The Gold Standard of Volcanology | Separating Source Depth from Volume Change in Surface Deformation

A displacement is a change in position relative to a specified reference and interval. Vertical uplift, horizontal motion and motion along a satellite’s viewing direction are related quantities, but they are not interchangeable readings of one number.

Begin with a fictional flat surface and one source directly beneath its coordinate origin. A station’s horizontal distance from the point above the source is r. Depth d is positive downward. Vertical displacement w is positive upward, while horizontal radial displacement u is positive away from the origin.

These conventions are part of the model. A negative vertical value would describe subsidence in this coordinate system. Reversing an instrument’s reported sign without updating the explanation can turn a contraction account into an expansion account on paper.

The first learning checkpoint is a labelled diagram described in words: a source at depth d, a surface station at radial distance r, and two displacement components. Nothing in a single reported ten-millimetre value supplies all those labels automatically. Define the component, reference, interval and geometry before interpreting its magnitude.

Contents · Next chapter

Use the small spherical-source, elastic half-space model commonly associated with Mogi deformation. It treats the surface as flat and the surrounding medium as homogeneous and isotropic. The source is small relative to its depth, and the response is represented elastically.

The University of South Florida model explanation provides the displacement equations and their assumptions. Our numerical records are independent fictional constructions using that mathematical framework.

Let ν denote Poisson’s ratio and ΔV the model’s source volume-change parameter. Define C = (1 − ν)ΔV/π. Then w(r) = Cd/(r² + d²)^(3/2), and u(r) = Cr/(r² + d²)^(3/2). Use consistent length units throughout.

The equations are forward relations: supplied parameters produce predicted displacements. Reading parameters back from observations is the inverse task. A neat forward formula does not guarantee that every inverse question has one answer. The available components, station geometry and independently known quantities determine what can be recovered within this restricted model.

Contents · Previous chapter · Next chapter

C has dimensions of volume. Multiplying it by d gives length to the fourth power. The denominator has length cubed, leaving a displacement measured in length. The horizontal equation has the same dimensional structure.

If depth and distance are entered in metres, C must use cubic metres to obtain displacement in metres. A ten-millimetre displacement is 0.010 metres. Substituting ten as though it were metres would make the inferred volume-change parameter one thousand times larger.

A dimensional check does not establish that the model fits a real system. It protects a more basic relationship: the calculation should at least describe quantities with compatible units. Correct arithmetic on incompatible units remains an incorrect explanation.

For practice, rewrite the model using kilometres consistently. The numerical volume unit becomes cubic kilometres, and displacement initially emerges in kilometres before conversion. The physical prediction should agree with the metre calculation. If it does not, inspect the conversion rather than adjusting a source parameter to absorb the error. A model fit can otherwise conceal a unit mistake behind apparently plausible parameters.

Contents · Previous chapter · Next chapter

At r = 0, the vertical equation becomes w₀ = C/d². Thus C = w₀d². If ν is known, ΔV = πw₀d²/(1 − ν).

This is already the central inverse relationship. A supplied w₀ constrains a combination of depth and source strength. Without another depth constraint, every allowed positive d has a corresponding C that produces the same on-axis uplift.

Use ν = 0.25 and w₀ = 0.010 metres for the teaching case. At d = 2,000 metres, C = 40,000 cubic metres and ΔV is approximately 167,551.608 cubic metres. These are exact model relationships followed by rounded numerical values.

Do not call C the measured chamber volume. It is a grouped parameter defined from a model volume change and an assumed material ratio. Nor is ΔV the complete original size of a cavity. The distinction between a source’s size and its change will matter later. A parameter’s convenient name should not make its physical interpretation broader than the equation supports.

Contents · Previous chapter · Next chapter

Keep ν = 0.25 and w₀ = 0.010 metres, but choose d = 4,000 metres. Then C = 160,000 cubic metres and ΔV is approximately 670,206.433 cubic metres.

Both models predict exactly ten millimetres directly above the source. Doubling depth requires four times the C and four times the ΔV in this on-axis comparison. The observation has not chosen between them.

