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Anelasticity is a delayed but recoverable strain response caused by relaxation of internal degrees of freedom. After a small imposed stress is removed, the strain can return to its original value over a finite time rather than instantaneously. Under oscillatory loading, this delay causes a phase lag and dissipation, measured by a loss modulus or loss tangent. Stress-induced redistribution of interstitial impurities in a metal gives the Snoek effect. This solid-state response is distinct from an approximation to fluid mass conservation carrying a similar name.

For the convention $e^{i\omega t}$, complex viscosity is the ratio of oscillatory shear stress amplitude to shear rate amplitude. A causal relaxation modulus gives the displayed transform. Its associated complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, where $G^{\prime }$ is the storage modulus and $G^{\prime \prime }$ the loss modulus.

An experimental condensed matter physicist who, with T. G. Mason, developed optical microrheology of complex fluids. Dynamic light scattering measures the random displacement of probe particles; a generalized Langevin equation relates those fluctuations to the fluid's storage modulus and loss modulus. Comparing the result with mechanical measurements showed that microscopic motion could probe bulk viscoelasticity over a broad frequency range.^[ref]

For strain $\gamma ={\gamma }_0\cos \omega t$ and stress $\sigma ={\gamma }_0(G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t)$, integrating supplied power $\sigma \dot{\gamma }$ over one period gives the displayed dissipated energy per unit volume. The storage term integrates to zero; passivity requires nonnegative loss modulus for positive frequency in this convention.

The dynamic modulus is the complex stress-amplitude to strain-amplitude ratio in steady oscillatory response. With the convention $e^{i\omega t}$, the real part is the storage modulus and the imaginary part is the loss modulus. For $ϵ={ϵ}_0\cos \omega t$, $\sigma ={ϵ}_0(G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t)$. The average supplied power is $\omega {ϵ}_0^2{G}^{\prime \prime }/2$, fixing the loss-sign convention. The elastic coefficient is a shear or extension modulus according to the deformation measured; it is unrelated to the complex modulus meaning absolute value of a complex number.

The loss modulus is the quadrature coefficient $G^{\prime \prime }$ in oscillatory shear. It measures viscous energy dissipation.

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

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

The printed diagram and the word “parallel” describe a Kelvin-Voigt model, not a series Maxwell fluid. In the depicted model both elements have displacement $x$ and their forces add:For $x=a{e}^{i\omega t}$ and the convention $G(\omega )=F_{\mathrm{amplitude}}/x_{\mathrm{amplitude}}$,The storage modulus and loss modulus here are element stiffnesses with units $\mathrm{N}{\mathrm{m}}^{-1}$; conversion to stress/strain moduli requires a geometric factor. If “response” instead denotes displacement per force, it is the compliance $1/(k+i\omega \mu )$. Specifying which ratio is used removes that convention ambiguity. The mean dissipated power is $\mu {\omega }^2|a{|}^2/2$, confirming the positive loss sign for $e^{i\omega t}$.

For completeness, the actual series Maxwell fluid analogue has a common force and additive spring/dashpot extensions. Thus $\dot{x}=\dot{F}/k+F/\mu$. With $\tau =\mu /k$,These different functions cannot both describe the printed parallel arrangement. The distinction also changes the elastic beam's response to a constant load.

An objective time derivative transforms as a tensor under every superposed time-dependent rigid translation and rotation of the observer. A constitutive prediction using it is therefore frame-indifferent. The upper-convected derivative, lower-convected derivative, and Jaumann derivative are objective; the componentwise partial time derivative, and generally the uncorrected material derivative of a tensor, are not objective under rotating observers.

The Johnson--Segalman--Oldroyd model uses the Gordon--Schowalter derivativeHere $\eta$ is the zero-shear viscosity, ${\lambda }_1$ is the stress-relaxation time, and ${\lambda }_2$ is the retardation time associated with the more rapidly responding viscous contribution. Passive Oldroyd-type materials normally have $0\le {\lambda }_2\le {\lambda }_1$.

A low-rate steady-shear experiment measures $\eta$. A step strain followed by stress relaxation determines ${\lambda }_1$ from the exponential decay. Creep and recovery, or a frequency sweep in small-amplitude oscillatory shear, then separates ${\lambda }_2$ from ${\lambda }_1$; fitting both storage modulus and loss modulus is especially direct.

