---
title: "Active Margins"
book: "University Earth and Space Sciences — Year 1"
subject: earth
language: en
chapter: 14
exercises: 12
source: https://one-course.com/books/earth/3/en/chapter/14-active-margins
license: CC-BY-NC-SA-4.0
credit: "One Earth and Space Sciences Book, One Course (one-course.com)"
---

# Chapter 14 — Active Margins

On 11 March 2011, at 14:46 local time, the plate boundary off north-east Japan broke over some $400\,\mathrm{km}$ along the trench and $150\,\mathrm{km}$ down its [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude). The fault slipped by as much as 50 to $60\,\mathrm{m}$, most of it [up-dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) of the hypocentre, towards the Japan Trench; on land, the coast moved horizontally and subsided. The plate that sinks there, the Pacific plate, goes down beneath Japan along a plane that earthquakes outline down to hundreds of kilometres; and where that plane lies about a hundred kilometres deep, a line of volcanoes runs parallel to the coast. This chapter reads such a margin from its shape, its gravity, its [heat flow](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) and, above all, its earthquakes.

**You already know.**

Oceanic [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) sinks back into the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) at subduction zones, marked by trenches and by inclined zones of earthquakes, and old [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) sinks under its own weight (Book 2, grade 11); water from the sinking plate lowers the melting temperature of the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) above it (Book 2, grade 12). Heat diffuses over a distance $L$ in a time of order $L^2/\kappa$ ([Chapter 10](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#ch-b1-heat-budget)); an infinite slab of density $\rho$ and thickness $h$ attracts with $2\pi G\rho h$ ([Chapter 11](https://one-course.com/books/earth/3/en/chapter/11-isostasy-and-gravity-anomalies#ch-b1-isostasy-gravity)); plates move about [Euler poles](https://one-course.com/books/earth/3/en/chapter/12-plate-kinematics-on-a-sphere#def-b1-plate-kinematics-euler-pole) ([Chapter 12](https://one-course.com/books/earth/3/en/chapter/12-plate-kinematics-on-a-sphere#ch-b1-plate-kinematics)); the [heat flow](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) of the ocean floor falls as the inverse square root of its age ([Chapter 13](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#ch-b1-rifts-ridges-margins)).

## 14.1 The anatomy of an active margin

**Definition 14.1 (Active margin, oceanic trench, forearc, back-arc basin).**

An *active margin* is a convergent plate boundary at which oceanic [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) sinks beneath another plate, either beneath a continent or beneath another oceanic plate. An *oceanic trench* is the long, narrow, V-shaped depression of the sea floor along which the sinking plate bends down beneath the other. The *forearc* is the part of the overriding plate between the trench and the line of volcanoes. A *back-arc basin* is a basin behind the volcanoes, on the side away from the trench, where the overriding plate is stretched and new [oceanic crust](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#def-b1-rifts-ridges-margins-oceanic-crust) may form.

Trenches are the deepest places of the oceans, and the deepest of all is the Challenger Deep of the Mariana Trench. Across the Kuril Trench the floor drops from about $5200\,\mathrm{m}$ on the abyssal plain to $8700\,\mathrm{m}$ at the axis, some $3.5\,\mathrm{km}$ lower, and the trench is only tens of kilometres wide (figure in [Section 14.2](#sec-14-2)). The convergent margins of the Earth measure about $55\,000\,\mathrm{km}$ in all; in a digital model of the plate boundaries the subduction boundaries add up to about $51\,000\,\mathrm{km}$, against about $67\,000\,\mathrm{km}$ of spreading ridges.

Seaward of the trench the plate rises slightly before it [plunges](https://one-course.com/books/earth/3/en/chapter/6-field-and-lab-methods-i#def-b1-field-lab-methods-1-trend-plunge), in a broad, low outer rise. There the plate is bent, and its upper part is stretched: earthquakes of normal faulting strike the outer rises. The width of the outer rise matches the bending of an elastic plate; this flexure of the [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) is studied in Book 5. The trench itself is the place where the weight of the dense sinking plate pulls its leading edge down.

**Definition 14.2 (Subducting slab).**

The *subducting slab* is the part of the sinking plate that has already gone down into the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers): a sheet of oceanic [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere), with its [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) on top, inclined beneath the overriding plate from the trench to depths of several hundred kilometres.

The overriding plate is cut into strips parallel to the trench (figure below): the [forearc](#def-b1-active-margins-active-margin), often with a pile of scraped-off sediments at its front and a sedimentary basin behind; the [volcanic arc](#def-b1-active-margins-volcanic-arc); and, behind some arcs, a [back-arc basin](#def-b1-active-margins-active-margin) and an older, extinct arc beyond it. Between the slab and the overriding plate lies a wedge of [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers), dragged down along the top of the slab.

![A subduction zone in section (schematic: lithosphere and slab drawn to the scale bar, relief above the sea floor greatly exaggerated). The slab dips gently beneath the forearc and more steeply below; the volcanic arc stands above the place where the top of the slab is about 100\, km deep. The cold slab drags its isotherms down into the mantle (dashed: one isotherm), and the mantle of the wedge is dragged down along it. Dots: earthquakes, on the plate boundary near the trench and inside the slab below. Water released by the slab rises into the wedge; melt rises to the arc.](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/fig-e8812626c64b.svg)

*A subduction zone in section (schematic: [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) and slab drawn to the scale bar, relief above the sea floor greatly exaggerated). The slab [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) gently beneath the [forearc](#def-b1-active-margins-active-margin) and more steeply below; the [volcanic arc](#def-b1-active-margins-volcanic-arc) stands above the place where the top of the slab is about $100\,\mathrm{km}$ deep. The cold slab drags its isotherms down into the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) (dashed: one isotherm), and the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) of the wedge is dragged down along it. Dots: earthquakes, on the plate boundary near the trench and inside the slab below. Water released by the slab rises into the wedge; melt rises to the arc.*

Summed over the same digital model, about $2.7\,\mathrm{km}^{2}$ of ocean floor go down the subduction boundaries each year, close to the $3.0\,\mathrm{km}^{2}$ made at the ridges ([Chapter 13](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#ch-b1-rifts-ridges-margins)): within the accuracy of such a model, which leaves out the slow deformation of wide plate-boundary zones, the ocean floor is destroyed as fast as it is made, and the surface of the Earth keeps its size.

## 14.2 Gravity and heat flow

Felix Vening Meinesz found the long, narrow belts of negative gravity anomalies along the trenches ([Chapter 11](https://one-course.com/books/earth/3/en/chapter/11-isostasy-and-gravity-anomalies#ch-b1-isostasy-gravity)). Satellites now map them everywhere at sea. Across the Kuril Trench (figure below) the [free-air anomaly](https://one-course.com/books/earth/3/en/chapter/11-isostasy-and-gravity-anomalies#def-b1-isostasy-gravity-anomalies) rises to about $40\,\mathrm{m}\mathrm{Gal}$ over the outer rise, falls to $-260\,\mathrm{m}\mathrm{Gal}$ over the trench, and is strongly positive again over the [forearc](#def-b1-active-margins-active-margin) and the islands of the arc.