The table is a constructive non-uniqueness example. It does not claim that these are the only models, that either describes a real volcano or that the parameters satisfy every possible alternative physical constraint. It shows that this one observation, under the stated framework, leaves a continuum of compatible depth-strength pairs. More decimal places on the ten-millimetre reading do not remove that structural ambiguity.

Contents · Previous chapter · Next chapter

Suppose an instrument reports the on-axis uplift with very small measurement uncertainty. That improves knowledge of w₀. It narrows the relationship C = w₀d² around the measured amplitude, but it does not independently supply d.

A perfectly known w₀ still permits the two exact models in the table. The missing information is not another decimal place on the same quantity. It is evidence with a different relationship to the unknown parameters.

This distinction is useful when reading parameter reports. A narrow numerical uncertainty obtained after holding depth fixed does not necessarily describe uncertainty when depth is allowed to vary. The conditional statement may be legitimate, but its condition must remain visible.

For a paper exercise, ask students to calculate C for three selected depths using the same w₀. Then repeat with w₀ slightly higher and lower. They should see a family of curves or relationships, not a single answer made inevitable by a precise observation. Identification asks whether different parameter choices remain observationally compatible. Precision asks how tightly an observed or estimated quantity is constrained within its stated model.

Contents · Previous chapter · Next chapter

Divide the vertical displacement at r by the on-axis value. The grouped amplitude C cancels, giving q(r) = w(r)/w₀ = [1 + (r/d)²]^(-3/2), provided w₀ is non-zero and both observations describe the same source change and reference interval.

The ratio depends on relative distance r/d. Within this model, it describes the breadth of the vertical deformation pattern independently of its amplitude. A shallow source produces a more concentrated pattern; a deeper source produces a broader one when the on-axis uplift is held fixed.

This statement is conditional on the supplied geometry and model. An unknown source centre changes r, and additional sources can change the pattern. A spatial ratio does not automatically separate every possible geological explanation.

The learning move is to identify what cancellation removes and what it leaves. C disappears because both measurements share it multiplicatively. Depth remains because it also shapes the spatial variation. Cancelling a common factor can create a useful discriminator, provided the measurements genuinely share that factor. Incompatible dates or reference corrections can destroy the relationship the ratio assumes.

Contents · Previous chapter · Next chapter

Place a second fictional station 2,000 metres from the point above the source. Model A has d = 2,000 metres, so r/d = 1. It predicts q = 2^(-3/2), approximately 0.353553, and w about 3.535534 millimetres.

Model B has d = 4,000 metres, so r/d = 0.5. It predicts q = 1.25^(-3/2), approximately 0.715542, and w about 7.155418 millimetres. The previously identical on-axis predictions now differ substantially at the second position.

This is why a changed observation geometry can be more informative than repeating the same amplitude many times. The models disagree on a quantity that the additional record can examine. The exercise concerns supplied hypothetical observations, rather than instructions for designing a real monitoring installation.

Ask the learner to predict which model has the larger off-axis uplift before calculating. They should refer to the broader depth-normalised shape while retaining the shared ten-millimetre centre value. A vague statement that deeper sources always create more uplift would miss the comparison’s condition: source strength was changed to keep the centre amplitude fixed.

Contents · Previous chapter · Next chapter

For r > 0 and a positive finite depth, the model ratio satisfies zero < q < 1. Rearranging q = [1 + (r/d)²]^(-3/2) gives d = r/√[q^(-2/3) − 1].

Substituting r = 2,000 metres and q = 2^(-3/2) returns d = 2,000 metres. Once d is identified under these assumptions, use C = w₀d² and then the stipulated ν to obtain ΔV.

The order matters. The ratio supplies a shape-based depth constraint; the amplitude then supplies a strength constraint. Treating both original displacements as independent direct measurements of volume would obscure that structure.

Values at or outside the allowed ratio range need investigation. A ratio above one cannot be explained by the model’s monotonic positive vertical pattern around its assumed centre. It could reflect noise, an incorrect centre, incompatible intervals or an unsuitable model. It should not be forced into the square root by deleting its sign or clipping the ratio without explanation. A failed inversion can reveal an assumption worth examining.

Contents · Previous chapter · Next chapter

Another shape description is the radius where vertical uplift reaches half its on-axis value. Set q = 1/2. Rearrangement gives r½ = d√[2^(2/3) − 1], approximately 0.766421d.

