YouTube what's up another game play this is actually mutt head-to-head Sometimes we get so ranked so high in salary cap it takes minutes and 30 minutes to go ahead and get get a game So when I play a little mutt head-to-head if you like any offense and defense that you see here The link is below for man turf to make sure you cop that offense the New Orleans Saints Also that defense is going to be the Detroit Lions playbook And this is mutt head-to-head a lot of my players are a little just base elites because of salary cap It's hard to keep you know the power-up players and also play salary cap this year. So it's a little bit different You see first play. I don't know who this guy is super shredded So, you know super sure it was a good back on his turtle. So he might be tough first play obviously coming out in the split Close I'm in dollar So we're gonna have to fight I had base elites on my d-line. Nothing crazy I think it's a Fletcher Cox clowny, and I'm Justin Houston Here we go, this is a play that he's he went to most all game It's the first time he hit me with it It was just had that wheel but I was playing hard flats And he was able to have enough time to hit the wheel route and the deep blue one with the post So I keep that in the back of my mind for the rest of the game That's one play he kept looking at to get and here he goes He fires a pick Sean Taylor. We try to go to deep end route in the back of the end zone with Sean Taylor So right now I'm thinking all right This guy is nothing special nothing crazy. Nothing I haven't seen before So here we go first play I'm gonna go ahead and go up top Randy boss He got by I'm going to spin moon nice little finish Sean Taylor You know he's gonna rip the ball out and that becomes a theme of the game and Ray Lewis picks it up And so pretty much him firing a pick you got it back from a fumble the strips are a little wild this year right now Especially if the computer is doing it for you Here goes running around with Matt Carlson once he has which is you know the best quarterback in real life Wouldn't say he's the best quarterback a mutt because of Vic came out But Carson once it's still very good Here we go. He ran a little bench there to get sacked. We hold him to three after the fumble. So that's okay I'm gonna get the ball back. We get the ball a half being down three nothing There's no big deal right now while I'm playing this game I really don't think much of this guy already so I feel like and the longer the game is the more I will surpass him. Here we go Vic looking a wide open Vernon Davis and he doesn't want to go catch the ball That's okay because we got second down second down here. That's a pretty good job Michael vick and looks like Brandon Graham actually just you know shoestring tackles and the quarterback tends to fumble on the shoestring tackles this year early in the year I hope it's something they fixed because nobody should have caught Vic let alone make a fumble right there So that's tough. So that's two about three players in two fumbles and the second play after the fumble He takes it into the crib with Leonard for net on a stretch got beautiful blocking Cam Chancellor actually got cut That's why I have it on strong safety. He got cut blocked and for that was able to get to the touchdown So two fumbles. He's got ten points. I'm still alright. It's a long game. It's the first quarter Just gotta get a couple possessions back. That's all I'm gonna go Michael big again got to get down He saw Brandy Graham was going for it again that they come from behind strip sack It was a little baby crowed back on my one. No problems over to fall right on the ground. There's nothing I can do there Dump it off here to mr. I missed a shady McCoy and he's gonna get cranked. I think that's by Ray Lewis The hips are pretty sure you're going to be pressed you want to press and we just got the diamond fully upgraded Old Del Beckham junior out wide right there able to burn him for touchdown So we get one of them touchdowns back. It's 10 to 7 Filling our eyes running ball a little bit right there in that there he is that's Daniel Hunter actually with a hit stick number 99 Here he goes dumping it on the knees Chantel coming up and bring or Jim Graham like fell right there in this play He just do right where I was running around and do it to me and it turned into a touchdown I'm like literally I was running the route with his wide receiver, but he decided to throw it Sometimes when you're lurking that player, you know, you don't expect somebody to throw it right at you But you're always looking at the other players to see where they want to go the ball and be ready to click on another player But he threw right at my face right there and then put his foot in the ground and went and got a touchdown So that's rough right now give up another easy score So back down 10 where I started last year. So let's go ahead and get some yards here He got crazy start man