 Thank you. First of all, let me thank the organizers for putting this wonderful conference together and for giving me the opportunity to meet Mike again. We haven't seen each other in a number of years and On the personal note, let me say that this is the first conference in person that I attend since the pandemic starts And I have been very I would say liberating for me to go back to something that I've been missing a lot which is meeting in person with friends and And it has been a little bit kind of crawling back to normality. I don't feel we are fully back in normality but getting there and I'm going to talk to you about an overview of recent work I've been doing with a number of friends, but First of all, let me justify. Why am I here? Of course, we have heard people talking about different stages of my career We have the great lack of having John to tell us about his PhD years and other collaborators that were collaborating with Mike Junior collaborators slightly more senior than him But I can offer you a complementary perspective because I was a PhD student of Mike And that's something that actually a number of times people have asked me So this is a typical conversation that I could have on a coffee break during a conference Someone tells me so who was your advisor and I say Mike Douglas And then a sudden silence appears in the coffee break people gather around and ask me Really? Yes How was that like? because you know my career of course skyrocketed from the very beginning and He's very accomplished, but for some people he was a bit intimidating. So they wanted to know what it feels every day Working with Mike when you are starting your own career So as a public service, let me offer you please follow me on a Memory trip down memory lane To tell you what is a bit what is a day in the life of a student of Mike And I remember very well the first time I met Mike I was a grad student that I just arrived fresh from the plane to Rutgers The golden day for Rutgers And I was offered to go to Mike's office. So I entered Mike's office and he tells me what do you know? And I say, oh, I know quantum field theory because I know the scholar field theory I know how to write the QCD Lagrangian. So I was very proud of myself. So what else is there to know about quantum field theory? So I say, oh, then sure you are ready for string theory Something you should know is that in the string palace as it was known those days There was no regular course on of perturbative string theory The running theory amongst the students is If you need Someone to explain to you perturbative string theory, perhaps you shouldn't be here So there were some courses Irregular courses on on deep brains and non perturbative stuff, but perturbative string theory is something that you learn by yourself So Mike told me, well, you know quantum field theory You will figure out string perturbative string theory for yourself. So here there is some paper You should go and read And I thought, oh, perfect. This is going to be a pedagogical gentle introduction to some current topic And everything will be explained in nice detail. So Perfect a gentle introduction to non perturbative stuff So that was the paper that Mike asked me to read You can say many things about this paper is Of course, it's a masterpiece. It's one of the top three. I just checked it this morning One of the top three types of papers by Mike. So that's a very tall order, of course But gentle and pedagogical are not the first words that come to mind Deep brains, quivers, alien symptoms. I thought, boy, I thought this was going to be in English, but it was Having trouble just decoding the paper. So it was a steep, it was a steep learning course But this was, they won and Little by little we developed kind of a routine. So let me tell you about my routine of working with Mike I Would get there early in the morning and start preparing for meeting with Mike And of course, this is my planning board. Everything was planned. I was going to do some heavy computations and think about clever arguments And think that he didn't understand. I would ask some questions So I'm Finally about to I'm ready to meet with Mike and you know, Mike, we know is deep Is wide-ranging but also very fast. If you have discussed with him, you do something at least way, way, way faster than me So I was thinking, okay, this is going to be the meeting that makes the difference. Finally, I'm going to catch up with Mike You see, he's going to be impressed by my computations I have very clever arguments to support my the way I see things And the things I don't understand my questions will be well, will be pointed and Hopefully we will figure out together the the answers. So you see at 12, I know because Mike's office and well same old same old Immediately when the meeting starts, Mike tells me, oh sure computation is okay but sure you could have guess the answer just by this immediate argument or well Everything will be most of the time not always but very often I would see that Mike was still ahead of me And that was the rest of the day trying to put back my pieces together start to work again and hopefully well Happily some days were better than others and eventually Mike and I were able to well Collaborate it and wrote three papers together. The common topic the common thread was the brains on Calabello manifolds I'm particularly Oh, sorry, that was not intended to happen. I'm particularly proud of this one Which again, I remember telling Mike Mike, do you think anyone is going to understand this paper where we start talking about the right categories and so on say Oh, no, no one is going to understand it at first, but eventually things will Kill cut and I have to agree with Frank when we wrote this paper Uh, I thought it was hard to read but then Mike wrote by himself a very very insightful night Neat and clear lecture notes on the topic Which is not an easy topic, but uh Again, Mike lectures on any topic he thought about are among the clearest and most insightful lecture you can find Or if you also want a gentle pedagogical introduction, you can no never mind Forget about it. Oh, this is a requirement. Damn it. Okay uh, so This was a nice, uh trip back in the past but It's important to revisit the past but you shouldn't get stuck there. So let's go back to the percent times And uh, what I'm going to I want to present now is some other view of what's done with