For d = 2,000 metres, the half-amplitude radius is about 1,532.842 metres. A depth estimate based on this restricted shape would therefore divide a supplied half-amplitude radius by approximately 0.766421. The width convention must be stated: a radius is not the same as the full distance across both sides of the profile.

This example helps prevent a visual shortcut. A broad plotted curve can appear narrow if its horizontal axis is rescaled. A physically meaningful width requires actual distances, a named amplitude level and a consistent centre.

Nor does identifying a half-amplitude point make the model universal. A non-spherical source, topography or material heterogeneity can alter the pattern. The width-to-depth relationship belongs to the stated forward model. A learner should be able to derive the coefficient from that model rather than treat it as an empirical rule applicable to every volcanic displacement map.

Contents · Previous chapter · Next chapter

A record directly above the source may be unavailable. Two different known radial distances can still constrain depth within the same restricted model. Let r₂ > r₁ and define q = w(r₂)/w(r₁) for compatible non-zero measurements.

Then q = [(r₁² + d²)/(r₂² + d²)]^(3/2). Write t = q^(2/3). Solving gives d² = (tr₂² − r₁²)/(1 − t). The numerator and denominator must support a positive finite depth.

For a fictional case, use r₁ = 1,000 metres, r₂ = 3,000 metres and t = 5/13. The calculation gives d² = 4,000,000 square metres and d = 2,000 metres. The corresponding displacement ratio is (5/13)^(3/2).

The model has not gained information merely because two numbers were supplied. Their distances must differ in a useful way, and the observations must describe a shared source and interval. Two duplicated positions provide the same shape information, even if their files have different names. Geometry, identity and compatibility remain part of the inference.

Contents · Previous chapter · Next chapter

Using the off-axis example with d = 2,000 metres, suppose the first station’s vertical displacement is approximately 7.155418 millimetres. Return to the original forward relation and solve C = w(r₁)(r₁² + d²)^(3/2)/d.

Keeping the unrounded supplied model value gives C = 40,000 cubic metres. With ν = 0.25, ΔV again equals approximately 167,551.608 cubic metres. Predicting the second station from these parameters gives about 1.706770 millimetres.

A useful verification has two stages. First reconstruct each observation using the inferred parameters. Then ask whether additional observations, not used in that reconstruction, also follow the predicted pattern. The first checks algebra and compatibility with the supplied points. The second can test aspects of the model beyond the fitted equations.

Neither stage proves a unique real geological mechanism. Other source geometries or material structures may reproduce similar records. The result should therefore be reported as parameters inferred under this model, together with its residuals and assumptions. A perfect fit to exactly enough supplied equations is a weaker model check than successful prediction of independent observations.

Contents · Previous chapter · Next chapter

The horizontal and vertical equations share the same denominator and grouped amplitude. At a station with non-zero vertical motion, their ratio is u/w = r/d. If r is independently known and both components are reliably referred to the same source change, d = rw/u.

At r = d = 2,000 metres, the model predicts equal positive radial and vertical components. With C = 40,000 cubic metres, both are approximately 3.535534 millimetres. This supplies another route to the same depth within the restricted framework.

At r = 0, horizontal radial motion is zero by symmetry. The ratio cannot provide depth there. This is a useful example of an additional measurement component that contributes no new identification information at a special geometry.

Do not interpret horizontal displacement magnitude as a radial component unless its direction has been resolved relative to the assumed centre. An east-west component can differ from the outward radial component at a station elsewhere around the source. The equations describe particular directions, not every number that happens to be reported in millimetres.

Contents · Previous chapter · Next chapter

A satellite line-of-sight observation combines displacement components according to the viewing geometry. It is not automatically a vertical uplift measurement. In a deliberately simplified two-component exercise, let the reported projection be l = 0.5u + 0.866025w.

If the actual model components satisfy u = w, then l = 1.366025w. Calling l the vertical displacement would overstate w by approximately 36.6% in this invented geometry. A different orientation or component sign changes the relationship.

The exercise does not describe a particular satellite, orbit or real monitoring record. It shows why measurement operators belong in the model. A projection may be very accurately measured while leaving its separate components unresolved.

For learning, supply l alone and ask students to construct two different u,w pairs satisfying the equation. Then add the restricted deformation relation and a known station geometry. They should explain which new assumption permits further separation. A data product’s familiar colour scale cannot replace the mathematical relationship between what was observed and the physical component being inferred.

Contents · Previous chapter · Next chapter

So far, the source’s horizontal position has been supplied. If it is unknown, each station’s radial distance depends on that additional parameter. A two-point depth calculation that treats guessed distances as measured facts can produce unwarranted certainty.

Imagine the same two surface stations with an assumed centre midway between them. Now shift the proposed centre. Their r values change, so the same displacement ratio can imply a different depth-strength combination. Additional spatial observations may help constrain the centre, but their usefulness depends on geometry and model adequacy.

This distinction explains why a clean algebraic inversion is conditional. It can be exactly correct after specifying the centre while leaving centre uncertainty unaddressed. Report the supplied geometric information rather than silently treating it as known.

For a paper exercise, give two competing fictional centre locations and ask the learner to recalculate the distances before attempting inversion. They should identify that the observation has not changed; the model geometry has. This is a practical way to separate measurement uncertainty from uncertainty about the configuration used to interpret the measurement.

Contents · Previous chapter · Next chapter

In the on-axis ratio formula, q approaches one as depth becomes very large relative to r. The inferred depth therefore grows without bound as q approaches one from below. A small change near one can produce a large depth change.

That is a sensitivity issue within a valid model, rather than proof that a deep source has actually been found. If measurement uncertainty permits q to reach one, the data may not supply a finite upper depth bound under this relationship.

A ratio near zero creates a different limit: the model depth becomes small relative to the fixed off-axis distance. The small-source assumption must still be assessed separately; allowing d to shrink without checking the source scale can leave the model’s intended regime.

Ask the learner to distinguish three statements: the ratio is mathematically admissible, its inverse is numerically sensitive, and the physical model is appropriate. None implies the other two automatically. This separation is useful far beyond deformation studies. An equation can have a solution that is highly unstable or physically unsuitable for the observation being interpreted.

Contents · Previous chapter · Next chapter

Suppose the centre and r are supplied, but q lies in a justified interval strictly between zero and one. The depth function increases with q. Applying the inverse to the lower and upper endpoints therefore gives conditional lower and upper depth bounds.

If w₀ is also uncertain, C = w₀d² must be evaluated with the supported joint information. A rectangular range assuming independent endpoint combinations can be unnecessarily broad or incompatible when the same observations determine both amplitude and ratio.

The distinction between a hard bound, a statistical interval and a scenario envelope remains important. The inversion does not convert one type into another. A probability interpretation needs a measurement and inference model that supports it.

For a practical note, retain the original component readings and their dependence, rather than only the derived ratio. Ratios can share a common denominator and correlated reference corrections. Another analyst should be able to reconstruct which uncertainty belongs to the readings and which belongs to the geological assumptions. A neat pair of depth endpoints cannot reveal those distinctions by itself.

Contents · Previous chapter · Next chapter

The spatial ratios in this restricted model remove C, so they do not identify ν separately from ΔV. Once C is estimated, the conversion is ΔV = πC/(1 − ν). An independently supplied ν is another model input.

For C = 40,000 cubic metres and ν between 0.20 and 0.30, the converted ΔV ranges from approximately 157,079.633 to 179,519.580 cubic metres. The earlier value at ν = 0.25 lies within that interval.

This is a conditional parameter range, not an uncertainty distribution or a measurement of material properties. It illustrates how a well-constrained displacement pattern can still leave uncertainty in the physical interpretation of its amplitude.

A learner may think that more surface points must eventually identify every parameter. But if ν and ΔV enter all these equations only through their product combination C, more observations of the same forward form do not separate them automatically. Additional information with a different dependence is needed. The model’s structure determines what the data can identify, alongside the number and quality of observations.

Contents · Previous chapter · Next chapter

In the small spherical-source representation, source volume change can be related approximately to radius a, pressure change ΔP and shear modulus G through ΔV = πa³ΔP/G. These are different parameters with different interpretations.