and everybody else. So we might have topped the moss and moss turned around and Got a little fluke it looked like a little mini rocket catch there able to score a touchdown in Randy Moss always near Randy Moss 18 I don't care if it's the 83 to 85 to 87 to 92 needs some version of Randy Moss But the height and speed is just unmasked by anybody else Here he goes firing the flats. We need to pick there But chanting it. Oh that jelly rings. You didn't want to go get it next play here He goes again to that wheel route to the running back But I puts that time is able to hit it Runs here on 31 gets the first down I want to get the ball back before the half here running again. Good job by cam chancellor there Bam Bam go ahead and cover in the flash there. He goes again rolls out looking for that wheel route But too much pressure on Carson once there. He was not able to hit it He goes runs the screen champ Bailey unblock tackle. I use my time. I'll let's get the ball back boys Let's go make a play downfield That's what it's about 47 seconds. We've got three timeouts Let's see what Tyree go can do and he gets tackled by I think there's juice juice check Stay able to make the play there.  That's a terrible play by me getting sad. They have to use a timeout 27 seconds left I draw up another play Get out of the pocket get some yards with Victor. We go 22 seconds to timeouts. Just a couple more dots I'll be able to get into the red zone. He's running coming for air and this is where I force a pass Then it had a spacing right a lot of that cloud flat They go pick off that post route if I was on a different hash a different part of the field It would be an easier pass But that was forcing something I could have got do a couple flat passes and got my feelable But the second half pick off Tyree kill We just find a little sliver and that's all Tyree killed needs to go to the crib It's just a small little sliver in between the defense great blocking by my guys nice little use of my turbo I didn't hit the turbo there right until I got into the hole So probably 40 yards down the field before I use the turbo button That's what made it so so easy to move around with Tyree kill It's hard to move when you're using the turbo if I can give you one point of advice Don't use turbo as much. You'll be a better user man. Here he goes again looking for that wheel route looking for that wheel route Champbelly swats it up. I feel like he could have jumped to pick that off But you know that's asking a lot for champ but champ costs a lot He goes down underneath to the curl cam chance. We're getting no time for that That's campaign champ and cam chance to where he's punting the ball 4-point lead but now fake punt You know you're playing a bazooka when they fake punch you Chris Harris He don't has time for the fake pun though picked it off. Here we go. Now that the ball we're up for what could go wrong I don't know we go here hit Odell over Milly holds on to a ball getting hit from Adrian Amos Well, that was on that's the diamond fully upgraded Odell Vic throws the ball into the stands. I appreciate it Mike They're getting one door underneath the Tyree killing crack Tyree kill fumbles the ball again. That's the third fumble of the game for me But that's okay. I mean at the end of the day Tyree kill got his belt run by Anthony Bar right there So when you put you know little receivers on the field, obviously, they're fast and they can run by everybody But they're not the best ones to be spinning into linebackers with and Tyree kill coughed it up there So I'm still at 4-0. I mean he still needs a touchdown. I feel fine He's really just taking what he can on the ground. You see 4-9 11 carries. He's got 59 yards Fletcher Cox fighting for me. Just try to give him no big plays. Here go the little PA slide Going up top Jalen Ramsey might bag this right, but no, he is volleyball season for Jalen Ramsey. So Get to a fourth and three play a little man coverage and this is to just kill the Prince hairs That's that's the new to I guess the Hall of Fame to I don't know what level it is But he definitely just killed his hair is there in man coverage Here he goes again This time I man up the running back because I told you I remember that play I remember play manned up the running back and he got cooked on a wheel route next play half back power He gets into the end zone. So now I'm down after all this all the drop pigs all the crazy plays all the fumbles I'm down three. It's some money drive time. No, this is what I do. I'm supposed to be professional man player I'm a pro money drivers. No big deal. I don't want him to touch the ball back So I'm gonna go ahead and just take what I got here. Oh here get out of bounds staying bounds Hopefully take all these classes two-minute warning Tyree kill right there. We got out of bounds there. So the clock will stop So we're gonna draw a little McCoy. This will take it to the I'm definitely on conservative now even though I Want to be able to spin and I want to go to make some moves I don't want to fumble and lose this game or fumble after three times already this game Here I have two people open But then make really good reads on this play I'm able to scream around they can find O'Dell late Throw it high run some more clock hit a draw. He has a crazy shed right there He's using his timeouts to try to get the ball back. I don't want him to get the ball back I want to end the game right here with a touchdown Arlo again. He's played good defense only rush three I don't know why that they're doing nothing past rush wise get out of pocket a bit gotta get down Like I said, I already fumbled three times. We can't make it four.  