The geomorphology of active normal faulting records a competition between accumulated tectonic displacement and the ability of erosion, transport and deposition to modify it. A hanging wall moves down relative to a footwall, producing a fault scarp, mountain-front relief, and accommodation space for a sedimentary basin. Repeated earthquakes can build a large front even though each individual surface break is much smaller.^[ref]

The principal controls and their observable effects are:

Useful observations therefore include scarp profiles, repeated offsets of dated surfaces, drainage organization, and continuity along strike. A high rugged front is not by itself evidence of a high present fault slip rate; conversely, a muted or buried scarp need not mark an inactive fault.

The $31.7^{\circ }$ result predicts initial failure in a uniform material under the assumed principal-stress orientations. The actual Figure 3 is a present-day NW–SE cross section through a thick, heterogeneous Alpine stack, extending to about 50 km depth. It contains near-horizontal basal stretches, curved ramps, tightly bent contacts, and steep structures in the internal zones around Viso and the Dora Maira complex. It does not show a single undeformed conjugate pair newly formed in horizontal uniform rock.

Several processes explain the departures:

Finally, a section shows apparent dip unless it is perpendicular to fault strike. For an ideal plane, $\tan {\delta }_{\mathrm{app}}=\tan {\delta }_{\mathrm{true}}\sin \alpha$, where $\alpha$ is the angle between the section direction and fault strike. Obliquity can make a plane look shallower, but cannot on its own explain an apparent dip steeper than its true dip. Do not infer that each black curve is a currently active brittle earthquake fault.

Assume an initially unfractured homogeneous isotropic material described by a linear Mohr-Coulomb failure criterion, with the given coefficient of friction ${\mu }_f=0.5$ representing the relevant internal friction angle. Use spatially uniform pore fluid pressure, effective stresses ${\sigma }_i^{\prime }={\sigma }_i-P_f$, and compression-positive signs. ThenThe Mohr circle in the ${\sigma }_1^{\prime },{\sigma }_3^{\prime }$ plane has centre $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and radius $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$. At first failure it is tangent to the failure envelope. The radius to the positive-shear tangent point is perpendicular to the line of slope $\tan \phi$, hence makes angle $90^{\circ }+\phi$ with the positive normal-stress axis.

If $\theta$ is the angle between the fault normal and ${\sigma }_1$, the stress transformation isThus $2\theta =90^{\circ }+\phi$, $\theta =58.2825^{\circ }$, and the angle of the fault itself to horizontal ${\sigma }_1$ isThe two conjugate thrust faults therefore strike approximately east–west and dip about $32^{\circ }$ north and $32^{\circ }$ south. This is the fault dip, not the $58.3^{\circ }$ fault-normal angle. Both sediments and basement have the same optimum orientation under these assumptions because their stipulated friction slopes are equal; their cohesion and absolute failure stresses could still differ.

The first two panels in solution give the labeled construction. Constant cohesion shifts the straight failure envelope without changing its slope, and uniform pore fluid pressure translates the Mohr circle in normal-stress coordinates; neither changes this tangency angle. In the illustrative cohesionless case, $R/C=\sin \phi$, giving ${\sigma }_1^{\prime }/{\sigma }_3^{\prime }=(1+\sin \phi )/(1-\sin \phi )\simeq 2.618$. Pre-existing weaknesses, a nonuniform stress field, anisotropic strength, or a nonlinear failure envelope invalidate the simple optimum-orientation prediction.^[ref]

A reverse or thrust fault earthquake is the expected end member: a hanging wall moves upwards relative to its footwall, accommodating north–south tectonic shortening. A plausible double-couple seismic source focal mechanism has an approximately east–west fault strike, a moderately north- or south-dipping fault plane, and up-dip fault slip. For example, right-hand-rule strike $090^{\circ }$, southward fault dip about $32^{\circ }$, and fault rake $+90^{\circ }$ specify a pure thrust mechanism.

In a lower-hemisphere equal-area plot, with shaded regions denoting compressional P first motions, this focal mechanism has a shaded central region and two unshaded outer lobes. The P axis is approximately horizontal north–south and the T axis approximately vertical for a symmetric idealization; for a single $32^{\circ }$ plane their axes are slightly tilted. The P and T axes of a double-couple seismic source need not exactly equal the regional principal stress axes. The nodal curves represent both the fault plane and its auxiliary plane; additional evidence is needed to choose between them.^[ref]

Use compression-positive principal stresses, ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$. For the intended thrust fault regime during approximately north–south tectonic shortening, the likely orientations areA locally horizontal traction-free surface motivates taking one principal direction as vertical, while the strong north–south compression selects the maximum horizontal principal stress. The expected fault strike is then approximately east–west.

This ordering assumes the two horizontal compressions exceed the vertical principal stress. North–south shortening by itself does not establish the magnitudes of all three: if the vertical principal stress were intermediate, strike-slip faulting would instead be possible. The answer above is the simple compressional end member used in the following parts.

Assume that the intrusion has a positive magnetic susceptibility contrast, that induced magnetization dominates its remanent magnetization, and that the local present-day geomagnetic field points roughly northward and downward, as in this northern mid-latitude setting. Writewhere $\mathbf{b}$ is the anomalous magnetic field and the last expression assumes $|\mathbf{b}|\ll |{\mathbf{B}}_0|$. A total-field magnetic anomaly measures a field projection, so it need not have the same sign everywhere above a high-susceptibility body.^[ref]

The 100 km east–west extent makes an approximately two-dimensional, east–west-striking body a useful model away from its ends. On a profile drawn north on the left, south on the right, the expected normally induced response has a subsidiary negative lobe on the northern side and a stronger positive lobe over the body, displaced towards the south. The northward horizontal field component makes the anomaly asymmetric; its downward component produces the dominant positive response. The field falls back towards the regional value away from the intrusion. Broad bodies can have edge shoulders or a broad central high, so the sketch represents a qualitative shape rather than one uniquely specified curve.

An exactly vertical inducing field would remove this north–south skew for a symmetric induced body. Real magnetic inclination and magnetic declination should be taken from a suitable geomagnetic field model for the survey date and location; no exact local value is supplied here. Strong oblique or reversed remanent magnetization can shift or reverse the lobes, and heterogeneous magnetic susceptibility, body dip and three-dimensional end effects can alter the profile. Higher magnetic susceptibility alone does not determine a unique amplitude or polarity pattern without these qualifications.

For the two negative gravity anomalies in the actual Figure 1, examine where the steepest flanks occur within the total fall into the trough. The solid upper intrusion profile has maximum slope relatively low on its flanks, close to the flat trough; its body widens downwards. The lower basin profile has maximum slope relatively high on its flanks, near the shoulders; its body narrows downwards. The dashed gradient traces and arrows mark these transitions. The signed derivative changes sign on opposite sides of a symmetric trough; a plot of gradient magnitude instead gives two positive peaks.

A shallow edge controls a relatively abrupt local change, whereas deeper geophysical density contrast spreads its effect over greater horizontal distance. Downward-narrowing geometry leaves relatively more of the anomaly associated with the shallow broad part, placing the strong flank change nearer the outer shoulders. Downward-widening geometry adds a broader deep contribution beneath the narrow outcrop, shifting the inflection towards the lower part of the trough. Thus the location of a flank's maximum slope relative to its shoulder and minimum is more useful here than the shared $-40\mathrm{mGal}$ peak alone.

A symmetric constant-density body with vertical sides gives a symmetric negative trough with paired rounded flanks; for a broad shallow body the main slope changes occur approximately halfway down those flanks. It is intermediate between the two end members, rather than a triangular anomaly or a rectangular discontinuity. Exact fractional inflection heights depend on width, depth, burial and rounding. The following independently calculated infinite-strike models illustrate the distinction without copying Figure 1.

This qualitative inference assumes approximately comparable simple bodies, uniform geophysical density contrast, and a properly removed regional field. Gravity inversion is non-unique: lateral density changes, multiple bodies, and a buried top can imitate changes of wall slope. Use the seismic reflection constraints to test the interpretation.

Use the same surrounding basement density inferred in solution, carrying its unrounded value through the arithmetic. The intrusion's geophysical density contrast isThe Bouguer slab thickness needed to produce the same negative peak is thereforeThe smaller absolute geophysical density contrast requires a thicker body. This is an approximate vertical thickness and, if the top is close to the surface as drawn in Figure 1, an approximate base depth below that top. It is not an independent estimate of the intrusion's burial depth. For an infinite slab the anomaly does not depend on top depth; a buried top would require additional geometry. Finite width and the downward-widening outline in Figure 1 make 8 km a slab-equivalent estimate rather than a uniquely determined geological depth.

For a wide horizontal Bouguer slab replacing surrounding basement, use the density contrast $\Delta \rho ={\rho }_{\mathrm{sed}}-{\rho }_{\mathrm{base}}$, not the absolute sediment density:Take downward attraction as positive. The negative anomaly therefore requires $\Delta \rho <0$. With $\Delta g=-40\mathrm{mGal}=-4.0\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$, $h=3000\mathrm{m}$, and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$,soThis is the requested Bouguer slab approximation: a flat observation surface, laterally infinite constant-thickness material, uniform average geophysical density contrast, and a regional anomaly already removed. The actual basin in Figure 1 narrows downwards and is finite. For a fixed maximum thickness, finite width reduces the central anomaly magnitude, so this calculation can underestimate the true magnitude of the average geophysical density contrast. It does not establish a unique basement density without the geometrical assumption.

A gravity anomaly is the difference between observed gravitational acceleration and a specified reference prediction after defined reductions. A free-air gravity anomaly, a Bouguer gravity anomaly, and a residual local gravity anomaly differ because they remove different reference contributions. State the reference ellipsoid, datum, reference density, and regional trend before interpreting an anomaly as a subsurface geophysical density contrast.

On land, tie gravimeter observations to a gravity base, calibrate them, correct instrumental drift and Earth tides, and determine station latitude and elevation accurately. Subtract latitude-dependent normal gravity $\gamma (\phi )$. Reduce to the reference elevation using a positive free-air correction for a station above that datum: about $+0.3086h\mathrm{mGal}$ for $h$ in metres. A simple Bouguer gravity anomaly then subtracts attraction of the intervening rock slab and adds the conventional terrain correction:with all acceleration terms expressed consistently. In SI units $1\mathrm{mGal}=10^{-5}\mathrm{m}{\mathrm{s}}^{-2}$.

At sea, also correct vessel-motion effects, especially the Eötvös correction from east–west vessel motion and Earth rotation, and vertical ship acceleration; average or filter the moving-platform data appropriately. GPS navigation and tide information specify position and measurement height. Sea-surface measurements give a marine free-air gravity anomaly after motion, latitude, tide and height reductions. A marine Bouguer gravity anomaly additionally allows for measured bathymetry and the replacement of low-density water by reference rock; the simple water-column replacement correction is positive, $2\pi G({\rho }_{\mathrm{ref}}-{\rho }_w)H$, relative to the sea-surface free-air value. A consistent three-dimensional bathymetric correction is preferable near irregular coasts. Remove a justified broad regional field if the intended quantity is a residual anomaly of the local body.

The three stable fields areThe bounding equalities give the stable two-mineral coexistences. In particular, the stable corundum region lies below the lower of the two oblique lines; treating every line extension as stable would mislabel parts of the mineral activity diagram.

A stability boundary in a mineral activity diagram must also be stable against the third mineral. Use the common-potential test for a three-mineral activity diagram. For the two corundum-producing directions,Negative values favour corundum. For the albite-to-nepheline conversion, ${\Delta }_{r}G/(RT\ln 10)=2x+11.03$, so low silica activity favours nepheline and high silica activity favours albite. These comparisons give:

All three coexist at the intersection. Pairwise mass-action equations alone do not distinguish the stable segments from their metastable extensions.

The net reaction quotient is $a_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}^2$ when the solid and water activities equal one. Therefore its equilibrium line isa vertical line through the intersection of the other two. There is no dependence on $y$, because the net reaction contains neither hydrogen ions nor sodium ions.

Subtract the nepheline–corundum reaction from the albite–corundum reaction and divide by two. In the albite-to-nepheline direction the resulting net reaction isAdditivity of the standard Gibbs energy of reaction makes the logarithms of the equilibrium constants combine with the same coefficients:This normalization is per one albite formula unit. The doubled reaction has ${\log }_{10}K=-22.06$, and reversing the reaction changes its sign.

Use base-ten logarithms and dimensionless thermodynamic activities at the stated temperature and pressure. Define $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{\mathrm{N}{\mathrm{a}}^{+}}/a_{{\mathrm{H}}^{+}})$. The concentration notation is an ideal dilute-solution approximation with a consistent concentration reference.

With pure solids and unit water activity, the two reaction quotients are $10^{2y+2x}$ and $10^{2y+6x}$. Equating them to the corresponding equilibrium constants givesThe slopes on the mineral activity diagram are $-1$ and $-3$. The lines meet at $(x_{\ast },y_{\ast })=(-5.515,8.955)$, within the negative silica-logarithm range.