 So it looks like I'm on the schedule for two hours. I think we'll talk for a little while and then have a break, stretch, maybe get a coffee and then put coffee and come back. I wanted to take our discussion yesterday in two different directions. There are many different directions we could go to expand on the very simple model that we derived yesterday, the so-called logi source model. What I've chosen to do is to address a question that came up yesterday, actually, what about different constitutive law? The logi source was for a spherical pressurized cavity in a fully elastic half-space. So the first part I want to discuss today is what happens if the region around the magma chamber is not behaving elastically. And so we'll sketch out how that works. And then hopefully we'll take a break around then and then I'll talk about some work that we've been doing in my group to couple the ideas of volcano deformation with some of the things that Michael talked about yesterday in terms about conduit models and what the physical properties of magmas are as they move from high pressure to low pressure towards the surface and how we can try to link these two paradigms together. So the motivation for this discussion about viscoelastic behavior is this is from a paper of Del Negro just looking at a steady state thermal distribution. And the inescapable fact is that we can't imagine that we have magma in some kind of reservoir at order of 1,000 degrees C that's up against rock that's going to behave in a perfectly elastic fashion. The rock is that hot. It's going to creep under stress. And there's been quite a lot of work on this done in the last decades or so. One of the things that we'll see is that when we and we're going to restrict our attention today to so-called linear viscoelastic material to find this a little bit more. Viscoelastic material is a nice kind of silly putty I often bring. And Michael had a silly putty here yesterday. So you all imagine that. That is under low rates of loading that it behaves like a fluid. It flows viscously. And when you deform it rapidly, if he had turned it into a ball and bounced it on the desk, it would have bounced so at high stressing rates it behaves in elastic fashion. We won't go through the full duration here, but it turns out that in contrast to the elastic material, we had a solution that said displacement was a linear function of pressure. So in elastic material, if the pressure in the magnet changes over time, the surface displacement will change in proportion to that pressure change. The details will, the shape at the surface will depend on whether it's a spherical source or a dyke or a sill or a stock. Somehow that shape will be influenced by the spatial pattern of deformation of the surface. But there'll always be this linear mapping between pressure and displacement. Now in a viscoelastic medium, what happens is we get a convolution in time. So the displacement at some position at some time will, in fact, we have to integrate over the pressure rate and some kernel. And that kernel will have in it a characteristic relaxation time, very similar to the relaxation time that Michael talked about in the glass transition, where we'll get a ratio of viscosity to shear modulus. He used eta for viscosity, as I do. He used g for shear modulus. I'm using mu for shear modulus. But the point is that the viscosity has units of Pascal seconds. Shear modulus has units of Pascal. So this is a characteristic time. And this is the characteristic relaxation time for so-called Maxwell material, which I will define in a moment. The challenge here has been that what we would like to do is learn something about how the pressure is changing within the magnet chamber. But to deconvolve, to solve this, we would have to know what the viscosity is a priori. And the viscosity structure within the earth, it's not going to be just one viscosity. That viscosity will vary as a function of position, how far away we are from the hot magnet chamber. So this inverse problem gets very complicated. We don't know the pressure history. That's what we're trying to solve. And we don't really know the viscosity structure either, although we may have some constraints that we can place on it. So this has been a challenge. And what I'm going to do today is look at a special case that I think can be instructive and it's illustrated schematically here. And that is where we look at the changes that follow an explosive eruption. Because a short-term explosive eruption is basically an instantaneous change. On the time scales that the crust relaxes, we have a very rapid depressurization of the magnet chamber. So although we don't know the magnitude of that pressure change, we at least know its shape. So that gives us some constraint and makes it easier to actually, I meant to back up to show you a figure. It is early in the morning. I'm still working. This is from a paper of Andy Newman and others showing some inferred pressure history. The details don't matter on the bottom. And these are different uplift responses depending on the different viscoelastic model. So the point is you could have the same input pressure. And for a different viscoelastic model, you get different displacement histories. And of course, we're trying to solve the inverse problem. And we measure the displacement of the surface, and we're trying to figure out what's going on at that. So you can see this is a tricky problem. And so again, we're going to try to exploit this idea that if we know something about the history, that is, we have a rapid depressurization. We study how the surface deforms in response to that. So the model I have in mind is shown pictorially here. We're going to go back to the idealized spherical magma chamber, that in this case, we imagine that there's a shell surrounding it that behaves in a viscoelastic way. Why a shell? Because the higher temperatures near the magma chamber would lead that rock to creep under stress. So we're going to imagine right before the eruption, there's a magma chamber. It's at some pressure of P0 minus, minus meaning just before the eruption. This magma chamber is connected to a deep mantle reservoir that we're going to assume is at fixed pressure. And if the pressure within the magma chamber is lower than this deep pressure, adjusted for the magma static gradient, the magma will flow into the magma chamber. So at some time, the eruption occurs. We assume this is an instantaneous eruption. Maybe it lasts for a few days, but that's effectively instantaneous. Mass is ejected. The pressure drops by an amount delta P. And two things are going to happen after that. One is that now the pressure is lower in the magma chamber. So new flow is going to come in from the mantle. It's going to recharge. But also this viscoelastic shell is going to deform. It's going to start creeping in response to the fact that now the stresses around it have changed. And so we're going to have these two effects going on simultaneously. We want to understand how that would affect what we measure at the surface, whether it's from INSAR, GPS, or TILT. So let's see. I think I can go ahead one or two slides before we go to the board. So we're going to treat the magma chamber in terms of its mass and pressure as a single lumped parameter. In other words, we're not going to look at the spatial distribution of pressure within the reservoir. We treat it as a scalar parameter. And conservation of mass would say that the flux, the mass flux in would be the time rate of change of mass within the reservoir, which would be its density times its volume. We just expand that differential. And we can write the pressure dependence in terms of a magma compressibility, all betas in my notes will be compressibilities and would be magma. So we have a term related to the change in pressure. And we also have a term related to the change in volume. Again, the change in volume for spherical magma chamber is just going to be the surface area times the radial displacement. But now that radial displacement is going to be time dependent because it's a viscoelastic medium. Just for my own notation, we're going to call the pressure within the magma chamber, p naught minus, remember, was the initial pressure just immediately before the eruption. And everything after that is a deviation in that pressure. We'll call delta p. And then this capital delta p naught is the difference between the equilibrium deep pressure, sorry, the equilibrium pressure within the magma chamber minus its initial pressure. So if this was zero, the magma chamber would be in magma static equilibrium with a deep source and there would be no flux. But if the magma chamber's slowly inflating, then this is a positive quantity. The equivalent pressure, equilibrium pressure, is slightly higher than the initial pressure. That means it's slowly refilling at the time of the eruption. So that would be the more general case where you might have some slow inflation going on. So then we just rewrite the mass balance from the previous equation. I do want to say that there's a very simple case to solve. And that is if we ignore viscoelasticity and have an elastic magma chamber, then we get a term which looks like the compressibility of the chamber. It's the volume change of the elastic reservoir per unit change of pressure. And I call that the chamber compressibility, where's that beta c? So beta m is the magma chamber compressibility. Sorry, beta m is the magma compressibility. This is the chamber compressibility. So for perfectly elastic medium, you get a first-order ODE, which has an exponential solution. So pressure drops, pressure's now lower, magma starts flowing in. The more the magma flows in, the lower the pressure gradient becomes. And therefore the slower the flux is. And so you get an exponential solution with a characteristic time constant, which depends on the volume of the magma chamber, its density, the compressibility, the combined compressibility, and this flux parameter, which I didn't define, but that's the proportionality between the mass flux and the pressure difference. And if you remember, I think I'll show this again later, but remember those tilt records I showed from Kilauea volcano, where there was rapid depressurization and then infilling inflation-deflation cycles. They had this sort of exponential character. In fact, they look a lot like this. And it's not uncommon to see inflation-deflation histories that look a lot like this. So it looks a lot like this elastic limit. But that doesn't include the viscoelastic response. So to do that, we're gonna do pretty much exactly the same thing that we did before, but now we're going to look at a model with a shell that surrounds the magma chamber that behaves in a viscoelastic fashion. So again, we have this flux in, it's proportional to the difference in pressure times this flux parameter, balancing the mass within the magma chamber. So there's a pressure change, here's the volume change, but what we need to now to solve is how this radial displacement, so it's four pi r squared times the displacement on the boundary, how that changes as a function of time in a viscoelastic meaning. And we're gonna make use of a solution by Dragoni and Manunensi that came in 1989 for a full space. So we do exactly what we did yesterday that is find the full space solution, right? And then fix up an approximate solution to account for the free surface exactly the way we did yesterday, okay? New questions so far? We will ignore gravity again. So again, on the time scales, the spatial scale. Yeah, I made a comment yesterday if it was viscoelastic and the stress is relaxed, I was thinking of if we have a distributed layer, then it will tend to sink over time if we add mass. But still I think we're here looking at time scales maybe on the orders of decades and so forth. So we're still probably at a limit where gravity is a small effect. Okay, so the first thing I wanna do is look at different viscoelastic reologies and I'm only gonna do one today, but once you get the general gist of this, it's pretty easy to expand. We're gonna look at what's called a Maxwell material and the Maxwell material is written as a dashpot which I'll define with viscosity eta in series with the spring with stiffness mu. So I realized a few years ago that not everybody knows what a dashpot is. This is like a damper in a shock absorber in your car or another one I like to think of. Anybody know those rowing machines? We still have those rowing machines where you pull and it dissipates energy as you're pulling on it. So that little piston thing is a dashpot. So we can put springs and dashpots together in parallel and series and make lots of complicated models that can get quite interesting but we're just gonna take the very simple case here. And the reason this is viscoelastic is you can imagine if I grab this end and I cycle it very rapidly, the viscous damper doesn't have time to move and so I'll just get a perfectly elastic response. I'll just pull this very rapidly and the spring will stretch and relax and stretch and relax. On the other hand, so it behaves like a solid or in other words, seismic waves would propagate through it because it's deforming very rapidly. On the other hand, if I grab this and pull it very slowly then the dashpot will have a chance to deform and it'll deform like a linear viscous material. Why am I insisting on linearity, linear viscous rather than power law because some math I'm gonna use requires linear equations. We could obviously generalize that to nonlinear behavior. So the way we develop constitutive laws for viscoelastic materials is to look at this and we say when we have them in series like this, the total strain that is the total stretch of this little body is gonna be the stretch of the spring plus the stretch of the dashpot. And the forces have to be equal. The stress has to be the same stress because we could just see if the stresses weren't the same something would be out of balance and it would have to accelerate. So we're gonna write the strain rate and the total strain rate as the sum of the elastic strain rate plus the viscous strain rate. So elastic strain rate is gonna be just stretching rate over modulus. So just to sort of look more like the tensor form which we'll get through momentarily, I'm gonna write stress rate over twice the shear modulus and the viscous strain rate is just stress over viscosity stress over two eta. All right, so the total strain rate depends on the stressing rate plus the stress. So we have this elastic part and we have a viscous part. If everybody see that I know in the back sometimes it's hard because of this. I think I'll move up here. Now the way I'm gonna write this now is in the form of just a linear differential operator acting on the stress. So I can write the stressing rate on maybe twice the stressing rate is equal to one over mu times time derivative plus over eta is all operating on stress. And so now I wanna exploit the linearity and make use of what are called Laplace transforms. So how many people are familiar with Laplace transform? Uh-oh, not as many as I'd hoped. Okay, we won't have time to go into great detail in Laplace transforms but it's an integral transform and I will define it here. So the Laplace transform of a function of time is integral from zero infinity of this f of t u minus st t. Sometimes I'll write this as f bar of s. So I have some function of time. I multiply it by an exponential function e to the minus st and then I integrate out time. So I get rid of time and I end up with this s variable which is our transform variable s. So are more people familiar with Fourier transforms? More people have seen Fourier transforms. The seismologists have all seen Fourier transforms. This is kind of like a Fourier transform. The difference is Fourier transform, we'd integrate from minus infinity to infinity and this would be an imaginary exponent here. We'd have say e to the minus i omega t rather than st. But that's basically the difference. So it's very much like a Fourier transform. Laplace transforms are useful when we have initial value problems. So the only feature that I wanna exploit here of Laplace transform is that the Laplace transform of a derivative turns out to be very simple. So if I take now the Laplace transform of df dt, I have zero to infinity df dt e to the minus st. And I won't go through the details here but I think you can see that I can integrate by parts very easily. So I integrate this one, I get rid of the time derivative, this becomes s. I differentiate this and I get a minus s in front. And then in the limits, one of those terms is gonna go away because e to the minus st as t goes to infinity will go to zero. So when all is said and done, I'm gonna get s times the Laplace transform of f of s minus, I'm gonna get the sign on here, s minus f of t equals zero. So again, those are familiar with Fourier transforms. Every time you take a time derivative, you get an i omega out front. Here every time you take a time derivative you get an s out front. It's exactly the same thing. But what that means is that we can go back to this expression and now go into the Laplace domain. So I didn't write this more explicitly. So I differentiate this, I get twice 2s Laplace transform, now I function the transform variables. Every time I have a time derivative, I get an s. And this is s over mu plus one over eta times sigma bar. And then I can just rearrange and write this as, let's see, 2s finally stress equals two. So I have in the Laplace domain, the stress and strain are proportional. That proportionality looks like a modulus, but that modulus is dependent on the transform variable. So where mu bar is equal to, let's see if I do this, s mu over. So again, we have this ratio of shear modulus to viscosity. This is like an inverse relaxation time. This is the so-called Maxwell time. There's everybody with me so far. So the really key thing and what's very cool about this is it says this just looks like Hooke's law. This looks like elasticity, except I'm now in this transform domain. So now the neat thing is, if I can solve an elasticity problem in this Laplace domain, all I have to do, I write down the solution and then I have to invert the Laplace transform, which I won't have time to go through today, but it's actually reasonably straightforward. Once I can do that, then I have my solution. And in many cases, those inverse transforms are very easy to do. In the cases we'll be concerned with today, they're actually trivially done. So it's really nice. This is called the correspondence principle. It has this name that any viscoelastic solution has a corresponding elastic solution once we go into Laplace domain. So this isn't the only way of solving viscoelasticity problems, but it's the most convenient. The only restriction that the constitutive law has to be linear, we're exploiting the linearity to use these integral transforms. Okay, any questions? Yes, I have to speak up a little louder. That's right, this is, I'm just right now assuming a fixed viscosity. Temperature dependent viscosity, which would be completely appropriate, is another layer of complexity. So what I'm gonna do now is solve a very simple problem and try to learn as much as we can from that simple problem. Yeah, but you're absolutely right. We would want a more general analysis would consider a temperature dependent viscosity. And of course, as Michael emphasized yesterday, you have, when you have work done, you have dissipation, so that feeds back into changing the temperature. We're not including any of that today. Okay, so this is the basic scheme and we're gonna follow this very nice paper by Dragoni and Maninensi, who did this solution in a full space. And so the problem is going to look like this. We have a magnet chamber of radius R1 with pressure delta P applied to the interior of the magnet chamber. We have a viscoelastic shell, R2 that has some viscosity eta and shear modulus mu. So this is the Maxwell dashpot with spring. And we're gonna call this region, we're gonna call the displacements here region one. And then outside in region two, it's just purely elastic. There's no viscoelasticity. Okay? So this shell out here is at higher temperature. In reality, of course, the viscosity would be radially dependent, but we're just gonna simplify that and say there's a shell of constant viscosity. Okay, so we already know how to do that problem because we did it yesterday. Wait, let me take a second to dry. But we found last time that the displacements could be written in the following form. In region one, so this is an interior region. We can write this as AAR over three plus B over R squared. This is our general solution. Yesterday when we were considering an infinite medium we set A equal to zero because we said as R goes to infinity the displacements have to be finite. But if we're in the interior region here, R never gets to infinity, it only gets as big as R2. So we have to keep this term, but in the exterior region for the same argument we don't have a term proportional to R. We only have the term proportional to one over R squared. So we're solving the corresponding elastic problem and then we're gonna inverse Laplace transform. So I have three unknown coefficients. Hopefully I have three boundary conditions. So can you help me decide what the boundary conditions should be? What about at R1? They must be the same in the interior or around the end of the body. What must be the same? The displacement. Okay, so the displacement at this outer boundary between region one and region two better be the same. If it's not then either a space is gonna open up and the material's gonna interpenetrate. So that's one boundary condition. The displacement has to match at R2. What about the stress? The radial stress has to be zero. What about sigma RR is region one versus sigma RR is region two at R equal R2. Better be equal, right? Otherwise I'm transmitting a net force across that boundary. So the displacement has to match R equals R2. Not gonna write everything out. So that's two boundary conditions, we need one more. Exactly, the one we had yesterday. So the radial stress on the interior has to equal the pressure or minus the pressure because of the sign. So sigma RR region one at R equal R1 is minus P. So we have three boundary conditions. Displacements and stresses at the edge of the shell have to match and the stress has to equal, the radial stress at the inner boundary has to equal minus the pressure, okay? This is actually, I repeat this analysis in the chapter, so you have all the notes for this. So we have three equations and three unknowns. There's nothing interesting that happens, it's a lot of algebra, okay? So to keep it short, we're just gonna look for the time being here at the displacements in the exterior region because once we put in the free surface, it's the outer region where we're gonna measure things. I'll show you what the results are in the shell as well. So now we go to Laplace domain and we'll say the displacements in the outer region are gonna be some function of S over R squared. And that function of S, forgotten is exactly what it is. So we'll go find it here again for a while, here we go. Okay, so here finally we have, I'm looking for it. So the displacement in the outer region is a function of radius and the transform variable, which is basically our time, it has units of inverse time, is the C, which is a function of S over R squared. And that C looks like this. It's the transform of the pressure, which is gonna tell us something about the pressure history. Okay, R one cubed, this is the radius of the chamber over four mu. We saw that yesterday, we had PA cubed over four mu, times S plus something over denominator. And that denominator has this ratio of stiffness to viscosity, but also depends on the ratio of the radii cubed. So this is the characteristic relaxation time for this problem, or it's inverse actually. Okay, so the characteristic relaxation time for this problem, which I'm gonna write is tr is three eta over mu plus mu. So this characteristic relaxation time always has the ratio of viscosity to stiffness, right? Pascal seconds over Pascal's, but it will also depend on geometry in some way. And here it ends up depending on the ratio of the radii to the cubed, which you don't know before you start the problem, but that's what falls out. So as I said, now we're almost done. We have the solution in the Laplace domain. All we have to do is inverse Laplace transform. And that's a step I won't show you unless people who have seen Laplace transforms, we can talk about it later. We're just gonna jump to the solution, and it has this wonderful form. So this is for a step change in pressure. So at time zero, the pressure changes from zero to P naught. That's a special case. So what do we have? Let's look at this. This is what we wanna, can everybody, is this big enough font or should I enlarge it? Enlarge it, okay, let's enlarge it. Because this is really kind of important. Is that better? So this is now in the exterior region. So this is the radial displacement in the outer elastic region as a function of radius and time. And let's look at two limiting cases. Let's look at t equals zero. So when t is zero, e to the minus zero is one, so we get one minus one is zero, that drops out. This just becomes one, okay? So at instantaneous time, we just have that. Pressure, radius cubed, radius cubed over four, sure, modulus times radius squared. That's exactly what we had yesterday. That's like the equivalent mogie solution in a full space, which it better be, right? Because at very short time, this behaves like an elastic medium. The viscous component hasn't had a chance to relax yet. So it better look like the elastic solution. In fact, that's a check that we're doing things correctly. Everybody see that? Pull it really fast that the dashpot doesn't have a chance to flow. But now let's look what happens when time goes to infinity and time goes to infinity relative to this relaxation time. So if this relaxation time is one year, then 100 years is pretty close to infinity. But when this goes, when time gets very large, this term is gonna go to zero. This term is gonna go to zero and we just get the ratio of the radii cube. But this R1 cube cancels out that R1 cube. And now it looks like a mogie source, but it has a radius of R2, not R1. So it looks like we started with a small magnet chamber and then as time relaxes, we end up with a big magnet chamber. Well, why is that? The reason for that, we didn't solve for the stresses, but I do that in the notes. And it's very interesting. What happens is that at short times, this is time over relaxation time of zero and this is the radial stress normalized by the pressure. One, this is distance, one on this scale is the magnet chamber boundary. And I picked the radius of the outer shell to be here. So at time zero, stress decays as one over R cubed from the magnet chamber. At infinite time, when this ratio gets to about 10, we can see that stress is decaying from like one over R cubed, but now from the outer boundary. And the stress within the shell is now everywhere uniform. So what's happened is that as time has gone on, the shear stresses within the viscoelastic region have relaxed and that's had the effect of taking that pressure boundary condition from the inner radius and transferring it to the outer radius. So it ends up looking like a source with a bigger radius, the same pressure. But because the displacement scales with pressure radius cubed, that means that the displacement will increase even though the pressure hasn't changed. R2 just has to be bigger than R1, but other than that, there's no constraint. At this point, the question was whether there's a constraint on the ratio of R2 to R1, okay? So now, all we've done so far, and this is the Dragone and Mananensi paper, is do this in a full space. Well, how might we imagine doing this and getting an approximate solution in an elastic half space where the outer region is elastic but there's a free surface? Well, we could do exactly what we did yesterday. And remember, we found out, almost magically, that we got the solution by taking the displacements, resolving them into tangential and vertical components, and it turned out that all we had to do is multiply by four, one minus nu, right? So that same thing applies, except now we're gonna have the restriction. Now here's where this restriction comes in. We have R1 and radius R2. Now we have to have the outer radius over the depth should be less than one. Although we saw from the McTighe analysis that in the elastic case, you could get away with ratios that were really rather modest and it still seemed to work. When I was first thinking about this, I was contacted by a colleague who was running finite element calculations, and he said, you know, I compare the analytical solution for full space with a half space solution that I get from a finite element calculation, and when I put in a possible ratio of a quarter, I find that with the free surface, the displacements are three times as large as they are for the full space solution. Four times one minus a quarter is three. So apparently this works recently well even in the viscoelastic case. So we can, it's still an approximate solution. We're not matching, we'll be matching the boundary conditions at the free surface exactly. We'll be only matching the stresses and the pressure on the magnet chamber boundary approximately. So that's the Dragroni and Manonensi analysis in the full space. Here's how we would extend it to a half space. Are there any questions? Because I'm gonna about to go back to the slides and show you what I think the consequences of this are. So remember, I started out by saying conservation of mass requires that the flux into the system be balanced by change in mass within the system, and I can change mass by having a compressible fluid by compressing that fluid or I could expand the boundaries, the outer boundaries, but that radial displacement rate now is a function of time because it's viscoelastic. So there's a bit of math here, which is not too terrible, but I just wanna outline it. So we're gonna do everything like I just said. We're gonna go into the Laplace domain so I can write those displacements at the boundary and then solve for pressure. And what happens when I do that is you can see I can write the denominator in terms of a cubic in the S. And for those of you that know about inverting Laplace transforms, it's the poles, the roots here that determine the effective relaxation times. You see some people nodding. So it turns out I'm gonna get pressure that behaves with two exponentials, either the S1T, either the S2T, where these roots, S, are roots of a quadratic and they depend on all kinds of stuff. They depend on compressibilities, they depend on this characteristic relaxation time and so forth. So I get pressure that has two characteristic times and therefore the displacements are gonna have two characteristic times, S1 and S2. So this looks like mogie. Remember we had r cubed over modulus times depth squared. There's this shape factor over here but now pressure has this complicated time dependence because pressure is changing due to two things. It's changing as magma flows in because the pressure's dropped and it's changing because as the magma chamber walls creep, they have the potential of pressurizing the fluid. When I drop the pressure, the walls wanna creep inward. And again, this is an approximation but we have some belief that this approximation might be interesting. So I just wanna point out that if you look at data from a number of places where there's good time dependent histories, and I just picked this one from Grim's voting volcano from this recent paper by Reverso, shows what this is the time history of a single GPS site. This is unfortunately under the ice cap so there's only a single GPS site showing how that behaves post-eruption and they found empirically that they required two characteristic times. They could fit this time history with exponential like solutions but it required two characteristic times. Now it's not unique that this is the interpretation but it could be that this comes from a behavior like this where we get two characteristic times related to the viscoelastic response. I do wanna emphasize, we always should, remember this is a highly idealized model. This is what some people think Long Valley Caldera and California might look like with collapse caldera fill, partially molten magma chamber, influx of basalt. We're modeling that by that. So we shouldn't forget, this is very idealized. Nevertheless, so is the picture of my shirt, of course. Absolutely. The beauty of analytical solutions and sometimes people say, you know, why can't we just go into finite element calculation and do this? Well, you can. The beauty of analytical solutions is you can explore those and see dependencies on parameters and ways that speak to you much more so than a numerical solution does. You'd have to run so many numerical solutions to get the insight for the single equation consult. So we're gonna push on this highly oversimplified analytical solution for a minute just to see where we go. And one thing that I found actually kind of surprising is what I call the no recharge limit. So let's look at the case where no flow comes in from the mantle. The only thing that's happening is we're getting the response due to deformation of this viscoelastic cavity. Well, it turns out then we can look at the infinite time response normalized by the instantaneous response. The instantaneous response is elastic. It's just what Moghi would do if you drop the pressure by a certain amount. But with time as a man with chamber creeps, we get something that has this kind of interesting time dependence. Oops, and I forgot to define this. This is a very important parameter, beta, which actually will be defined on the next slide. So maybe I'll just jump ahead. Beta is this, or capital B, is the ratio of compressibilities. It's the chamber compressibility divided by the total compressibility of the system. So there are two key limits here. One is if the magma is completely incompressible. If the magma is incompressible, so if this goes to zero, then this quantity goes to one. If the magma is infinitely compressible, then this goes to zero. So this becomes infinity, the ratio goes to zero. So zero is perfectly compressible, magma. One is incompressible magma. So now we're gonna look at what happens for different case, well it says that right here. Perfectly compressible versus incompressible. On the left-hand side, we have the pressure change within the magma chamber. And on the right-hand side, we have the displacement. So let's think about this. When it's perfectly compressible, we drop the pressure, so this is the instantaneous pressure drop in the eruption, and it stays constant. For compressible fluid, the pressure will stay the same, regardless of how much goes in and out, or what the walls of the boundary are doing. And in that case, these are displacements. The inner boundary is dashed. The outer elastic region is in the solid curve. So what we would see in terms of GPS, we would see an instantaneous deflation during the eruption. And post- eruptively, we would see continued deflation. Walls would just keep moving inward because the pressure's dropped, and the pressure stays constant. That's actually the DeGroni and Maninensi result, the constant pressure. But what happens if it's incompressible? If the magma's incompressible. If the magma's incompressible, the volume, and there's no recharge coming in, the volume of the magma chamber's fixed, that magma chamber can't creep inward. The relaxation occurs outward. So as it tries to creep, so this is B equals one, the pressure actually increases. Because the magma chamber's trying to creep in, pushing on this incompressible fluid, it causes the pressure to go up. And what's actually kind of at first counterintuitive is you can get partial re-inflation. So you get an instantaneous deflation, and then the surface would kind of rebound a little bit, not because there's any new magma coming in, but just because you're trying to compress this incompressible magma. And interestingly enough, there's somewhere in between where the surface displacement would be exactly zero, it would be time independent. You'd get instantaneous deflation during the eruption, and then nothing would happen. But there's all this relaxation going on, it just cancels out. And that turns out when, going back to this expression, when that ratio of B depends on Poisson's ratio by this expression, and it turns out for Poisson's ratio of a quarter, that's five ninths. So for this ratio of five ninths, you get no displacement. So that's the first key thing we've learned, that you can actually get partial re-inflation at the surface with no recharge, if the melt is sufficiently incompressible. And I think that's important. So that was the case of no inflow. What if I relax that restriction now and allow inflow to occur? Then we're gonna get a boundary between inflationary, post-eruptive responses and deflationary responses. Perhaps it's easier if I look here. There are cases where the displacement will get an instantaneous deflation during the eruption. In some cases, it will continue deflating for a while until magma starts flowing in and causes it to begin re-inflating. And there are other cases where it will instantaneously begin re-inflating. We'd like to know what controls that boundary because that's something we could go out and observe easily in data. We could see, does it keep deflating after explosive eruption, does it immediately turn around and begin re-inflating? And it turns out that boundary is given by this expression, which I've plotted here. So the inflationary field is on this side. The deflation is to the left of the colored lines. The colored lines are for different ratios of the outer radii to inner radii. This axis is that ratio of compressibilities. So this is perfectly compressible. This is incompressible over here. And this is the ratio now that characteristic relaxation time for the refilling time. The refilling time is this characteristic time for the pressure to recover in a perfectly elastic system. So we're looking at the ratio relaxation time to this refilling time. And the no recharge limit that we looked at on the previous slide is when this characteristic refilling time becomes infinite. So that was along this axis here, remember? And the boundary was at five ninths, which is exactly here. So on this side we got, when it was incompressible, we got inflation. And on this side we got deflation, continued deflation. But now it's more complicated because it depends on the ratio of these characteristic times. So if I pick, as I apparently did here, this ratio of 0.2, for example, and I have the ratio of the radii of 1.5, so that would be the green curve, we could go over here and we would say that ratio of those characteristic times is about 1.5. So if it's less than that, if it's less than 1.5, say 0.5, we would predict continued deflation after the eruption. So instantaneous deflation during the explosive eruption continued deflation until magma starts flowing in and then it recovers. But if the characteristic relaxation time is longer than that, say 4.5, we get into instantaneous reinflation. And in exactly 1.5 you see the time derivative there is zero at the limit. So instantaneously it's neither inflating or deflating, all eventually it turns around to deflates. So why is this interesting? What's interesting to me is that if we could constrain this parameter, and we could constrain the ratio of the radii, that we might be able to constrain this characteristic relaxation time and say something about the effective viscosity. There's a bunch of ifs, right? You'd have to constrain some of these parameters to get at the other ones, but at least this is something we can easily go out and measure. Does the post-eruptive response look like continued deflation or does it immediately turn around and start reinflating? So there are other things that we can look at actually that I think are kind of interesting. One is how long does it take the pressure to recover in the magma chamber? And the other is how long it takes the mass of magma to recover? What's the characteristic times for mass recovery and for pressure to recover? So in the perfectly elastic system, that is easily determined, so I call this T star, tau is the characteristic refilling time. This is the pressure drop, the instantaneous pressure drop during the eruption and delta P naught was how far away we were from magma static equilibrium at the time the eruption started. That was a long time ago in one of the earlier slides. Remember, so I'm asking the question here, I should graph this, I'm asking the question, this is pressure, there's some initial pressure at time zero, let's call it here, this is delta P zero minus, the pressure drops during the eruption and then it's gonna recover in some way. I wanna say how long does it take to get back to where the pressure was just before the previous eruption? That doesn't guarantee that an eruption can occur but maybe it's a necessary condition. You think that you at least have to recover the pressure that you lost previously. And remember the pressure recovered exponentially, so if we were at equilibrium, I call this point here where pressure was such that there was no flow coming in, then that would be infinitely long. So if that delta P naught goes to zero, it takes infinitely long time to recover because it's an exponential recovery. But we'll assume that we're in a state where at least there's some very slow inflation going on before the eruption, so we're not at that case. Anyway, so it's easy to find what that is for the pressure to recover. In the elastic system, the mass recovery is proportional to pressure. In the elastic system, everything's proportional. So pressure recovery and mass recovery are proportional, that's not true in the viscoelastic case. And you can show that for very short relaxation times in that limit, the characteristic recovery time for pressure is different and it has this factor out front which is greater than one. So in other words, viscoelasticity can delay the recovery of pressure. So rather than showing you equations, let me just show you a couple examples. It's sort of not necessarily immediately intuitive how this works. So we have pressure recovery in red and mass change in blue. This is time on this axis normalized by this characteristic recharge time. And so the left axis is either pressure or mass and zero by definition is where we were immediately before the eruption occurred. And this is for a case where the relaxation time is short. It's one-tenth of the characteristic refilling time for the elastic system. So you can see that there are two characteristic times in this problem. There's a short relax, there's a short time and then there's a long time. But the interesting thing that happens at least in this particular example, and I picked this because it shows this interesting effect, as the magnet chamber starts creeping in, this B-parameters 0.8, so it's pretty incompressible, it drives up the pressure, but that means you reduce the pressure gradient that's driving recharge. So you actually slow down recharge by driving the pressure up. So the first thing that happens is the pressure recovers pretty quickly, but now the gradient driving flow in from the mantle is very reduced, so it takes a long time for the pressure to recover. This is where it would be in an elastic system. So the viscous relaxation has delayed recovery of the pressure. The mass is shown here in blue, they both recover at the same time, and that was given by that simple expression, but that's not always the case. Here's when that ratio of relaxation time to refilling time is 10, and you can see the pressure recovers pretty quickly, but it takes the mass to recharge the magnet chamber a really long time before that recovers. I think I will skip this. I have a bunch of plots that look like this. The interest of time, I'm going to show you this one. As volcano geodesists, we use surface deformation as a proxy for what's happening in the magnet chamber. So what this shows is the displacement, how much of the deflation, the coerruptive deflation has recovered, at the time the pressure recovers. A lot of these plots are going to have this parameter B on this axis, so this is perfectly compressible, this is incompressible, and this is the log of the ratio of the relaxation time to recovery time. If this was zero, it would mean all the deflation had recovered at the time the pressure recovered. But you see there's a part of the parameter space where this is considerably less than zero, it's negative, it's about 0.3, so that means only 70% of the deflation had recovered at the time the pressure recovered. You say that again. You've got instantaneous deflation, you're tracking your GPS receiver, you recovered 70% of the deflation has been restored, but now you're already back to the case where the pressure in the magnet chamber is equal to what it was prior to the last eruption. In an elastic system, tracking displacement is exactly proportional to what the pressure is doing. When the material starts behaving in a time-dependent manner, you no longer have that correspondence. I think I wanna make one more point and then we'll maybe take a break. That is, it's also interesting to look at the stresses. So we'd like to think about what are the conditions when the next eruption might occur? Well, you might say recovering the pressure in the magnet chamber to what it was before, maybe that's an important condition. But if the physical mechanism for the eruption is a dike propagating away from the magnet chamber, then it's not just the pressure, the stresses at the boundary of the magnet chamber will also matter. So we have these poop stresses, these sigma theta, theta's. They can go into compression, which would make it difficult for a dike to propagate away from the magnet chamber. So maybe we should look at the stress as well as the pressure and think about how it's changing. Well, here's something very interesting happens. Now I'm plotting the stresses as a function of time. Both the radial stresses and the poop stresses as a function of time for three different ratios of relaxation time to refilling time. Now, what happens? Let's think about this. The pressure in the magnet chamber decreases. The magnet chamber moves in. That's a radial tension. It's pulling radially, so it's an effective tension. And then as the pressure recovers, it's gonna get more and more and more compressive over time. If we look at the hoop stress, the stress is going around the magnet chamber, which is what the dike has to, will be facing as it tries to propagate away. When the magnet chamber shrinks, it moves in, those hoop stresses become compressive. So what we see is the radial stresses are shown here dashed. So at the time of the eruption, they come up and become tensile, and then they relax away with time and go back to zero. But the hoop stresses do this interesting thing. They start out being compressive. Magnet chamber shrinks, goes into compression, but then as the magnet chamber relaxes, these actually change sign. They shoot up and become tensile. This is relative tension. This is the no recharge limit. So in the limit of very fast relaxation, it will approach this no recharge limit, and then they decay back away to zero. So you have this sign change with fast relaxation, which actually causes the hoop stresses to go from being inhibiting dike propagation to actually assisting dike propagation, which is something that I didn't anticipate when we started this. So if we plot what you could call an excess pressure, that is the sum of the magnet chamber pressure and the hoop stress, again, is a function of time normalized by relaxation time. This is fast relaxation, intermediate, and slow relaxation. You can see that you can get back to the case, and zero here is just what it was prior to the eruption. We're just measuring where we were prior to the last eruption. That can happen pretty quickly. But that doesn't guarantee that a dike could actually get very far away. It just says you reach a condition where you could potentially open the dike. Let me just go to this one here. So I'm gonna plot this effective, this excess pressure as a function of distance. This is within the shell. This is outside of this elastic shell at three different times. And you have to remember what those times are. The first is when the excess pressure recovers at the magnet chamber boundary. So this is saying the sum of the hoop stress plus the magnet chamber pressure is back to where it was at the time of the last eruption. It's recovered. And the interesting thing is when that recovers at the boundary, it's recovered everywhere within the viscoelastic shell. And it turns out that can happen way before the pressure recovers in the magnet chamber system. The green curve is when the magnet chamber pressure recovers. So you can see here that the excess pressure is positive within the viscoelastic shell, but it's still negative outside of the elastic region. Now, what controls dike propagation is much more complicated than this. I don't mean to oversimplify it, but you might say that at least you have to have pressure and stress conditions to allow the crack to open. That may be a necessary, but not sufficient condition for the dike to grow. For the dike to actually propagate, things come in like thermal effects. The dike has to be able to propagate fast enough that it doesn't freeze before it gets very far. We're not worrying about that. We're just looking at the pressure condition. You can, the third curve here is the blue curve is to say, when is the excess pressure recovered at the boundary of the viscoelastic shell? And there it's either zero or positive everywhere. But I think the take-home message from this is that it could be in the systems where you have rapid relaxation that you get to a stage where it's possible for incipient dikes to start propagating into the viscoelastic shell. They may not be able to get very far, but they would serve to transfer heat into this outer region, this viscoelastic region, keep it hot and keep it viscoelastic. So there may be the potential for incipient diking at relatively early times before the pressures even built back up to where it was before. And as I showed you, you can't be guaranteed that the surface displacement that you measure from GPS has recovered the deflation even from the last eruption. It will take longer for the states to recover to the point where you can probably propagate a dike all the way to the surface and initiate another option. That's probably beyond this simple-minded analysis. But I think we've learned a couple of things from this. Idealized, very idealized analysis. I'll just run through them. I think it's probably somewhat unintuitively that post-eruptive deflation, oh, that should say deflation, sorry, inflation can occur without recharge. If the mag is sufficiently incompressible, even if there's no new mag flowing in, you can get a partial recovery of the deflation whether or not in the more general case you go into deflation and continue deflation or immediate reinflates is gonna depend on the material properties and the viscoelastic timescales. Fast viscoelastic relaxation does delay recovery of the chamber pressure. So it can sort of stall the next eruption if this material relaxation occurs. And again, you're not guaranteed to have the surface displacements that we measure from GPS are not tracking directly what the pressure's doing. I showed you for short relaxation times that the hoop stresses go through this sign change so it can actually favor the conditions for incipient diking at relatively early times within the viscoelastic region that may act to keep that viscoelastic region hot, transfer heat to the surroundings. I think this is something that is worth trying to test how we would do this. I'm not quite sure at this point but that's something we might try to look for. And right, excess pressure can exceed that at the time the previous eruption. In some cases, even before the deflation has recovered it. So I've shown you a bunch of things that you can deduce from a very simple model. And I think it's been an hour and a quarter. That's a pretty long time. Maybe we should take some questions and then have a quick stretch.