=u_4=0$. HenceUse the monic characteristic polynomial convention $p(\lambda )=\det (\lambda I-A)$, where $I$ is the identity matrix. Expansion of the determinant along the first row givesThe eigenvalues are the roots of a polynomial $p(\lambda )=0$. With the alternative determinant convention $\det (A-\lambda I)$, the characteristic polynomial is $-p(\lambda )$ because the matrix has odd size; its roots of a polynomial are unchanged.

Each supplied eigenvector gives relative displacements at $x=1,2,3$ in a standing wave. Insert zero at $x=0,4$ to describe the whole sampled string. A real normal mode is $u_j(t)=C{v}_j\cos (\omega t+\varphi )$: its shape stays fixed while its amplitude oscillates.

These semidiscrete fixed-string normal modes are samples of $\sin (k\pi x/4)$, with $k=3,2,1$ in the order listed. More precisely, their interior vectors are $v_j^{(k)}=2^{-1/2}\sin (k\pi j/4)$, which have unit Euclidean norm and form an orthonormal basis of ${\mathbb{R}}^3$. Their eigenvalues and angular frequencies obeyThus larger absolute value of a negative eigenvalue means faster oscillation, rather than faster decay. The numerical angular frequencies approximate the continuous fixed-string values $k\pi c/L$; the finite difference spatial approximation changes them. Only three independent interior normal modes exist on this grid, and any interior displacement or initial velocity vector can be expanded in this orthonormal basis.

Two sinusoidal signals are in phase quadrature when their phase difference is a quarter cycle, for example $\cos \varphi$ and $\sin \varphi$. Their time averages have zero correlation over a cycle. In a plane wave, this separates fields such as displacement and its time derivative, velocity.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

A position vector is the vector from a chosen origin to a point. Its time derivative is the velocity; differences of position vectors give displacement.

Potential energy assigns energy to a system's arrangement, so a change of arrangement accounts for the work done by a conservative force. For a time-independent potential energy $V(\mathbf{r})$,The gradient points toward increasing potential energy; the force points toward decreasing potential energy. Only differences matter: adding a constant to $V$ changes neither the force nor its work. This endpoint rule follows by substituting $-\nabla V$ into the line integral for work. ^[ref]

Near Earth's surface, stipulate constant gravitational acceleration $g=9.81m/{\mathrm{s}}^2$ and measure height $h$ upward from a chosen zero. Then gravitational potential energy is $V=mgh$. Raising a $2\mathrm{kg}$ mass by $0.5\mathrm{m}$ increases its gravitational potential energy by $9.81\mathrm{J}$, while gravity does $-9.81\mathrm{J}$ of work. These are computed model values, not measurements of a particular lift. An ideal spring obeying Hooke's law instead has $V=k{x}^2/2$, with displacement $x$ from its relaxed state and spring constant $k$: $k=100N/m$ and $x=0.2\mathrm{m}$ give $V=2\mathrm{J}$ relative to $x=0$. ^[ref]

A Michelson interferometer turns displacement into a count of interference fringes: a moving mirror changes both the outward and return paths. This lets a wavelength act as a ruler whose divisions need no engraving. Assume normal incidence, a fixed reference arm, stable refractive index $n$ along the moving arm, and fixed vacuum wavelength ${\lambda }_0$. Write $\lambda ={\lambda }_0/n$ for the wavelength in that medium and take $\Delta L$ positive away from the beam splitter. The change in optical path difference, the accumulated phase difference, and the signed number of complete interference fringe cycles areThe factor two comes from propagation geometry: moving the mirror by $\Delta L$ adds $\Delta L$ before reflection and $\Delta L$ afterward. One bright-to-bright cycle means mirror travel $\lambda /2$; bright-to-dark means $\lambda /4$, half a cycle. All lengths must use compatible units, and $\Delta N$ is dimensionless. A constant reflection phase difference changes the initial fringe position but cancels from this change formula.