The size of the trench anomaly tells how the trench is held. Suppose the trench, $3.5\,\mathrm{km}$ deeper than the floor around it, were not compensated at all: where the trench is, $3.5\,\mathrm{km}$ of rock of density about $2700\,\mathrm{kg}\,\mathrm{m}^{-3}$ is replaced by sea water. By the Bouguer slab ([Proposition 11.9](https://one-course.com/books/earth/3/en/chapter/11-isostasy-and-gravity-anomalies#prop-b1-isostasy-gravity-slab)) gravity there would be lower by

$$
2\pi G\,\Delta\rho\,h = 2\pi \times 6.674 \times 10^{-11} \times 1700 \times 3500
  \approx 2.5 \times 10^{-3}\,\mathrm{m}\,\mathrm{s}^{-2} = 250\,\mathrm{m}\mathrm{Gal} .
$$

That is the size of the observed low. A trench is therefore not in isostatic equilibrium: it is a depression held down, by the strength of the bent plate and by the pull of the slab, against the buoyancy of the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) below. (A trench only tens of kilometres wide attracts less than an infinite slab, so the water in it alone gives somewhat less than this; the rest of the deficit lies deeper, where the whole bent plate, its light [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) included, is pushed down into the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers): exercise [14.1](#exo-b1-active-margins-1).)

![Gravity and heat flow across a subduction zone. Top: the free-air gravity anomaly measured by satellite across the Kuril Trench, along the section of the figure in : 38\, m Gal over the outer rise, -260\, m Gal over the trench, strongly positive over the forearc and the islands. Bottom: heat flow, a schematic profile: about 43\, mW\, m-2 on the old ocean floor, lower over the forearc, 70 to 85\, mW\, m-2 on average over the arc and the back-arc. Dotted: the trench and the volcanic front. Data: Scripps Institution of Oceanography marine gravity model (Sandwell and others, 2014).](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/fig-717b38a07c8b.svg)

*Gravity and [heat flow](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) across a subduction zone. Top: the free-air [gravity anomaly](https://one-course.com/books/earth/3/en/chapter/11-isostasy-and-gravity-anomalies#def-b1-isostasy-gravity-anomalies) measured by satellite across the Kuril Trench, along the section of the figure in [Section 14.3](#sec-14-3): $38\,\mathrm{m}\mathrm{Gal}$ over the outer rise, $-260\,\mathrm{m}\mathrm{Gal}$ over the trench, strongly positive over the [forearc](#def-b1-active-margins-active-margin) and the islands. Bottom: [heat flow](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow), a schematic profile: about $43\,\mathrm{mW}\,\mathrm{m}^{-2}$ on the old ocean floor, lower over the [forearc](#def-b1-active-margins-active-margin), 70 to $85\,\mathrm{mW}\,\mathrm{m}^{-2}$ on average over the arc and the back-arc. Dotted: the trench and the volcanic front. Data: Scripps Institution of Oceanography marine gravity model (Sandwell and others, 2014).*

The [heat flow](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) tells the same story from the other side. The old ocean floor that comes to a trench loses little heat: about $43\,\mathrm{mW}\,\mathrm{m}^{-2}$ by the law of [Chapter 13](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#ch-b1-rifts-ridges-margins) for the Kuril plate, $122\,\mathrm{Myr}$ old. Over the [forearc](#def-b1-active-margins-active-margin) the flow is lower still: the cold slab beneath it takes up heat from above, and the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) under the [forearc](#def-b1-active-margins-active-margin) is cold and stagnant. Over the arc and the back-arc it is high, 70 to $85\,\mathrm{mW}\,\mathrm{m}^{-2}$ on average, twice the flow of the old interiors of continents (about $40\,\mathrm{mW}\,\mathrm{m}^{-2}$). An active [volcanic arc](#def-b1-active-margins-volcanic-arc) needs a [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) wedge at about $1200\,{}^{\circ}\mathrm{C}$ some $100\,\mathrm{km}$ down; behind it the whole back-arc is hot, with a [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) only 50 to $60\,\mathrm{km}$ thick and its [Moho](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-discontinuity) at 800 to $900\,{}^{\circ}\mathrm{C}$.

Why does the slab stay cold? It sinks faster than it can warm up. The Kuril slab of [Section 14.3](#sec-14-3) enters at about $82\,\mathrm{mm}$ a year and [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) at $45{}^{\circ}$: it reaches $600\,\mathrm{km}$ after a path of $600/\sin45{}^{\circ} \approx 850\,\mathrm{km}$, in about ten million years. By the conductive time scale ([Proposition 10.12](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#prop-b1-heat-budget-timescale)), heat from the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) crosses the $50\,\mathrm{km}$ to the middle of a slab $100\,\mathrm{km}$ thick in

$$
\tau = \frac{L^2}{\kappa} = \frac{(5 \times 10^{4}\,\mathrm{m})^2}{10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}}
  = 2.5 \times 10^{15}\,\mathrm{s} \approx 80\,\mathrm{Myr},
$$

about eight times longer. The core of the slab arrives at $600\,\mathrm{km}$ almost as cold as it was at the trench, and the isotherms are dragged down with it, as the section of a subduction zone above shows.

## 14.3 Earthquakes and the shape of the slab

**Definition 14.3 (Wadati–Benioff zone).**

A *Wadati–Benioff zone* is the inclined zone of earthquakes that starts near a trench and deepens beneath the overriding plate, down to as much as about $700\,\mathrm{km}$: it outlines the [subducting slab](#def-b1-active-margins-slab).

Its earthquakes are of two kinds. Down to some tens of kilometres they happen on the plate boundary itself, the gently dipping fault along which the two plates slide past each other; the largest earthquakes on Earth, such as that of 2011 off Japan, break this fault. Deeper, they happen inside the slab, which alone is still cold enough to store stress. At the greatest depths even the slab should flow rather than break, and how deep earthquakes start is still debated. The deepest earthquake of the catalogue, of magnitude 4.2, struck beneath the Vanuatu region in 2004 at $735.8\,\mathrm{km}$. In every subduction zone examined, the zone is double at intermediate depths: two layers of earthquakes, further apart in older plates.

The Kuril Islands, between Hokkaido and Kamchatka, sit above a slab that earthquakes outline from the trench down to more than $600\,\mathrm{km}$. There the Pacific plate, about 120 million years old where it reaches the trench, sinks beneath the Okhotsk plate (map below). The earthquakes of magnitude 5 or more since 1976 draw the slab in plan: shallow ones along the trench, intermediate ones beneath the islands, deep ones under the Sea of Okhotsk. On 24 May 2013 an earthquake of magnitude 8.3 struck $598\,\mathrm{km}$ beneath the Sea of Okhotsk, in the same slab farther north; it was felt as far as Moscow, some $6400\,\mathrm{km}$ away.

![The Kuril subduction zone: earthquakes of magnitude 5 or more, 1976–2025, coloured by depth, and the volcanoes active in the last ten thousand years. The trench (teeth on the overriding plate) and the band of the section shown in the next figure (dashed, 150\, km on each side of the section line). Data: USGS earthquake catalogue; Smithsonian Global Volcanism Program; trench from Bird (2003); coast from Natural Earth.](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/fig-5f89367ac08d.svg)

*The Kuril subduction zone: earthquakes of magnitude 5 or more, 1976–2025, coloured by depth, and the volcanoes active in the last ten thousand years. The trench (teeth on the overriding plate) and the band of the section shown in the next figure (dashed, $150\,\mathrm{km}$ on each side of the section line). Data: USGS earthquake catalogue; Smithsonian Global Volcanism Program; trench from Bird (2003); coast from Natural Earth.*

To see the slab in section, keep the earthquakes within $150\,\mathrm{km}$ of a line that starts at the trench and runs perpendicular to it, towards the overriding plate, and plot each at its distance $x$ along the line and its depth $z$ (figure below). The points form a band; the straight line that follows it best is found by least squares.