Like I said extend the game I'm not playing the superstar man player the longer the game goes the more I have a chance to win Here goes a pat Pete out the back of the hands. Oh, oh, oh, that was a clutch tackle right there I don't know who shed it right there, but it looked bad. It definitely looked bad And here we go, this is the end of the end of the first half I'm gonna chuck it up to will but you know Patrick Peterson the best part of the game isn't having that He's gonna return this and I thought he's gonna make a play but he got his sticks by the running back So we're going over time. I'm playing super shredder. So I'm not really that afraid of him I want the game to go as long as possible.  They have kind of a Pink tint to him because of the red lights. So look, maybe that's why I'm playing so bad and coughing the ball up. Don't wear pink Well, here we go. We're stopping around. I said if we stop the run will be alright and here We blitz the crib. This is exactly looking back on this game. That was just a terrible defensive call by me to be that aggressive Give up a big player to the tight end got a slowdown for that now. He definitely can't run in one now They were getting actually Melvin anger getting a strip animation now kind of me one of my fumbles back would be nice This is looking bad Bobby Wagner with the hit stick. He's at the 12-yard line first and 10 Nothing he rolls out left.  He went for the PA slide post round the back of the end zone Great play by tell them Smith now. We just need a fuel Vic. It's just fighting for me. I'm on conservative. They didn't really let me slide on conservative You will set up a big play here. We got beautiful pass protection Lane Johnson on a tiger. He'll get to the sideline And what what another fumble dank the Lord it went out of bounds. So right now I'm tripping I'm like, you know what that's that this can't be that changed. Nope. I'm on conservative still fun But I'm telling you it's the pink glare from the red lights on these white jerseys, but I'm pretty much in in Fogel range now feeling good. Nothing's really open here. I keep it safe. Throw the ball away I'm gonna try to pop a run here. Once I get this extra 10 yards. It feels good about my fillable Feel perfect. I'm running gale McCoy make it even easier. Really. I'm on conservative shouldn't fumble So I'm gonna go ahead and take this football he caused time out Which is the right thing to do now. There's a bunch of different ways You can avoid ice and all these things but but for me It's like, you know pretty pretty much down the line somewhere You're gonna have to kick it ice feelable and you got might as well get used to it now. So I'm gonna go ahead Knocked this field go through barely. It looked like it went right, but I knocked it through to win the game It was a tough game. I had to fight three fumbles drop picks You know, he played a good he plays smart he ran a ball kept it short But I said this this pink tent on my jerseys really did not do the right thing for me Good game to super shredder Like I said, if you like any of the offense any of the defense it didn't look amazing this game But you can check out man turf the link is below a premium membership will give you everything throughout the whole year I got tons more gameplay coming for you guys I'm gonna post the next gameplay when this one gets to 200 likes I appreciate all y'all checking out the videos and you know come through for even more


about exploring big AP physics with co-gases. Thank you, Marcello. Hello, everyone. I'm very happy to be here again after the colloquium of yesterday. And so the topics of my two talk, actually, will be a build-up about the colloquium. I will try to go not as fast as yesterday. I will try to really set the ground for this Berezinski-Kosare-Tauris physics. Explain to you why two dimension is not the same as three, whereas we lose long range order, but we gain other things. So I will go relatively slowly. Probably for those of you who are very educated in this field, you will not learn so much at least today. But I hope that I will take everyone to the next level. Just before I start, I just uploaded my slides on my own website. So if you have a tablet or a computer and if you want