uh, these collaborators Uh, like Arlera, Fred Gerchkovitz, Sohar Komagorski, Aytolio Kovic, Jero, Jero, Martin Montoya, Alan Ryoth Fackleman, and Janis Dorens And as I said, Mike and I haven't seen each other in a number of years So I have news for you, Mike Some of them are great. Some of them before this workshop. I thought they were not so great, but uh Now I have rethought uh, the men perhaps are also great So the first some news is that uh, I given you four great children. Okay Those are three more than I gave my parents. So believe me, my mother is not happy about the comparison, but well, that's my cross to bear and Actually, I totally covered his half A son of mine because he started during his undergrad year to do research with me, but eventually He went to work famously with Juan Maldacena. So of course, that's your grandson that I I kept for a while, but uh He of course moved away But the not so great news is that well, you see I was hoping that they would have a career in theoretical physics and that well, Alan is taking a now a postdoc So that's great. But the other three have left physics. So I thought that was very sad So let me for those of you who are not String theories in the audience. Let me tell you about one of the big threats that We have in theoretical physics, which are black holes. And of course, uh, you should know that whenever we discuss black holes You have to be careful of the conventions. There are west coast conventions East coast conventions and in other convention, there is something called the even horizon We know that once at least classically in the classical picture once you cross an even horizon You disappear from the exterior So depends on the conventions, but In these conventions the even horizon is called wall street. You enter wall street as a physicist. You disappear forever On the west coast, this is your typical west coast physicist Wow Sorry I was thinking somewhat a little bit north perhaps around stanford Once you enter silicon valley, that's the name of the even horizon there. You disappear So I don't know what happened. I my three of my kids turned out to be a bunch of hippies. I was trying to Raise them properly, but they all went into machine learning computer vision things like that Never to uh, and I thought well, this is the end. They are not coming back but now of course, uh, I see that uh, this was way too classical way of thinking because as we learn from mike Uh, and we have heard many times, uh, recently about black holes That picture is way too classical if you find wormholes you can happily go from one place to another so The idea that there is an even horizon and once you get stuck. I don't know in physics finances or Computer science is way too classical. So we follow the elite of mike and think well You don't you don't know what the future is going to prepare for you Okay, so Find the wormholes keep this idea in mind because it will come back later Uh, so finally, let me go back to the talk. Uh, I wanted to present The motivation and before I get to the specifics is the following of course, uh, radiation in electromagnetic radiation is one of the key conceptual and practical Particle most important things in the whole of electromagnetism And the typical questions, uh You can you can explain are what is the total related power or a more detailed question of angular distribution of radiation Okay And also, uh radiation has played a very important role recently in gravity So the first question you could ask is what happens in more generic, uh, gauge theories or more generic quantum field theories Even if they don't have a gauge description And in this talk, I'm going to focus on a very particular family of sorry. Sorry. That was not supposed to happen No the other way I'm going to focus on Uh Here it is. Sorry about that of conformals Yeah, this is the one conformal field theories Both because the question is very can be very clearly formulated for conformal field theories And because the technical tools we have let let us address it more clearly And there have been new developments main chiefly holography and supersymmetric localization that allow us to explore the strong coupling regime of conformal field theories And in particular to address this question And when people way before I started to work on this, uh, started exploring radiation in conformal field theories Some of the results that the encounter were unexpected were not not very different from Well were unexpected in different ways so This is not going to be the topic of my talk these papers on holographic cfts, but uh By the way, this is a very very beautiful paper by liu radiogolpallon collaborators where they studied holographic cfts and radiation and the the The results that they found were the following first of all the radiative energy density Only depends on retarded time. Of course, that's a result that we know For uh, free cfts and masswalk, of course, we teach that and undergrad courses But if you look at the computation that is a string moving in five-dimensional ads 5 and then you have to see what is the The metric that it creates and how it propagates to the boundary and then you read off from that the The 10 the 10 the stress and urethane sorry in the boundary theory the fact that it depends on retarded time is not Something obvious at all. Okay Furthermore, they notice that the relative energy density is not positive definite There are regions that are where the density energy density is negative and they said of of course Well, that is we are in the deep quantum regime very strong coupling So There is nothing that forbids in the quantum theory at very strong coupling that the relative energy density is not positive definite Furthermore, the relative power is not Lorentz invariance. That's not a disaster But again, that's different from what happens in a Maxwell theory and also it depends on the derivative of the acceleration That's something that we don't encounter either in Maxwell theory. Okay So some of the authors for experts Particularly in the second paper. So these are our results, but we don't believe them This probably is an artifact of super gravity. We are doing something wrong. We don't know what but uh, we shouldn't trust these results okay so In particular, let me stress the first point that the second authors were very critical of that was their expectation expectation is once you are in a deep quantum regime Of course, you could get something coming from the retarded time would be here But also there will be off-shell gluons to use their language that uh, they are time-like and also affect the point of observation But their computation is gives an answer that depends on the retarded time exactly like in a free theory would happen Okay so uh For those of you with a more formal bending actually I want to understand CFTs, uh, that might not be a non-lagrangian for instance I might Encounter the argyrus Douglas point that is a non-lagrangian field theory and you on those CFTs Lagrangian or not you can have a defect and in the question I'm asking in the present of a time-like defect Time-like line The stress and retention can develop an expectation value. How much is it fixed just by symmetry considerations? Okay So this is the plan of the talk Actually, the question can be very nearly separated into questions one is angular distribution But is the angular distribution is qualitatively and quantitatively similar to say Maxwell the one we are more familiar with or not And then the the question this question factorizes into a second one What is the coupling dependence as I crank up the coupling? Can I explore the medium and very strong coupling regime of the theory supposing that the theory has some couplings? Okay So let me start discussing the angular distribution of radiation And well the classical procedure I'm going to start with free free CFTs And we know the procedure from Maxwell. So we just Give an arbitrary trajectory everything I'm going to say is for completely arbitrary trajectories You solve the equation of motion evaluate the stress energy tensor for those fields and keep the relative part the one that decrease like 1 over r square And of course from that you can define Following the standard textbook notation the angular distribution and the total related power And you do that for Maxwell just a reminder. This is the stress energy tensor This is the angular distribution, which is positive definite and the total related power We know it's Lorentz invariant. So so far so good And we can repeat that for a free scalar And when we repeat that for a free scalar This is the way they did it in in those beautiful papers I was telling you that they did the holographic computation and they also did the field theory computation to compare And again, you get a purely positive definite angular distribution at the classical level and a Lorentz invariant Rated power you see that there is a factor of two that they say well, it's the polarization of the photon. So everything is fine right well not so remember we want to study conformal field theories And court and friends told us long ago that well Perhaps if you want not not perhaps if you want to study a conformally Unconformable scalar that was not the right Lagrangian in particular. You see that The strain tensor was not traceless So you have to add an improvement term to the stress energy tensor And that improvement term involves second derivatives of the fields and the second derivatives are going to play a very important role They change qualitatively the results So this is a very computation to do now once you realize that you should have Introduced this improvement term in the stress energy tensor something that I think it is Clear but it was mentioned is that the equations of motion don't depend on this Parameter that tells you that you are using conformally scalars But the stress energy tensor does and once you evaluated You immediately see that the angular distribution now at the classical level is not positive definite This is not a quantum effect. This is pure classical Okay, no surprise here Beckenstein already told us in the 17s and I don't know if in in Collins paper was already mentioned that Not minimally coupled fields violate energy conditions even at the classical level in minkowski space. So I'm not saying anything new. I'm just Putting a giving you an example in a in a context that I hadn't seen before but this this is something new And the rate of emission Has now this term that depends on the Derivative of the acceleration. Why the derivative of acceleration? Well, remember the improvement term has second derivative So you are going to derive once more and that gives you the derivative of the acceleration And the related power is non-laurician variant precisely because of this term Okay So to recap We have found that at the classical level no holography here pure classical computation The relative energy density is not positive definite The total rated power is not laurician variant and the derivative of acceleration appears in the formula Sure, no problem. Yeah, look at the definition of the rated power. I have it Sorry, I'll come back in a second Here look how we generally define the rate power we define this four vector the rate of emission And when we take the zero component And divide by coordinate time not proper time This is not a manifestly laurence invariant definition So why is true that in Maxwell theory? It happens to be laurence invariant result In Maxwell theory this four vector is proportional to the four velocity So the zero component is a gamma Laurence gamma and when you usually switch from proper time to coordinate time to lap time There is another gamma they cancel out and so happens that in Maxwell theory because it's proportional to the four velocity You get a laurence invariant result However Here we have an extract term that is proportional to the four accelerated the derivative of the four acceleration So this is laurence covariant And it had to be if you will buy construction But no one told you that uh, no one says and it's not true that in general for generic theories This has to be proportional to the four velocity. It's not And therefore this term is going to violate uh, when you construct this construction No one guaranteed that it was going to be laurence invariant and it's not in this case so, uh Maxwell in a sense was an exception and it was perhaps giving a giving you A wrong expectation that every time this rate of emission is going to be proportional to the four velocity It so happens that it's not That this that is answer your question Okay, sure Uh, so I told you that the qualitative agreement with the Holographic computation was good, but let's compare actually the details and this here, uh, I'm not expecting you to Check that all the details, but this is the a result of a long painful masterful very difficult holograph computation again It's a work of art, uh to derive this result. I'm really impressive This is a large a lambda and large n, okay This is the result of a one afternoon less Good afternoon of a free theory computation You see that the agreement on the special dependence is exact For arbitrary trajectory, this is not a bps quantity. This is not arbitrary trajectory. It's not a straight line or sharp or uh hyperbola So this is also striking So we have found, uh, I would say two surprises so far First is that when you compute this, uh, expectation value of the stress intensity tensor depends on retarded time holographically Of course, and the classical level is no surprise and that the algorithm angular distribution is exactly the same in the holographic regime and in the free theory regime One can argue and I'm going to skip the details, but one can argue that, uh, if you assume the first property The follow on the second one follows from conformal symmetry A supersymmetry that cannot play an important role because I said these are not bps quantities Okay, so let me sketch the idea. The idea is that by dimensional analysis If you say that everything depends only on retarded time, these are the same type of terms that you could expect And, uh, let me consider two particular kinematical configurations I want to describe the genetic situation, but I'm going to claim that with two particular kinematical configurations I will need all the ingredients. I will find all the ingredients I need First, uh, let me consider just a static probe And Kapustin argued that in this case there is just, uh, the space time dependent part is fixed And then you need only to compute a coefficient of the possible marginal couplings of the theory that I'm calling G here Then, of course, by Lorentz and Barian you know the expectation value at arbitrary constant velocity And now we use, uh, heavily the properties of a conformal field theory. In a conformal field theory We can relate the relative field with the columbic terms because we have special conformal transformations So a static probe or a probe moving constant velocity Is related by special conformal transformation to a Rindler trajectory And that gives me some information about certain kinematical situation where we have radiation So that's a piece of information that we are going to use And the second kinematical configuration I want to study goes back if you look at a Mike's tree that we saw yesterday to JJ Thonson, which considered the following situation You consider a static probe, then you kick it And the particle changes velocity and in this acceleration that happens here, there is also radiation That, uh, Kasb Wilson lined that, uh, polygons started in the 80s And more recently, Maldesena collaborators from that extracted a second function of the marginal couplings That is also related to radiation in this particular case So we were able to argue that if you assume that's something that we have to, we assume to, uh, To move forward, that the stationary tensor depends only on retarder time The full angular distribution is fixed in terms of these two functions One coming from the Rindler trajectory, one coming from the, uh, Bernstrahlung From kicking a particle at rest, everything else is given by these two functions Okay So, uh, to recap this first part, we have argued that these terms, uh, are universal But the acceleration, the relative part, the one that decayed like 1 over R squared Are theory dependent and depends on these two functions that we still have to determine So I think this is a good spot to, uh, pause for a second If there are any questions about this first part of the angular distribution If not, uh, I can move on to the discussion on the coupling dependence Okay So now we are going to switch gears to coupling dependence and as you will see, uh, although We started with a very question, uh, separated from things that Mike had been doing Eventually, all wrote to Mike and we will encounter things that Mike did in the past Okay So the coupling dependence, I told you that in the previous part of the talk That we need to determine two functions, one that comes from the Bernstrahlung And that, uh, the another one from a Rindler observer And the first observation is that for half-VPS probes, these two functions are actually related Okay This is a statement that is true even for null Lagrangian CFT This is a statement that is true for half-VPS probes of n equals to super conformal through theories It doesn't matter if the theory has a Lagrangian description or not Okay So an intuitive way to think about it is that, uh, for arbitrary probe, arbitrary CFT You have scalars and you have gluons and you are reading both scalars and gluons And these two things are independent, okay, so that's why you have two functions This, uh, a knife way to understand this counting However, for half-VPS probes, they are sitting in the same multiplex of supersymmetry Forces you that there is a relation between scalars and gluons They have no longer independent, so these two functions have to be one and the same So we have reduced the problem to computing just one function on the marginal couplings And then the full stresses in the tenson depends on a single function And this is something that we observe at least for n equals to 4 When we compare the free theory computation to the holographic result Okay, that would explain why we encountered that the angular distribution was exactly the same And everything on dependent coupling sits in front of the full expression So, uh, we have to still compute one function And let me tell you what's the procedure that we are going to follow The strategy is to reduce this, uh, function that is of a four-dimensional CFT To a matrix model computation Okay, so we are going to go from four to zero dimensions And then once we have done that, we still have to solve the matrix model Some of the matrix model, we will rewrite it in a way that is perhaps not so well known But actually this rewriting of the matrix model will allow it to solve it in the planar limit Okay, with full effort you can solve it beyond the planar limit But the answer is not enlightening at all for us Okay, so this is the strategy and this is the way we are going to proceed So, let me start discussing, uh, how to The different steps that go into the computation The first one is that, uh, we are concerned about n equals two theories And you can consistently put them on a sphere and also on the former sphere This is the word of Haman Hosomichi You can, there is a way to, uh, place these theories on a squash, uh, a force And you can also define how VPS will some loops on those spheres And the first ingredient that goes into computing this coefficient Is that we were able to argue with Efrat and Sohar That, uh, because you have the theory on a sphere If you vary a little bit the background You go from a, you squash a little bit the force sphere This squashing, this changing the background metric Is amount to introducing the stress in the tensor That's, so that's why you are computing the one point function of the stress in the tensor Okay This is the precise formula we argued We gave some perturbative evidence that was not, Didn't amount to a proof but luckily these guys, uh, Some time, some years later were able to completely prove And rigorously prove this formula again Without assuming that the theory has a Lagrangian formulation So this is an exact result for any n equals two Super conformable theory, Lagrangian or not So fine, we have reduced the problem of computing this coefficient To compute it, uh, the expectation value of a particular Wilson loop on a sphere That does might not seem like a lot of progress But now we invoke a very impressive result by Peston Before he moved also, he jumped into his own black hole of machine learning And, uh, we can use the work of supersymmetric localization Because in this work Peston was able to show that the half VPS circular Wilson loop Is given by the following intimidating matrix model Is intimidating but is now is a zero-dimensional problem And why we care about circular Wilson loops? Well, because the circular Wilson loop is the analytic continuation Of the ring load trajectory, which is a hyperbola Okay So, because this is the analytic continuation of hyperbola It immediately relates to that coefficient I was telling you Of the accelerations in by, uh, an observating probe In proper constant acceleration And because we only need one coefficient to compute, we compute that one We are done Why do you call it a matrix integral? Sorry? Why do you call it a matrix integral? Because there is a matrix integral I'm not, I'm not Where is the matrix? Oh well, sorry, yeah, I mean These are eigenvalues of a, uh, uh, matrix of, uh, some Lie algebra Because there is an edge group, uh, that I'm going to keep G And these are the eigenvalues, the A's So, for UN you would have, this is a Hermitian matrix model It's not, because your measure is completely different from the measure For the integral of the eigenvalues of the matrix Z1 loop, Z instanton, it's Oh well, sure, sure It's a matrix integral with a lot of insertions, yeah Anything is a matrix integral with a lot of insertions I will, I will, uh, repackage it in a way that might look more familiar But this is, uh, you would agree that this is the, the expression or in the, in the piston paper But it's not a matrix integral Okay Fine So the usual approach to deal with this integral is to reduce it, oh sorry To reduce it to the, uh, cartons of algebra That's the story that most of you are, are much more familiar than I am with And this generates a Van der Mond determinant And you use, uh, you write that as an effective potential And you can solve it with different techniques It's always the integral of a cartons of algebra Huh? It's always the integral of Yeah, yes, yes, I'm saying, sorry The way I'm presenting here is the usual approach here The result of, I'm going to take a step back Which I'm not going to give you the details here But you can rewrite it in a different way than this one I agree So something I learned from Mike is the pleasure of taking, uh, to following the road was taken And I'm not going to try to solve the matrix, uh, the integral as it's presented here You can rewrite it in a way that, uh, instead of going to cartons of algebra You integrate over the fully algebra And the advantage of that is that you don't generate the Van der Mond determinant Which is something that was a hindrance to solve it fully, uh, for this type of, of integrals And, uh, as a warm up Let me consider the case of n equals to 4 where this, uh, one loop at instanton terms are absent So the integral is Gaussian And you can do that on a case-by-case basis What I mean by that is you have to choose the gauge group You could say u n or s u n or s o Whatever gauge group you want to consider And you have to choose the representation of the probe The probe, uh, has a character representation Typically you could choose the fundamental but You could choose a probe in any other representation And if you do that, the Wilson loop, uh, was already computed exactly by Drucker and Gross Uh, following word by Ericsson Seminoff and Sarenbo This is the answer, this is a Laguerre polynomial times an exponential And from that, you can immediately get this Berenstrahlum function Which, in this case, for a fundamental probe in the u n group Is a ratio of Laguerre polynomials Okay, so here is, for instance, an exact, uh, answer of what is this Berenstrahlum function It's a, in this particular example, it's a portion of Laguerre polynomials And you can do that in other representations for other gauge groups But you have to go on a case-by-case basis But, uh, another approach, the other approach I was telling you Instead, uh, allows you to consider in a single computation all An arbitrary gauge group, an arbitrary representation And gives you the exact answer at finite g and finite n So I'm not taking the planar limit in n equals to 4 We can solve it at finite g and finite n The Wilson loop is just given by this infinite sum Where the coefficients are fully symmetrized, uh, traces Of the generators of the gauge group, or the algebra And once you have the Wilson loop, which is defined as expression You can take the log and you are done So the question of what is the coupling dependence of radiation In n equals to 4 for arbitrary gauge group And arbitrary representation of the probe Is solved essentially, uh, at any finite g and finite n Just taking the log of this expression Okay, and in this paper we discuss a little bit Some simplifications, because there is the non-novelian Sponization theorem Tell you that the log is going to be simpler than the Wilson loop But those are details that you can find in the paper So this is the whole story for n equals to 4 supergen mills What happens when now you move to the most, the more difficult And the richer case of n equals to 2 theories Now, both the one loop and the instanton contributions in the integral are non-trivial I'm going to set the instanton contributions to 1 That's something, uh, we don't have anything new to say So I'm going to focus on the zero instanton solution You should keep in mind that the finite g, finite n result Has instantonic contributions No question about that But we are going to work in the zero instanton sector And, uh, we were able to argue, following word by video and collaborators That in that case, this one loop action, uh, you can rewrite it As an effective action With an infinite number of single and double traces Okay So, uh, again, I emphasize that only single and double traces appear No, no higher term, no triple or four or Products of four traces But the number of them is infinite Okay And in the large, uh, this is a result, a set result, uh, finite n But in the large n, uh, this simplifies And we are just left with an infinite number of double traces in this effective action So, uh, going back to the 90s, uh, these matrix models with double trace interactions Were studied in the context of 2D quantum gravity And something that Mike, of course, knows very well And although I think he never, I don't know if you wrote any paper with double traces But, uh, uh, you certainly made very important contributions to model with single traces Can you explain, I mean, where does this come from, uh, the previous one? Yeah So, what is this effective action? So, you can write, uh, this, when you go back to this, uh, here This set one loop, you can write as the, uh, sorry, I keep messing up This set one loop, yes You can write as the exponential of an effective action How do you, can you do that? Well, uh, this set one loop comes from, uh, there are roots of the different, uh, Ways of the different representation, the adjoining, the matter is a quotient So, you have products of the squares of eigenvalues, uh, some products of eigenvalues And you can say, oh, if I have the sum of all the squares of the eigenvalues I can write that as a trace of the matrix And because the matter that you are, uh, concerned about If it's a conformal or asymptotically free theory, has at most two indices That's why you get at most two, uh, products of two traces Okay So, you have, uh, you think of the eigenvalues as coming from, uh, This sums and products of eigenvalues as coming from having taken the trace of the full matrix Okay Well, if I can comment, so the result of Peston of 2007 Was that the sphere partition function with Wilson loop Is the integral of the absolute value square of my partition functions who are float So first sphere has two poles, no, the south pole And my partition function has a perturbative part and an instanton part The perturbative part is the double, it's a product of Bern's gamma functions That's the contribution of W bosons They depend on differences of eigenvalues of this scalar and vector multiple So you have a function which is the, you can expect, it's low, you can expand this But let me show, uh, things like A i minus A j to some even power So that you can expand this product Perfectly And, uh, these double traces in the 90s were thought as, uh, wormholes I told you that we would hear about wormholes again Connecting different blobs of, uh, you will, you will geometry built by single trace terms And these models of course, uh, can be solved with ordinary techniques Here you will find it, uh, how to solve these models But, uh, what we found is a different way to tackle them And we found a purely combinatorial, non-recursive solution Which in some cases is much simpler and much more compact So let me try to give you an idea on how that works Imagine that you have a term that goes like a cube, uh, of, uh, a single, a double traces So the large hand counting tells you a number of things First tells you that if you have m pairs of double traces You have, uh, the term that contributes has the right large hand counting Is given by a product of m plus 1 connected correlators This is large hand counting You can, uh, go and replace it, uh, easily And furthermore, because remember we are, we are one to compute, um, uh, Things like the free energy we are going to take a look If they are going to survive the look And there is a further restriction is that no correlator that appears in this product Should contain traces coming from the same pair So now this is a purely combinatorial question And here is the example Imagine that you want to distribute these three pairs of double traces into four correlators Such that no, uh, no pair sits in the same correlator You start playing and you convince easily That, uh, these are the only types of configurations Which are given in this particular case by these three graphs By a graph I mean that every node is a connected correlator And every edge connects, uh, pairs of double traces that are sitting By construction in different nodes Okay, so actually this is the incarnation of the wormholes That people were talking in the 90s, uh, because they are coming These edges appear every time you have, uh, a pair of things of double traces Sitting at different nodes Okay, and you can go and convince yourself that this is the general picture So the plan, uh, our results We encounter it in the context of n equals to four, n equals to two theories But it's true for any matrix model, Hermitian matrix model Such that the potential is given by several double traces In the planar limit, uh, because I'm, I use the large incontent to do this In the planar limit, it's a combinatorial sum Where, uh, all the terms are given by three graphs Okay So, for instance, uh, here is the planar free energy of n equals to four, n equals to two, uh, large chain QCD, okay This is the theory that has the number of, of, of Matter multiplets, twice the number of colors And you see, well, I'm not going to get into detail But it's purely some, it's just summing over all possible trees And an important factor, because we want to get explicit result Is you want to know the explicit form of the planar correlators in the large chain limit And that's a problem that was solved by, uh, Mathematician Thieu in the 60s And in modern, in the language that we are more familiar with By Ngopal, Kumar and Pius Okay With the restriction that you need all the double, all the traces to be even We don't, there is no known formula for odd traces Okay, uh, so that was for the free energy For the, um, Wilson loop, which is what we are after, uh, the, the story is very similar But now the position, the insertion of the Wilson loop, uh, distinguishes one of the notes in the tree So that's what mathematicians call rooted trees So again, we have a full expression for the Wilson loop And the coefficient that appears in the radiation But now the sum is over rooted trees So you have to draw again all the possible trees And find the different ways to, uh, select one of the notes And here is the expression again For the, uh, Wilson loop And you just take the log of this expression I'm not claiming this is nice, but it's not going to get any nicer We try it and there are no further simplifications So in a sense, this gives you the coupling dependence of radiation In the planar limit for arbitrary, uh, gauge theories Okay, and equals to gauge theories on a probe that is half EPS So to recap, we have managed to compute this coupling dependence of radiation for arbitrary In the planar limit, 12 orders in the hoof coupling For any Lagrangian theory And equals to super conformal theory with a single gauge group As a welcome bonus, something that we were not after at the beginning But, uh, it was easy enough, but that we also did it We were able to compute also the planar free energy for these theories Something that I haven't discussed today, but, uh, most of you know You know that in general quantum field theories, the perturbation theory is asymptotic But in the Lagrangian limit, there is a drastic reduction of the diagram you have to consider And then for finite theories, there are generic arguments That the radius of convergence is finite in the planar limit It's hard to combine two with the specific samples We don't have a mathematically rigorous proof But I think we have overwhelming or very, very, uh, strong evidence That for any observable, any observable that is captured by a super symmetric localization In equals to theories, the radius of convergence is pi square In the, this is the hoof coupling So, we are able to argue both numerically and with some analytic approximations That the radius of any observable that you capture by localization is pi square We also extended this to cool work theories And finally, although the techniques we developed with n equals to super conformal theories inside I was very attentive once you encountered a hammer to look for nails And there are other potentials that we can give very compact results for the planar free energy For instance, this is something that Casacov on Mike and collaborators Studied a potential we have to put the restrict of only single traces But in our case, we have restricted to even traces just because we don't know the formula Of the planar correlator for odd traces And this is the very compact expression of the planar free energy that we have encountered in the literature I'm sure that experts know how to derive this in other ways, but we never saw it published And for instance, if you go back to these models with double trace interactions And you just consider a single term We, uh, people of course knew how to compute these two arbitrary orders The planar free energy is recursively, but you didn't see any pattern But with our methods, we immediately see, it's a three-four line computation That the full planar free energy of this model is given by partial Bell polynomials And we cannot also add single traces The expression is a bit more, it's uglier But we can also solve it to all orders in perturbation theory in the planar limit So I think I wanted, this is everything I wanted to tell you There is one more thing Is happy burden, Mike, whatever you do keeps floating in the landscape So any question? So the effect is coupling, how does it grow? Can you, did you show us? Ah, uh Lambda is, this is a conformal theory, it's a parameter Well I thought lambda was the top Yes, and because it's a conformal theory, it's a parameter, I can vary it Yeah, yeah, but how does the coupling grow? I mean, how much more do I radiate a strong coupling than a twin coupling? Oh, yeah, I didn't, the plot is in the paper, it's not very eventful, how to say Really, it's a bit more, but there is no nice striking feature that I can tell you But it grows It grows, but again, there is nothing that, I wouldn't say featureless, but close to it There is nothing striking to say But there's a way of formula Done in the holographic of gravity picture, right, where you have an explicit formula Oh, yeah, yeah, it goes like a square root of lambda in the fundamental In the fundamental, I can be more precise because it goes like a square root of lambda But that, you didn't need the computation because that square root of lambda is ubiquitous Because it comes from the Nambu Goto action Yeah So if you want to, you can explore The, how do you go from small lambda to large lambda in the fundamental in other representations Yes, thanks sir, but yeah, thank you But can you reproduce, I don't know, the coefficients in front? Sure, sure, sure, sure Yeah, yeah, I don't know, that we checked, all the numbers come precisely, yeah Not only the square root of lambda, but the numerical coefficient we also derived, yeah Good What happens at lambda equals pi squared? Do you have a cut? I don't know, I don't know If it happens to be already divergent or if it will diverge, I don't know that As I said, we don't have mathematical, I don't know the mathematical tools to prove that Frequently or to approach it, simply that We have very good numerical evidence that that's the, for different observables I should mention, there are other observables that I haven't discussed Like, extremal correlators and other things that you can use these techniques And again and again, you see that it's pi squared But I would need, I mean, we would need a mathematician to actually prove it rigorously But rather convinced that it would take a lot of effort for me to be able to prove that So when you say you're rather convinced, why? I mean, it's true, true, right? Aren't you just computing like in the expansion of the coefficients and you have these formulas You can compute whatever coefficient you like and then you can compute the radius of convergence No, no, it's harder than that because you have to sum over trees And of course there are, if you ask, if I ask, you give me all the trees with 400 nodes You have to generate them, okay So there are routine, there are very efficient ways to do it But at some point, I mean, we use Mathematica, we generate them, but eventually it runs out So the way we did it, I can give you For instance, if you, ah, sorry, you consider a model With arbitrary k, you could see, for this type of model, that with the right of With the behavior that these coefficients have in the particular case often equals to theories The way they grow, these tend to a universal limit for large k And because you are going to have infinite terms And that's the pi, in our language, the pi square Then some other groups, beyond collaborators, studied with Paddy resummonation and so on, and also they converge with very different methods to pi square In the particular, in some particular cases where you can do simple questions, you can use this You can derive pi square So these are all complementary evidence that this pi square, but The sum over all possible trees at a given order is hard to tackle analytically Because imagine, what are the, you can count the number of trees And there are asymptotic formulae to tell you how many trees are there So that's the kind of games we play Assuming that all the trees give you more or less the same But that's out of desperation, we don't know how to do better than that Okay, thank you These lovely formulas were all for the case of basically uniform acceleration No, no, no, no, what about the kick case where we're talking? Yeah, what I emphasize is that the angular distribution is determined For arbitrary trajectories determined by a single function in the case of an equals to theories So for arbitrary CFTs without supersymmetry, there are two functions And then the angular distribution of an arbitrary CFT is determined is, let me see, is here In terms of these two functions, the kick and the ringler one, if you will So I'm claiming if you know, if you compute the kick and you compute ringler This is for arbitrary trajectory So you look at these two particular kinematic cases and then you know everything In the particular case of supersymmetry, things are even simpler If you know one of them because there is How do infrared divergences fit into this? I mean, I was just thinking back to the old days, right? I mean, so radiation of a particle that scatters, right? If you do the calculation naively, you get an infrared divergent result And then Block and Nordseek show you how if you ask the right kind of question Then we watch the total energy radiated into some, you know, net energy summed overall possible Plays of getting it by emission of photons You get a finite result, but it depends on the measuring apparatus And, you know, this is a question completely outside the framework of your talk But I was just wondering No, I haven't thought about it I would imagine, but I don't know Of course, I'm heavily using conformal symmetry here Which say in QD you don't have it at one look But yeah, my whole expectation is that Well, this is already doing the taking care of the infrared divergences But I haven't done the exercise of trying to go from Or what you're saying for a purely conformal theory But if you only compute the energy momentum in terms of Do you get this in theory? Probably not I guess not in some sense Yeah, of course I haven't done that yet Could it be that, you know, the soft photons which you normally have to take in detail Here, they don't, they don't try again Because a symmetry creates a supersymmetric trajectory Which you can see, right? So you want a supersymmetry only at the moment of the kick Well, but this result is for an arbitrary trajectory You don't care about how many photons you admit I mean, you only compute the inclusive Right, but no, that's in a conventional way Which is, you know, that's the essence of solving the infrared divergence problem idea So, and the source has to be a half BTS state For this formula, no For this, this is not, there is no supersymmetry in this formula This is for an arbitrary CFT How many examples I know of Where I can write a Lagrangian of an interacting CFT that is not supersymmetric? No, but this formula we derive But there is an important assumption here That is, we are assuming that it depends on the retarded time Which could be a too strong an assumption And the only evidence we have That this is, that we think, we gave us to think about it Is the holographic computation The holographic computation shows by explicit computation That in the holographic regime, it only depends on retarded time It could be an artifact of the holographic regime It could be, I mean, it would be perverse But it could be true that this is the result in the holographic regime This is the result in the free theory limit They look very similar But at finite and finite coupling, things get, I don't know It depends on the integral of the whole war line up to retarded time Because this affects that I was showing in this picture at the very beginning So it could be that these things switch off in the free theory limit And the holographic regime, but are present at finite and finite coupling At this point, I cannot argue against that possibility Okay, but I think it's remarkable that in the holographic regime at least It depends only on retarded time So those are questions Do you think that there could be a kind of a gross written idea Based transition at lambda equals pi squared So the value of this scale is spreading over I haven't thought about that It could very well be, but I don't know I don't have the tools, certainly, to study the thing So this one loop contribution for the round sphere Is kind of more singular than it is for the general absurd And so since it involves gamma functions So there are poles or zeros And when you take a logarithm, they lead to cuts So your eigenvalues get spread over Because if your coupling is too strong, the suppression is small And so they can reach up to the point when the gamma function will have a singularity And that's the point when, I don't know, w-buzzles become massless Okay, and you have an understanding in your way of thinking You take out instant dots Okay Okay No more questions or questions for Langeberg It's okay, so thank you