 All right. Good morning. Can everyone hear me? Yes, good so Sounds like a lot of you are interested in volcano deformation, which is one of the areas I work in and so what I had planned to do this morning was to give a little bit of background motivation and Then talk a little bit about the steps we go through for this very famous source model called the mogi source which is probably the most widely used model to describe volcano deformation and The emphasis is going to be a little bit on what the assumptions are that goes into this model so we understand what the basis is and then I'll be talking again tomorrow and I just want to invite you first of all again to ask questions and Also, if there are particular things that that you want to hear about and many of you know, I've written a book on earthquake and volcano deformation of a chapter on volcano deformation meant some of these are a little bit hard to go through the some of these topics in a short amount of time, but if there's something Specifically that's of interest. Let me know and we'll see if we can squeeze in a little bit or I can talk to you individually So again, I want to start with just a little bit of motivation. I'm not going to spend too much time going through a lot of examples, but So I assume most people are aware of the sort of basic idea that is we could measure how the earth surface deforms as magma rises from the mantle ultimately being stored in crustal reservoirs and As pressure within a magnet chamber increases this tends to cause the ground surface to swell We call that inflation and that can be measured with a variety of methods nowadays mostly GPS and insar It seems like there are a lot of insar folks in the audience today Ultimately when the conditions are such that either a dike can propagate from the magma chamber or magma can force its way to the surface We can get an eruption that will cause the pressure within the magnetic system to decrease And we end up with deflation. So in many volcanoes, we see these inflation deflation cycles This is from Kilauea volcano during the early part of the current eruption episode all the way back in 1983 and This shows tilt record with up being inflation So tilting away from the volcano as pressure increased as magma was stored and then the sudden tilt Drops these deflationary events were associated with eruptive phases during this early part of the eruption So the measurements we make either with GPS insar tilt or a variety of other methods Are things that we want to compare we want to understand what's going on in the subsurface And we need mathematical models to describe the relationship between what happens in the magnetic system with what we can measure at the surface and So this this famous Mogi model is it actually an approximate representation of a spherical pressurized cavity within a fully elastic medium Now whenever you I see this kind of picture and think about what maybe real magma tumors look like This is somebody's cartoon of what maybe a more realistic magma chamber geometry might be Clearly this is not what we think a real Magma chamber looks like in fact it always reminds me of the joke about you know You've all heard versions of this that ends with the punch line of postulate a spherical cow or something like that So this is like postulate a spherical magma chamber. Why do people use this model other than it's been around for a long time? Two reasons I would say one is that you end up with very simple expressions for both the vertical and radial displacements at the surface That will see end up being related to the volume change That is the amount of either intruded or extruded volume within the magmatic system They're very simple expressions and actually many cases they fit data reasonably well this this very simple model And so this is a radar interferogram from Darwin volcano in the Galapagos Many years ago, and this is the best-fitting Mogi source model. So you can see it it represents the data very well More generally what we can learn from these kinds of measurements It turns out that the shape of the magma system the magma chamber Does a control the ratio of horizontal to vertical displacement in fact sill like sources They're relatively flat tend to push up and exert a lot of relative vertical deformation relative to horizontal Whereas things like stocks or pipes push outward more and so you get less vertical relative to the radial so We can use that then information To learn a few things in summary What we can get from these kinds of measurements usually is something about the depth of the source the pressure source Some limited information about the geometry again, whether it tends to be more sill like or more stock like The exception being dykes dykes have very distinctive deformation patterns Something like this because as the crack moves into the crust it pushes the walls aside And that gives rise to very characteristic both vertical and horizontal displacements We could get as I said information about the volume change how much magma has been added or subtracted from the system Not the total volume although we'll talk about that So the some of the things we can't get that are the absolute chamber volume How much magma is there from a volcanological spec perspective? That's something we very much like to know how much eruptable material is actually there Turns out to be also difficult to get at pressure in an absolute sense and of course these models are essentially kinematics So they don't have predictive capability. You know come back to that tomorrow So now I want to go through just a few examples showing deformation leading to eruptions This is from the Galapagos Sierra Negra volcano Deformation there was first found with insar measurements and then campaign GPS and ultimately continuous GPS You can see very rapid changes here uplift of the floor of the caldera leading to interruption Back in when was it Cindy? I don't remember exactly, but we can focus in on the little bit more on the details here showing Actually very interesting sequence of faulting events that preceded the eruption But dramatic changes in the deformation leading up to the eruption. So Magma was filling the system pressure was increasing going towards the eruption a similar thing for The eruption in Iceland we all call IA because it's too hard to pronounce the full name So again interferograms GPS time series the option is somewhere over here Rather dramatic changes indicating that magma was being stored in the in the near surface prior to that eruption As I mentioned dykes are rather different. This is interferogram from an intrusion into the upper East River zone of Kilauea volcano and I love this butterfly pattern here because it very much shows these two uplift lobes as the Dyke is is intruding into the crust it's pushing aside by the Poisson effect you get uplift on both flanks and That's shown in the calculation here, and you can see those two uplift lobes on either side of the roof zone there again These signals precede eruptions are their precursory in that sense There's a lot going on on this slide some of this is tilt data some of this is GPS But we see tilt downward here as magma started leaving the summit magma chamber intruded into the rift zone Propagating earthquake swarm going out these two GPS sites here eventually started moving apart That's what this time series. So first deflated up near the summit meaning magma moved out intruded into the rift zone and ultimately these two stations started moving apart as the dyke propagated Down rift I think in the interest of time. I'll skip this one last example, and then just go now and talk about The derivation of this bogey source model, okay, and that's what we'll spend the rest of our time on so this is as I said is a highly simplified model and It looks something like this. We're going to imagine an elastic material with a free surface here. So the tractions on the surface vanish and we have a spherical cavity of radius a whose center is at depth D within an elastic half-space and There's some pressure applied to the boundary of this This cavity will call P so it could be a Change in pressure But we're gonna imagine whatever the initial stress state is itself equilibrated So there's some equilibrium stress that that equilibrates all the body forces that are acting on the system And now we add pressure P but because this is an elastic system Everything is linear and we can superpose different solutions. So we're only looking at the change as masses added or subtracted to the cavity Okay, so We're gonna imagine that outside here. This is fully elastic it's linear and Just to keep everything simple. We're gonna assume it's homogeneous and isotropic It means its properties don't vary as a function of position. So The governing equations here are again, we always have is as Michael emphasized We have conservation of mass momentum and energy. We don't have an energy equation in this case We're assuming everything is isothermal. We're not worrying about heat transfer and We're gonna then write the equilibrium equations in the standard way the divergence of the stresses So if people aren't familiar with the tensor notation, you might check with me Afterwards, we'll talk about this just saying the divergence of the stresses plus body forces vanishes So I don't have an inertial term in this equation Why not there's no Yeah, okay, that's exactly right. So we're gonna assume that these are so-called quasi static That these displacements these deformations occur slowly enough that we can neglect the inertial terms in the equations of motion So there actually should be a term over here. It's density times the The total derivative of velocity with respect to time, but we're gonna look at the limit where this is approximately zero So these are taking place slowly. We cannot in this calculation Therefore look at radiated waves elastic waves for example from this assuming the deformations occur slowly So let's think about the boundary conditions here The boundary conditions acting on the wall of this cavity suppose we erect a spherical coordinate system that's centered on the chamber We would have that the radial stresses evaluated on the boundary Would be equal to minus the pressure Why minus? Because I use attention positive stress convention So push here pressure P induces a compressive stress sigma rr and This this fluid magma that's within the magma chamber is static So there are no viscous stresses being exerted on the walls and therefore the shear stresses are theta And our fee are exactly zero This is on r equals a and then as I mentioned here we have This is a free surface. This is representing mathematically the surface of the earth So all the tractions that act here are zero. So in a in a Cartesian system Sigma I3 equals zero Z equals zero where I is one two three So the the vertical normal stress vanishes well, it's atmospheric pressure, but that's effectively zero And there are no shear stresses acting on this Planes equals zero. Okay, so this this turns out to be to solve this problem Exactly is actually rather complicated and you can think about it because We'd like to be in a spherical coordinate system here to solve this part of the problem But here we have this Cartesian boundary over here That we need to match the free surface boundary conditions And so that would be sort of awkward in a spherical coordinate system So even this highly simplified model for an elastic homogeneous medium is actually difficult to solve analytically and the Mogi solution is actually an approximate solution and that Approximation is going to turn out to be at accurate in the limit that the radius over the depth is small compared to zero so a over d is Small sorry small compared to one Now the derivation I'm going to outline is one that was presented in a paper by Dave McTigue back in