Lecture 21 — Isostasy & Lithospheric Flexure

ESS 314 — Introduction to Geophysics
University of Washington · Spring 2026

Marine Denolle · Wed May 6, 2026 · JHN 111

By the end of this lecture, you should be able to…

  • [LO-21.1] Apply Archimedes — compute crustal-root thickness (Airy) and column density (Pratt).
  • [LO-21.2] Distinguish local (Airy/Pratt) vs. regional (flexural) compensation.
  • [LO-21.3] Interpret a Bouguer profile as evidence for compensation; describe the isostatic-residual.
  • [LO-21.4] Estimate post-glacial rebound timescale and connect to mantle viscosity.
  • [LO-21.5] Use the flexural parameter α\alpha to predict forebulge wavelength.

Course objectives addressed: LO-1, LO-2, LO-3, LO-4.

A puzzle from Lecture 19

The Bouguer anomaly over a mountain range is broadly negative, not positive.

→ The mountain is not an uncompensated excess mass.

Beneath every large topographic feature there is a mass deficit at depth — a root of light material — that almost exactly balances the topographic excess.

The mountain is floating.

Archimedes for mountains

Hydrostatic equilibrium at the depth of compensation:

ρcolumn,1gdz  =  ρcolumn,2gdz\int \rho_{\text{column,1}} \, g \, dz \;=\; \int \rho_{\text{column,2}} \, g \, dz

Below the depth of compensation, the asthenosphere flows freely on geological timescales.
Any pressure imbalance there drives lateral flow that restores equilibrium.

→ Two end-member ways to satisfy this: Airy and Pratt.

Airy and Pratt — two end-member compensation styles

Airy and Pratt isostasy compared

  • Airy: same density, variable thickness. High mountain → deep root.
  • Pratt: same depth, variable density. High mountain → low-density column.

Both satisfy hydrostatic equilibrium but make different deep-Earth predictions.

Airy — the root formula

ρch=(ρmρc)rr=ρcρmρch\rho_{c} h = (\rho_{m} - \rho_{c}) r \quad\Rightarrow\quad r = \frac{\rho_{c}}{\rho_{m}-\rho_{c}} \, h

For ρc=2700\rho_{c} = 2700, ρm=3300\rho_{m} = 3300 kg m⁻³:

r/h=4.5r/h = 4.5

A 1-km mountain → 4.5-km root.

The Tibetan Plateau (h5h \approx 5 km, Moho 70\approx 70 km) is approximately Airy.

Pratt — the density formula

ρcHref=ρblock(Href+h)ρblock=ρcHrefHref+h\rho_{c} H_{\text{ref}} = \rho_{\text{block}} (H_{\text{ref}} + h) \quad\Rightarrow\quad \rho_{\text{block}} = \rho_{c} \, \frac{H_{\text{ref}}}{H_{\text{ref}} + h}

For Href=35H_{\text{ref}} = 35 km, h=5h = 5 km: ρblock=0.875ρc2362\rho_{\text{block}} = 0.875\,\rho_{c} \approx 2362 kg m⁻³ (12% lower).

Mid-ocean-ridge crests, hot spots, and oceanic plateaus are approximately Pratt.

Real Earth: a mixture

Neither model is solely correct. Earth applies both.

  • Airy dominates large-scale crustal features (continents, plateaus).
  • Pratt-type density variations dominate at the oceanic-continental boundary and near hot spots.

→ The choice is a scale and setting question.

The isostatic-residual anomaly

Δgiso=ΔgBouguerΔgcompensation prediction\Delta g_{\text{iso}} = \Delta g_{\text{Bouguer}} - \Delta g_{\text{compensation prediction}}

  • Zero residual → load is fully compensated.
  • Positive residual → uncompensated excess mass; load supported elastically (e.g. recent volcano).
  • Negative residual → uncompensated deficit; load not yet equilibrated (e.g. active glacier).

Transient residuals are diagnostic of active geodynamic processes.

Time-dependent compensation — post-glacial rebound

Post-glacial rebound

Ice sheet removed → lithosphere rebounds toward equilibrium.
Rate set by mantle viscosity:

τ4πηρmgL\tau \sim \frac{4\pi\eta}{\rho_{m} g L}

For Fennoscandia (L1000L \sim 1000 km, η1021\eta \sim 10^{21} Pa s): τ4 ⁣ ⁣5\tau \approx 4\!-\!5 ka — uplift continues at ~9 mm/yr today.

Inverse problem: get viscosity from rebound

Two pieces of data:

  • Total post-glacial uplift (Fennoscandia: 300\sim 300 m).
  • Present-day uplift rate (9\sim 9 mm/yr).

→ Ratio gives τ\tau → equation gives η\eta.

This single inverse problem has driven mantle-viscosity research for 50 years (Whitehouse 2018 review, CC BY 4.0, https://doi.org/10.5194/esurf-6-401-2018).

