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Alfred Wegener was a meteorologist and geophysicist who developed continental drift into a coordinated scientific argument. He presented the proposal publicly in 1912 and published its first book-length treatment in 1915. His investigations of the atmosphere grew alongside his polar fieldwork: observations from Greenland expeditions contributed to his work on atmospheric thermodynamics.^[ref]

The matching continental outlines were a starting clue. Alfred Wegener assembled evidence from geophysics, geology and paleontology, including related rock formations and past organisms on opposite sides of the Atlantic. He also used isostasy to challenge the idea that vanished land bridges explained these connections: the density contrast between continents and their substrate mattered to his physical argument. He attempted to test motion through geodesy in Greenland, but the inferred displacement rates were much too large. This distinguishes the broad evidence for continental drift from an individual attempted measurement of it.^[ref]

In the linear flat-elastic plate model, let downward deflection be $w$, applied downward pressure be $q$, flexural rigidity be $D$, and the restoring geophysical density contrast be $\Delta \rho >0$. The static balance isFor a sinusoidal Fourier mode of horizontal wavenumber magnitude $k$, applying the fourth spatial derivative multiplies its amplitude by $k^4$. HenceThe displayed dimensionless ratio compares bending resistance with buoyant restoration. Large $\chi$ means substantial elastic support and suppressed deflection. Small $\chi$ gives nearly local isostasy. The associated length $L_f=(D/\Delta \rho g{)}^{1/4}$ makes the load-width dependence explicit; some conventions define a flexural parameter with an additional factor $4^{1/4}$. At zero wavenumber, bending contributes nothing: total support remains buoyant even when the spatial distribution of subsidence changes.

Cooling oceanic lithosphere becomes denser and subsides under isostasy. A simple calculation explains the square-root age law. Let depth $z$ be measured below a fixed cooling surface, take constant thermal diffusivity $\kappa$, and write the half-space cooling modelFor mantle reference mass density ${\rho }_m$ and small constant volumetric thermal expansion coefficient $\alpha$, the cold-column density excess relative to hot mantle is $\delta \rho ={\rho }_m\alpha (T_m-T)$. Its integral per unit horizontal area isThis uses ${\int }_0^{\infty }\mathrm{erfc}udu=1/\sqrt{\pi }$.

At a common compensation depth, additional water depth replaces the dense upper column by seawater of mass density ${\rho }_w$. Balancing column masses gives $({\rho }_m-{\rho }_w)[d(t)-d_r]=\Delta \Sigma$, henceThe relevant geophysical density contrast is therefore mantle versus seawater, and $t$ is seafloor age measured from the ridge reference depth $d_r$. The model neglects sediment loading, dynamic support and changes of material properties. A finite oceanic plate cooling model and basal heating limit indefinite square-root deepening at old ages.

Use the intended hot-ridge state and a linear law of thermal expansion about mantle density ${\rho }_m$ at $T_1$. Let $s$ be depth below the old seafloor. Its excess density isand its excess column mass is ${\int }_0^a\delta \rho ds={\rho }_m\alpha \Delta Ta/2$. Isostasy requires equal pressure at a compensation depth below both columns. Lowering the old seafloor by $h$ replaces rock by water, reducing column mass by $({\rho }_m-{\rho }_w)h$. Therefore the isostatic ridge elevation in a plate model isHere $\alpha \Delta T=0.052$; the result is a first-order estimate and neglects sediments, crustal-composition differences and flexural support. If one instead insists on the literal printed zero-age profile, its column is denser than the old plate and the formal relative elevation reverses sign, $-3.9\mathrm{km}$ to this order. That profile cannot describe the elevated hot ridge posited by the question.