**Method 14.4 (Fitting the slab by least squares).**

Given $n$ earthquakes at $(x_i, z_i)$ inside the slab:

1. compute the means $\bar x = \frac1n\sum x_i$ and $\bar z = \frac1n\sum z_i$ ;
2. compute $S_{xx} = \sum (x_i - \bar x)^2 = \sum x_i^2 - n\bar x^2$ and $S_{xz} = \sum (x_i - \bar x)(z_i - \bar z) = \sum x_i z_i - n\bar x\bar z$ ;
3. the slab line is $z = a + s x$ with $s = S_{xz}/S_{xx}$ and $a = \bar z - s\bar x$ ; its [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) is $\delta = \arctan s$ .

The line minimises the sum of the squared vertical misfits $F(a, s) = \sum_i (z_i - a - s x_i)^2$, a quadratic function of $a$ and $s$ that is smallest where both partial derivatives vanish. $\partial F/\partial a =
-2\sum (z_i - a - s x_i) = 0$ gives $\bar z = a + s\bar x$: the line passes through the centre of the points. Substituting $a = \bar z - s\bar x$, $\partial F/\partial s = -2\sum x_i\,[(z_i - \bar z) - s(x_i - \bar x)] = 0$; since $\sum \bar x\,[(z_i - \bar z) - s(x_i - \bar x)] = 0$, this is $\sum (x_i - \bar x)[(z_i - \bar z) - s(x_i - \bar x)] = 0$, that is $S_{xz} = s S_{xx}$.

Beneath the central Kuril Islands the band holds 599 earthquakes, 121 of them deeper than $70\,\mathrm{km}$, inside the slab. The shallower ones are of little use for the fit: they mix the plate boundary and the bending plate, and 147 of the 478 are listed at the same round depth of $33\,\mathrm{km}$. The 121 deeper ones give

$$
z = -52.7 + 0.996\,x \quad\text{(kilometres)},
$$

a [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) of $44.9{}^{\circ}$, about $45{}^{\circ}$, down to the deepest earthquake of the band, $632\,\mathrm{km}$. The line does not pass through the trench: it reaches the surface $53\,\mathrm{km}$ landward of it. The slab is not straight: it leaves the trench at a gentler [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) and steepens below the [forearc](#def-b1-active-margins-active-margin), as the shallow earthquakes above the line show.

![The Kuril slab in section, without vertical exaggeration: the earthquakes of the band of the map above at their distance from the trench and their depth. The least-squares line through the 121 earthquakes deeper than 70\, km dips 45. The first volcano of the band, 164\, km from the trench, stands above the place where the line is 111\, km deep. Data: USGS earthquake catalogue; Smithsonian Global Volcanism Program; trench from Bird (2003).](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/fig-55236564314b.svg)

*The Kuril slab in section, without [vertical exaggeration](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-section): the earthquakes of the band of the map above at their distance from the trench and their depth. The least-squares line through the 121 earthquakes deeper than $70\,\mathrm{km}$ [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $45{}^{\circ}$. The first volcano of the band, $164\,\mathrm{km}$ from the trench, stands above the place where the line is $111\,\mathrm{km}$ deep. Data: USGS earthquake catalogue; Smithsonian Global Volcanism Program; trench from Bird (2003).*

A cold slab is denser than the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) around it, and its weight pulls on the plate behind it: this is the slab pull of Book 2, grade 11. Its size follows from the depth of the ocean floor.

**Proposition 14.5 (The excess weight of a slab).**

Under the hypotheses of the [depth–age law](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#thm-b1-rifts-ridges-margins-depth-age) ([Theorem 13.7](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#thm-b1-rifts-ridges-margins-depth-age)), a plate of age $A$ carries, per unit area, the excess mass

$$
m = (\rho_m - \rho_w)\bigl(d(A) - d_r\bigr)
    = \frac{2}{\sqrt\pi}\,\rho_m\,\alpha\,(T_m - T_0)\sqrt{\kappa A}
$$

over the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) at $T_m$. A slab of this plate, of length $L$ along its [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $\delta$ and not yet warmed, has the excess weight $F = g\,m\,L$ per unit length of trench, of which $F\sin\delta$ acts along the slab.

**Proof.** A layer $\mathrm dz$ of the plate at the temperature $T$ has the mass $\rho_m\bigl(1 + \alpha(T_m - T)\bigr)\mathrm dz$ per unit area, that is $\rho_m\alpha(T_m - T)\,\mathrm dz$ more than the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) at $T_m$. Summed over the plate, this excess is $\rho_m\alpha\int_0^\infty (T_m - T)\,\mathrm dz$, which the proof of the [depth–age law](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#thm-b1-rifts-ridges-margins-depth-age) showed to equal both $(\rho_m - \rho_w)(d - d_r)$ and $(2/\sqrt\pi)\rho_m\alpha(T_m - T_0)
\sqrt{\kappa A}$. A slab of length $L$ has, per unit length of trench, the area $L$ and so the excess mass $mL$, whose weight is vertical; its component along a line dipping at $\delta$ is $F\sin\delta$. ∎

The ocean floor thus weighs its own excess mass: each metre it has subsided below the ridge stands for $2300\,\mathrm{kg}$ of excess per square metre. For the Kuril plate, $122\,\mathrm{Myr}$ old, the half-space law ($C = 393$), which for floor this old somewhat overestimates the subsidence ([Chapter 13](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#ch-b1-rifts-ridges-margins)), gives $d - d_r = 393\sqrt{122} = 4340\,\mathrm{m}$ and $m = 2300\times4340 \approx 1.0 \times 10^{7}\,\mathrm{kg}\,\mathrm{m}^{-2}$, ten thousand tonnes on every square metre. Down to the deepest earthquake, $632\,\mathrm{km}$, the slab is $632/\sin45{}^{\circ} = 894\,\mathrm{km}$ long, so

$$
F = 9.81 \times 1.0 \times 10^{7} \times 8.94 \times 10^{5} \approx 8.8 \times 10^{13}\,\mathrm{N}
  \text{ per metre of trench},
$$

$6.2 \times 10^{13}\,\mathrm{N}\,\mathrm{m}^{-1}$ of it along the slab. This is an upper bound: the slab warms at its faces, and much of its weight is held by the viscous [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) it pushes aside rather than passed to the plate at the surface. Still, since the analyses of the driving forces of all the plates in the 1970s, the excess density of the sinking [lithosphere](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-lithosphere) is thought to provide most of the power that moves the plates.

## 14.4 Accretionary and erosive margins

The sinking plate brings its sediments to the trench. Some are scraped off and stay; others go down with the slab, and the slab may even take with it rock from the overriding plate.

**Definition 14.6 (Accretionary prism, tectonic erosion).**

An *accretionary prism* is a wedge of sediments, with slices of [oceanic crust](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#def-b1-rifts-ridges-margins-oceanic-crust), scraped off the sinking plate and added to the front of the overriding plate at a trench. *Tectonic erosion* is the removal of rock from the overriding plate by the sinking plate, which carries it down: from the front of the overriding plate, or from the base of its [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers).

A prism grows in two ways. At its front, sediments are bulldozed off the sinking plate and stacked as slices that [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) away from the trench; the youngest slices are the outermost, and the older ones farther back are lifted and steepened. At its base, sediments carried down along the plate boundary are stuck to its underside, a growth called underplating. The Franciscan rocks of California are an old prism whose underplated parts are now chaotic mixtures of blocks in a sheared matrix, mélanges. A prism, like a heap of sand pushed by a bulldozer, keeps a constant taper: once the angle between its surface and its base has reached a critical value, it grows only as a larger copy of itself (exercise [14.11](#exo-b1-active-margins-11)).