to download them, tell me if it doesn't work, but it should work in principle. So, OK. So everyone, some time to know that. So fees.ins.fr, tilde dali bar, without small letters, trieste dali bar 1 PDF. Tomorrow it will be 2. OK, so let me start. And I start by, again, speaking about two dimensions. This will be the topic of today. And before going to fundamental physics, I would like first to say how important it is to understand what happens in two dimension, how important it is from the applied point of view. There are many devices in our world, technological world now, which are based on two dimensional physics. And here I have just given three of them, but there are many others. The three of them that I'm pointing out here are quantum wells, which are everywhere in electronic devices.  Another place where two dimensional physics appears is in high Tc superconductor. It is now, I think, widely accepted that this superconductivity occurs because the cooper pairs move in two dimensional environment. And another very famous example of 2D physics is graphene, this set of carbon atoms, these sheets of carbon atoms, which have very interesting properties for application. So this motivates to study two dimensional physics. Now there is also a fundamental motivation, which is, as I mentioned yesterday, the fact that in 2D, you do not expect to have the same type of phase transition as in 3D. And this was understood first by Piers in 1935. And the question that Piers was asking at that time was, if we live in a 2D world, would physical objects like crystals or magnets, would they exist? And the answer that Piers brought and that we will review thoroughly in the next slides is the following one. If you take a non-zero temperature, there is no true crystalline order in dimension 1 or 2. And the reason for that is that if you take a possible arrangement of a crystal, say a square lattice like that, if you look at the displacement of this atom that I label 0 here, I call this u0, this one I call it qj. And I assume that I am at a non-zero temperature. So I look at the variance of the displacement of 0 versus j, so uj minus u0 squared. I average over thermal fluctuation. Then the result that you get in 2D, and again, we will show it rigorously in the following. But I am just announcing the result here. This variance is proportional to temperature times the log of the distance between atom 0 and atom j. This means that if I take rj large enough, the log is going to tend to infinity. And therefore, even if I know perfectly well where the atom 0 stands, I will have no information where the atom j stands within a unit cell. Because this quantity here will be larger than a square, where a is the size of the unit cell. And therefore, I will lose completely the long range order knowing where an atom is sitting. It doesn't give me any information where another atom far enough is sitting. And this was understood, as I said, by Payor very early, just after the beginning of quantum mechanics. And it was generalized 30 years later by Mermin, Wagner, and Hohenberg for any system with short range interaction. Then this gentleman showed that you cannot have at a non-zero temperature a phase transition involving a breaking of a continuous symmetry. The continuous symmetry here being simply the formation of a periodic lattice of atoms. And this, as we will see again today, also applies to the case of both ion condensation, where the symmetry to be broken when you both condense a system in 3D is that your macroscopic wave function acquires a definite phase. So you want symmetry for the phase. However, as I explained yesterday, which will be the topic of the two lectures, there is still room for unconventional phase transition. And this was understood by Kossel and Thaoules in 1933 in a famous paper, which was called Ordering Metastability and Phase Transition in Two-Dimensional Systems. And last year, Kossel and Thaoules got the Nobel Prize, shared with Duncan Eldane. So this is what we want to study. This is the outline of the lecture of today. So I would like to review the argument by Piers. Again, it may be something that some of you already know, but I think it's good to see this kind of argument because it will play a role in the following. So I want to show that there is indeed no long range order in 1D or 2D and take the example of a crystal. Then we will generalize this to the case of both ion condensation. I gain 1D or 2D. In the first part, I will deal with an ideal gas to interaction. And in the second part, this may be after the lunch break, we will deal with an interacting both gas. We are there, we will be more careful by separating what is called phase fluctuation and density fluctuation. And we'll show actually that interactions bring some new feature respect to ideal system because it brings this idea of a quasi long range order, not the true long range order, which is forbidden because of the MAM invigner