Oceanic thermal subsidence. Newly formed oceanic lithosphere is hot and buoyant. Cooling increases its column density, causing subsidence by isostasy as it moves away from a ridge. The half-space cooling model has profile isWith volumetric thermal expansion coefficient $\alpha$, the excess column mass relative to hot mantle isCompensation with seawater density ${\rho }_w$ givesThus young-to-intermediate-age seafloor generally deepens approximately as the square root of age. At old ages, finite thermal-plate thickness and basal heat supply flatten this trend; indefinite half-space deepening is inappropriate. Sediments, variable crust, volcanic swells and dynamic topography also perturb bathymetry. Ocean-floor depth and heat flow jointly constrain cooling models.^[ref]

Effective elastic thickness. This is the thickness of a homogeneous elastic plate with the observed effective flexural rigidity, not necessarily the crustal or thermal lithospheric thickness:Infer it by modelling deflection around known loads, such as seamounts or trenches, or fitting gravity-topography admittance and gravity-topography coherence across wavelengths. A simple Fourier balance is $(D{k}^4+\Delta \rho g)w_k=q_k$ for a signed transverse load. The transition from buoyancy-dominated to bending-dominated behaviour constrains $D$; then assumed Young's modulus $E$ and Poisson's ratio $\nu$ give $T_e$.

Good inference includes load geometry and density, buried interfaces, sediment and internal loading, boundary conditions and data resolution. These can trade off with rigidity. Plastic yielding, faulting, mechanical layering and creep make $T_e$ an effective, timescale- and stress-dependent quantity. Older, colder lithosphere is often stronger, but age at loading and subsequent relaxation matter; a single universal age-to-$T_e$ conversion is unjustified.

Using the preceding conductive estimate, warmer incoming crust is less dense. Interpret the given $\alpha =10^{-5}{\mathrm{K}}^{-1}$ as the volumetric coefficient entering $\delta \rho /\rho =-\alpha \Delta T$. The representative contrast givesFor ${\rho }_s\simeq 3000\mathrm{kg}{\mathrm{m}}^{-3}$, that is only about $10.5\mathrm{kg}{\mathrm{m}}^{-3}$. Let $L$ be the vertical thickness of the anomalous slab material in a compensated column. Its mass deficit per unit area is approximately ${\rho }_s\alpha \Delta {\overline{T}}_{c}L$. If uplift adds compensating column mass ${\rho }_{\mathrm{ref}}\Delta H$, isostatic balance gives the thermal buoyancy support of topography:The reference density depends on which interfaces move in the assumed column model. For approximately equal densities and a near-horizontal $7$ km crustal layer, this gives $\Delta H\simeq 25$ m; using ${\rho }_s=3000$ and ${\rho }_{\mathrm{ref}}=3300\mathrm{kg}{\mathrm{m}}^{-3}$ gives about $22$ m. A modest dip changes the vertical intersection to approximately $h_c/\cos \theta$ and does not change this order of magnitude. Different graphical flux choices and uncertain geometry change the number; they do not justify replacing a $7$ km crustal anomaly by a whole $100$ km lithospheric anomaly without additional information.

Thus the crust-only thermal contrast can supply tens of metres of isostatic relief, not kilometres. Its sign would favor greater thermal support from the warmer Juan de Fuca crust, so it cannot by itself explain why much of the central Andes stands higher than Cascadia. Crustal shortening and roots, whole-slab thermal structure, phase transformations, hydration, magmatic addition and dynamic stresses can dominate. If the stated expansion coefficient instead meant a linear solid expansion coefficient, the first-order density coefficient would be about three times larger; the estimate would still be well below kilometre-scale mountain relief.

To test isostasy, jointly measure surface topography, free-air gravity anomalies, Bouguer gravity anomalies and crustal thickness from receiver functions and seismic refraction surveys. An Airy isostasy interpretation predicts crustal roots beneath high terrain and a corresponding regional negative Bouguer signal; thermal or compositional compensation predicts different density structure. Compare gravity and geoid observations with compensated-column predictions over a range of wavelengths, and use gravity-topography admittance to test local versus flexural isostasy. A small long-wavelength free-air anomaly can support compensation, but is not a sufficient test by itself. GNSS uplift, leveling and dated marine or river terraces constrain transient or sustained vertical motion; seismic tomography and rock-density estimates constrain deeper buoyancy. Residual relief or gravity after these corrections may require dynamic topography or tectonic support. An actively deforming subduction margin need not satisfy instantaneous local isostatic equilibrium, even if the estimate above is a useful scale comparison.

The volume alone does not determine a unique change of elevation. A local isostatic response to an added surface layer depends on its thickness and density, so the Deccan Traps' footprint must be specified or retained as a parameter.