Whether a margin builds a prism or loses rock depends on the sediment that reaches the trench and on the speed of convergence. Accretionary margins, such as Cascadia, the Aleutians and the southern Peru–Chile Trench, receive plenty. Erosive margins, such as the northern Peru–Chile Trench, Tonga–Kermadec or the Marianas, receive little, and the sinking plate gnaws at their front and at the base of their [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers). Erosion prevails where the plates converge faster than about $6\,\mathrm{cm}$ a year and the sediment cover is less than $1\,\mathrm{km}$ thick; accretion where the convergence is slower or the trench sediment thicker than $1\,\mathrm{km}$.

## 14.5 Volcanic arcs and back-arc basins

**Definition 14.7 (Volcanic arc, arc–trench gap).**

A *volcanic arc* is a belt of volcanoes on the overriding plate of a subduction zone, parallel to the trench and convex towards the sinking plate, fed by magma from the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) wedge above the slab. The *arc–trench gap* is the horizontal distance between the trench and the volcanic front, the line of the volcanoes nearest to the trench.

Arc volcanoes stand where the top of the slab is about $105\,\mathrm{km}$ deep on average, though the depth beneath different arcs, averaged over stretches $500\,\mathrm{km}$ long, ranges from 72 to $173\,\mathrm{km}$. The magma is made where water released by the slab, as its hydrous [minerals](https://one-course.com/books/earth/3/en/chapter/7-rock-forming-minerals#def-b1-rock-forming-minerals-mineral) break down, lowers the melting temperature of the hot [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) of the wedge (Book 2, grade 12; the magmas of arcs are studied in the Year 2 volume). Beneath the central Kurils the volcanic front lies $164\,\mathrm{km}$ from the trench, where the Kuril slab line is $111\,\mathrm{km}$ deep.

**Proposition 14.8 (The arc–trench gap).**

Suppose the top of the slab [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) at $\delta_1$ from the trench down to the depth $z_1$, then at $\delta_2$ below, and that the volcanic front stands above the place where the top of the slab is at the depth $H > z_1$. Then the [arc–trench gap](#def-b1-active-margins-volcanic-arc) is

$$
g = \frac{z_1}{\tan\delta_1} + \frac{H - z_1}{\tan\delta_2} .
$$

For a slab of uniform [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $\delta$, $g = H/\tan\delta$. For given $z_1$, $H$ and $\delta_1$, the gentler the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $\delta_2$, the wider the gap.

**Proof.** On the first segment the slab sinks by $z_1$ while it advances horizontally by $z_1/\tan\delta_1$, since the tangent of the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) is the depth gained per unit of horizontal distance. On the second it sinks by a further $H - z_1$ over $(H - z_1)/\tan\delta_2$. The front is above the end of the second run, at the sum of the two horizontal distances from the trench. With $\delta_1 = \delta_2
= \delta$ the sum is $H/\tan\delta$. Finally $\tan\delta_2$ grows with $\delta_2$ on $(0, 90{}^{\circ})$, so the second term, and $g$, decrease as the slab steepens. ∎

A slab of $45{}^{\circ}$ with the volcanoes $111\,\mathrm{km}$ above it would put the front $111\,\mathrm{km}$ from the trench if it were straight; the Kuril front is at $164\,\mathrm{km}$ because the slab [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) more gently beneath the [forearc](#def-b1-active-margins-active-margin) (exercise [14.6](#exo-b1-active-margins-6)).

Arcs are curved, and their curvature follows from the sphere. A plate that goes down into the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) must bend without being stretched, for a plate tens of kilometres thick cannot be stretched by tens of percent. On a sphere this fixes how a bent plate can look.

**Proposition 14.9 (A bent spherical shell).**

Reflect the cap of a sphere cut off by a [small circle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-spherical-triangle) of angular radius $\rho$ in the plane of that circle. The reflected cap has the same shape and size as the original one, so the surface is bent without being stretched; and along the circle it is bent through the angle $\theta = 2\rho$. A plate that sinks at the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $\theta$ along such a [fold](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-fold) can do so without stretching along a [small circle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-spherical-triangle) of angular radius $\theta/2$.

**Proof.** A reflection is an isometry: the reflected cap has the same lengths and angles as the cap, and it meets the sphere exactly along the circle, which the reflection leaves in place. Let $O$ be the centre of the sphere and $P$ the centre of the cap, on the sphere; the plane of the circle is perpendicular to $OP$. At a point $M$ of the circle the angle $\widehat{MOP}$ is $\rho$; the tangent plane of the sphere at $M$ is perpendicular to $OM$, so it makes the angle $\rho$ with the plane of the circle, which is perpendicular to $OP$. Crossing the circle towards $P$, the sphere therefore rises above the plane of the circle at the angle $\rho$; the reflected cap, its mirror image, goes down below that plane at the angle $\rho$. Where the unbent sphere would have gone on at $\rho$ above the plane, the bent surface goes on at $\rho$ below it: it is bent through $\rho + \rho = 2\rho$. ∎

![A sphere cut through its centre O and the centre P of a cap of angular radius (here 40, to make the angles visible). Reflected in the plane of its circle, the cap keeps its shape and size; at M the surface, which went on rising at above that plane, now goes down at below it: it is bent through 2.](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/fig-caf80eba318d.svg)

*A sphere cut through its centre $O$ and the centre $P$ of a cap of angular radius $\rho$ (here $40{}^{\circ}$, to make the angles visible). Reflected in the plane of its circle, the cap keeps its shape and size; at $M$ the surface, which went on rising at $\rho$ above that plane, now goes down at $\rho$ below it: it is bent through $2\rho$.*

The Kuril slab [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $45{}^{\circ}$, so the proposition predicts an arc of angular radius about $22.5{}^{\circ}$. The PB2002 positions of the Kuril Trench, from Hokkaido to southern Kamchatka, are best fitted by a [small circle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-spherical-triangle) of radius $21.9{}^{\circ}$, centred near $58.9{}^{\circ}$ N, $123.8{}^{\circ}$ E, in eastern Siberia, on the side of the overriding plate. The agreement is close here, but the model is rough: the central part of the same trench alone gives $18.1{}^{\circ}$. The idea is Charles Frank’s, in 1968: the Kuril–Kamchatka, Aleutian and Sumatra–Java arcs, above slabs dipping about $45{}^{\circ}$, all have radii of about $20{}^{\circ}$, about $22{}^{\circ}$ measured to their trenches.

![Sarychev Peak, a volcano of the Kuril arc on Matua Island, in an early stage of eruption on 12 June 2009, seen from the International Space Station through a hole in the clouds. Grey ash, probably a pyroclastic flow, runs down from the summit, and a smooth cap of cloud, a pileus, sits on the rising column; ash from the eruption was detected more than 2400\, km away. Photograph: NASA, public domain (Wikimedia Commons).](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/img-daab8e805e1f.jpg)

*Sarychev Peak, a volcano of the Kuril arc on Matua Island, in an early stage of eruption on 12 June 2009, seen from the International Space Station through a hole in the clouds. Grey ash, probably a pyroclastic flow, runs down from the summit, and a smooth cap of cloud, a pileus, sits on the rising column; ash from the eruption was detected more than $2400\,\mathrm{km}$ away. Photograph: NASA, public domain (Wikimedia Commons).*

The [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) beneath arcs is thickened both by shortening and by the magma added from below; beneath the arcs built on continents, such as the Andes, it is among the thickest on Earth.

