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Authors: Francis Nimmo, Ulrich H. Faul, Edward J. Garnero.

Journal of Geophysical Research 117, E09005. Tests solid-state anelasticity against lunar dissipation and Love numbers; mantle melt is not required by every constraint, but the model has a tidal frequency-dependence discrepancy.

I would not assign the lunar mare an automatic terrestrial 100–125 km compensation level. For present-day mare basins, a local equal-column-mass model may have no physically valid compensation depth; first test whether lithospheric flexure supports the load. For ancient thermally driven subsidence that did relax, choose a common datum beneath the cooling layer in deformable lunar mantle appropriate to that epoch.

The Moon is smaller than Earth and has lost a larger fraction of its original heat. Its cold, mechanically strong outer shell can be much thicker relative to its radius than the terrestrial oceanic lithosphere. Lunar mare basalts filled pre-existing impact basins: their loading, basin excavation, mantle uplift, cooling and the evolving strength of the lithosphere all matter. A thick elastic lithosphere can support lateral stress differences, so equal column mass need not hold even after a long time. The strong gravity anomalies of lunar mascons, measured by GRAIL, are evidence that the present mare-basin system is not everywhere locally compensated.^[ref]

For an explicitly assumed ancient relaxed thermal model, estimate its thermal penetration depth from thermal diffusivity and the elapsed cooling time, infer the contemporaneous mechanical lid from gravity-topography admittance or basin deformation, and place $z_c$ below that layer where lateral pressure anomalies relax. For example, a hypothetical several-hundred-kilometre-thick lid requires a deeper datum than that lid, not the shallow crust-mantle boundary. The paper supplies neither an epoch-specific thermal model nor a load wavelength, so a unique kilometre value is not determined. If the chosen datum is a substantial fraction of the lunar radius, spherical geometry and radial variation in gravity must replace the flat, constant-gravity column approximation.

Does the Moon have an asthenosphere? Not a demonstrated shallow, globally connected Earth-like layer underlying mobile plates; but “no weak or partially molten mantle anywhere, ever” is too strong. Apollo lunar seismology indicates a cold outer interior and more attenuating deep regions. Reanalyses support a deep low-velocity or partially molten region near the lunar core, while lunar laser ranging and tidal Love numbers constrain deep deformability and tidal dissipation. Such a deep weak region is sometimes called a lunar asthenosphere; it is not automatically the correct local compensation surface for a mare load.^[ref]

The inferred melt fraction and distribution remain model dependent: Dissipation at tidal and seismic frequencies in a melt-free Moon by Francis Nimmo and others demonstrates that much of the dissipation can also be explained by hot solid-state anelasticity, although its model does not reproduce every frequency-dependent constraint. The Moon was hotter during mare emplacement and could relax loads differently then. Choose the datum from the relevant thermal and mechanical history, or use a flexural model when local isostasy fails.

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.

The Snoek effect is anelasticity caused by stress-induced redistribution of mobile interstitial defects among differently oriented sites in a crystal. An applied stress favors orientations whose local distortion best accommodates it. At low frequency the atoms can track that bias and provide additional relaxed compliance; at high frequency their populations are nearly frozen. Dissipation is largest between these regimes. For a weak single relaxation, the loss tangent has the approximate shape $\Delta \omega \tau /[1+(\omega \tau {)}^2]$, with $\tau ={\tau }_0{e}^{E_a/(k_{B}T)}$ for thermally activated hopping. The peak is near $\omega \tau =1$. This describes a standard mechanism for interstitial impurities in body-centred cubic metals, not every damping mechanism in every alloy. ^[ref]

Ulrich H. Faul coauthored lunar models using a laboratory-calibrated description of anelasticity in melt-free olivine. This connects small-scale mineral deformation with measurable planetary energy loss.^[ref]

Such a comparison tests whether solid-state relaxation can explain observations before an additional molten layer is required.