Let the deposited basalt have uniform thickness $H=V/A$ and density ${\rho }_b$, where $V=10^{15}{\mathrm{m}}^3$ and $A$ is the loaded area. Let $w$ be downward displacement of the original land surface, and let ${\rho }_m$ be the density of displaced mantle. Assume a subaerial, laterally extensive deposit, negligible air density, unchanged thickness and density of the pre-existing crust, and complete local isostatic adjustment. Take the magma supply to be external to the local compensation column, and neglect cooling, erosion, magmatic underplating and any deep thermal or source-depletion anomaly.

Choose a compensation depth beneath all the displaced interfaces. The added column mass per unit area is ${\rho }_{b}H$. Downward translation of the pre-existing crust replaces a thickness $w$ of mantle within the column, reducing its mass by ${\rho }_{m}w$. Equality of hydrostatic pressure at the compensation depth therefore givesThe newly formed top of the basalt stands $H$ above the depressed original surface. Consequently the change of the region's top-surface elevation isThus the original substrate subsides, but the top of the deposit rises if ${\rho }_b<{\rho }_m$. Distinguishing these two surfaces prevents reporting the large downward displacement as the change in plateau elevation.

For an explicitly illustrative choice $A=10^{12}{\mathrm{m}}^2$, ${\rho }_b=2900\mathrm{kg}{\mathrm{m}}^{-3}$ and ${\rho }_m=3300\mathrm{kg}{\mathrm{m}}^{-3}$, the deposit is 1000 m thick, the substrate subsides about 879 m, and the final surface rises about 121 m. This is a conditional estimate, not a number fixed by the paper. Half that footprint would double all three heights under the same assumptions. For a nonuniform deposit, local isostasy gives the same relation pointwise with $H(x,y)$; the total volume still cannot specify a unique local elevation.

Elasticity lets the lithosphere share a load with neighbouring regions through lithospheric flexure. Beneath the centre of a localized load, subsidence is generally smaller than the local isostasy estimate, so the deposited surface stands higher. The surrounding plate also bends down, and a more distant flexural forebulge may rise. The extra support is regional bending resistance, rather than removal of the load's mass.

The controlling comparison is load width against the lithosphere's flexural length, which increases with flexural rigidity and decreases with the relevant buoyant geophysical density contrast. Equivalently, the flexural support ratio becomes large for short-wavelength loading. A very broad nearly uniform load produces little curvature in its interior, so even a strong elastic lithosphere has a nearly local isostatic adjustment there. Bending matters most near its edges. A narrow load requires more curvature and can be supported appreciably by the surrounding elastic lithosphere.

The Deccan Traps form a broad continental large igneous province. The volcanic edifices of the Hawaiian Islands are more localized loads on an old oceanic lithosphere, for which regional bending and a flexural moat are prominent. Under the usual broad-Deccan versus narrow-Hawaiian-load comparison, elastic strength matters more for the Hawaiian Islands. The more coherent mantle-dominated strength of an old oceanic plate can reinforce this contrast. It cannot be deduced simply from “oceanic” versus “continental”: a cold Indian craton can be strong, and a site's effective elastic thickness depends on its thermal and loading history.

For an initially submerged Hawaiian load, replacing seawater changes both the load and buoyant restoring contrast. At fixed deposited thickness, the added load relative to the initial water column uses the rock-minus-water density, and the restoring term uses the mantle-minus-water density. An emerged island requires separate treatment of its subaerial and submerged portions. These corrections change the local isostasy reference but do not remove the importance of load width for lithospheric flexure. The broad Hawaiian swell should also be distinguished from the narrower volcanic load: deeper thermal or dynamic support of the Hawaiian swell is not the same mechanism as elastic support of the volcanic ridge.

Finally, over long periods material creep, yielding and thermal modification can reduce elastic support. Without measured load geometries, ages and effective elastic thicknesses, the ranking is a justified qualitative expectation rather than a theorem valid for every continental and oceanic load.

Cooling makes the oceanic lithosphere denser and contracts it; isostatic adjustment therefore lowers the seafloor. The cooling layer includes substantial mantle, not just the approximately fixed-thickness igneous oceanic crust.