Behind some arcs the overriding plate is pulled apart. Dan Karig explained these marginal basins of the western Pacific within plate tectonics, in 1971: when the slab sinks steeply and rolls back, the trench and the arc migrate towards the sinking plate, and the plate behind them is stretched until it splits, sometimes leaving an extinct remnant arc on the far side. In the plate model used above, back-arc spreading runs at about $32\,\mathrm{mm}\,\mathrm{yr}^{-1}$ in the Mariana Trough near $18{}^{\circ}$ N, and at more than $110\,\mathrm{mm}\,\mathrm{yr}^{-1}$ in the north of the Lau Basin, behind Tonga. Where the slab is shallow, on the contrary, it drags the overriding plate, whose [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) is shortened and thickened behind the arc, as behind the Andes.

**In the field — Warning of a tsunami.**

A tsunami warning centre has minutes. In 2011 the first warning for the coast of north-eastern Japan came about three minutes after the earthquake, from a magnitude based on the short-period seismic waves; for earthquakes of magnitude 8 or more such a magnitude is underestimated, and so was the height of the tsunami. The moment magnitude, which does not saturate, took about fifteen minutes: the broad-band seismometers of Japan had been driven beyond their range. Since 2013, when the size of so large an earthquake is uncertain, the first warning assumes the largest earthquake possible there and describes the tsunami in words, such as “huge”, rather than with a number, and the forecasts draw on offshore measurements of the sea itself, from buoys positioned by satellite and pressure gauges on the sea floor, which see the wave before it reaches the coast.

![Duty officers in a tsunami warning centre, before screens of computed wave fronts and sea-level records (AI-generated illustration).](https://one-course.com/images/onecourse/chapters/earth-3/b1-active-margins/img-c04c6e821b33.jpg)

*Duty officers in a tsunami warning centre, before screens of computed wave fronts and sea-level records (AI-generated illustration).*

## 14.6 Exercises

**Exercise 14.1 ★.**

A trench lies $3\,\mathrm{km}$ below the ocean floor around it. (a) If it were not compensated, by how much would the missing rock, of density $2700\,\mathrm{kg}\,\mathrm{m}^{-3}$, replaced by sea water of density $1000\,\mathrm{kg}\,\mathrm{m}^{-3}$, lower gravity there? Use the Bouguer slab, with $G = 6.674 \times 10^{-11}\,\mathrm{m}^{3}\,\mathrm{kg}^{-1}\,\mathrm{s}^{-2}$. (b) Compare with the observed anomaly over the Kuril Trench, about $-260\,\mathrm{m}\mathrm{Gal}$, where the trench lies $3.5\,\mathrm{km}$ below the floor around it: redo the calculation for that depth. (c) Why does the water in a trench only tens of kilometres wide give less than the slab formula, and where must the rest of the deficit lie?

**Solution of Exercise 14.1.**

(a) $2\pi G\,\Delta\rho\,h = 2\pi\times6.674 \times 10^{-11}\times1700\times3000 =
2.14 \times 10^{-3}\,\mathrm{m}\,\mathrm{s}^{-2}$: gravity would be lower by about $214\,\mathrm{m}\mathrm{Gal}$. (b) For $3.5\,\mathrm{km}$, $2\pi\times6.674 \times 10^{-11}\times1700\times3500 = 2.50 \times 10^{-3}\,\mathrm{m}\,\mathrm{s}^{-2}$, $250\,\mathrm{m}\mathrm{Gal}$: the size of the observed low, so the trench is held far from isostatic equilibrium. (c) The slab formula assumes the missing rock extends without limit on every side; a trench only tens of kilometres wide removes much less mass around the point of measurement, so its water alone gives less than $250\,\mathrm{m}\mathrm{Gal}$. The rest of the deficit lies deeper: the whole plate is bent down beneath the trench, and its [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers), lighter than the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers), takes the place of [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) at its base.

**Exercise 14.2 ★.**

The Pacific plate enters the Kuril Trench at $82\,\mathrm{mm}\,\mathrm{yr}^{-1}$. (a) What length of plate goes down in 10 million years? (b) If the slab sank along a straight path dipping $45{}^{\circ}$, how deep would its leading edge then be? (c) How long does it take to reach $600\,\mathrm{km}$?

**Solution of Exercise 14.2.**

(a) $82\,\mathrm{mm}\,\mathrm{yr}^{-1} = 82\,\mathrm{km}\,\mathrm{Myr}^{-1}$, so $820\,\mathrm{km}$ in $10\,\mathrm{Myr}$. (b) $820\sin45{}^{\circ} = 580\,\mathrm{km}$. (c) The path to $600\,\mathrm{km}$ is $600/\sin45{}^{\circ} = 849\,\mathrm{km}$, covered in $849/82 = 10.3\,\mathrm{Myr}$.

**Exercise 14.3 ★.**

Volcanoes stand above the place where a straight slab is $110\,\mathrm{km}$ deep. Compute the [arc–trench gap](#def-b1-active-margins-volcanic-arc) for a slab dipping $30{}^{\circ}$, $45{}^{\circ}$ and $60{}^{\circ}$. Which arc lies farthest from its trench?

**Solution of Exercise 14.3.**

$g = 110/\tan\delta$: $191\,\mathrm{km}$ for $30{}^{\circ}$, $110\,\mathrm{km}$ for $45{}^{\circ}$, $64\,\mathrm{km}$ for $60{}^{\circ}$. The arc above the gently dipping slab lies farthest from its trench.

**Exercise 14.4 ★.**

In the digital plate model of the chapter, about $2.7\,\mathrm{km}^{2}$ of ocean floor go down the subduction boundaries each year, along about $51\,000\,\mathrm{km}$, and about $3.0\,\mathrm{km}^{2}$ are made along about $67\,000\,\mathrm{km}$ of spreading ridges. (a) Compute the mean convergence rate and the mean [full spreading rate](https://one-course.com/books/earth/3/en/chapter/12-plate-kinematics-on-a-sphere#def-b1-plate-kinematics-spreading-rate), in $\mathrm{mm}\,\mathrm{yr}^{-1}$. (b) Over all its boundaries the model destroys as much area as it makes. Where does the rest of the new floor go?

**Solution of Exercise 14.4.**

(a) Convergence: $2.7 \times 10^{6}\,\mathrm{m}^{2}\,\mathrm{yr}^{-1}/5.1 \times 10^{7}\,\mathrm{m} =
0.053\,\mathrm{m}\,\mathrm{yr}^{-1}$, about $53\,\mathrm{mm}\,\mathrm{yr}^{-1}$. Spreading: $3.0 \times 10^{6}\,\mathrm{m}^{2}\,\mathrm{yr}^{-1}/6.7 \times 10^{7}\,\mathrm{m} = 0.045\,\mathrm{m}\,\mathrm{yr}^{-1}$, about $45\,\mathrm{mm}\,\mathrm{yr}^{-1}$. Along a shorter total length, the subduction boundaries consume ocean floor faster than the ridges make it, yet consume less in all. (b) The other $0.3\,\mathrm{km}^{2}$ or so is taken up along the other converging boundaries of the model, such as continental collisions and convergent zones within the oceans, where the plates are shortened rather than subducted; and the ridges are not the only diverging boundaries: [continental rifts](https://one-course.com/books/earth/3/en/chapter/13-rifts-ridges-and-passive-margins#def-b1-rifts-ridges-margins-rift) and some transforms also open a little.