theorem, but quasi long range order. And I will finish the lectures of today by discussing the case of a harmonic entrap system and speak of something which at first sight looks very strange, which is both ion condensation of photons. But I will try to convince you that this is indeed possible and has been seen experimentally. And tomorrow, in the lecture of tomorrow, we will learn more about the role of vertices and we will really understand the BKT transition point. And I will discuss a nice feature that occurs in two-dimension, which is not present in 3D, which is a notion of scaling variance. So let's start with today's lecture and this argument by Payals that indeed there is no long range order in 1D or 2D. So the argument of Payals, let's start with one dimension where it is the simplest. The argument of Payals goes in the following way. Let's suppose that I have my atom, which are interacting by some short range interaction and I model this short range interaction by a potential u of x, which is like this, so attractive at a long tail, which is attractive at long, for long distance, a local minimum here for distance a and then a short range repulsion. So if I am at temperature zero, it's clear that all atom will make perfect crystal, classically, and the distance between the two atom will be equal to the two little a where a is the position of this minimum. So at zero temperature, classically, there is no fluctuation, the answer is easy. If it was a quantum description, then I would have quantum fluctuation and it would be a bit more difficult, but let's do it classically for the moment. So I expect perfect crystal at zero temperature. What happens at non-zero temperature? So I will excite some fluctuations because of thermal fluctuation and this fluctuation will be characterized by the spring constant here. If I replace this by an approximation which is a parabolic potential, I have the strength of the oscillator is this kappa, which is the second derivative of u with respect to x taken in x equal a. So I will now model my chain at a non-zero temperature and the best way to do that is to use periodic boundary conditions. So instead of having a chain which is between minus infinity to plus infinity, I put my particle on the circle and I put n particle for a circle which has a length n times a. So at temperature zero, again, I have one atom every a, everything is perfect. Now, if I'm at non-zero temperature, the atom one may move by a quantity u one, the atom two may move by a quantity u two, and so on. And if I look at the energy of this system, I have both some kinetic energy and some potential energy due to the interaction of the atom between the atoms. And so the kinetic energy is here. It's promised to the uj square. So uj, again, is a displacement with respect to the equilibrium position. And the potential energy is just the proportional, so distance between the atom one and atom two, atom two and atom three. And it would be minimum if the distance would be a, and if the distance is not strictly a, it's proportional to the square of the uj plus one, the displacement of atom j plus one minus uj square, and all this is sum over j. And the hypothesis here is that I am allowed to expand the potential between the atoms, so uj plus one minus uj has to be small compared to a, but I don't assume that uj is small compared to a, that is, the atom j may move by a lot as long as its neighbors move by nearly the same amount. Okay, so that's the hypothesis. Okay, so how does one treat that? Well, I suppose that you have all seen this in your lecture. You can write the equation of motion for the variable uj, and uj double dot double time derivative is coupled to itself and to the two neighbors, j plus one and j minus one. And a simple way to solve this equation, excuse me, is to go to Fourier space, to Fourier transform the equation. So I introduce uq with a hat, which is just a Fourier transform of uj. So q is a wave number here, but that I choose in the Brillouin zone. Emmanuel has described this in his lecture, so I'm not going through that in detail, but q goes from minus pi over a to plus pi over a. And when you are in Fourier space, you get normal modes and you get n independent equation if you have n atoms, and each equation writes uq double dot plus omega q square uq equals zero, where the frequency of the mode q is just given, well, by sine qa, so again the wave number here, times coefficient, which involves the spring constant and the mass of an atom. So again, this is something I think you have all seen once, so I'm going pretty fast, but if you have any question, as for the other speakers, don't hesitate to raise your hand. So at this stage, I suppose everyone is happy. Yes, okay. So then we look for thermal equilibrium for this system. So we have n independent harmonic oscillator because this is the equation of an harmonic oscillator. So at thermal equilibrium, a given oscillator is decoupled