New plate near a spreading ridge is hot and relatively buoyant. As it moves away, conductive cooling penetrates more deeply, and material formerly in the hot asthenosphere becomes part of the colder thermal boundary layer. To first order,where $\alpha$ is the volumetric thermal expansion coefficient, ${\rho }_m$ a reference mantle density, $T_m$ the hot reference temperature, and $t$ plate age. A positive depth-integrated density anomaly makes the column negatively buoyant relative to the ridge column. Isostasy is restored by downward displacement and mantle redistribution; cold rock also occupies less volume.

At a fixed sufficiently deep datum, deepening the seafloor by $\Delta d$ removes an equivalent thickness of rock from the column and replaces it with less dense seawater. In the usual constant-property approximation this giveswhere ${\rho }_w$ is seawater density and $z_c$ lies below the relevant thermal anomaly. This explains why contraction alone and water-loaded seafloor subsidence are not identical numerical quantities.

For a half-space cooling model,so integration yieldsHere $\kappa$ is thermal diffusivity and $z$ is measured downward from the surface of the thermal model. The leading oceanic thermal subsidence is proportional to $\sqrt{t}$. A finite oceanic plate cooling model limits this growth at old ages. Sediment loading, dynamic topography, anomalous crust thickness and hydrothermal heat transport can produce departures from the simple relation.^[ref]

Choose a common horizontal compensation depth beneath the deepest cooling-induced density anomaly, within mantle that can relax pressure differences over the timescale considered. In a normal oceanic plate model, this means below the old plate's thermal lithosphere, not at the base of the approximately 7 km thick oceanic crust.

The reason is that cooling changes density through both crust and lithospheric mantle. A datum at the Mohorovičić discontinuity would omit most of the changing column mass. A different datum for each age would also compare unlike columns. Once the common datum is sufficiently deep, extending it farther into laterally uniform mantle adds the same mass per unit area to each column and does not change the predicted relative subsidence.

There is no unique number fixed by the question. For an illustrative oceanic plate cooling model with a thermal thickness about 100–125 km, a common datum beneath that layer, of order 150–200 km below sea level, is a defensible choice provided the model has no substantial lateral density anomaly there. These are model choices, not measured universal boundaries. In a half-space cooling model, use the oldest age $t_{\max }$ to choose a practical cutoff: $z_c\gtrsim 4\sqrt{\kappa t_{\max }}$ makes the remaining temperature anomaly less than about 0.5% of its surface contrast. For $\kappa =10^{-6}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $t_{\max }=100$ million years, this is about 225 km. A half-space model has no mathematically sharp finite thermal base.

Finally, equal column mass is the local, approximately constant-gravitational acceleration representation of equal pressure at the datum. A weak asthenosphere permits pressure relaxation through slow solid-state material creep; it need not be liquid. Short-wavelength loads, lithospheric flexure or dynamic topography can violate local column-by-column isostasy, requiring a more complete mechanical model.

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.

The addition of sediments as a mechanical load, contributing to subsidence through isostasy and lithospheric flexure.

Suppose a compensated vertical column contains thickness $L$ of material warmed by $\Delta T$. For a small anomaly, the volumetric thermal expansion coefficient gives $\delta \rho \simeq -{\rho }_s\alpha \Delta T$. Its column mass change is $\delta M\simeq -{\rho }_s\alpha \Delta TL$. Define ${\rho }_{\mathrm{ref}}\Delta H$ to be the additional column mass associated with the permitted uplift and interface motion in the chosen isostasy model. The mass balance ${\rho }_{\mathrm{ref}}\Delta H+\delta M=0$ proves the displayed scaling.

The model must specify ${\rho }_{\mathrm{ref}}$: translating an upper column of fixed internal thickness above mantle compensation gives a mantle-density scale, while fixing an internal interface and changing only surface-layer thickness gives that surface layer's density scale. If their ratio is near one, $L=7$ km, $\Delta T=350$ K and $\alpha =10^{-5}{\mathrm{K}}^{-1}$ imply only about 25 m of relief. Extending this to a whole lithosphere requires its depth-dependent temperature anomaly, not only the mean temperature anomaly of the crust. Dynamic topography and flexural stresses lie outside a simple local-column calculation.