**Exercise 14.5 ★★.**

Four earthquakes of a section lie at $(x, z) = (200, 150)$, $(300, 250)$, $(400, 330)$ and $(500, 430)$, in kilometres. (a) Fit the slab line by least squares and give its [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude). (b) Where does the line reach the surface? (c) Compare with the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) of the line through the first and last earthquakes only.

**Solution of Exercise 14.5.**

(a) $\bar x = 350$, $\bar z = 290$; the deviations of $x$ are $-150$, $-50$, $50$, $150$ and those of $z$ are $-140$, $-40$, $40$, $140$, so $S_{xx} = 2(150^2 + 50^2) = 50\,000$ and $S_{xz} = 2(150\times140 + 50\times40) = 46\,000$. Then $s = 0.92$, $a = 290 - 0.92\times350 = -32\,\mathrm{km}$, and the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) is $\arctan 0.92 = 42.6{}^{\circ}$. (b) $z = 0$ at $x = 32/0.92 = 34.8\,\mathrm{km}$ from the trench. (c) Through the end points the slope is $280/300 = 0.933$, a [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) of $43.0{}^{\circ}$: close, but it uses only two of the four earthquakes.

**Exercise 14.6 ★★.**

A model slab [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $25{}^{\circ}$ from the trench down to $50\,\mathrm{km}$, then $45{}^{\circ}$. The volcanic front stands where the slab is $110\,\mathrm{km}$ deep. (a) Compute the [arc–trench gap](#def-b1-active-margins-volcanic-arc). (b) Compare with a slab dipping $45{}^{\circ}$ all the way. (c) The deep segment, extended upwards, reaches the surface: where? Compare with the Kuril line, which reaches it $53\,\mathrm{km}$ landward of the trench. (d) What gap would a deep segment of $60{}^{\circ}$ give?

**Solution of Exercise 14.6.**

(a) $50/\tan25{}^{\circ} + 60/\tan45{}^{\circ} = 107.2 + 60.0 = 167.2\,\mathrm{km}$. (b) A slab of $45{}^{\circ}$ all the way gives $110\,\mathrm{km}$: the gentle shallow segment widens the gap by $57\,\mathrm{km}$. (c) The deep segment is $z = 50 + (x - 107.2)$; it reaches $z = 0$ at $x = 57.2\,\mathrm{km}$, landward of the trench, as the Kuril line does at $53\,\mathrm{km}$: a gentle shallow segment explains the Kuril offset. (d) $107.2 + 60/\tan60{}^{\circ} = 107.2 + 34.6 =
141.9\,\mathrm{km}$.

**Exercise 14.7 ★★.**

(a) By the bent-shell proposition, what angular radius, and what radius in kilometres ($R = 6371\,\mathrm{km}$), should an arc have above a slab dipping $45{}^{\circ}$? $30{}^{\circ}$? $70{}^{\circ}$? (b) Which arcs are the most tightly curved? (c) The Kuril slab [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) $44.9{}^{\circ}$ and the Kuril Trench fits a circle of $21.9{}^{\circ}$: compute the relative difference from the prediction.

**Solution of Exercise 14.7.**

(a) The radius is half the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude), and $r = R\,\theta$ with $\theta$ in radians: $22.5{}^{\circ}$ and $6371\times0.3927 = 2502\,\mathrm{km}$ for $45{}^{\circ}$; $15{}^{\circ}$ and $1668\,\mathrm{km}$ for $30{}^{\circ}$; $35{}^{\circ}$ and $3892\,\mathrm{km}$ for $70{}^{\circ}$. (b) The arcs above gently dipping slabs. (c) The prediction is $44.9/2 = 22.45{}^{\circ}$; the fit, $21.9{}^{\circ}$, is $0.55/22.45 = 2.4\,\%$ smaller.

**Exercise 14.8 ★★.**

Take a conductivity of $2.5\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$ and a surface temperature of $10\,{}^{\circ}\mathrm{C}$. (a) Compute the surface temperature gradient beneath a [forearc](#def-b1-active-margins-active-margin) losing $35\,\mathrm{mW}\,\mathrm{m}^{-2}$ and beneath a back-arc losing $80\,\mathrm{mW}\,\mathrm{m}^{-2}$. (b) Extrapolating these gradients linearly, estimate the temperature at $30\,\mathrm{km}$ in each. (c) Back-arc [Mohos](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-discontinuity) are measured at 800 to $900\,{}^{\circ}\mathrm{C}$. Why is the linear estimate too high?

**Solution of Exercise 14.8.**

(a) $q/k$: $0.035/2.5 = 0.014\,\mathrm{K}\,\mathrm{m}^{-1} = 14\,\mathrm{K}\,\mathrm{km}^{-1}$ beneath the [forearc](#def-b1-active-margins-active-margin), $0.080/2.5 = 32\,\mathrm{K}\,\mathrm{km}^{-1}$ beneath the back-arc. (b) $10 + 14\times30 = 430\,{}^{\circ}\mathrm{C}$ and $10 + 32\times30 = 970\,{}^{\circ}\mathrm{C}$. (c) Part of the surface [heat flow](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) is produced by radioactivity within the [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) itself, so less heat crosses the deeper levels and the gradient decreases with depth: a linear extrapolation of the surface gradient overestimates the temperature at depth.

**Exercise 14.9 ★★.**

In the plate model of the chapter, the Mariana Trough opens near $18{}^{\circ}$ N at a full rate of about $32\,\mathrm{mm}\,\mathrm{yr}^{-1}$. (a) Express this rate in $\mathrm{km}\,\mathrm{Myr}^{-1}$, and give the half-rate if the spreading is symmetric. (b) How long would the trough take to open by $100\,\mathrm{km}$ at this rate? (c) Give two reasons why the age of a [back-arc basin](#def-b1-active-margins-active-margin) cannot simply be read from its width and its present rate.

**Solution of Exercise 14.9.**

(a) $32\,\mathrm{mm}\,\mathrm{yr}^{-1}$ is $32\,\mathrm{km}\,\mathrm{Myr}^{-1}$; the half-rate is $16\,\mathrm{mm}\,\mathrm{yr}^{-1}$. (b) $100/32 = 3.1\,\mathrm{Myr}$. (c) The rate need not have been constant; part of the width may come from the stretching and rifting of the arc before spreading began; and the axis may have jumped or spread asymmetrically.

**Exercise 14.10 ★★★.**

A slab $100\,\mathrm{km}$ thick sinks at $82\,\mathrm{mm}\,\mathrm{yr}^{-1}$ along a path dipping $45{}^{\circ}$. Take $\kappa = 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$. (a) How long does it take to reach $600\,\mathrm{km}$? (b) How far does heat diffuse into it in that time, by $\sqrt{\kappa t}$? (c) How long would heat take to reach its middle? (d) Conclude about the temperature of the slab at $600\,\mathrm{km}$ and about its earthquakes.

**Solution of Exercise 14.10.**

(a) The path is $600/\sin45{}^{\circ} = 849\,\mathrm{km}$, covered at $82\,\mathrm{km}\,\mathrm{Myr}^{-1}$ in $10.3\,\mathrm{Myr}$, that is $3.27 \times 10^{14}\,\mathrm{s}$. (b) $\sqrt{10^{-6}\times3.27 \times 10^{14}} = 1.8 \times 10^{4}\,\mathrm{m}$, about $18\,\mathrm{km}$. (c) $(5 \times 10^{4}\,\mathrm{m})^2/10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1} = 2.5 \times 10^{15}\,\mathrm{s}$, about $79\,\mathrm{Myr}$. (d) Heat has reached only the outer $18\,\mathrm{km}$ of each side of the slab: its core arrives at $600\,\mathrm{km}$ almost as cold as it left the trench, colder and stronger than the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) around it, which is why it can still break in earthquakes at that depth.

**Exercise 14.11 ★★★.**

A model [accretionary prism](#def-b1-active-margins-accretionary) has a triangular section: its surface slopes at $\alpha = 4{}^{\circ}$ towards the trench and its base [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) at $\beta = 6{}^{\circ}$ beneath it, both meeting at its toe; it keeps these angles as it grows. (a) Show that its area is $A = \frac12 L^2(\tan\alpha + \tan\beta)$ for a width $L$, when the angles are small enough for the height at the back to be $L(\tan\alpha + \tan\beta)$. (b) Sediment $1\,\mathrm{km}$ thick arrives on a plate converging at $40\,\mathrm{mm}\,\mathrm{yr}^{-1}$, and all of it is accreted: compute the width after 1, 4 and 10 million years. How does $L$ grow with time? (c) What if half the sediment goes down with the slab?

**Solution of Exercise 14.11.**

(a) At the distance $L$ behind the toe, the surface stands $L\tan\alpha$ above the level of the toe and the base lies $L\tan\beta$ below it: the back of the prism is $L(\tan\alpha + \tan\beta)$ high, and the triangle has the area $\frac12 L\times L(\tan\alpha + \tan\beta)$. (b) $\tan4{}^{\circ} + \tan6{}^{\circ} =
0.175$. The prism receives $1\,\mathrm{km}\times40\,\mathrm{km}\,\mathrm{Myr}^{-1} =
40\,\mathrm{km}^{2}\,\mathrm{Myr}^{-1}$ of section, so $A = 40t$ and $L = \sqrt{2A/0.175}$: $21.4\,\mathrm{km}$ after $1\,\mathrm{Myr}$, $42.8\,\mathrm{km}$ after $4\,\mathrm{Myr}$, $67.6\,\mathrm{km}$ after $10\,\mathrm{Myr}$. $L$ grows as $\sqrt t$: four times as long gives twice the width. (c) $A$ is halved and $L$ divided by $\sqrt2$: $15.1\,\mathrm{km}$, $30.2\,\mathrm{km}$ and $47.8\,\mathrm{km}$.

**Exercise 14.12 ★★★.**

(a) Show that the least-squares line passes through the mean point $(\bar x, \bar z)$. (b) A catalogue lists every depth $10\,\mathrm{km}$ too deep. How do the slope and the intercept of the fitted line change? (c) The distances $x$ are all measured from a trench point placed $20\,\mathrm{km}$ too far landward, so that every $x$ is $20\,\mathrm{km}$ too small. How do the slope and the intercept change? (d) Does either error change the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude)?

**Solution of Exercise 14.12.**

(a) $\partial F/\partial a = -2\sum(z_i - a - s x_i) = 0$ gives $n\bar z = na + sn\bar x$, that is $\bar z = a + s\bar x$. (b) Every $z_i$ and $\bar z$ grow by $10\,\mathrm{km}$, so the deviations $z_i - \bar z$ and $S_{xz}$ are unchanged: the slope is the same, and $a = \bar z - s\bar x$ grows by $10\,\mathrm{km}$. (c) The deviations $x_i - \bar x$ are unchanged, so the slope is the same; $\bar x$ decreases by 20, so $a$ increases by $20s$. (d) Neither: the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) depends only on the slope. Errors shared by all the points move the line without turning it.

## 14.7 Problem: Reading a subduction zone from its earthquakes

**Problem 14.1.**

Weekend problem — the Kuril slab from eight earthquakes, how fast it goes down, what its gravity and heat flow say, and where its volcanoes should stand

Eight earthquakes of the Kuril slab, all deeper than $70\,\mathrm{km}$, are listed by their distance $x$ from the trench along the section line, towards the Sea of Okhotsk, and their depth $z$, both rounded to the kilometre.

| event | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- |
| $x$ ($\mathrm{km}$) | 155 | 222 | 305 | 319 | 431 | 533 | 578 | 609 |
| $z$ ($\mathrm{km}$) | 96 | 169 | 245 | 272 | 389 | 489 | 518 | 573 |
| magnitude | 5.5 | 5.1 | 5.0 | 5.4 | 5.3 | 5.3 | 5.2 | 6.4 |

**Part I — The slab line.**

1. Plot the eight earthquakes at the same scale on both axes, depth downwards.
2. Compute $\sum x_i$ , $\sum z_i$ , $\sum x_i^2$ , $\sum x_i z_i$ and the means $\bar x$ and $\bar z$ .
3. Compute $S_{xx}$ and $S_{xz}$ , then the slope of the least-squares line to three decimals and its intercept to one decimal.
4. What is the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) of the slab?
5. Where does the line reach the surface? What does this say about the shape of the slab near the trench?
6. Compute the misfit $z_i - (a + s x_i)$ of each earthquake. Which lie farthest from the line, and how far?

**Part II — How fast the slab goes down.** At the start of the section the PB2002 model gives a convergence of $87.4\,\mathrm{mm}\,\mathrm{yr}^{-1}$: relative to the Okhotsk plate, the Pacific plate moves towards [azimuth](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) $300{}^{\circ}$. The section line, perpendicular to the trench, runs towards [azimuth](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) $320.9{}^{\circ}$.

7. Compute the angle between the convergence and the section line, then the components of the convergence perpendicular and parallel to the trench.
8. What length of plate goes down the trench in 10 million years?
9. Using the [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) of question 4 (rounded to $45{}^{\circ}$ ), how long does the slab take to carry a point of its top from the trench down to the deepest earthquake of the band, $632\,\mathrm{km}$ ? Neglect the gentler [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) near the trench.
10. The plate is about $122\,\mathrm{Myr}$ old at the trench. How old is the sea floor that is now $632\,\mathrm{km}$ deep?
11. By $\sqrt{\kappa t}$ , with $\kappa = 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$ , how far has heat diffused into the slab during that descent? Compare with its thickness, about $100\,\mathrm{km}$ .

**Part III — Gravity and heat flow.** Take a conductivity of $2.5\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$.

12. The old ocean floor in front of the trench loses about $43\,\mathrm{mW}\,\mathrm{m}^{-2}$ , the back-arc about $80\,\mathrm{mW}\,\mathrm{m}^{-2}$ ; take $35\,\mathrm{mW}\,\mathrm{m}^{-2}$ for the [forearc](#def-b1-active-margins-active-margin) . Compute the temperature gradient near the surface of each.
13. Explain why the [forearc](#def-b1-active-margins-active-margin) loses less heat than the old ocean floor in front of it.
14. The [free-air anomaly](https://one-course.com/books/earth/3/en/chapter/11-isostasy-and-gravity-anomalies#def-b1-isostasy-gravity-anomalies) over the Kuril Trench is about $-260\,\mathrm{m}\mathrm{Gal}$ . What thickness of rock, $1700\,\mathrm{kg}\,\mathrm{m}^{-3}$ denser than water, would a Bouguer slab need to give this deficit? Compare with the depth of the trench below the floor around it, about $3.5\,\mathrm{km}$ .
15. Suppose now that the whole plate is pushed down by $3.5\,\mathrm{km}$ : water takes the place of rock at its top, and its [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) takes the place of [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) at its base, so that, in all, [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) of density $3300\,\mathrm{kg}\,\mathrm{m}^{-3}$ is replaced by water. Compute the Bouguer slab anomaly. Why does the real trench, only tens of kilometres wide, give less?
16. The trench and the outer rise are not in isostatic equilibrium. What holds them out of it?

**Part IV — Where the volcanoes should stand.** Use the line of question 3, with its rounded slope and intercept.

17. The first volcano of the band is $164\,\mathrm{km}$ from the trench. How deep is the slab beneath it?
18. Is this depth within the range found beneath arcs, 72 to $173\,\mathrm{km}$ ? How far is it from the average, $105\,\mathrm{km}$ ?
19. At what distance from the trench is the slab $100\,\mathrm{km}$ deep?
20. By the bent-shell proposition, what angular radius should the Kuril arc have, and what radius in kilometres?
21. The Kuril Trench fits a [small circle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-spherical-triangle) of $21.9{}^{\circ}$ . Compare.
22. Arc volcanoes stand, on average, where the slab is about $105\,\mathrm{km}$ deep. At what distance from the trench does the slab line reach that depth?
23. Compare this distance with the observed $164\,\mathrm{km}$ : by how many kilometres and by what fraction does it differ?
24. To the nearest kilometre, what [arc–trench gap](#def-b1-active-margins-volcanic-arc) does the Kuril slab line predict for volcanoes standing above the slab at $105\,\mathrm{km}$ ?

**Solution of Problem 14.1.**

**1.** The eight points lie close to a straight line dipping about $45{}^{\circ}$, from $96\,\mathrm{km}$ deep at $155\,\mathrm{km}$ from the trench to $573\,\mathrm{km}$ deep at $609\,\mathrm{km}$.

**2.** $\sum x_i = 3152$, $\sum z_i = 2751$, $\sum x_i^2 = 1\,442\,910$, $\sum x_i z_i = 1\,290\,548$; $\bar x = 394.0\,\mathrm{km}$, $\bar z = 343.875\,\mathrm{km}$.

**3.** $S_{xx} = 1442910 - 8\times394^2 = 201\,022$ and $S_{xz} = 1290548 - 8\times394\times343.875 = 206\,654$. Then $s = 206654/201022 = 1.028$ and $a = 343.875 - 1.02802\times394 =
-61.2\,\mathrm{km}$: $z = -61.2 + 1.028\,x$.

**4.** $\arctan1.028 = 45.8{}^{\circ}$.

**5.** At $x = 61.2/1.028 = 59.5\,\mathrm{km}$ landward of the trench. A slab that left the trench at this [dip](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) would reach the surface there; it reaches it at the trench, so it [dips](https://one-course.com/books/earth/3/en/chapter/4-geological-maps#def-b1-geological-maps-attitude) more gently near the trench and steepens below.

**6.** The misfits are $-2.1$, $2.0$, $-7.3$, $5.3$, $7.1$, $2.3$, $-15.0$ and $8.1$ $\mathrm{km}$. Event 7 lies $15\,\mathrm{km}$ above the line, event 8 $8\,\mathrm{km}$ below it; the others lie within about $7\,\mathrm{km}$ of it, a fraction of the thickness of the slab.

**7.** $320.9{}^{\circ} - 300{}^{\circ} = 20.9{}^{\circ}$. Perpendicular to the trench: $87.4\cos20.9{}^{\circ} = 81.6\,\mathrm{mm}\,\mathrm{yr}^{-1}$; parallel: $87.4\sin20.9{}^{\circ} = 31.2\,\mathrm{mm}\,\mathrm{yr}^{-1}$. The convergence is oblique.

**8.** $81.6\,\mathrm{mm}\,\mathrm{yr}^{-1}$ is $81.6\,\mathrm{km}\,\mathrm{Myr}^{-1}$: $816\,\mathrm{km}$ in $10\,\mathrm{Myr}$.

**9.** The path is $632/\sin45{}^{\circ} = 894\,\mathrm{km}$, covered in $894/81.6 = 11.0\,\mathrm{Myr}$.

**10.** $122 + 11 = 133\,\mathrm{Myr}$.

**11.** $t = 10.95\,\mathrm{Myr} = 3.46 \times 10^{14}\,\mathrm{s}$ and $\sqrt{10^{-6}\times3.46 \times 10^{14}} = 1.86 \times 10^{4}\,\mathrm{m}$, about $19\,\mathrm{km}$ from each face: far less than the $50\,\mathrm{km}$ to the middle of a slab $100\,\mathrm{km}$ thick, whose core is still cold.

**12.** $q/k$: $17.2\,\mathrm{K}\,\mathrm{km}^{-1}$ beneath the old ocean floor, $14.0\,\mathrm{K}\,\mathrm{km}^{-1}$ beneath the [forearc](#def-b1-active-margins-active-margin) and $32.0\,\mathrm{K}\,\mathrm{km}^{-1}$ beneath the back-arc.

**13.** Under the [forearc](#def-b1-active-margins-active-margin) the cold slab lies only a few tens of kilometres down; it takes up heat from the rock above it, so less heat reaches the surface.

**14.** $h = 2.60 \times 10^{-3}\,\mathrm{m}\,\mathrm{s}^{-2}/(2\pi\times6.674 \times 10^{-11}\times1700)
= 3647\,\mathrm{m}$, about $3.6\,\mathrm{km}$: as much as, or a little more than, the whole depth of the trench below the floor around it, about $3.5\,\mathrm{km}$. The trench is missing mass that nothing compensates.

**15.** The density difference is $3300 - 1000 = 2300\,\mathrm{kg}\,\mathrm{m}^{-3}$, and $2\pi\times6.674 \times 10^{-11}\times2300\times3500 = 3.38 \times 10^{-3}\,\mathrm{m}\,\mathrm{s}^{-2}$, $338\,\mathrm{m}\mathrm{Gal}$. The slab formula counts the missing mass out to any distance; the real trench is bent down only over a band tens of kilometres wide, so its deficit is smaller, about $260\,\mathrm{m}\mathrm{Gal}$, between the two slab estimates.

**16.** The strength of the bending plate, which holds the trench down and the outer rise up as an elastic plate would, and the pull of the dense slab sinking into the [mantle](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers).

**17.** $-61.2 + 1.028\times164 = 107.4\,\mathrm{km}$.

**18.** Yes: $107\,\mathrm{km}$ lies between 72 and $173\,\mathrm{km}$, and only about $2\,\mathrm{km}$ deeper than the average, $105\,\mathrm{km}$.

**19.** $x = (100 + 61.2)/1.028 = 156.8\,\mathrm{km}$.

**20.** $45.8{}^{\circ}/2 = 22.9{}^{\circ}$, and $6371\times22.9\times\pi/180 = 2546\,\mathrm{km}$.

**21.** The fit, $21.9{}^{\circ}$, is $1.0{}^{\circ}$ smaller, about $4\,\%$: the bent shell describes the curvature of the Kuril arc well.

**22.** $x = (105 + 61.2)/1.028 = 166.2/1.028 = 161.7\,\mathrm{km}$.

**23.** $164 - 161.7 = 2.3\,\mathrm{km}$ nearer the trench than the observed front, $2.3/164 = 1.4\,\%$: the Kuril volcanoes stand above a slab very slightly deeper than the average, as question 18 found.

**24.** To the nearest kilometre, the Kuril slab line predicts an [arc–trench gap](#def-b1-active-margins-volcanic-arc) of **$162\,\mathrm{km}$** for volcanoes above the slab at $105\,\mathrm{km}$ (observed: $164\,\mathrm{km}$).