In small-amplitude oscillatory shear, expand the measured shear stress asComparison with the definitionof the storage modulus and loss modulus givesand henceThus the measured phase directly gives the loss tangent.

At low angular frequency, a standard viscoelastic fluid has time to relax and responds mainly as a viscous liquid: stress is approximately in phase with strain rate, so $\delta \to \pi /2$ as $\omega \to 0$. At high frequency it cannot relax during one cycle and responds mainly as an elastic solid, so stress is approximately in phase with strain and $\delta \to 0$ as $\omega \to \infty$.

For a material dominated by one viscoelastic relaxation time $\lambda$, the crossover between these regimes occurs when $\omega \lambda$ is of order one. Measure the frequency ${\omega }_c$ at which $G^{\prime }=G^{\prime \prime }$, equivalently $\delta =\pi /4$, and estimateA broad or multi-peaked crossover instead indicates a spectrum of relaxation times.

Here the complex viscoelastic modulus is a stress-to-strain response coefficient, rather than a complex number's absolute value. Use strain $\gamma (t)=\mathrm{Re}({\gamma }_0{e}^{i\omega t})$ and stress $\tau (t)=\mathrm{Re}[G^{\ast }(\omega ){\gamma }_0{e}^{i\omega t}]$, withThe storage modulus $G^{\prime }$ measures the elastic component; the loss modulus $G^{\prime \prime }$ measures dissipation. The mean power per volume is $\omega {\gamma }_0^2{G}^{\prime \prime }/2$, and the energy dissipated per cycle is $\pi {\gamma }_0^2{G}^{\prime \prime }$, fixing the sign convention.

A Hookean solid has $G^{\ast }=G_0$, while a Newtonian fluid has $G^{\ast }=i\omega \eta$. The Maxwell model interpolates with $G^{\ast }=G_{0}i\omega {\tau }_R/(1+i\omega {\tau }_R)$, giving $G^{\prime }=G_0(\omega {\tau }_R{)}^2/[1+(\omega {\tau }_R{)}^2]$ and $G^{\prime \prime }=G_0\omega {\tau }_R/[1+(\omega {\tau }_R{)}^2]$. It is fluid-like at low frequency and elastic on times short compared with relaxation time ${\tau }_R$.

A rheometer imposes small-amplitude oscillatory shear and resolves the torque amplitude and phase relative to the imposed motion. The torsional measurement of a complex shear modulus makes the conversion from torque and twist to $G^{\ast }$ explicit. A frequency sweep locates relaxation scales; keeping the strain small tests the linear-response assumption.

For the convention $\gamma =\mathrm{Re}({\gamma }_0{e}^{i\omega t})$, the complex viscoelastic modulus is $G^{\ast }=i\omega {\int }_0^{\infty }G(t)e^{-i\omega t}dt$. ThusThe storage modulus exceeds the loss modulus for $\omega \tau >1$, so the response is predominantly solid-like above the crossoverAt low frequency $G^{\ast }\simeq i\omega \eta$; at high frequency $G^{\ast }\to G_0$.

A polymer melt has an observation-time-dependent response because chains need time to rearrange. Short compared with their terminal relaxation time, entangled chains constrain one another and can store elastic free energy. Long compared with that time, chains escape their original constraints by reptation and the uncrosslinked melt flows. A permanently crosslinked elastomer instead retains a long-time elastic modulus.

A single-time approximation is the Maxwell fluid: a spring of shear modulus $G$ and a viscous dashpot of viscosity $\eta$ in series. Their common stress and additive shear strains give $\dot{\gamma }=\dot{\sigma }/G+\sigma /\eta$, with relaxation time $\tau =\eta /G$. For sinusoidal deformation proportional to $e^{i\omega t}$, the storage modulus and loss modulus areAt low frequency $G^{\prime \prime }\simeq \eta \omega$ dominates; at high frequency within this coarse-grained model $G^{\prime }\simeq G$ dominates. Real melts have a spectrum of relaxation times and additional segmental or glassy regimes, so the single-time model does not describe every frequency. A step-strain experiment measures the decay of stress, whereas oscillatory rheometry separates stored energy from dissipated energy through the phase of the response.

