Thank you for the invitation. It's a pleasure to be here. And so I indeed, I'm gonna talk about a coherent charge dynamics and phonons. So let's start with conventional charge dynamics and phonons just to remember a few things. So phonons, right? So electrons are charged and they can emit phonons. And so phonons are emitted by particle whole pairs, right? So electrons coming along, emits a phonon. So how can that happen? So the kinematically, the relevant quantities are the spectral weights of particle whole pairs. So I'm not sure what should I point at? If I point at this thing. So now I can't, anyway. Okay, so this is omega and K. And so at low temperatures, the particle whole pairs have the Lindhard continuum, right? And so this is where the particle whole pairs are. Perfect. Okay, thank you. Is there a pointer here? Perfect, okay, thank you. So the particle whole pairs live, this is where they live. And then the phonon has some dispersion. So for example, here I've plotted an acoustic phonon. It doesn't matter what it is, right? And so the sound speed is much less than the thermal velocity. So the acoustic phonon mode is pretty flat on electronic time scales. It just goes along the bottom. And so this region here is where, from here to here is where the particle whole pairs are kinematically able to emit the phonon, okay? And so if we want to see how strongly electrons couple to the lattice, there's gonna be some electron phonon coupling and you're gonna have to integrate the electron phonon coupling along this line, which is where the particle whole pairs are able to emit the phonon, okay? So there's an interplay between two kinematics, the particle whole kinematics, which is this band and the phonon kinematics, which is this line, okay? And that determines how they can interact. That makes sense. So pick your favorite phonon dispersion and that'll be some line on this curve. Now, that's all fine. And that's been understood for almost 100 years. So however, in strange metals, it seems clear I'll review some experiments in a minute that the density spectral function, so that was this blue curve, it just does not have the Lintard form and particularly does not have this sharp. So this Lintard form has a sharp singularity at small k and at 2kf, right? At the two sides of the Fermi surface, right? So that band comes from the small k at low energy. You can either be close or you can go all the way to the other side of the Fermi surface and that's responsible for this band, right? This is 2kf. Here, these are low energy fermions doing that. And so this structure is very much based around having coherent excitations that live on the Fermi surface. So as I'm gonna tell you, that does not, by definition almost, is not the case in the incoherent metal. And so if the spectral density is not of this form, then obviously that's gonna make a big difference on the electron phonon coupling. So an example of what an incoherent spectrum might look like. So these somewhat recent experiments on bisco, these are these M eels experiments. So what was measured is this quantity, but how many charged excitations there are as a function of omega and k. This k squared is pretty much fixed by kinematics. And then what the point here is there's hardly any k dependence. So these things basically just flash as a function of k, all right? And so in particular, there is not a 2kf. There are not these sharp features. So this plot is at quite high frequencies. So this is a line sort of like this, at quite high omega. But what is missing are these features where you would notice where you see the sort of sharp edges of the Fermi surface, okay? Another example is from this cold atom simulation of the Hubbard model. So I think this is actually pretty interesting. I'll tell you how the experiment goes is you set up, you have your Hubbard model in a trap, right? You engineer your cold atoms to have Hubbard model Fermi Hubbard model interactions. You, and you set up a profile, a charge profile of the atoms which is modulated in space and it has some wave vector of k, okay? And then you let go and you watch it decay, right? And so the amplitude, this is the amplitude. So as a function of time, this amplitude decays, okay? Then you vary k, you change the modulation, and then you plot the amplitude again for a different k, all right? And so this plot here is the decay of the amplitude against time for different values of k, right? The different wave vectors. And so the point here, when k have a very long wavelength, sorry, let me get this the right way around. Yeah, the long wavelength one is decay, but for shorter wavelength is actually a beginning of an oscillation and so on. And so what this is is a direct, a very direct position space measurement of this quantity, because you're measuring how a charge is decaying as a function of omega and k. And so their results fit very nicely to this form, which is sort of almost the simplest form you can imagine. If this term were not here, this would be purely diffusive, right? And so this tau is a boundary between, so if this term wins, then here you have a linear dispersion between k and omega, and that's responsible for these oscillations, okay? So there's a crossover as a function of k between a diffusive form and an oscillating form in this function, okay? And this is actually, I think this function also has a very respectable pedigree. It's sort of the simplest, if you're gonna write down the collective form for the charge dynamics, this is the first function you would write down, okay? So the upshot here is that there's again, there are no 2KF singularities in sight, and instead there's a very collective form for the charge dynamics, okay? Very quite high. That's like at some about one-third of the Fermi energy to a few times the Fermi energy. Yeah, thank you. These are quite high. So this is in the bad metal regime with the Hubbard model at quite high temperatures. And this is 2D or 3D? 2D, okay. So all of this is to motivate. Suppose the electronic dynamics are, suppose the electronic dynamics give you some incoherent non-Fermi liquid, take that system and now couple it to phonons. What happens? So what is the formalism in which we can add phonons to a already existing, strongly incoherent electronic system, okay? Another motivation for doing that, came from something I noticed a couple of years ago in Cooperates. And so let me ask the following question. So in Cooperates, there's some phase diagram and so on. And so let's say, let's be at room temperature, okay? Now, sorry, before Cooperates, let's think about copper for a minute. So on copper, we know very well that the resistivity is a function of temperature. Looks something like this, right? And it has this linear form. When is the linear resistivity of copper kicking? It kicks in from about one-third of the divide temperature, okay? So above the divide temperature, the phonons are classical and they scatter with linear and T. And this process starts at about one-third of T-debye. In Cooperates, T-debye is maybe 600 Kelvin, something like this. So one-third of T-debye is, you know, down somewhere down there, roughly. I'll just call it T-phonon. I'll just take phonon scale. So here, as is well known, the resistivity is linear, okay? And most people believe that has nothing to do with phonons, okay, in Cooperates. However, the phonons are there. And so you could ask the question at room temperature in Cooperates, do phonons contribute to scattering or not? So does this T-linear resistivity of the Cooperates have a contribution, which is scattering off classical phonons? The classical phonons are definitely there. Do the electrons see them or not? That's the kind of question that you need a formalism to couple incoherent electrons to phonons, okay? That's the kind of question I would like to have a formalism that even asking that question, right? So suppose you believe that the electrons here are described by the Haber model, solve the Haber model and then couple it to phonons and ask, do phonons contribute to the scattering or not, okay? So one way to think about this question, let's not start here, let's start over here, okay? At some very high doping, let's say 0.33, just in LSEO. So that's beyond the superconducting dome. Here at low temperatures, the resistivity is T squared, okay, at low temperatures, that's this plot. This is LSEO at 3.3, okay? Well, way beyond the superconducting dome, at low temperatures, and what's being plotted here is the derivative of the resistivity, okay? The derivative, that's why it's a bit noisy, right? So take a bunch of points, you just take the differences, that's the derivative, okay? At low temperatures, this is a straight line with intercept, basically zero, right? And so the derivative is T squared with intercept zero, sorry, if the derivative is T with intercept zero, then the resistivity is T squared with no T linear components, okay? So it's the statement I think hopefully familiar to many of you, if you're down here, away from anything interesting at low temperatures, the resistivity is purely T squared, okay? Clearly electrons. But if you go up, when you cross this scale, which is about 150 or 200 Kelvin, which is roughly a third of the phonon scale, there's sort of a kink, okay? And a simple, I mean, this is just data, right? A simple interpretation of this kink is that the line changes and now there's an intercept. This is the derivative. An intercept in the derivative means a T linear term is turning on in the resistivity, okay? So around here, so up here, the resistivity is T plus T squared above this point. There may be many things happening at this scale, but one thing that is happening is that scattering of classical phonons is possible, okay? So a possible interpretation of this kink is the onset of phonon scattering. It's like that, so. It's exactly what you should expect because the mass gets renormalized also when you go through the phonon scale. Yeah, so two things happen. So phonons do two things, so it's quite, so this is temperature and there's some characteristic scale. Above this temperature, the phonons scatter efficiently. But they don't renormalize because they're too slow. Because above this temperature, the electrons are much faster than the phonons. Below the phonons don't scatter, but they do renormalize the mass of the phonons. And so a kink in the T squared is very reasonable that it happens at the same scale. So okay, I have no ideological commitment to this being the phonons, but that is at least a simple possible explanation for that. Okay, let's suppose that we found some phonon scattering, which has every right to be there. There are phonons in the system. It's a Fermi liquid. You go above the phonon scale. Why should you not see phonon scattering? Okay, but then we can ask what happens to these phonons as you go across towards optimal doping. So here the resistivity is purely T linear. There's no T squared components, right? And so if you keep going, you can do these plots. And actually what happens is that this kink survives. It remains at roughly the same scale at the same temperature, but it gets smaller, okay? And that's because at low temperatures, a T linear term starts appearing, okay? And so that's why at low temperatures, it goes up and eventually the kink disappears around this doping. So to put it on a face diagram that you might have seen before, there's a famous plot from this Cuperdall paper. And so this feature I'm talking about, by the way, these log derivative plots are very obscure, a lot of things, okay? But actually, there's a little bit of a bending. So this feature I'm talking about is here and it continues until about at this point, the kink disappears and it merges into a completely continuous T linear all the way. Okay, by the way, one of the mysteries, right? That the Cuperdall is that this T linear scattering doesn't have kinks in it, okay? It goes straight from low temperature to very high temperature, well, there is a kink if you're away from optimal doping, but it disappears as you go here, okay? So what this led me to think is that it's really urgent to try to disentangle the ball of electrons and ponons in strange metals, okay? And so to do that, I thought it might be nice not to consider the resistivity, but a different process where electron-phonon interactions are really essential. Resistivity is complicated, they're electrons, spins, ponons, collective modes. I wanna consider now a different process which absolutely crucially depend on electron-phonon interactions. I'm gonna think about that process, which I'll tell you what it is in a second and ask how to apply that to systems that are incoherents, okay? All right, so that was motivation, right? So two motivations, there exist incoherent metals where the spectral density is not the lint-hard form and it's plausible that phonons are playing an interesting role in systems that we care about. So we need a formalism that couples phonons to incoherent metals, okay? So the process, the observable that I'm gonna think about is joule has to do with joule heating. And so let me tell you how this about this. So really the electrons that we like, our carbon model lives here, but in reality it's coupled where there's the environment, the big environment. And then there's sort of the immediate environment which is the lattice in which the electrons live, okay? So the electrons are coupled to themselves, right? They have electronic interactions that are coupled to the lattice. They're also coupled to the environment and the lattice is also coupled to the environment. So I'm gonna consider, I need some structure I don't want to distill the most general case of anything. So the idea is gonna be suppose you apply, supply some energy to the electrons. For example, you apply an electric field, right? Or you just whack the system with a laser. I'm gonna apply some energy to the electrons and this energy has to go somewhere, right? Otherwise the electrons are just gonna keep heating up all the way to infinite temperature. I'm gonna consider situation where they mainly go is into the lattice, okay? And I'm gonna consider situation where this step is slow compared to this step, okay? So what that means is that if you perturb the electrons, they quickly reach an electronic thermal equilibrium but they're not in equilibrium with the lattice. And then there's a slower process where the electrons are thermalized with the lattice. That should be very reasonable in these incoherent metals where the electron interactions are very, very strong. But if you believe that the Hubbard model is the underlying essential physics of the cuprates, that means you believe that electronic interactions happen much faster than interactions in the lattice. Yes? So they even emerge on metal, right? Yeah, so perfect. If you're happy, perfect. If you believe this in conventional metals, you should only be even more happy in unconventional metals, that's all, yeah. Okay, so, yeah. Why is that so intellectual? So, okay, I think it might depend on the temperature. And for example, as I'll review in a second, the conventional theory of heating does assume that the electric, it doesn't, okay, let's just stick around conventional metals where it's more likely to be true. So, and what's this gonna buy me is this hierarchy of time scales is gonna be the theoretical input into what I do, okay? So if they don't, very good. So there's the simplest model of dual heating is this two-temperature model where the electrons have a temperature, the lattice has a temperature and they thermalize. In reality, you have to consider more complicated things where they're sort of both thermalizing in parallel, okay? But there's the simplification. A common simplification is one where the electrons and the lattice thermalize separately and then that interacts. Okay, so let's put some equations that go with this picture that I just showed you. So suppose the electric, so it's applying electric field, okay? And so that pumps energy into the electronic system. So this thing we're about to show you here, I think it's very well known, but it's not so easy to find the equations written down somewhere. So maybe this is helpful. So when you apply the electric field, that starts heating up the electronic system, that's dual heating, right? And so the energy in the electronic energy starts increasing by E dot J, right? E dot J is the rates at which you're pumping energy into the electronic system. If the electronic system were a closed quantum system, the temperature would just go off to infinity. So to have a steady state, it's crucial that there's some bath where the electrons can dump their energy, okay? That I'm gonna imagine the case where that's the lattice that is dumping the energy and there's a timescale tau associated with this process, right? So that tau was this process here, right? The rates at which energy is going out of the lattice. So in a steady state, we have a steady, what you actually measure, right? You apply electric field, you measure a steady current. You want this time derivative to be zero. And so you have to balance these two terms. If you balance these two terms, you see that the electric field leads to a shift in the energy which causes a shift in the temperature, right? The lattice is gonna heat up a little bit, right? By balancing these two terms. So the shift in the temperature is the shift in the energy divided by the specific heat. And now the energy, according to this formula, is given by E dot J. And now at this point we use Ohm's law, say that J is sigma E. And so we get this. And so the electrons are a little bit hotter than they would be proportional to the electric field squared. So the electric field heats up the electrons a little bit but because the heat is also constantly going out, it's stable, okay? But this means that this electric field shift of the temperature causes a non-ohmic heating, right? So J has the original temperature of the lattice times E, but the shift in temperature causes a shift in this temperature. And so in fact, there's an E cubed term because the temperature depends on the electric field, right? And so this is the simplest physics of this time scale, okay? Where this time scale appears, is it determined of course, there may be many other sources of non-ohmic conduction, but one of them, the universal one, is due to dual heating, okay? And so something that we can see in this, I find this formula very useful. So if tau goes to, if we turn off the bath, the coupling to the bath, tau will go to infinity and linear response will break down. But for any finite coupling to the bath, as long as tau is finite, if you send the electric field to zero, it will recover Ohm's law, okay? So a non-zero coupling to the bath is crucial to get linear response, but we can take most of the time, we don't care about that, because you can take the electric field to zero and then this is sub-leading, right? So tau to infinity and electric field to zero don't commute. Okay, very good. Another place where this time scale appears is when you is in pump probe spectroscopy and there you actually measure the temperature of the electrons as a function of time, okay? So you whack the electrons with some laser. Because the electrons have a lower specific heat than the lattice, they heat up much more than the lattice does, right? So most of the energy, most, so you whack the system that hit, that the energy goes into everything. The electrons have a lower specific heat so they heat up more. And so, and then what, so you whack it with one laser and then you start probing it with another laser and you wanted to see how the electron temperature decays over time, okay? So this one version of this is time-resolved opposite, where these lasers are actually spatially resolved, then you can zoom in on the Fermi surface and you can get plots like this one, which actually was in Bisco. And so here, let me get it right, that's right. So this is basically the width of the distribution of fermions at the Fermi surface as a function of time. And so at early times, the electrons have been heated up to some very high temperature. So this is broad and as the electrons cool down, it narrows, right? That this is the sort of the width of the Fermi direct distribution. And so one way to directly measure the temperature of the electrons is to measure the width of the Fermi direct distribution on the Fermi surface. And that's what this is doing. And so now you can plot, and so this is a plot of the electronic temperature, the red and the black are two different initial temperatures. So here the temperature you see is getting, it gets whacked up and the decay is down. And this decay is not a trivial thing, but the initial decay is exponential. And that also defines this timescale tau, okay? So this tau due to coupling to the lattice is a very physical thing, okay? It determines dual heating. It determines this kind of process. And clearly it could not exist without the lattice, right? So the reason I'm talking about this is to identify a process where electron-phonon interactions are not optional, they're crucial, okay? Resistivity exists without phonons. This does not exist without phonons, okay? The energy needs to leave the system, okay? So these kind of probes, I think the kind of thing that we should be doing, not me, but other people should be doing in these strange metals to really directly see what the electron-phonon coupling is doing, okay? All right, very good. So now what's the theory of this timescale? So in a conventional metal, this existed for a long time and it goes back to the Soviet paper of Kaganov-Lyschitz and Tanatorov. And so for non-interacting electrons, they just coupled to phonons, they obtain this timescale. There's a simple formula for it and the result looks like this. So this is for non-interacting phonons coupled to, sorry, non-interacting electrons coupled to phonons. And so this is the, this rate, this rate, this is, I chose an acoustic phonon, but it doesn't matter. This is the temperature over the Bloch-Ruhneisen temperature, which remember that is that. So if this was VF, it would be the Fermi energy, but it's V sound, okay? And this is the natural scale in electron-phonon coupling, okay? So if it has this form, there's some big peak. So a couple of interesting things. So this plot is very intuitive. The reason it goes to zero here is because there are not many phonons there, okay? At low temperatures. This, the reason it goes to zero at high temperatures is more interesting. That's because above the Bloch-Ruhneisen scale, the electron, the scattering is elastic from the electrons point of view. The electrons are much more energetic than the phonons, right? Because the electrons have energy T, but the phonons have energy below the scale. And so because the scattering is elastic from the electrons point of view, they don't lose a lot of energy in any scattering. And because they don't lose a lot of energy, this thermalization becomes slow, right? The rate at which it loses. And so this decrease at high temperature is not because the scattering is weak, it's because the scattering doesn't transfer energy efficiently out of the electronic system. And so there's a maximum, and curiously enough, if the electron-phonon coupling is this value, which is a typical value in a non-superconducting metal, this maximum is actually this Planckian scale. All right, so as a first, so, okay, very good. So we will rewrite this in a second. So what I'm gonna talk about now is in this paper with Paolo Glorioso, who's up, yes. Bye-bye, what's your bye? Yeah, I'm assuming that that is not the dominant factor. Indeed, so that would be, which I think it isn't, but I'm not, yeah, I say something, that's kind of like this, right? Where you go straight into the environment, and I'm just not, I haven't thought about it. Well, one should think about it. I agree. Good, so what we did in this paper is get a formula, a tau, for this rate at which you lose energy to the lattice that holds for any form of the electronic spectral density as long as the time scale at which you lose energy is much longer compared to electronic timescale. So the assumption is that the electrons thermalize quickly and then slowly lose energy to the lattice. In that limits, you can sort of treat the energy loss as the perturbation, right? And you can get the formula, which I'll show you. And so the formula, oh, sorry, no. Let me quickly tell you about where the formula comes from. And so it's a method that I think should be more widely used in general. And this is, which is this sort of, so there's this thing called the memory matrix methods and which are based around Cooper formulae. And they're useful precisely when there's one slow process that you want to think about. So for example, so an example is the Druda formula. Right, so when you have something like this for the conductivity, this is not normally exact, right? This is that there are many timescales in a metal, right? You have no right to have the Druda formula. When is the Druda formula exact? The Druda formula is exact when momentum lives much longer than anything else, okay? So say you have a translation in variant metal, okay? Then the connectivity would be infinite. And so to get a finite connectivity, you have to break translation in variants. The interactions of everything else can be very strong. But if you break translation variants weekly, then momentum is much longer lived than all other operators. It decays much more slowly, right? When momentum decays much more slowly, then this time scale in the Druda formula is what Druda originally thought it was, which is the momentum relaxation rates, okay? In a conventional metal, it's not the momentum relaxation rates, okay? But if the metal is very, very clean, then it's the momentum relaxation rate. Then this formula can be directly related to a single slow quantity, which is the momentum. Here, the slow quantity is gonna be the electronic energy, right? So the electronic energy in a closed system, the electronic energy is conserved, right? The dots is zero, right? Now I'm gonna imagine the couplet that lattice is weak. So the electronic energy is not conserved. Of course, the full energy is conserved, not the electronic energy, and it's much longer lived than typical electronic processes that decay quickly. So when you have one long lived quantity, it's possible to get a formula for this time scale that depends on the two-point functions quantity. So that makes sense, okay? So whenever you have one operator that lives much longer than all other operators, it's decay rate controls everything, okay? So for example, so here is the formula you get in this case is at this time scale, right? This rate of decay is given by the spectrum, the two-point function of E dots, right? So E is the electronic energy. Another way to say it is that the electronic energy now obeys a Drude-like formula, right? So normally this is like a Drude formula, but not for momentum, but for the energy, right? The energy decays slowly. It has a time scale, okay? And so if the electronic energy obeys this formula, right? Which means which it will if it decays slowly, then you can extract this one of a towel using this, right? This order of limits are designed to extract well, this towel from this formula. So if you take this formula and you plug it into here, you'll get this towel, okay? Now, but it's interesting so where E dot just means that you multiply by omega squared, okay? And E dots are used for quantity because E dot's gonna be small, okay? So what is E dot? So E dot is the commutator of the Hamiltonian, the full Hamiltonian, including the phonons with the electronic energy, okay? The only part of the Hamiltonian that degrades the electronic energy is the electron phonon part of the Hamiltonian. So this is the commutator, the electron phonon part of the Hamiltonian with the electronic parts. But the electron phonon part of the Hamiltonian is some electron phonon coupling times the density. So I'm gonna assume this is just the model which we can simply change that the phonons couple to the density, right? And this is the simplest way. So then here, this just becomes the electron phonon interaction B of K and the electronic density N. At the Hamiltonian with the electronic density, that's just N dot, right? Because this is the electronic Hamiltonian, okay? So this formula just becomes a two-point function for N dot, yeah? Which is a purely electronic quantity. And if this time scale is very long, one evaluates this to, so this E dot depends on the electron phonon coupling times, but it's an electronic operator. And so if this time scale is very long, you can evaluate this quantity in the theory without the electron phonon coupling. So I think this is an important point. So suppose you have the Hamiltonian is some electronic Hamiltonian as a small number times the electron phonon Hamiltonian, all right? Then what I'm calculating is the rate, some Green's function for the rate at which this decays this electron phonon coupling, right? What I'm showing you here is that this is related to some Green's function of N dot. But this will have an epsilon squared in front, right? Because H dot comes from the electron phonon term. So it's electron squared. So now this Green's function to leading order, I evaluated in the epsilon equals zero theory because I already have the epsilon squared out the front. The trick of this memory matrix is to use these time derivatives to bring out the factors of the small parameter. And then once you have left, you can just evaluate in the theory without the phonons because I already have the epsilon squared, okay? And so now I just need to calculate. Does that make sense? Maybe, okay. So I might not be able to solve the electronic dynamics, but as long as I can measure this quantity in the electronic system, I now know how to couple it to phonons. I don't have to consider the couple, the electronic phonon system and non-perturbitably. I'm sorry, take it. Phonon is completely perturbed. Yes, in this thing, that's right. It's being added on perturbedly to some. And but what the point of doing all this is that the electrons can be as strongly coupled as you like, but the only information I need will be this, this single Green's function. That's why it's useful, okay? So I'll show you, I think this will become clear in a second. So for elucidative purposes, not essential, we can consider the simplest, an acoustic phonon that has this deformation potential coupling. So that means that the electron phonon coupling is acoustic, right? So it's linear in K. And then, okay, but like this is the textbook coupling of electrons and phonons, right? So this is what you get, right? So I now take this electron phonon coupling and I calculate this Green's function for an arbitrary electronic system. The electrons can be as strongly coupled as you like. And this formula is very intuitive, okay? So it's a formula for this decay rate. So there's an overall constant, which is just a strength of the electron phonon coupling. You integrate over all possible, we're calculating the rate to which we emit phonons, right? So you integrate over all possible weight vectors of the phonon. There's a K squared, that's because it's an acoustic phonon and low K acoustic phonon don't like to couple to electrons. There's some typical thermodynamic factor. What this does is tell you that if the phonon that you emit has an energy that's bigger than the temperature, that's exponentially unlikely, okay? All this form does is say you can only emit phonons with energy less than the temperature. It's very reasonable, right? And then this tells you, this is the NN, this is the spectral weight of the Green's function. It tells you how many electronic excitations do you have that can emit a phonon with momentum K and energy omega K, right? So how many, and this is where all the electronic dynamics just has the height, it gives you this quantity, right? It tells you how many excitations are there in phase space that can emit phonons. And then the rest is just kinematics. So you then need to get this quantity from the electronic sector. So as an example, you could take this formula and plug in the free fermion version of the GNN, which is this guy. And you take the, this is the same plot I showed you before. So now you do the integral along the, so this integral here is the integral along the orange line. You do the integral along the orange line of this formula you put in the Lindhard form and you'll get this, the Soviet formula on the nose. Okay. All right, so just I'm just trying to convince you that this is a correct formula, right? If you put in free fermions, you'll get the textbook formula for this one of a town. Yeah. Okay, but now we can just, I showed you two experimental examples of this quantity that were not Lindhardt functions. Okay, say you solve the Hubbard model in the cold atom simulator and then you say, well, how does it couple the phonons? Well, you have to take, you have to take this quantity from your cold atom experiment and plug it into this formula. Obviously the cold atoms does not have phonons in it. All right, but the cooperates do have phonons in it. And so if you believe the cooperates of the Hubbard model, you simulate the Hubbard model, then you put in the dispersion, the phonon dispersion of the cooperates and you plug it into this formula. Okay. Yes. I said this formula was obtained for high temperature. For what's over? This cold atom formula was obtained for high temperature. High temperatures, high temperature. Yeah, sorry, in principle, that's right. Thank you, sorry. So the data at the moment, that's right, thank you very much. At the moment, data on this formula in the physically most relevant regime in metals is a bit sparse, right? So the eels experiment is at quite high frequencies and the cold atom experiment is at quite high temperatures. Okay. And so I hope that this will change over the next few years. All right. So once you have this formula, I'm telling you what you should do with it. That's right. So for purely illustrative purposes, I'm gonna, that's right, thank you. Although that said, let's go back to here. When these systems are whacked with a laser, the temperature actually goes up quite a lot. In the electronic temperature can go above 1,000 Kelvin in these pump probe experiments. And so high temperatures may not be totally irrelevant. But I certainly hope there will be data at lower temperatures and lower frequencies soon. But so then, so these two examples I'm showing you are just illustrative.  It's just to get some sense of what happened. So for example, we can just take this form. Okay. As experimentally reported. So A is, the big A is known, the omega C is known. Okay. These things are known. These are for, it's for frequencies above 0.1 EV. Okay. So it's quite, quite high frequencies, right? But just for illustrative purposes, we can take this and we can couple this to an optical phonon at 0.1 EV. Okay. And let's assume the electron phonon coupling is order one just as an estimate. So you take this, you plot with these numbers are unknown. You plug it into the formula and you get one picosecond for this time scale. Okay.  Okay. These processes do happen on that kind of time scale. Which actually is not, is actually a bit annoying because it means that, oops, sorry. So it means that actually the fact that the spectrum is incoherent, it doesn't mean that you don't couple the phonons because the coupling, remember it was this integral along this line, okay? And so really it's sort of the integrated spectral weight that ends up mattering, okay? So even though the spectral weight is incoherent, the time scale that you get is quite comparable to what you would get in a conventional system. Okay. So it has to be a precision tool, right?  So a first lesson, maybe it's not surprising is that incoherent spectrum can dissipate into phonons at comparable timescales to conventional metals, okay? Which is bad if you want the phonons not to do anything in the cupboards. Okay. Yeah. All right. So before I, I'm gonna talk about the cold element. How am I doing? How much longer do I have? Okay. Am I at my negative time? Okay. Yeah. I started late. Okay. Yeah. Okay. So just I'll make, I'll make two comments. So something that, that this made me realize when I was thinking about this. So I mean, I'll do the cold atoms very quickly. But I think this point is interesting. So you could ask, this was the Lindhardt continuum? Okay. Pre-framing. So what do interactions do? Imagine adding small interactions, perturbative and electronic interactions to this system. So interactions will introduce a mean free path, right? And so let's put that mean free path with, and in a good metal, this mean free path is very long, right? And so one over the mean free path, which has units of K is very small. So one over the mean free path might be here and one over the lifetime might be there. Okay. And so the effect of interactions is only to change. So if you're at wave, if you look at wavelengths bigger than the mean free path, sorry, lengths bigger than the mean free path then the particle undergoes several scatterings before that length. And so the interactions are important. That's really here. And so this integral, okay, that you do to see the rate of phonons, it doesn't actually care too much about these interactions. But is it, I'm saying, yes, please subject. Yes. So the temperature scale is here. I can say that the unit of the gap where it's walking will probably wait a second. So this is- It's not sharp enough. No, wait, so I'm good. No, no, so this is a very conceptual. So I'm saying, so for example, a diffusion. So charge has to diffuse on the longest time scales, right? And so GNN is gonna have this diffusive form as K goes to zero. Okay. At what K does this kick in? This will happen, but K less than one over the mean free path, right? That's where this diffuses from. So this scale, forget this axis. Along this K axis, the natural scale of interactions is the mean free path. If your distance is much shorter than the mean free path, you don't know the interactions are happening. And so- No, then it's just this, right? No, no, no, no, no, no. So too much less than here. That's fine, yeah. That's fine. You already put on the grid or various elements. Yeah, so, no, that's right. But this very, right, but on that scale, this line is way out there, right? Because KF, this is KF. Temperature is negligible in this, yeah. I'm trying to say that the gap you could draw, the little spectrum gets smaller, the little spectrum gets the finite features. It's a spectrum that's there. Where do you want me to be here? You're drawing it on the line, which is through the spectrum, yeah. Mm-hmm. It's still trying to do things like, I'm saying that this one will be non-zero. Of course, yeah. Yes, that- It's not determined by interest, again, people are determined by chemistry. No, no, that's fine. This is, no, no, that's fine. But all those broadening effects are gonna be in this formula. What I'm asking, what are the effects of interactions? Okay, so I'm saying interactions are important here.  So I make it, this is a preliminary step, sorry. So interactions, so the, by the way, just for phonons, we're never up here, right? The phonon line is very, very flat. So this region is not relevant for phonons ever. So the phonons are basically at zero as far as the electrons are concerned. So when you calculate the rate of decay into phonons, most of the decay comes from large wave vector electrons because you have to do the, they all, all the electrons emit all the particle-hole pairs. And so most particle-hole, most transitions that emit phonons are these large wave vector ones and they don't see the interactions so strongly to the right of the here. However, oh, the point I wanted to make in a bad middle, the definition of a bad middle is when the mean free path becomes of order one of a KF, okay? In a bad middle, that means that the scale, let's say two KF, just to make it more fun. That means that in the, this area where the interactions are important, goes all the way up to the, up to the two KF, okay? So a bad middle is by definition, it's normally defined in terms of the mean free path. I think a better definition is when interactions eat the whole Lindhart continuum because the Lindhart continuum goes to two KF. And so the mean free path is comparable to two KF. That means the whole thing is collective, okay? You need to consider interactions. And that's kind of what this cold atom experiment sees that the diffusive form eats everything. It's not some small thing, it's everything. So diffusion is normally an effective long wavelength theory, but in the bad middle, diffusion is microscopic. It goes all the way up to KF, okay? And that's this red plot here. Okay, that's just a tangential point. Maybe I'll skip this. So you can take this formula that they measured, you can plug it into our formula and you can do a comparison. So if you take the parameters from the Hubbard model that they used and you stick it into the Kagan-Ovedale formula, this is what you get. And if you use our formula, this is what you get and it's a factor of four. It's not a auto-magnitude difference, but there is a quantitative difference, all right? So to summarize, so the main point I tried to make is that we have a Kubo formula for the rate at which electrons dump energy into the lattice. That's a measurable quantity. And I illustrated the use of this formula on two measured incoherent densities, but these are really estimates. And when these things are measured better, we could do something more precise. A final comment. So suppose as many people believe that phonon scattering is irrelevant for strange metals, okay? I'm very sympathetic to that, probably true. So then you could measure or calculate this electronic green function with no phonons, right? Those phonons are irrelevant. You can then measure this tau quantity and that will determine how strong the electric phonon coupling strength is using our formula, right? Because our formula has tau, electron phonon, and this quantity. If that electron phonon coupling is not small, okay? If it turns out to be large, then your starting belief is not true and you should have considered the phonons from the beginning, okay? So what I'd like to head towards is really trying to disentangle electrons and phonons in strong interacting systems, okay? Thank you. All right, we are running a little bit late, but maybe a couple of questions. If it's a couple, can I have two? All right, so I have two, but I'll ask one. Your assumption when you wrote down your equivalent, if you want for the Allen formula, is that you can get tau because the relaxation to the phonons is happening much faster than everything else. Slower, slower. Yeah, sorry. And usually there is an assumption, an additional assumption, which is that diffusion is irrelevant and that's why you first get the decay when the lattice and the electrons equilibrate and then there's some slower relaxation process by diffusion. This is actually, I think, an Allen's paper. But if you have incoherent metals, can you really neglect the fact that your electrons are diffusing heat themselves to the boundaries rapidly? Very good. So, right, so I think the question has to do with the, that's the same as, very good. An equivalent way to state this is that the phonons have been treated as free, as non-interacting in this, yes, I think, well, I'm not sure. I mean, ideally one should model, actually people do in conventional systems just model the electrons, the phonons and the coupling and really do everything properly. It's more complicated. I should say that even to the time scale for this phonon diffusion is quite a bit slower than the electronic one.  So, I think it may be reasonable, but what one, yeah, thanks.


We have the biggest number in the first row and calculate from right to left starting with the units But oh dear. We cannot subtract 8 from 3 Yes, you know it. We just borrow one of the tens The tens are now only 5 but our units are 13 Remember that we took 10 units Okay, our new number is now 13 in the units 13 minus 8 is 5. We now start subtracting the tens and we can subtract 5 from 5 Which leaves 0 We move next to the hundred. We see that again. We cannot subtract 1 from 0 So we go borrow from the thousands The thousands remain now with 4 but the new number in the hundred is now 10 10 minus 1 is 9 We write a comma in front of the hundreds Place value now we subtract the thousands and 4 minus 2 is 2. The difference between the numbers is 4,905 times Viewers and learners it was so nice to have your attention for this lesson Thank you. And from the hard work educational multimedia project. We wish you success in your mathematics studies Before we go who just recap We have seen that there are three main units of mass namely Tons the biggest which is abbreviated at T kilograms is most often used in Day life and it's written shortly as KG and Gramps is the smallest it is written as G each of this unit is a Thousand times heavier than the other before there are 1000 grams in kilogram and 1000 kilograms in one tone Remember that when we compare measurements we must make sure to use the same units This means that sometimes we must convert everything to a smaller unit before we do calculations The four main operation used in master division multiplication addition and subtraction


Rick Admiral Thomas Rowden, Director of Surface Warfare Division, recently spoke of the importance of new technology and surface warfare capabilities like the DDG-1000 platform. I think the capabilities of DDG when the 1000 bring to the fleet are going to be absolutely tremendous. We have a lot of new technologies that we're putting into the ship, obviously a very different hull form than we've seen in the past, but the fact that we have integrated power on the ship, a lot of the missions that we're going to ask it to do are of tailing into the missions that we've executed previously, but from the standpoint of being able to support forces ashore, certainly DDG-1000 is going to be tremendous. Later this month, the Zumwatt will begin testing to see how the ship reacts to how winds and stormy seas. From the Defense Media Activity, I'm Petty Officer Tyrell Morris.


Ukrainian forces defending Bakhmut are facing increasingly strong pressure from Russian forces, British military intelligence set on March 4, with intense fighting taking place in and around the eastern city. Explosions could be heard in Reuters footage showing Ukrainian soldiers digging trenches on Bakhmut's outskirts. Ukraine is reinforcing the area with elite units, while regular Russian army and forces of the Russian private military Wagner group have made further advances into Bakhmut's northern suburbs, the British Defense Ministry said in its daily intelligence bulletin on Twitter. Two key bridges in Bakhmut have been destroyed within the last 36 hours, it said, adding that Ukrainian-held resupply routes out of the city are increasingly limited. One of those bridges connected Bakhmut to the city's last main supply route from the Ukrainian-held town of Chasev Yar, about 13 kilometers to the west, it said. Ukraine's military command said Russia was still trying to surround Bakhmut but added that over the past day Ukrainian forces had beaten back Russian attacks in the city. The Ukrainian briefing note also said Russian attacks had been foiled in the villages of Ivanovsk and Bodnivka, both of which lie less than 8 km west of Bakhmut city center he captured of those villages, which flanked the crucial Bakhmut Chasev Yar road on either side, would leave the city on the cusp of total Russian encirclement.


Okay. Welcome everyone. I'm Chan-Sin Chinpasa and my colleague, Khom Kitt when we say. So we are from Nipah Khao, Thailand, Bangkok. So today we are going to talk about the performance tuning and resource management for Valiant, Dedicated, and Share Call favors. So, okay. Agenda for today. The first one is how we decide a favor for public car customers. Share call tuning guys, Dedicated call tuning guys, and Conclusion Future Works. In our company, we have a marketing team to talk our customer and get their requirement. My marketing team concludes our customer needs. The main factor that customer want to use a public car in in Thailand is high already or H.A. And the second is a cost and discount convenient sale service and pay as you go. So for the high already in the last year we have a session talk at the Berlin with Dr. Apisak is talk about how we decide the open stack in H.A. That in the picture we decide the concept like push the controller in the T side with the all active and we decide a zone for do the like a NOVA sale controller and we are using the tungsten fabbing instead of open with which you can you can go to watch is again in YouTube. You can sort of how we decide an open stack public car as a care with from big networking. So today we are talking about the second factor is a cost and discount.  After that, we need to decide the compute note hardware spec for each type. We talk to customer. We can grouping the requirement in the two parts. The first part is a non-critical application like low traffic web servers, CMS, blocks, depth test servers or just more database or someone used to keep a coding repository. This section I'm called a non-critical application. It can down, it can not work at high performance. And the second one is a mission critical or performance VM. For example, it's a high performance SQL in memory caching, indexing, memcache, ladies or JVM video streaming and video coding. So here is an example on our customer. They move from on premise to our cloud. They have Kubernetes cluster on the for the depth server and production server. So the customer requirement is how to optimize cost for the non-mission critical service and how to maximize the performance and stability for Kubernetes production. Because this customer will provide the POS and vending machine for it in the Thailand. So after we discuss to customer, we know how to save cost for them. We decide the two favor group. The first one is a check call favor and second one is a dedicated call favor. The first one check call favor, it can be the Kubernetes development cluster, locking server, Git server open search and monitoring. And the dedicated call favor is will be the Kubernetes production, database node, gateway or VPN side to side. So we are public car, right? We need to calculate the cost. So on the left-hand side is cost effective. It's focused to the low medium workload application, possible CPU, lower cost as possible. And the CPU and LAM will over commit. And so on the right-hand side is performance. We are focused to the medium high workload application, sustained CPU performance and minimized CPU and LAM latency. And it will be dedicated CPU and LAM. So we decide favor and dedicate node together for the check call. The spec of compute node that we use is 32-core 64-trade and 768 gigabytes of LAM. The CPU and LAM will show will be high. It's around 1 to 10. And the dedicated call favor is the CPU pinning. So there are no over commit, right? CPU and LAM will show will be lower. It's around 1, 2, 3, 4. The spec of a dedicated compute node that we are using is a 68-core 96-trade with two sockets. It's a total 192-trade. And memory is 500 gigabytes of LAM. So when you have a check call and dedicate call compute node in the same cluster, you need to create a host aggregate for grouping that compute node. So here is an example we are creating. The first line is a grouping of a dedicated node. And the second line is a grouping of a shared node. After that, we are using the feature of Noah's scheduler named aggregate in-stand extaspec. So we create a host aggregate, right? And we create a favor for a shared call and dedicate call. And how to map the favor to host aggregate, yeah. You need to add the favor metadata named aggregate in-stand extaspec to favor metadata and add into the host aggregate. When user launches the in-stand with match the value of the favor metadata, it will launch into the compute node correctly. For the resource type we offer, we provide to type of favor, one share call favor and dedicate call favor. We cause a lot our concept form hyperscaler cloud. If we look at example of hyperscaler, we can see that event with the same specification. The same specification. Performance testing score by gig band are not same. So we calculate the price per performance ratio. You will see which one is valuable. Yeah, because customer ask customer with provider, we want four call, eight gig, right? But they don't care about the performance that we get. So here is an example on the address. Yeah, we just test on the gig band of all the favor and calculate cost and performance ratio. Yeah. Our public cloud user can adjust the resource type to optimize cost or service to focus on performance with use residing. Yeah, here is a benefit for do the share and dedicate in the same cluster because a user can reside by their sales form share to dedicate and dedicate to share. Four share call over commit ratio is one to eight by default. That means if we have 16 call in our inventory, we can utilize up to 128 vCPU. When creating a share call favor, we specify the option. HWCPU policy equal share. When create favor or set favor. Actually default open stack favor that you create is a share call. Yes. And important aspect for share call is that we have to impose what is known as a CPU quota limit. If we don't do this, the problem that appear is unequal CPU uses high congestion, performance not stability. Giving the change for some VM to use too much CPU, which impact other VM since each VM share CPU uses. This example that we use in setting CPU quota limit when creating share call favor. This is an example. Next, let's see how creating favor that dedicate different form share. The difference is that we add the option CPU policy equal dedicate. It's different form share when creating favor or set parameter favor. For dedicate call when dumping the revert HTML, we will see is the mapping between each compute CPU and each VM CPU is one by one. For the share call, it do like a pool of CPU range. And the VM CPU can run on any CPU in that pool. But in the dedicate call, the VM CPU will map one to one to the hypervisor CPU. So it's quite guarantee and latency is good. Let's look at how configuration know the difference between share call and dedicate call. The important parameter relate to share call are the configuration of over commit and CPU share set. For dedicate call, the important parameter is CPU dedicate set. When we conflict as mentioned, the inventory collection of resource in paceman service we use with CPU for share call, PCPU for dedicate call. The number of CPU share set is CPU call ID. If you have 128 call, you can add up to 128 call. You can mix the share call and dedicate call in the same compute. But please do not overlap the CPU range. This slide talk about the huge page. Last slide, we talk about the CPU, right? And now we need to improve the memory also on the dedicate call favor. In the Linux system, there are page table. The page table is for store mapping, virtual memory updates to the physical memory updates. The virtual memory updates are used by the application VM instance. The default of Linux page is 4KB. So you mean more page, you have more time that you can find the page mapping into the physical memory, right? The concept of pure page is increase the page size from 4KB to 2 megabyte or 2 gig. Here is an example. If you use the default page and VM required to use the 2 gigabyte of RAM, you will have about 500,000 pages. It is a lot. But if you use the huge page, the page will be increased to 1,000 pages. So let me conclusion about the benefit of pure page. So page size will be increased, right? And number of page will be decreased. And the little page table walking is faster. So it's no swapping and no case swap operation. And overhead operation will be less. So two things you need to do. The first one, you need to enable the huge page on the cloud parameter like this. Just add parameter here. Default huge page, 1 gig, 1 gig, it means page size will be 1 gig. And huge page, 400, it means you will have page, 400 pages. And you need to update and reboot the server. And the second, you need to add the favorite metadata. HW main page size, large. It tells NOVA to use huge page on this instance. I will provide example to build a favor for dedicate call with huge page. We will need specify the option HW, CVU policy equal, dedicate, and HW main page size equal, size of huge page. This example, we use 1 gigabyte for huge page. Here is an example for resort provider for both dedicate call and share call on the same compute, on the same compute. Yeah. Observe that PCPU is eventually PCPU, eventually for dedicate CPU which does not have an allocation ratio or over commit, PCPU does not allocation ratio or over commit. And PCPU is eventually for share CPU. You will notice there is an allocation value which is over commit value. It's a PCPU and have allocation. A placement service will not multiply for us if we working on capacity planning. Okay. Let me conclusion. We decide the share call favor and dedicate call favor for customer. We separate the compute node for share call and dedicate call. So we use the host aggregate and favor to mapping by the aggregate in standard aspect. Yeah. So future work is, we plan to upgrade to the latest version of NOVA because we are interested, new feature is about the strategy to reduce the CPU power consumption when unused. Yeah. So we haven't tried yet but it can reduce the power consumption of dedicated node. Okay. That is it. Thank you. Thank you. So do we have any questions? So you mentioned in order to have dedicated physical CPUs for virtual CPUs, you ping virtual CPUs on certain physical CPUs. So my question is, do you need to live migrate or cold migrate these virtual machines? And if they do, how does this CPU pinning work for VM migrations? Because if your hypervisors have different CPU models, different numbers of cores, it may work on, you know, the pinning may work on one hypervisor but not another. Right. So I wonder if you'll encounter that and how that works in your case. Actually the CPU here, right? Actually NOVA compute will map this one. When we live migrate, NOVA compute will recalculate the CPU. So nobody can live migrate from another compute. So for your hypervisors, do they have the same number of CPUs? No. Same number of cores? No. So the pinning is just a range and NOVA will figure out the details, you mean? NOVA compute will find the available CPU of that compute and really mapping again. Okay, so NOVA will find another hypervisor with the same physical CPU setup and that's how VM migration works. Okay. Thank you. Any more questions? Thank you. One more here. Sorry, only one question. Why did you combine the shared and the dedicated compute together? Why not do one compute with all the dedicated CPUs and one compute with all the shared CPUs? I didn't share in my production. I just told about it can be possible. But in our production, we split it. Thank you. Okay, thank you. Thank you.