Beyond local compensation — flexure

Real lithosphere has elastic strength. A vertical load at one point produces regional deformation that extends 100s of km.

The thin-plate equation:

Dd4wdx4  +  (ρmρw)gw  =  q(x)D \, \frac{d^{4}w}{dx^{4}} \;+\; (\rho_{m}-\rho_{w}) \, g \, w \;=\; q(x)

with flexural rigidity D=ETe3/[12(1ν2)]D = E T_{e}^{3} / [12(1-\nu^{2})]
and flexural parameter α=[4D/(ρm ⁣ ⁣ρw)g]1/4\alpha = \bigl[ 4D / (\rho_{m}\!-\!\rho_{w}) g \bigr]^{1/4}.

What flexure looks like

Lithospheric flexure under a line load

Load → central depression + forebulge at xπαx \approx \pi\alpha + zero crossing.
Increasing TeT_{e} widens and reduces the deflection.
Inverting the observed forebulge geometry → estimates TeT_{e}.

When does each model apply?

Feature Width Compensation
Continent / plateau α\gg \alpha Local (Airy)
Mountain range \sim a few α\alpha Mixed
Single seamount α\ll \alpha Regional (flexural)
Subduction trench bend in plate Flexural

A continental craton has Te80T_{e} \approx 80 km; an oceanic spreading centre has TeT_{e} near zero.

Worked example — Mount Olympus (PNW)

Olympic Mountains: h2.4h \approx 2.4 km. Treat as Airy-compensated.

Predicted root: r=(ρc/(ρmρc))h=4.5×2.4=10.8r = (\rho_{c}/(\rho_{m}-\rho_{c}))\, h = 4.5 \times 2.4 = 10.8 km.
Predicted Bouguer (slab approx of root):

Δgroot2πG(ρmρc)r272 mGal\Delta g_{\text{root}} \approx -2\pi G (\rho_{m}-\rho_{c}) r \approx -272 \text{ mGal}

Free-air anomaly should remain small (<50<50 mGal) — root and topography roughly cancel.

→ The Olympics sit on the Cascadia accretionary prism — active loading. The isostatic residual is the signal of that active process.

Course connections

  • Backward: L19 (Bouguer construction) and L20 (regional vs residual).
  • Forward: L22 — rheological structure of the lithosphere; viscoelasticity.
  • Forward: Module 7 — dynamic topography, vertical motion driven by mantle convection.
  • Cross-link: present-day GIA is a key correction to GPS-measured vertical land motion → tide gauges → sea-level reconstruction.

Research horizon

  • Whitehouse (2018) — GIA modelling review (CC BY 4.0). Key open question: depth profile of mantle viscosity.
  • Forte & Rowley (2022) — much of long-wavelength Bouguer / geoid is dynamic topography, not local isostasy.
  • Antarctic / Greenland mass balance — disentangling present-day ice loss from underlying GIA is the largest single uncertainty in 21st-century sea-level projections.

Societal — Cascadia subsidence and the tide gauge

PNW tide gauges show slower sea-level rise than the global mean.

This is not measurement error — it is inter-seismic uplift from Cascadia locking.

In the next M9 megathrust event:

  • Coastal Washington will drop 0.5–2 m in minutes.
  • Tide-gauge baseline resets.
  • Locally, "sea level" is redefined.

→ Vertical land motion (this lecture's mechanics, on a transient timescale) directly informs Cascadia hazard.

USGS Cascadia resource (public domain): https://www.usgs.gov/programs/earthquake-hazards/science/cascadia-subduction-zone

AI Literacy — Prompt Lab

Three prompts that test whether AI handles this lecture's reasoning.

P1. Compare Airy and Pratt for the Tibetan Plateau. Which fits observations and why?
→ Good: derives both, computes amplitudes, notes empirical mixture. Poor: qualitative-only.

P2. Estimate mantle viscosity from Fennoscandian uplift (9 mm/yr; total 300 m).
→ Good: τ33\tau \approx 33 ka → η1021\eta \approx 10^{21} Pa s. Common error: confuses rebound amplitude with load amplitude.

P3. Why do oceanic seamounts produce forebulges but individual continental peaks rarely do?
→ Good: invokes TeT_{e} contrast and notes that continental peaks usually sit within larger compensating structures. Either side alone is incomplete.

Concept Check

  1. Two ranges of identical surface elevation produce Bouguer anomalies of 100-100 mGal and 200-200 mGal. Which is more nearly locally compensated, and what does the difference suggest about lithospheric strength?
  2. A continental margin shows a forebulge 250 km offshore of a deltaic load. With ρmρw=2270\rho_{m}-\rho_{w} = 2270 kg m⁻³, E=70E = 70 GPa, ν=0.25\nu = 0.25estimate TeT_{e}.
  3. Fennoscandia: 9\sim 9 mm yr⁻¹ uplift, total 300\sim 300 m post-glacial. Estimate τ\tau and η\eta. Are they in the canonical range?