Choose complex amplitudes proportional to $e^{i\omega t}$ and define $\tilde{\sigma }=G^{\ast }(\omega )\widetilde{ϵ}$. The Maxwell model givesRationalizing the denominator yields the storage modulus and loss modulus:For $\omega \tau \ll 1$, $G^{\prime \prime }\simeq \eta \omega$ dominates and $G^{\prime }\simeq G_0{\omega }^2{\tau }^2$ is small: the material has time to flow. For $\omega \tau \gg 1$, $G^{\prime }\to G_0$ and $G^{\prime \prime }\simeq G_0/(\omega \tau )$: the dashpot scarcely deforms during a cycle and the response is mainly elastic. At $\omega \tau =1$, $G^{\prime }=G^{\prime \prime }=G_0/2$, and $G^{\prime \prime }$ is maximal. The stress leads strain by $\delta =\arctan (G^{\prime \prime }/G^{\prime })$, ranging from nearly $\pi /2$ to nearly zero.

For a fixed strain amplitude ${ϵ}_a$, dissipation per cycle is $\pi G^{\prime \prime }{ϵ}_a^2$. A peak in $G^{\prime \prime }$ therefore concerns that normalization. At fixed stress amplitude ${\sigma }_a$, it is instead $\pi {\sigma }_a^2/(\eta \omega )$, so it has no finite-frequency maximum in this ideal model. Reversing the Fourier time convention reverses the imaginary-part sign, while the physical dissipated energy remains positive.

A viscoelastic material stores part of the deformation energy and dissipates part of it. In linear viscoelasticity, a small imposed shear strain produces a stress determined by its history, $\tau (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. A shear step probes stress relaxation; a stress step probes creep compliance. A purely elastic solid has persistent stress after a held strain, while a Newtonian fluid relaxes that stress and creeps continuously under constant stress.

For a Maxwell model, a spring of shear modulus $G_0$ and a dashpot of viscosity $\eta$ are in series:For the convention $\gamma \propto e^{i\omega t}$,The storage modulus dominates at short times or high frequencies; the loss modulus dominates at long times or low frequencies. The energy dissipated per cycle per volume is $\pi G^{\prime \prime }{\gamma }_0^2$. A Kelvin-Voigt model places spring and dashpot in parallel and instead gives bounded creep $J(t)=(1-e^{-G_{0}t/\eta })/G_0$: a simple illustration of a viscoelastic solid. Real materials can require a spectrum of relaxation times rather than one element.

A small-amplitude oscillatory rheometer measures stress amplitude and phase relative to imposed strain, giving $G^{\prime }=|{\tau }_0/{\gamma }_0|\cos \delta$ and $G^{\prime \prime }=|{\tau }_0/{\gamma }_0|\sin \delta$. First sweep strain amplitude to identify the linear range; control temperature, sample ageing, wall slip and instrument inertia. Frequency-dependent phase lag alone is insufficient if these apparatus effects are unresolved.

A viscoelastic material stores part of the work of deformation and dissipates part. In the linear regime its stress depends on the strain history, not just the instantaneous strain or shear strain rate. For simple shear, a causal relaxation modulus $G(t)$ gives $\sigma (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. A sudden strain then probes stress relaxation, while a constant imposed stress probes material creep; these are different experiments.

For steady sinusoidal strain use the convention $\gamma (t)={\gamma }_0\cos \omega t$ and $e^{i\omega t}$ complex amplitudes. The complex viscoelastic modulus isso $\sigma (t)={\gamma }_0[G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t]$. The storage modulus $G^{\prime }$ is the in-phase elastic component. The loss modulus $G^{\prime \prime }$ multiplies the component in phase with shear strain rate; stress leads strain by $\delta =atan2(G^{\prime \prime },G^{\prime })$. The loss tangent is $\tan \delta =G^{\prime \prime }/G^{\prime }$ when $G^{\prime }>0$.

Multiplying stress by shear strain rate and averaging gives $⟨P⟩/V=\frac{1}{2}\omega G^{\prime \prime }{\gamma }_0^2$. Hence the dissipated energy per oscillatory shear cycle isThe elastic contribution returns its work over the cycle. A passive material has nonnegative dissipation, which fixes the sign of the loss term in the chosen Fourier convention. This complex viscoelastic modulus is a response function, not the mathematical absolute value of a complex number.