from the other because they are independent, so the equation is not coupled. So I know that several average of uq prime will be zero if q is not equal to q prime, and if I take q equal to q prime in this equation, then I have the equipartition of energy because I'm doing classical statistical physics. So one half of m omega q square uq square just equal to one half kT, which tells me that uq square is proportional to kT. This is the important result that actually will be valid even in 2D or 3D. So now let's go back to the problem that Pyre's had risen. Let's look at the fluctuation of the distance between two atoms. So I take the atom zero, which moved by a quantity u0 because of fluctuation. I take the atom j, which moved by a quantity uj, and I look at this quantity, uj minus u0. I plug here the Fourier expansion that I had written before, where hj here is the position of the atom j. The atom zero is position zero, so I've got a one here, and I remind you that now I have to take the thermal average of this, knowing that uq square is proportional to temperature. It involves the frequency omega q, and this omega q is itself partial to sine qa. So here I take the square of this. Well, first I look at the average value of the quantity itself, but this is going to be zero because all the uqs are zero in average. So in average, the atom stay at the position I am interested in the fluctuation. So I take the square now, and I take the thermal average. So the thermal average of the uqs get out. This is this kbt over m, which is here, and I'm left with a sum over all modes of the system of sine q, sine square of qxj over two, divided by this frequency omega q square. This is what needs to be evaluated. And this is when evaluating that, that we will see whether pi r is correct and how one goes from one d to two d to three d. So in order to evaluate this, I'm going to make an approximation, which will be validated later on. I will assume that the physics of, at low dimension, the physics of Berencico-Sarris-Sarris is dominated by low momenta, or if there are large wavelengths and you will see why it is like this in a moment. So if q is, I can assume that q is small, then it simplifies the algebra. I can say that q is small compared to pi over a, which means that I can linearize a sign here and I get simply omega q proportional to q, which just means that I'm looking here at phonons, at sound waves only, and these are the excitation which dominates the physics. Okay, so the omega q square, which is here, I will replace it by cq square. And the second step that I have to do is to replace this discrete sum, which are never easy to evaluate by an integral. And the integral over q here runs from zero to the maximal value of q, which is pi over a. So I get here in the integral, this sine square, which is not modified, and the omega q square, which I had here, which has been replaced by c times q, so I get the q square and the denominator. So keep this in mind because this is really the structure that I will discuss, both in 1D, 2D and 3D. And the coefficient which is in front is kBt over something which involves the parameters of the problem, the spring constant kappa and the distance between two sides, little a. Okay, so let's try to evaluate that if there are no questions at this stage. Let's try to evaluate that. So it's an integral, so if I replace the upper bound by infinity, actually I can calculate the integral exactly. And the result that I get in 1D for this integral is something which is kBt again, this was proportional, and the integral of sine square over x over x square integral from zero to infinity is equal to pi over two. So I get finally this result here. And I get the fact that first I can look where are the modes which contribute the most here, and I get from this integral that the mode which contributes the most are the q on the order of pi divided by xj, that is if I take my two atom, one atom in zero, one atom in j, the modes which contribute the most to the fluctuation of the distance between these two atom are the modes which have a wavelength on the order of this distance. Okay, so since I'm interested in long distance behavior, indeed these modes have a very low q, this validates the approximation that I was looking at only sound waves. And more important, and again this is what Piles showed, when you look at that, you see that this function here will get larger and larger when xj gets larger and larger, which is what I said before, that is even if I know perfectly well where the atom zero is sitting, that I know perfectly the value of u zero, if I take j far enough, I will have no information where the atom j is sitting within a lattice spacing because this whole quantity here will be larger than little a, or little a square, so I will not have a true crystal, I will not have any long range order, no long range order in one d at non-zero temperature. Is that clear for everyone? Everyone? