---
title: "What a complex Love number can identify"
description: "Derive a shell–solid degeneracy, build a weighted fit from frozen responses and separate tidal power from a conditional thermal budget."
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last_modified: "2026-10-28"
---

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

AC12 · L3

# What a complex Love number can identify

Sandbox Physics · 2026-10-28

A precise complex response can constrain combinations of material properties without uniquely revealing a liquid layer. In our model the response depends on geometric stiffness times material modulus. That product lets a thin stiff shell and a soft solid agree exactly, even at several frequencies. We will derive the ambiguity and then ask what extra assumptions change it.

## Before you begin

Use the preceding response derivation. Weighted residuals, elementary parameter fitting and steady spherical heat conduction are developed here.

## Find the combinations the data actually see

Write the response using two effective quantities: geometric stiffness times elastic modulus, and the Maxwell time. The first sets the strength of the restoring resistance, while the second sets how it varies with frequency. Neither expression separately contains thickness and elastic modulus. This is an identifiability statement about the equations, before any noisy data are introduced.

Choose any other shell thickness and scale both modulus and viscosity by the ratio of old to new geometric stiffness. Their ratio, the Maxwell time, stays fixed. The product of new stiffness and new complex modulus also stays fixed. The radial and gravity Love numbers therefore agree at every frequency, not just at the physical orbital frequency.

For the default fifty-kilometre shell, a fully solid sphere with elastic modulus about 0.214 gigapascals and viscosity about 19.4 trillion pascal-seconds gives the same external response. Its internal strain field differs. The model is not saying those material values describe real solid ice: independent material constraints could reject that alternative. It says the response alone cannot reject it within this family.

$$
A=z_h\mu_E,\quad \tau_M=\frac\eta{\mu_E},\quad k_2(\omega)=\frac{3/5}{2/5+\dfrac{A}{\rho gR}\dfrac{i\omega\tau_M}{1+i\omega\tau_M}}
$$

The observations constrain two effective combinations: stiffness and relaxation time.

$$
\mu_E^\prime=\mu_E\frac{z_h}{z_h^\prime},\quad \eta^\prime=\eta\frac{z_h}{z_h^\prime}\quad\Longrightarrow\quad k_2^\prime(\omega)=k_2(\omega)
$$

This transformation also preserves h because the equal-density example fixes k/h. Density stratification changes that proportionality and the independent information in the two responses.

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

## Fit observations with their stated errors

The public record contains three synthetic harmonic probes, each with a real and an imaginary gravity response. The frequencies are one-half, one and two times the chosen orbital frequency. These are teaching probes, not three Cassini measurements or a claim about the tidal spectrum of a named moon. Independent Gaussian errors have standard deviations 0.002 and 0.0016 on the two components.

Subtract each predicted component from the corresponding observation to form a residual. Divide by its standard deviation, square the standardized residual, and sum all six terms. Minimize that sum over the logarithms of the two positive effective parameters. Logarithms maintain positivity; they do not add an observational prior. The supplied solver uses a coarse starting search followed by a damped two-parameter Gauss–Newton step.

The saved fit gives effective stiffness about 825 million pascals, Maxwell time about 88,602 seconds and squared-residual sum about 4.004 for six numbers and two fitted parameters. Inspect the six individual residuals as well as their total. A deliberately nonpassive record with a strongly positive imaginary response fails the declared passive model under our sign convention.

$$
\chi^2=\sum_j\left[\frac{(k_{R,j}^{\rm obs}-k_{R,j}^{\rm pred})^2}{\sigma_{R,j}^2}+\frac{(k_{I,j}^{\rm obs}-k_{I,j}^{\rm pred})^2}{\sigma_{I,j}^2}\right]
$$

This diagonal-error expression assumes independent real/imaginary errors. Actual correlated observations require the full covariance inverse.

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

## A narrow thickness interval can come from a strong assumption

First allow modulus and viscosity to change with thickness. The output family includes shells and a fully solid sphere, all with exactly the same fitted response and squared-residual sum. Some thin-shell rows require moduli beyond the Lab’s display range. They are mathematical members of the synthetic family; their material plausibility must be assessed separately.

Next fix elastic modulus exactly at 3.5 gigapascals. For each thickness, minimize over viscosity only. The saved quarter-kilometre grid selects 50.5 kilometres; points with a profile increase of at most one span 49.5 to 51.75 kilometres. This is a local one-parameter Gaussian profile approximation conditional on an exact modulus and every other fixed world property. It is not an unconditional ocean-depth uncertainty.

The contrast between a flat unrestricted family and a narrow conditional interval is the lesson. Better measurement precision shrinks errors perpendicular to the degenerate family. It cannot by itself eliminate the direction that the equations leave invisible. Add independent material knowledge, a genuinely different observable, or a richer structural model before interpreting a unique layer boundary.

$$
\chi^2_{\rm prof}(d)=\min_{\eta>0}\chi^2(d,\eta\mid\mu_E=3.5\,\mathrm{GPa}),\qquad \Delta\chi^2_{\rm prof}=\chi^2_{\rm prof}-\chi^2_{\rm prof,min}
$$

The conditioning bar records which material statement creates the thickness information.

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

## A heating prediction needs an energy source and an exit

The power formula uses a prescribed eccentric orbit. It is sustained work supplied to the deforming body, balanced by irreversible internal loss. The browser does not evolve the orbital and spin reservoirs that ultimately provide that energy. In a complete orbital model, tides can reduce eccentricity unless another interaction maintains it. A fixed driver cannot be read as a prediction of unlimited long-term heating.

For a simple conductive reference, assume a spherical shell with constant thermal conductivity 2.5 watts per metre per kelvin, a liquid-boundary temperature of 273 kelvin and a surface temperature of 100 kelvin. Steady radial heat flow has the same total power through each sphere. Integrate the temperature gradient between the two radii to obtain the expression below.

At fifty kilometres this reference carries about 24.5 gigawatts, while the chosen tidal model supplies about 100 gigawatts. That mismatch poses a thermal evolution problem. The specified boundary temperatures and constant conductivity determine this reference; convection, latent heat and radiogenic heating would add further terms. Its thickness sweep stops before the no-liquid solid limit, where the specified inner liquid boundary would lose its meaning.

$$
Q=-4\pi r^2\kappa\frac{dT}{dr},\quad T_i-T_s=\frac{Q}{4\pi\kappa}\left(\frac1{R-d}-\frac1R\right),\quad Q_{\rm cond}=4\pi\kappa(T_i-T_s)\frac{R(R-d)}d
$$

The independently specified conductivity and boundary temperatures define this steady conductive reference.

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)

## What a feedback calculation would have to add

A warm region can become less viscous. On one side of the loss peak that raises heating; on the other side it lowers it. Conductive or convective cooling also changes with temperature. To discuss stability one needs a specified heat capacity, a temperature-dependent material law, heat transport and the location of the dissipation. The derivative of net heating around an equilibrium then determines its local thermal tendency.

Beuthe’s thermal examples use specified transport and rheological assumptions; their equilibrium crust thicknesses are not consequences of Love numbers alone. The 2025 Titan analysis couples a stratified interior interpretation to a thermal model of high-pressure ice. These are useful next readings because they add the missing links explicitly, while also exposing new parameter dependence.

Liquid water, heat and a large tide each answer a different question. Habitability further depends on chemical energy, composition, exchange and persistence. Our synthetic response neither rules an ocean in nor rules life out. It gives a solvable example of how mechanical evidence reaches an interior claim only through a model.

$$
C\dot T=P_{\rm tide}(T,d,e)+P_{\rm other}-Q_{\rm out}(T,d),\qquad \left.\frac{d(P_{\rm in}-Q_{\rm out})}{dT}\right|_{T_*}<0
$$

With positive heat capacity this derivative is a local stability condition for a specified single-temperature model. It supplies a next step beyond the prescribed-material response calculated in the Lab.

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.

| Noiseless matching solid material | $\mu_E=0.214321\,\mathrm{GPa},\quad\eta=1.93641\times10^{13}\,\mathrm{Pa\,s}$ |
| --- | --- |
| Noisy fitted effective material | $z_h\mu_E=8.25176\times10^8\,\mathrm{Pa},\quad\tau_M=8.86025\times10^4\,\mathrm{s}$ |
| Squared residuals and residual degrees of freedom | $\chi^2=4.00403,\quad\nu=6-2=4$ |
| Conditional best grid thickness | $d=50.5\,\mathrm{km}\quad(\mu_E=3.50\,\mathrm{GPa}\ \mathrm{fixed})$ |
| Fixed-property conductive reference | $Q_{\mathrm{cond}}(d=50\,\mathrm{km})=24.4573\,\mathrm{GW}$ |

The fitted rows use six frozen synthetic components and two effective fit parameters. The thickness result fixes the modulus exactly and scans in quarter-kilometre steps; its local profile-increase-of-one grid interval is 49.5–51.75 km. The public Python script independently evaluates the published geometric factor. The conductive reference assumes fixed boundary temperatures 100 and 273 kelvin, constant conductivity and no convection.

## [Compare a hidden interior at fixed external response](https://sandboxphysics.com/labs/tidal-locking/)

1. Choose the shell and rescaled-solid presets and export both records. Compare complex Love numbers, material values and inner radius.
2. At fixed material, inspect the thickness data row. List the material and temperature feedbacks needed to turn its conductive reference into a self-consistent equilibrium.

## Exercises with explanations

1. Would measuring radial displacement at the same frequencies break the exact degeneracy of this equal-density example?

   Read the explanation

   No. Its gravity response is always three-fifths of its radial response, so the extra response is algebraically dependent. A realistic density-stratified structure changes that relationship; the answer for a real mission must be calculated in that richer model.

2. Why is a small conditional thickness interval compatible with a completely flat shell–solid family?

   Read the explanation

   Fixing modulus removes the scaling freedom. The narrow interval answers a different conditional question. If modulus is released again, changes in thickness can be compensated by modulus and viscosity, restoring the flat family.

3. A heating and cooling curve cross. What additional information is needed before calling the crossing stable?

   Read the explanation

   Specify how each curve changes with the actual temperature state, including the viscosity law and heat transport. For a positive heat capacity in a single-temperature model, net heating must decrease with temperature near the crossing. Evaluating that derivative requires a temperature-dependent model in addition to the displayed thickness scan.

[Previous: A finite ice shell, from boundary tractions to Love numbers](https://sandboxphysics.com/stories/ocean-worlds/derive-a-finite-thickness-love-response/) [Next: Fit the response, then expose the shell–solid family](https://sandboxphysics.com/stories/ocean-worlds/rerun-a-shell-and-solid-response-fit/)

## 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.