the 1980s and I like it because it illustrates Sort of conceptually how we can think about building this and allowed him to go to higher order approximation The Mogi solution, but we'll walk through it even though this is historically not how This equation was actually derived in fact there were there were papers written before Mogi a whole series of Very important papers and yet it was Mogi's name that got attached to the solution So that the procedure is going to be as follows What we're going to do is we're going to derive a solution for the full space Pretending that the free surface isn't there at all Okay, so step one is going to be the full space solution and then step two where we're going to correct For free surface So we're going to solve this problem in spherical polar coordinate system centered on the magnet chamber pretending the surface is not here They were going to find out that they're actually stresses both share a normal stresses that act on the plane z equals zero Then we'll say well, let's add equal and opposite Stresses to that boundary now will satisfy the boundary conditions here But of course when we do that will now violate the boundary conditions on the surface of the magnet chamber And that that's why it's approximate We could imagine going back and forth and back and forth fixing up the boundary conditions here ruining them over there Fixing them here ruining them over here and just iterating back and forth But we'll only do one one step with that Okay, so this equation here and by the way, we're also going to ignore body forces For the time being because we imagine that the initial stress state Equilibrates all the body forces in the system Now it may seem paradoxical that we can ignore gravity In this calculation, but if someone wants to raise that at the end would come back and talk about why we can neglect gravity as a first approximation So this equation represents conservation of momentum or balance of forces and the continuum It alone is not sufficient to give us a solution. We need more information. What information are we missing at this stage? Sorry Elastic properties we need first of all more generally we need a constitutive equation We need something that relates stress and strain Or stress and strain rate if that were the case We're assuming this is elastic. So we're going to write that the stresses sigma ij Mu is the shear modulus Lambda is the other lame constant and in this case with a full-space solution We have radial symmetry and so there are only two strains That is the radial strain epsilon rr Which is the derivative of the radial displacement with respect to r and the hoop strains the theta theta And phi phi which are equal by symmetry Spherical symmetry so that two directions are the same and it turns out those displacements Sorry, those strains are u r over r So we can substitute this Into hooks law and get the stresses in terms of gradients and displacement The spherical symmetry this is expanding out. So the only non-zero displacements are u r There are no displacements in the theta or phi directions Okay, so If you start with the equilibrium equations the general equilibrium equations We can show that they reduce with it for spherical symmetry in general. These are how many equations? Here It's a tensor system. So these are how many? Three right, so these are three equations because we have One free subscript. We're summing over j the repeated indices are summed over So this represents Balance of forces in the Cartesian system and you know x y and z three different directions Because of symmetry there's only one Non-trivially satisfied equilibrium equation and that is Looks like this so There are two other equilibrium equations, but because of the symmetry they're automatically satisfied So this is the only one that's non-trivially satisfied We take hooks law and these equations substitute into here and we end up with a differential equation In our and I'm going to leave it to you as an exercise to deduce this and then from these equations get to here Total are you using total differential now because you know all the derivatives with respect to theta and phi are trivially satisfied Okay This is Above the last yeah, these are ours Rr minus Stress sigma rr stress minus a sigma theta See these are all in the notes if you don't want to be right Okay So we've taken an equilibrium equation in terms of stress through the constitutive law. These are kinematic relationships That relate strain and gradients and displacement This is a constitutive equation for an elastic meeting saying stress and strain are linearly proportional And from these we can derive a single Equilibrium equation in terms of displacements now so this is a first-order ODE we can integrate that and I won't let you again Let you do that and we end up with the radial displacement having two Okay, it's a second-order equation we end up with two constants a and b and now we have to figure out what a and b are so What about a? Do we notice what about this term? Should be zero why the displacement has to be finite as our goes to infinity Because this is an infinite medium So we have to set that term equal to zero now if it was not an Infinite medium then we might have to worry about that But in this case we were looking at the solution in the full space Extends infinitely far away from the source So we're still left with one constant how we're going to determine that constant. It's the boundary condition, right? so we have a boundary condition the boundary condition is in terms of Stress the radial stress has to match the pressure on the cavity wall, so I skipped some steps here, but we can write the the the radial stress so sigma rr is Twice the shear modulus D u r dr, so we take the vertical we take the derivative here with respect to radius We're going to get minus four B over r cubed This should be equal to minus p sigma rr at r equals a minus p And this is minus B over a cubed if I did that right minus cancels minus And B should be equal to P a cubed over four u and Then we can put that back in to here so the radial stress Should be equal to minus p a over r Cube and the radio and the radial displacement, which is B over r squared would be minus four P a cube over four u r squared All right, so let's think about this Dimensionally Does this make sense? On this side, we have stress On the right hand side, we have pressure so I have units of stress a over arc is Non-dimensional so at least dimensionally this makes sense, which is a sanity check And we find that just that the stress is decay as one or r cubed as we move away from the magnet chamber So the farther away we go the stresses are going to decay quite rapidly Let's look at the displacements now So we have units of length on this side pressure over modulus so stress over stress a Cubed over r squared has units of length. So I have units of length on this side units of length on that side At least it passes the check of having the right units. Okay. Hopefully we did the algebra, right? so Everybody any questions about this part so far Exactly right so far we have this centered on the source So this is the full space solution now we we've done this this step So the good news is that's pretty straightforward. The bad news is we're done with the easy part now It's gonna get harder, but we won't Do everything Just outline how to do that So the next thing we want to do then is we want to say Okay, what are the stresses acting on this z equals zero plane? due to this This magnet chamber here expanding And if I can erase this We can see that there's that the stresses are radial in the full space. There's only a radial stress so this is pushing out in a radial sense and so we get a radial displacement and we were interested in decomposing that into a vertical displacement in a tangential horizontal component and I'm going to call this direction the row direction And so we want you row which would be that the horizontal displacement on the free surface And that will have cylindrical symmetry right so the displacement in and out of the board will be the same as Parallel to the board or any other horizontal direction, but that's different from the vertical Okay, now we can see we kind of have similar triangles here. We have a triangle here. It's depth Here's the radial distance r and then we have a triangle here with radial displacement vertical and tangential So that I Can get the displacements the vertical displacement use z Okay over the radial by a similar triangle is so that's this over that is Proportional to the depth over the radial distance, so that's equal to d over r By similar triangles So all I'm doing is I'm taking the radial displacement decomposing. This is still in a full space Just decomposing it into vertical and tangential Well, you are For the full space we have over here so that's pi a cubed over 4 mu and Then we get d over r cubed but r By by Pythagoras r squared is d squared plus row squared So this is pi a cube over 4 mu D over d squared plus row squared Three-halves power this is the vertical So again pressure over modulus is non-dimensional we have Length cubed over length cubed times distance, so this is dimensionally correct and By the way the the tangential component is just row over d cubed so This over that instead of the vertical over that so that we have now the vertical and the tangential components This is still in a full space. We've just taken the full space solution and decomposed it into two components in a a Cylindrical coordinate system centered now Over here, so we haven't fixed the free surface yet. Is that clear to everyone? I've just taken this and done a coordinate transformation gone from spherical coordinates to cylindrical polar coordinates, so the the step now would be to compute What the stresses acting on the surface are? And there I'm not going to write them out, but they're going to be a Vertical stresses sigma zz and then they're going to be shear stresses So these are normal stresses you can see we're pushing up. So that's going to give compression here and There'll also be shear stresses. This would be on z equals zero and they're both shear stresses Rho z Theta, etc. and So resolving those stresses the full space solution stresses, which are these It's just a transformation of coordinate system again, just like we did here But now there's stresses, so we're dealing with a tensor instead of a vector and What McTighe did in the solution then is to apply equal and opposite stresses to this boundary And a rather remarkable thing happens when you do that that the details are a little bit complicated, but the In terms of the displacements it turns out and this is actually rather a general result is you just multiply by four one minus new so the effect of Of negating the tractions that vertical and shear tractions that act on this surface in terms of the displacements on z equals zero is You just take these solutions and multiply by four one minus renew as Poisson's ratio It's not obvious. So don't expect to see that intuitively That's just something that works out through the map So in fact then we end up with Equals that the four cancels the four in the denominator here and we get one minus new pi a cubed over shear modulus D over rows and similarly with that the tangential displacements you just replace D with row And this is the mogie solution so to be clear what we have done is Exactly now match the boundary conditions on the surface. They are traction free But when we apply Equal and opposite tractions on the planes equal zero we no longer exactly satisfied The radial stress is equal to minus the pressure. So that's and that's the reason this is an approximate solution McTighe went one step farther and Corrected those Stresses the stresses that we've added negated those But you see what's going to happen is those stresses are going to decay again like a over r cubed So by the time they reflect back up here. They're now proportional to a over r to the sixth power So you skip two powers in the ratio of a over r a over depth And it actually turns out even though this solution is approximate formally as a over D should be much less than one It turns out to be pretty accurate even when a over D is on the order of say for example a half It's not too terribly bad of an approximation Because the stresses decay as you go away from the spherical cavity like a over r cubed They decay so rapidly that you can actually and maybe if we have time. I'll show you a couple figures to show that So you can generate that What happens so Michael's talking about an image we could reflect by symmetry and put an image source a Distance D above the surface and that would cancel the normal stress But we'd still be left with sheer stresses So getting rid of the sheer stresses is a little trickier. Oh They're really complicated. Yes, you could do in the in the limit that you represent this by a point force Then they're there could their series of images their ways of constructing these as images, but it's not trivial I should say and we won't go through this the way. This was actually first arrived is to approximate And we could do that By what's called a center of dilatation In elasticity theory there's something called the greens function, which just gives us displacement At some point say here Due to a concentrated force acting in the medium you call that a greens function, right? So this is a concentrated force It gives rise to displacement at some point Now you can construct higher order sources from superimposing point forces So for example, I could get a double force to equal and opposite forces in a so-called center of dilatation Which goes back a long way in elasticity theory would be three orthogonal force couples so they push out equally in all directions and In the limit that the radius of this cavity gets small you can represent this as a super position of double forces with no moment This sort of interesting result turns out to be Somewhat more general, but it only applies for Sort of volumetric strains. It doesn't have to be spheres But if you had uniform volumetric strain over some region you could represent it that way So I think the important point that I want to emphasize what are the approximations linear elastic So we're imagining that even as you get very very close to this magnet chamber where temperatures could be extremely high This behaves in a linear elastic fashion. There's no viscous flow in the medium going on We could and we will tomorrow include viscoelastic effects tomorrow, yeah But for now we would assume that this is small strain linear elasticity We've assumed that the materials homogeneous It's isotropic neglected the effects of gravity we've neglected poor structure the fact that there could be fluids water within the pores of the rock Multi-phase fluids and We've assumed that we have this limit where formerly the radius of the magnet chamber is small compared to its depth So it's very idealized I will show Mind me before we leave Just how good the approximation is for The the depth getting relatively small, but before we do that. Let's look at this structure here So displacement pressure over modulus radius cube So pressure remember this is really the pressure increment This is the change in pressure because we assumed we were in some Self-equilibrated state whatever happened before the pressure increased was in equilibrium. It wasn't accelerating off into space And a cube is proportional to volume for the sphere So in some sense this displacement looks like pressure times volume So if all we're doing is measuring the displacements If we wanted to figure out the volume of the magnet chamber We would have to know independently what the pressure change was or conversely if we wanted to know how much the pressure was changing We'd have to know what a was so this is a feature of the solution That's kind of unfortunate because we'd like to know independently what these two things are that would be very useful to us But in fact, we only get them occurring as a product And that turns out to be true for more general shapes that could be ellipsoidal shapes or things like that You're essentially insensitive to the total volume Perhaps we can estimate a independently If we could that would be useful Maybe you have some ideas about how we might do that Maybe from seismic tomography. We might be able to figure out how big the magnet chamber was If we could then we could say something about what the pressure increment was Maybe we can put bounds on the pressure because we have physical constraints But the pressure can't be so high that it you know, it would break the rock or something like that And that would tell us well the radius would have to be at least so large We could put sort of physical constraints on what it would be but from the displacements alone. We can't say Independently what those are Now sometimes we see the mogie solution written in a different way and in fact I have it over here in terms of volume change So let's just show very quickly how that comes about So again, if I go back to the full-space approximation How would I relate the volume change? to Knowing the displacement on the boundary you are evaluated at r equals a Ignoring the free surface just in the full-space Change in volume would be I know Marco But generally speaking it would be the integral of the displacements over the surface That would be the first-order approximation of the volume change. I would take the outward pointing displacement integrate over the surface Here we have Symmetry so it makes that integral particularly easy. So the surface area would be 4 pi a squared Right times the radial displacement You are evaluated at r equals a In the full-space solution, which we had over here. We get 4 pi a squared times pi a cubed over 4 Mu a squared a Cancels out a squared So delta V equals pi P a cube Over mu pressure over modulus is dimensionless volume a cube So where I see this P a cube over mu. I can just replace that with delta V over pi so similarly this could be written as minus mu over pi Distance cubed over length cubed over length cubed displacement partial to distance So either way these are two equivalent representations. I can relate the displacements to the volume increment the displacement at the surface Related to how much volume either moved into or out of the magma chamber Or I can relate it to this product of the pressure change Times the radius cubed either of those is equivalent But we've seen that I can't get uniquely at pressure or volume independently so when you look in the literature or those of you that that Fit data you'll often see the data will satisfy be satisfied with a volume change of X number of million cubic meters of magma either added or subtracted to the chamber any questions about that Now I do want to say one other thing about volumes and and that is that There are other volumes that one could compute And by the way, these are what the displacements look like so the vertical displacement decays in the following way and so typically What you can do then is to this is the scaled radial distance the distance over the depth And so typically what you do is you say go to where it falls to half the height You can work out exactly what that be that will give you some information about the depth how deep the sources The radial displacement starts at zero increases to a maximum goes up of course by symmetry It has to go to zero right above the source It's going to say there are two other volumes that come in one is You could actually and I don't know why you do this But people have done the literature you could take the vertical vertical displacements and integrate them over the entire surface and that will give you the delta V of the uplift and Paradoxically it turns out that delta V of the uplift can be larger than the delta V of the magma chamber I'll let you ponder that The other volume that that people talk about is the The volume of the mass volume of the magma either added to or subtracted to the magma chamber If the magma were incompressible those two volumes would have to be equal but magma's have gases dissolved in them and those gases will form bubbles and the bubbles tend to be the gas phase Tends to be highly compressible So due to the compressibility it turns out that very often times if you add up all the magma that's erupted onto the surface Even the dense frock equivalent will be more Than the delta V of the magma chamber from which it came because the magma itself is compressible So you can't necessarily just eat the mass right we have conservation of mass. There's no conservation of volume loss conservation Dense frock equivalent means you just take out the pores hmm of the erupted products so Volcanologists, patrologists call that the dense rock equivalent. So you may have a deposit that's 10 million cubic meters But if it's 20 percent porosity, that's not You know 10 million cubic meters of okay, so this is a figure from David tides paper showing the The this is the vertical displacement and radial displacement I call this order epsilon cube epsilon here is a over D the ratio of the radius to depth And This is for that ratio being 0.5. So this is not very very small compared to one. It's a half That's less than one of course has to be less than one It's not very very small and you can see the difference in the displacements is rather modest This plots the maximum uplift as a function of that radius. So you can see at Radius of 0.5, you know, you're talking a 15% correction Given all the assumptions we've made about homogeneity Isotropy flat free surface neglecting a lot of effects That's not a huge Correction factor and again it comes because the next higher order correction Actually comes in at this ratio out of the six power because things drop off as a over D cubed each time you skip two powers in that ratio so if we were to look at Then the stress state around the magnet chamber. This is including the next order of fact this is showing the The deviatoric stresses the shear stresses and these are the mean normal stresses It's actually an interesting thing that happens in the full space solution This expanding source is actually a pure deviatoric pure shear stress source. I Didn't emphasize that but we if we we calculated the the radial stress It's compressive If we had computed the hoop stresses the theta theta stresses They would be I'm just going to write it is minus one half sigma rr so The cavity expands outward the radial stresses are tensile It's being stretched as you expand out the hoop stresses are tensile and They're half in magnitude and opposite sign But because they're two of them there's a theta theta in the Phi Phi So when you add them all together, there's zero So it's sort of paradoxical this expanding source Actually generates a pure shear stress. There's no mean normal stress change You can see that here in the contours of constant deviatoric stress are nearly spherically symmetrical Although the free surface does introduce a little bit of an effect This is the mean normal stress this the trace of the stress tensor You see it's completely due to the free surface in a full space. It would be zero everywhere. So this is the All right, and this is just showing that you can write this as an equivalent So-called center of dilatation these this triaxial point source Representation is also equivalent to logis source And this was the one that was actually derived first and is convenient to be used for example If you want to look at layered medium or the elastic properties are spatially varying Sometimes this representation is is very useful. Although it is exactly equivalent in the limit to what we just derived any last questions lunchtime