The same ΔV can arise from different combinations. Doubling a while dividing ΔP by eight preserves a³ΔP when G is fixed. The displacement equations then cannot distinguish those combinations solely through their shared ΔV parameter.

This algebra does not establish that arbitrary combinations are physically appropriate. The source must remain small relative to depth, the elastic approximation must remain suitable, and additional geological information can constrain the possibilities. The example identifies a model trade-off rather than a licence to choose any desired pressure.

A particularly important reporting distinction is between original source size, inferred volume change and erupted material. They are not interchangeable quantities. The present equations do not say that the model’s ΔV will erupt, that it equals a complete reservoir volume or that a pressure change predicts an eruption. Keep the parameter’s mathematical role and the larger geological claim visibly separate.

Contents · Previous chapter · Next chapter

The idealised vertical displacement can be integrated across the flat surface. For a positive source change, the mathematical uplift volume is the integral of 2πr w(r) over radius from zero to infinity. Substitution gives 2πC.

Because C = (1 − ν)ΔV/π, the surface integral is 2(1 − ν)ΔV. At ν = 0.25 it is 1.5ΔV, not ΔV. The deforming elastic medium connects the two quantities through the model.

For C = 40,000 cubic metres, the integrated surface uplift volume is approximately 251,327.412 cubic metres. Our source volume-change parameter is approximately 167,551.608 cubic metres. Both values can be correct for their different defined quantities.

A finite map captures less than the infinite mathematical integral. Within radius R, the fraction is 1 − d/√(R² + d²). At R = d it is about 29.3%; at R = 3d it is about 68.4%. These are properties of the idealised profile, not completeness factors for a real deformation map. Defining the integrated quantity and map boundary is essential before comparing volumes.

Contents · Previous chapter · Next chapter

A fitted point-source model can leave systematic residuals: observations may differ by direction, distance or time in ways the restricted model does not explain. Those patterns deserve examination rather than automatic removal as inconvenient noise.

Timothy Masterlark’s 2007 deformation study examines sensitivity to assumptions about the surrounding medium. It supplies primary research context for a key limitation: inferred source parameters can depend materially on the physical model, even when observation uncertainty is small.

Our teaching calculations do not reproduce that study or infer its case parameters. They isolate mathematical identification inside one supplied model. Model uncertainty is a further layer, addressed through appropriate independent evidence and comparison of plausible physical explanations.

The existing crystal-diffusion record guide examines another inverse problem with different observations and assumptions. Compare the reasoning habit rather than transferring its coefficients: identify what the record directly constrains, then state what extra conditions are required to infer the desired hidden quantity.

Contents · Previous chapter · Next chapter

Use a new fictional geometry with w₀ = 6 millimetres, ν = 0.25 and a second station 3,000 metres from the known centre. Its supplied ratio is q = 2^(-3/2). Find depth, C and ΔV before reading the answer.

The ratio identifies r/d = 1, so d = 3,000 metres. Convert w₀ to 0.006 metres. Then C = 0.006 × 3,000² = 54,000 cubic metres, and ΔV = π × 54,000/0.75, approximately 226,194.671 cubic metres.

The second station’s vertical displacement is approximately 2.121320 millimetres. Its radial displacement equals that value because r = d. Reconstruct both components from the original forward equations to check the result.

Now remove the second station’s ratio. The six-millimetre centre observation alone again permits a depth-strength family. The numerical amplitude changed, but the identification issue did not. This changed-context task checks whether the learner recognises which observation supplied depth, instead of treating a familiar-looking uplift number as if it contained every source parameter.

Contents · Previous chapter · Next chapter

A useful record identifies component, reference, observation interval, station geometry, source-centre assumption, units and missing observations. It then states the forward model, fixed inputs, inferred parameters and uncertainty interpretation.

Separate reconstruction from validation. Reproducing the fitted observations checks compatibility. Predicting additional independent observations tests more of the explanation. Comparing alternative source geometries or material models examines assumptions that a narrow parameter interval may leave untouched.

For repair, begin with displacement units and the on-axis depth-strength family. For stabilisation, predict the same two models at another distance. For extension, remove the centre observation, change the viewing direction or make the source centre uncertain. Ask which relationship must be rebuilt each time.

