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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.

For elements in series the stress is common and the strains add. If ${\widehat{ϵ}}_K$ and ${\widehat{ϵ}}_1$ are the Kelvin element and added spring amplitudes, thenThe compliances therefore add, yieldingThis is a standard linear solid model. DefineWriting $x=\omega \tau$, its storage modulus and loss modulus areThus $G^{\prime }$ increases from the low-frequency plateau $G_L$ to the high-frequency plateau $G_H$. The loss modulus has low-frequency slope $+1$ and high-frequency slope $-1$ on a log-log plot, with maximum $\Delta G/2$ at $\omega =1/\tau$. At that frequency $G^{\prime }=(G_L+G_H)/2$. There need not be a crossing of the two curves for arbitrary spring constants.

A Kelvin-Voigt model has no finite instantaneous elastic compliance: the viscous element prevents a sudden deformation under a finite stress step. Its creep compliance isConversely an ideal strain step requires an impulsive stress,After the impulse the held-strain response is simply constant. There is no finite stress-decay interval, and the loss modulus grows without bound as $\omega \to \infty$. This fails to represent a material with both a finite instantaneous stiffness and delayed stress relaxation.

The added series spring resolves this particular defect. At the instant of loading the Kelvin element cannot move but the added spring can. The standard linear solid model consequently hasandIts instantaneous stiffness is $G_H=G_1$, its long-time stiffness is $G_L$, and it exhibits genuine stress relaxation as well as delayed creep. The creep retardation time ${\eta }_0/G_0$ differs from the stress-relaxation time $\tau ={\eta }_0/(G_0+G_1)$. These properties also follow directly from its constitutive equation,It remains an idealized solid with one relaxation mode: it cannot describe unrestricted long-time viscous flow, nonlinear response, or a broad spectrum of relaxation times.

For a standard linear solid model with $0<G_R<G_U$, put $x=\omega \tau$. Its storage modulus is $G^{\prime }=G_R+(G_U-G_R)x^2/(1+x^2)$ and its loss modulus is $G^{\prime \prime }=(G_U-G_R)x/(1+x^2)$. The latter peaks at $x=1$, whereas its loss tangent isDifferentiation puts the loss-tangent peak at $x=\sqrt{G_R/G_U}$, with height $(G_U-G_R)/(2\sqrt{G_R{G}_U})$. A loss-modulus peak and a loss-tangent peak therefore need not occur at the same frequency. For weak relaxation, $G_U/G_R$ is close to one and their frequencies are approximately equal.

A standard linear solid model has a finite instantaneous stiffness, a nonzero long-time stiffness and one exponential stress-relaxation mode. A Kelvin-Voigt model with spring $G_0$ and viscosity $\eta$ in series with spring $G_1$ givesThese expressions follow by adding series strains under common stress. Its storage modulus is $G_L+(G_H-G_L)x^2/(1+x^2)$ and its loss modulus is $(G_H-G_L)x/(1+x^2)$ for $x=\omega \tau$. Its relaxation modulus is $G(t)=G_L+(G_H-G_L)e^{-t/\tau }$, while its creep compliance is $J(t)=G_1^{-1}+G_0^{-1}(1-e^{-G_{0}t/\eta })$. Thus creep retardation and stress-relaxation times differ. This solid has bounded creep and cannot represent terminal fluid flow.