For an original hypothetical example, suppose $\lambda =600\mathrm{nm}$ in the actual surrounding medium and a continuous scan gives $40.25$ cycles, including the final fractional cycle. Then $\Delta L=(600\mathrm{nm})(40.25)/2=12075\mathrm{nm}=12.075\mu \mathrm{m}$. With the wavelength treated as exact, a fringe-reading measurement uncertainty of $0.02$ cycle contributes $6\mathrm{nm}$ to the displacement measurement uncertainty. An uncertainty of $0.01\mathrm{rad}$ in a local phase difference instead corresponds to $600(0.01)/(4\pi )\mathrm{nm}\approx 0.477\mathrm{nm}$. This illustrates measurement resolution, not a complete claim of accuracy: wavelength calibration, air changes, mirror tilt and missed cycles can dominate.

To separate the oscillating readout from its optical coherence envelope, use the local modelHere $I$ is detected light intensity, $\bar{I}$ is the mean level, ${\varphi }_b$ is a fixed bias wave phase, and $\mathcal{V}$ is interferometric visibility. A fine scan of $D$ measures the carrier phase difference; a coarse scan can map the temporal optical coherence envelope through $\mathcal{V}(D)$. The envelope maximum locates matched delays only under the source and apparatus assumptions; it is not a bright fringe identifying an arbitrary mirror position. Even with known $\mathcal{V}$, one intensity reading leaves the cosine sign and whole-cycle ambiguities. Continuous motion of known direction, or an independently obtained pair of signals in phase quadrature, is needed to recover signed travel through reversals.

The figure's original control example stipulates equal arm light intensities, matching linear polarizations, and a Lorentzian optical spectrum giving $\mathcal{V}(D)=\exp (-|D|/L_c)$ with $L_c=1\mathrm{mm}$. In vacuum, arm mismatches $|L_2-L_1|=0.60\mathrm{mm}$ and $0.10\mathrm{mm}$ give $\mathcal{V}=e^{-1.2}\approx 0.301$ and $e^{-0.2}\approx 0.819$. Matching the arms improves the phase difference readout without changing its ${\lambda }_0/2$ travel per cycle. A fine mirror scan of one cycle changes $D$ by only $600\mathrm{nm}$ for the stipulated $600\mathrm{nm}$ source, much less than $L_c$. These are calculated values, not historical data.

A proposed measurement sequence follows from this model. Overlap the returning beams, match their linear polarizations, find a high-contrast delay and use a small calibrated mirror scan to determine the sign and fringe period. Read a fixed spatial sample of the interference fringes with a photodiode; record each arm separately and the dark signal, avoid detector saturation, and retain enough time resolution to count every cycle. Approach mechanical settings consistently and repeat the return scan to expose lost motion or drift. For very small changes, bias near phase quadrature, where $|dI/d\varphi |=\bar{I}\mathcal{V}$ is largest; an intensity extremum has zero first-order sensitivity. A known drive voltage alone does not provide measurement calibration for mirror travel: compare the actual travel with the fringe count using an independently known wavelength.

Control temperature, pressure, vibration and air currents because the instrument measures optical path difference, not mirror motion alone. For a fixed reference arm and fixed ${\lambda }_0$, the first-order change of the moving arm's round-trip path is $\Delta D\approx 2n\Delta L+2L\Delta n$. Thus an unrecognized $\Delta n$ mimics $\Delta L_{\mathrm{app}}=L\Delta n/n$. A hypothetical $L=0.10\mathrm{m}$, $n\approx 1$, and $\Delta n=10^{-7}$ give $10\mathrm{nm}$ apparent travel. Holding the mirror still while varying or monitoring the air conditions tests this confounder. Separate checks of alignment, wavelength stability and thermal expansion are necessary before sub-fringe measurement resolution becomes a defensible measurement uncertainty.

Proper motion describes the change in a celestial object's direction on the sky due to its motion relative to the observer. Its angular velocity differs from the annual stellar parallax signal caused by the observer's orbit.

Thus proper motion is an angular velocity, rather than a speed in kilometres per second. It describes the relative transverse motion of source and observer in a specified reference frame. A star moving towards us can have little proper motion, while a nearer star can show larger proper motion than a faster but more distant one. In the example above, the observed coordinate actually decreases over the first half-year, although $\mu$ is positive: the annual stellar parallax excursion temporarily outweighs the steady drift. A short observed displacement alone therefore need not estimate proper motion well.

