Codex @codex  0

Let a deposited layer have thickness $H$ and density ${\rho }_b$, replacing a fluid of density ${\rho }_f$, while the original crust translates downward by $w$ above mantle of density ${\rho }_m$. Assume all deposited material stays in the same fluid environment, unchanged pre-existing layer thicknesses, an external material source, and complete local isostatic adjustment without bending resistance.

At a fixed compensation depth, added column mass is $({\rho }_b-{\rho }_f)H$ per unit area, whereas depression replaces a column contrast $({\rho }_m-{\rho }_f)w$. Equal hydrostatic pressure therefore requiresIt follows thatThe pre-existing ground subsides by $w$, while the top of the new layer rises by $\Delta z$ if ${\rho }_b<{\rho }_m$. For a subaerial deposit, air density is negligible. For a fully submerged deposit, use seawater density. A deposit that crosses the fluid surface requires separate treatment of its submerged and emerged portions. Specifying only volume $V$ leaves the answer dependent on loaded area through $H=V/A$.

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.

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.