A Maxwell model places a spring of shear modulus $G$ in series with a dashpot of viscosity $\eta$, so the strains add at common stress:The Maxwell fluid dynamic modulus isAt $\omega \tau \ll 1$ the response is predominantly viscous, $G^{\prime \prime }\simeq \eta \omega$ and $G^{\prime }\simeq G(\omega \tau {)}^2$. At $\omega \tau \gg 1$ it approaches the elastic plateau $G^{\prime }=G$. The two moduli cross, and $G^{\prime \prime }$ reaches its maximum, at $\omega \tau =1$. A Kelvin-Voigt model instead has the spring and dashpot in parallel, giving $G^{\ast }=G+i\eta \omega$ and no fluid-like zero-frequency terminal flow. Real materials often need a spectrum of relaxation times rather than one Maxwell time.

Oscillatory rheometry measures stress amplitude and phase while imposing a small strain. An oscillatory-shear amplitude sweep first identifies a linear regime; an oscillatory-shear frequency sweep then measures $G^{\prime }$ and $G^{\prime \prime }$. Repeat at controlled temperature, suppress evaporation, check wall slip and sample aging, and correct instrument inertia at high frequency. Frequency-dependent structure and a nonlinear response cannot be inferred reliably by fitting one spring and dashpot to one point.

The requested model is the standard linear solid model: a spring of modulus $E_0>0$ in parallel with a Maxwell branch consisting of a spring $E_1>0$ in series with a dashpot of viscosity $\eta >0$. Both parallel branches have the same strain; their stresses add. In the Maxwell branch the spring and dashpot have the same stress and their strains add. The equilibrium spring prevents indefinite flow at long times.

For harmonic strain $ϵ(t)=\mathrm{Re}[{ϵ}_0{e}^{i\omega t}]$, define the dynamic modulus by $\tilde{\sigma }=E^{\ast }(\omega )\widetilde{ϵ}$. The Maxwell branch has compliance $1/E_1+1/(i\omega \eta )$. Defining $\tau =\eta /E_1$ therefore givesThe same mechanical model can represent a shear modulus if the appropriate shear stresses and strains are used consistently.

Writing $E^{\ast }=E^{\prime }+i{E}^{\prime \prime }$, its storage modulus and loss modulus areWith this time convention the physical stress is ${ϵ}_0[E^{\prime }\cos \omega t-E^{\prime \prime }\sin \omega t]$ and leads the strain by the loss angle. The dissipated energy per unit volume per cycle is $\pi E^{\prime \prime }{ϵ}_0^2\ge 0$. Thus $E^{\prime }$ describes elastic storage and $E^{\prime \prime }$ describes dissipation.

At low frequency, $\omega \tau \ll 1$, the dashpot flows during a cycle and relaxes the Maxwell spring, leaving $E^{\prime }\to E_0$, $E^{\prime \prime }\sim E_1\omega \tau$. At high frequency, $\omega \tau \gg 1$, the dashpot hardly deforms within a cycle, so both springs carry elastic load: $E^{\prime }\to E_0+E_1$, while $E^{\prime \prime }\sim E_1/(\omega \tau )$. The loss modulus is maximal at $\omega \tau =1$, with peak value $E_1/2$.

The equivalent constitutive equation isAfter a sudden strain ${ϵ}_s$ is imposed and held, the stress relaxes as $\sigma (t)={ϵ}_s[E_0+E_1{e}^{-t/\tau }]$. It falls from the instantaneous modulus $E_0+E_1$ to the nonzero relaxed modulus $E_0$. This single-time relaxation is the simplest improvement over a purely elastic solid or a Maxwell fluid with zero long-time modulus.

The loss tangent is $\tan \delta =E^{\prime \prime }/E^{\prime }$. Its maximum need not coincide with the loss-modulus maximum: the peak frequencies in a standard linear solid give $\omega \tau =\sqrt{E_0/(E_0+E_1)}$ for the loss tangent. The model captures one viscoelastic crossover between two finite elastic plateaux, with dissipation concentrated near the relaxation frequency.

Oscillatory response. A material with viscoelasticity stores some deformation energy and dissipates some over a loading cycle. In small-amplitude oscillatory shear, represent the strain by $\gamma (t)=\mathrm{Re}({\gamma }_0{e}^{i\omega t})$. The shear-stress amplitude is related to the strain amplitude by the complex viscoelastic modulusHere $G_1=G^{\prime }$ is the storage modulus, the component in phase with strain, and $G_2=G^{\prime \prime }$ is the loss modulus, the dissipative quadrature component. Both have units of pressure. The stress leads the strain by an angle $\delta$, withFor this $e^{i\omega t}$ convention, a passive material has $G_2\ge 0$ at positive frequency. Integrating $\tau \dot{\gamma }$ over one period gives dissipated energy per unit volume $\pi {\gamma }_0^2{G}_2$, and the average input power per unit volume is $\omega {\gamma }_0^2{G}_2/2$. Thus the loss tangent compares the dissipative and elastic parts of the response. The frequency dependence reflects which internal relaxation processes can follow the imposed deformation.