The lesson is complete when a learner can read an unfamiliar deformation claim and identify what was measured, what was fixed and what was inferred. They should also state one observation that could distinguish competing explanations. That is more useful than repeating that volcanoes swell before some eruptions without understanding the measurement or its limits.

Contents · Previous chapter · Next chapter

Suppose the instrument record adds the same unknown vertical reference offset b to every station: observed w(r) = true w(r) + b. The spatial ratio of the observed readings no longer cancels only C, because the additive term remains in numerator and denominator.

For Model A, add a fictional two-millimetre offset. The centre reading becomes twelve millimetres and the station at r = 2,000 metres becomes approximately 5.535534 millimetres. Their ratio is no longer 2^(-3/2), despite the unchanged underlying deformation.

Applying the offset-free inverse to these altered readings would infer a different depth. The formula has not failed algebraically; its observation model is wrong. A common additive effect behaves differently from a common multiplicative amplitude.

Differences between compatible stations remove a truly common offset. Ratios of such differences can then remove a common amplitude, provided enough distinct spatial observations and a valid geometry are supplied. That is a new inverse problem, with its own sensitivity and dependence. It should not be described as automatically solving every reference error.

This distinction gives a useful practical question: which corrections are shared, which vary by station and which change between observation dates? Preserve that information before interpreting a normalised map. A reference choice is part of how position changes become data, rather than a label that can safely disappear once the plot looks smooth.

Contents · Previous chapter · Next chapter

Return to Model A with a fixed known centre and d = 2,000 metres. Suppose a second compatible interval has twice the vertical displacement at every supplied position: twenty millimetres at the centre and about 7.071068 millimetres at r = 2,000 metres.

The normalised ratio is unchanged. Within this same single-source model, it supports unchanged depth while C doubles. With ν fixed, the volume-change parameter for that interval doubles too. The two intervals must describe comparable reference conditions for this account to hold.

Now suppose only the centre value doubles and the off-axis record is unavailable. A larger C at unchanged depth is one explanation. A changed depth with another C is also compatible with the centre amplitude. The apparent time trend has not supplied the missing spatial information.

For a learning check, ask whether a stable ratio proves that the actual geological source never changed. It does not establish that broad claim. It constrains depth only within the supplied geometry and physical model. Other mechanisms or compensating model changes can require a wider interpretation. Tracking amplitude and normalised shape separately keeps the observed pattern useful while avoiding an unsupported story about the entire subsurface system.

Contents · Previous chapter · Next chapter

Does ten millimetres of uplift identify source volume change?

Not alone. In the restricted on-axis model it constrains C/d². Depth and material assumptions are needed for the stated conversion.

Can a second observation always solve the problem?

No. It must add informative geometry or direction and satisfy the shared-source assumptions. Duplicated or incompatible observations need another account.

Is a line-of-sight measurement vertical motion?

Not generally. It is a projection whose components depend on viewing geometry and displacement direction.

Does a precise fitted depth prove a unique geological source?

No. Its precision can be conditional on a chosen model and fixed inputs. Alternative physical models and missing effects need separate examination.

Can these examples predict an eruption or define safe access?

No. They are fictional model exercises. Current decisions about a real volcano require the responsible observatory and local authorities’ information.

Contents · Previous chapter · Next chapter

Understand the volcanic evidence chain

Return to Volcanology and the existing volcano-monitoring guide for complementary observations. The gas concentration and emission-rate guide separates another pair of commonly confused quantities.

Compare different geological records

The crystal-diffusion guide concerns time-history inference. The mapped ashfall-volume guide concerns deposit geometry and unresolved map boundaries. Neither supplies a conversion from our fictional deformation parameter to erupted volume.

Read the model sources and test transfer

The USF displacement-model resource, credited to its authors in 2020, provides the equations and idealisations. Masterlark's 26 June 2007 paper develops the consequences of modelling assumptions. Sources checked on 23 October 2026. Continue through the Gold Standard education parent or How X Works with one task: identify the next independent observation that would distinguish two explanations of the same supplied uplift.

Contents · Previous chapter · Return to Volcanology