---
title: "Material memory turns a tidal cycle into heat"
description: "Derive Maxwell’s complex shear modulus, the sign of passive loss and the work per cycle, with a saved dimensional example."
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language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/ocean-worlds/from-a-maxwell-material-to-tidal-lag/"
last_modified: "2026-10-28"
---

[Ocean worlds from material memory to tidal heat](https://sandboxphysics.com/stories/ocean-worlds/)

AC12 · L1

# Material memory turns a tidal cycle into heat

Sandbox Physics · 2026-10-28

Viscoelastic heating comes from the timing between stress and strain. A spring and a viscous element in series give a model that is simple enough to solve, yet already distinguishes stored energy from irreversible work. We will derive its harmonic response before putting it into a self-gravitating world.

## Before you begin

Small-strain elasticity, a time derivative, elementary complex numbers and the mean of a sinusoid squared.

## Define a shear experiment before introducing complex numbers

Use a small simple shear. Engineering shear strain is the relative sideways displacement of two faces divided by their separation. It is dimensionless. Shear stress is force per area, measured in pascals. An elastic element obeys stress equal to shear modulus times engineering strain. A Newtonian viscous element obeys stress equal to viscosity times strain rate. Its viscosity has units pascal-seconds.

In the Maxwell arrangement the two elements are in series. They carry the same stress, while their strains add. Differentiate the elastic strain, add the viscous strain rate, and obtain the first equation. The characteristic time is viscosity divided by elastic modulus. A short load compared with this time looks mostly elastic; over a much longer load the viscous element can flow.

This is a linear small-strain model with one relaxation time. Ice in a real moon can have depth-dependent temperature, grain size and several relaxation mechanisms. The model is useful because each prediction follows from a stated material law; assigning it a plausible viscosity does not make it a complete ice rheology.

$$
\dot\gamma=\frac{\dot\sigma}{\mu_E}+\frac{\sigma}{\eta},\qquad \tau_M=\frac{\eta}{\mu_E}
$$

Engineering shear strain is used here. In tensor notation the shear stress is twice the modulus times the shear-strain tensor component.

Reading: Beuthe, Tides on Europa: the membrane paradigm, arXiv:1410.4735v1

## Solve one oscillation and fix its sign

Represent a real sinusoid as the real part of a complex amplitude multiplied by a positive imaginary exponential. Differentiation then becomes multiplication by imaginary angular frequency. Insert that rule in the material equation and solve the stress-to-strain ratio. Multiplying numerator and denominator by the complex conjugate separates the storage and loss terms.

Under this convention the imaginary part of the modulus is positive for a passive material. For imposed cosine strain, stress contains a negative sine term. Multiply stress by strain rate and average over a cycle. The cosine-sine cross term vanishes, while sine squared averages to one-half. The resulting mean work density is positive.

An inverse response has the opposite imaginary sign. When force is specified and displacement is the response, the compliant body lags the force and its response has negative imaginary part. This explains why our Love-number plots use minus the imaginary part as a positive loss magnitude. Changing the exponential convention reverses both signs without changing the physical heat.

$$
\mu^*(\omega)=\mu_E\frac{i\omega\tau_M}{1+i\omega\tau_M}=\mu_E\frac{x^2+i x}{1+x^2},\qquad x=\omega\tau_M
$$

The real part stores energy; the positive imaginary part dissipates it under the declared time convention.

$$
\left\langle\sigma\dot\gamma\right\rangle=\frac{\omega}{2}\operatorname{Im}\mu^*\,|\widehat\gamma|^2\geq0
$$

This expression is for a single engineering-shear component, per unit volume.

Reading: Beuthe, Tides on Europa: the membrane paradigm, arXiv:1410.4735v1

## A dimensional worked example

Take elastic modulus 3.5 gigapascals and viscosity approximately 316 trillion pascal-seconds. Their ratio is about 90,351 seconds, or 25.1 hours. The synthetic world’s orbital period is 178,720 seconds, or 49.6 hours, so its angular frequency times Maxwell time is about 3.18. Angular frequency equals a full turn divided by the period; using the inverse period would miss that factor.

For an imposed engineering strain amplitude of one millionth, the material law predicts mean work density about 17.6 nanowatts per cubic metre at this frequency. That is a local laboratory-style example, not the moon’s volume-averaged heating: the actual tidal strain varies with position and includes several tensor components. The next chapter integrates the appropriate deformation field.

Now double the viscosity while keeping the elastic modulus and imposed strain fixed. The material becomes more elastic at this frequency, and its loss modulus decreases because the original dimensionless frequency is already above one. At lower starting viscosity the same change could increase loss. A monotonic statement such as “more viscous means hotter” omits the timescale comparison.

$$
\tau_M=9.0351\times10^4\,\mathrm{s},\quad n=3.51565\times10^{-5}\,\mathrm{s^{-1}},\quad n\tau_M\simeq3.18
$$

The saved default uses viscosity ten to the power 14.5 pascal-seconds and the stated synthetic orbit.

$$
\mu^*=(3.18439+1.00251i)\,\mathrm{GPa},\qquad\langle q\rangle=\frac{3.51565\times10^{-5}}2(1.00251\times10^9)(10^{-6})^2=1.7622\times10^{-8}\,\mathrm{W\,m^{-3}}
$$

Insert the angular frequency, loss modulus in pascals and engineering strain amplitude before squaring; this gives 17.6 nanowatts per cubic metre.

Reading: Beuthe, Tides on Europa: the membrane paradigm, arXiv:1410.4735v1

## Connect material lag to a gravitational response

A moon is not a single sheared block. Its own gravity resists deformation and its geometry distributes strain. A radial Love number compares surface displacement with the external tidal potential divided by surface gravity. A gravity Love number compares the induced surface potential with the applied potential. Both are dimensionless, and both can be complex for a harmonic tide.

For the default shell, the radial response lags by about 16.9 degrees, much less than the phase of a freely creeping element would suggest. The self-gravity and finite shell thickness enter that answer. Even the viscosity at maximum global loss is not obtained by simply setting Maxwell time equal to the inverse angular frequency. The complete response denominator shifts the maximum.

Cassini’s measured quantity in the 2025 Titan analysis is a gravity response inferred from radio tracking. Mapping it to viscosity requires a much richer interior model than the one introduced here. Keep the order of reasoning: first the observed response and its uncertainty, then the material and structural assumptions, then the inferred interior.

$$
\widehat u_r(R)=h_2\frac{\widehat W(R)}{g},\qquad \widehat\Phi_{\rm induced}(R)=k_2\widehat W(R)
$$

Potential has units square metres per square second, so the displacement relation has units of length.

Reading: Beuthe, Tides on Europa: the membrane paradigm, arXiv:1410.4735v1 · Petricca et al., Titan’s strong tidal dissipation precludes a subsurface ocean, Nature 648 (2025)

Saved numerical example, units and method

Synthetic incompressible equal-density world. The parent mass is five hundred trillion trillion kilograms and the orbital semimajor axis is three hundred thousand kilometres. A four-condition elastic traction solve fixes the shell displacement; complex Maxwell correspondence fixes lag and heating. These values are predictions, with no observational noise except in the explicitly labelled fitted rows.

| Elastic modulus and viscosity | $\mu_E=3.50\,\mathrm{GPa},\quad\eta=10^{14.5}\,\mathrm{Pa\,s}$ |
| --- | --- |
| Material relaxation time | $\tau_M=9.03508\times10^4\,\mathrm{s}$ |
| Physical orbital period and frequency ratio | $P=1.78720\times10^5\,\mathrm{s},\quad n\tau_M=3.17642$ |
| Radial response phase lag | $\delta=16.8764^{\circ}$ |

The displayed scene removes the steady tide and magnifies the cyclic radial shape. Its gain is normalized separately at the reference eccentricity for each material and thickness; physical metre-valued amplitudes remain available for absolute comparisons. Independent external-work and volume-dissipation integrals test the power.

## [Separate bending from dissipation](https://sandboxphysics.com/labs/tidal-locking/)

1. Predict what ideal elasticity will do to the displacement curve and loop. Toggle it and explain both changes.
2. At the default shell thickness, sweep viscosity in the advanced controls. Find the loss peak, then compare with the maximum-loss preset and the frequency plot.

## Exercises with explanations

1. For a passive displacement response under our convention, should its imaginary part be positive or negative?

   Read the explanation

   Negative. A cosine force produces a displacement with a positive sine term when it lags. Writing that sinusoid as a complex amplitude times the positive exponential requires a negative imaginary amplitude. The modulus is the inverse relation and has the opposite sign.

2. Why does the ideal elastic preset have a changing shape but zero average heating?

   Read the explanation

   Stress and strain are in phase. Energy entering the elastic store during one part of the cycle is returned later; the closed force-displacement path encloses zero area. Motion itself is not evidence of dissipation.

3. Transfer the material model to a forcing period ten times shorter. Which dimensionless quantity changes, and can you infer the heating direction without the initial regime?

   Read the explanation

   Angular frequency and its product with Maxwell time increase tenfold. The loss modulus rises below its peak and falls above it; imposed strain and imposed force also produce different power laws. The full forcing and response must be specified.

[Previous: An icy world can warm as it bends](https://sandboxphysics.com/stories/ocean-worlds/how-tides-warm-an-icy-world/) [Next: A finite ice shell, from boundary tractions to Love numbers](https://sandboxphysics.com/stories/ocean-worlds/derive-a-finite-thickness-love-response/)

## Sources and calculation scope

1. [Beuthe, Tides on Europa: the membrane paradigm, arXiv:1410.4735v1](https://arxiv.org/html/1410.4735v1)

   Fixed 17 October 2014 preprint; Icarus publication DOI 10.1016/j.icarus.2014.10.027. Appendix A equations 105–116 give the finite-thickness homogeneous incompressible shell; Appendix C gives Maxwell correspondence; Appendix E derives work and power; Appendix G fixes the eccentricity/libration tide. Figures 8–9 and 12–13 expose approximation and thermal-model dependence. Full relevant sections and appendices were read. Our uniform-density Maxwell example is much simpler than a real moon.

2. [Petricca et al., Titan’s strong tidal dissipation precludes a subsurface ocean, Nature 648 (2025)](https://www.nature.com/articles/s41586-025-09818-x)

   Version of record, 17 December 2025, DOI 10.1038/s41586-025-09818-x, CC BY 4.0; full publisher-deposited text and Methods, PMC12711566, read. Figure 1 reports gravity-response estimates, Figure 2 compares conditional interior posteriors and Figure 3 is an inferred-interior schematic. Figure 1 error bars are three-sigma; Figure 2 ellipses are one-sigma. Positive reported imaginary response denotes their loss convention; our positive-time-exponential convention uses negative imaginary Love numbers for passive loss.