Torsional measurement. Use $\rho$ for radial position inside the rod, reserving $r$ for its outer radius. A small angular displacement $\theta$ of one end relative to the fixed other end gives shear strainIn harmonic amplitude notation, the shear stress is $G^{\ast }\rho \Theta /\ell$. Its restoring torque is obtained by multiplying by the lever arm and integrating over the circular section:The factor $J={\int }_A{\rho }^{2}dA=\pi r^4/2$ is the polar second moment of area; it is not the disc's mass moment of inertia $I$. The torsional spring constant becomes the complex stiffness $K^{\ast }=J{G}^{\ast }/\ell$. Angular Newton's second law therefore gives the paper's compact representationIts precise interpretation is a torsional measurement of a complex shear modulus. At a specified frequency, with complex angular amplitude $\Theta$, the driven response obeysSetting the external torque to zero supplies the free-mode condition, whose damped frequencies are generally complex. A dissipative free oscillator does not execute a sustained nonzero oscillation at a real frequency without driving. In real time, a general frequency-dependent complex viscoelastic modulus represents a causal memory response, rather than a complex instantaneous force on a real displacement. The derivation assumes a uniform slender circular rod, small twist, negligible rod inertia relative to the disc, and no additional bending, slip or external damping.

Frequency regimes and sketches. A permanently crosslinked elastomer has a low-frequency rubbery plateau $G_R>0$. Slow chain rearrangements can follow the strain, so the network is relatively compliant but still elastic. At high frequency the relevant segmental motions are frozen, giving a much larger unrelaxed or glassy stiffness $G_U$. Between these regimes the motions lag behind the loading and dissipate strongly.

A standard linear solid model supplies a consistent illustrative sketch of this process. Put $x=\omega {\tau }_r$, where ${\tau }_r$ is the stress viscoelastic relaxation time, and $\Delta G=G_U-G_R>0$. ThensoAt very low frequency, $G_1\to G_R$, $G_2\sim \Delta Gx$ and $\tan \delta \to 0$. Through the relaxation region $G_1$ rises, $G_2$ has a maximum and $\tan \delta$ has a loss peak. At high frequency, $G_1\to G_U$, $G_2\sim \Delta G/x$ and $\tan \delta \to 0$ again. Real polymers usually have a spectrum of relaxation times, so these features are broader and can have further structure.

The two kinds of loss peak need not coincide. The peak frequencies in a standard linear solid areThe first follows by differentiating $x/(1+x^2)$; the second by differentiating $x/(G_R+G_U{x}^2)$. This is why the sketch marks the peaks separately. If the polymer is an uncrosslinked melt rather than a permanent network, there is an additional ultimate low-frequency terminal-flow regime, with $G_1\propto {\omega }^2$ and $G_2\propto \omega$; a permanent elastomer does not have that zero-frequency fluid limit.

Interstitial atoms in a metal. A suitable mechanism is the Snoek effect, a form of anelasticity. An interstitial defect distorts its local crystal structure anisotropically. Different site orientations become unequally favorable under an applied stress, so hopping atoms redistribute towards the preferred orientations and supply additional strain. The hopping time is thermally activated, approximately ${\tau }_a={\tau }_0{e}^{E_a/(k_{B}T)}$, with activation energy $E_a$.

At low frequency, $\omega {\tau }_a\ll 1$, the populations nearly equilibrate during each cycle. The defect contribution to strain is nearly in phase with stress, giving a relaxed compliance and little dissipation. At intermediate frequency, $\omega {\tau }_a\sim 1$, redistribution lags behind the stress and energy dissipation is largest. At high frequency, $\omega {\tau }_a\gg 1$, the populations are effectively frozen; the extra relaxation strain cannot develop, and the response approaches the stiffer unrelaxed elastic limit. For weak single-mode relaxation, the loss tangent is approximately proportional to $\omega {\tau }_a/[1+(\omega {\tau }_a{)}^2]$. The metal remains a solid in the low-frequency regime: the extra delayed strain is recoverable, not unrestricted viscous flow.