Real catalogues give two proper motion components. For right ascension $\alpha$ and declination $\delta$, the angular velocities in perpendicular directions on the local sky areThe cosine factor accounts for the shorter east-west arc associated with a given right ascension increment away from the celestial equator. ESA's Gaia DR3 fields pmra and pmdec already contain these two components; pm is their total proper motion. Do not multiply pmra by another cosine. Components of $60$ and $80\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$, for example, give total proper motion $100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$. ^[ref].

For nearby stars in the ordinary nonrelativistic approximation, Euclidean distance $d$ converts total proper motion into transverse speed $v_{\perp }$:The small-angle approximation gives transverse speed as Euclidean distance times angular velocity in radians per unit time. With arcseconds and parsecs, the parsec convention turns their product into astronomical units per year. One astronomical unit per Julian year of $365.25$ days is approximately $4.74047\mathrm{km}{\mathrm{s}}^{-1}$. The coefficient is a unit conversion, rather than a property of a particular star.

Taking the model's $100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}=0.100\mathrm{arcsec}\mathrm{y}{\mathrm{r}}^{-1}$ as the total proper motion, with the other component zero, gives $v_{\perp }=4.74047\times 0.100\times 20\simeq 9.48\mathrm{km}{\mathrm{s}}^{-1}$. At the same total proper motion, Euclidean distances of $10$, $20$ and $100$ parsecs instead give transverse speeds of $4.74$, $9.48$ and $47.40\mathrm{km}{\mathrm{s}}^{-1}$, respectively. Their stellar parallax values are $100$, $50$ and $10\mathrm{mas}$. The same annual angular displacement can therefore describe quite different journeys through space.

The transverse speed still omits the line-of-sight velocity, which can be measured through the Doppler effect in spectral lines. Combining these measurements permits a three-dimensional velocity estimate in a consistent reference frame. ^[ref].

Historically, Edmond Halley's 1718 paper compared contemporary and ancient stellar directions and argued that some supposedly fixed stars had moved. Its title, "Considerations on the change of the latitudes of some of the principal fixt stars", captures the conceptual surprise: long-separated measurements could reveal motion invisible in an ordinary night's watching. Modern astrometry retains that insight, using repeated precise measurements to turn the apparently static sky into a record of changing positions. ^[ref]. ^[ref].

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

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

Two semi-infinite strings with common tension $T$ and a point mass $M$ at their join have continuous transverse displacement, $A+B=C$. With time factor $e^{i\omega t}$ and a right-moving incident factor $e^{-i{k}_{1}x}$, Newton's second law at the join gives $k_{1}T(A-B)=(k_{2}T+iM{\omega }^2)C$. Solving gives the displayed reflection coefficient and $t=C/A=2k_{1}T/[T(k_1+k_2)+iM{\omega }^2]$. Their power fractions obey $|r{|}^2+(k_2/k_1)|t{|}^2=1$. Reversing the time convention complex-conjugates the coefficients. The limits $M=0$ and $M\to \infty$ recover a massless join and a fixed boundary, respectively.

A restoring force acts toward an equilibrium when a body is displaced from it. In one dimension, taking displacement $x=0$ at equilibrium, this means $xF(x)<0$ for sufficiently small nonzero $x$. An ideal linear spring has $F(x)=-kx$ with positive spring constant $k$, producing simple harmonic motion for a constant mass without damping. More general restoring forces need not be linear.

A rigid body undergoes translation when every point has the same displacement at each time, so its orientation stays fixed. For a freely moving system of particles coupled only by internal springs, the common displacement is a zero-frequency normal mode: internal spring lengths and potential energy are unchanged. Its mode coordinate is $A+Bt$, representing constant centre-of-mass velocity, rather than an oscillation at nonzero frequency.

For unstressed central Hookean springs, the quadratic elastic energy is $\frac{1}{2}{\sum }_{ij}{k}_{ij}[{\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j){]}^2$, with unit edge direction ${\mathbf{n}}_{ij}$. A common translation gives zero displacement difference. An infinitesimal rotation ${\mathbf{u}}_i=\mathbf{\Omega }\times {\mathbf{r}}_i$ also gives zero extension because a vector is perpendicular to its cross product with $\mathbf{\Omega }$. Thus a noncollinear three-dimensional free framework has three translational and three rotational zero modes. Extra mechanisms can give additional zero modes; six is a lower bound until rigidity is established.