= Solution
\b[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 $\gamma(t)=\operatorname{Re}(\gamma_0e^{i\omega t})$. The shear-stress amplitude is related to the strain amplitude by the <complex viscoelastic modulus>
$$
\boxed{G^*(\omega)=G_1(\omega)+iG_2(\omega)},\qquad
\tau(t)=\gamma_0\bigl[G_1\cos\omega t-G_2\sin\omega t\bigr].
$$
Here $G_1=G'$ is the <storage modulus>, the component in phase with strain, and $G_2=G''$ is the <loss modulus>, the dissipative quadrature component. Both have units of <pressure>. The stress leads the strain by an angle $\delta$, with
$$
\boxed{\tan\delta=\frac{G_2}{G_1}}.
$$
For this $e^{i\omega t}$ convention, a passive material has $G_2\geq0$ at positive frequency. Integrating $\tau\dot\gamma$ over one period gives dissipated energy per unit volume $\pi\gamma_0^2G_2$, and the average input power per unit volume is $\omega\gamma_0^2G_2/2$. 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.
\b[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 strain
$$
\gamma(\rho)=\frac{\rho\theta}{\ell}.
$$
In harmonic amplitude notation, the shear stress is $G^*\rho\Theta/\ell$. Its restoring <torque> is obtained by multiplying by the lever arm and integrating over the circular section:
$$
\mathcal T=-\int_0^r \rho\frac{G^*\rho\Theta}{\ell}\,2\pi\rho\,d\rho
=-\frac{\pi r^4}{2\ell}G^*\Theta.
$$
The factor $J=\int_A\rho^2dA=\pi r^4/2$ 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 $K^*=JG^*/\ell$. Angular <Newton's second law> therefore gives the paper's compact representation
$$
\boxed{I\ddot\theta+\frac{\pi r^4}{2\ell}(G_1+iG_2)\theta=0}.
$$
Its precise interpretation is a <torsional measurement of a complex shear modulus>. At a specified frequency, with complex angular amplitude $\Theta$, the driven response obeys
$$
\left[-I\omega^2+\frac{\pi r^4}{2\ell}G^*(\omega)\right]\Theta=\mathcal T_{\mathrm{external}}.
$$
Setting 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.
\b[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 $\Delta G=G_U-G_R>0$. Then
$$
G^*=G_R+\Delta G\frac{ix}{1+ix},\qquad
\boxed{G_1=G_R+\Delta G\frac{x^2}{1+x^2},\quad
G_2=\Delta G\frac{x}{1+x^2}},
$$
so
$$
\boxed{\tan\delta=\frac{\Delta Gx}{G_R+G_Ux^2}}.
$$
At very low frequency, $G_1\to G_R$, $G_2\sim\Delta Gx$ and $\tan\delta\to0$. 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 $\tan\delta\to0$ again. Real polymers usually have a spectrum of relaxation times, so these features are broader and can have further structure.
\Image[/past-exam-of-the-soft-condensed-matter-course-of-the-university-of-cambridge/2018-scm-viscoelastic-spectrum.png]
{title=Storage modulus, loss modulus and loss tangent across a viscoelastic relaxation}
{description=Original standard-linear-solid illustration with a nonzero rubbery modulus and a larger unrelaxed modulus; frequencies and moduli are normalized, not fitted to a particular polymer.}
The two kinds of loss peak need not coincide. The <peak frequencies in a standard linear solid> are
$$
\boxed{\omega_{G_2}\tau_r=1,\qquad
\omega_{\tan\delta}\tau_r=\sqrt{G_R/G_U}}.
$$
The first follows by differentiating $x/(1+x^2)$; the second by differentiating $x/(G_R+G_Ux^2)$. 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.
\b[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 $\tau_a=\tau_0e^{E_a/(k_BT)}$, with <activation energy> $E_a$.
At low frequency, $\omega\tau_a\ll1$, 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, $\omega\tau_a\sim1$, redistribution lags behind the stress and energy dissipation is largest. At high frequency, $\omega\tau_a\gg1$, 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.
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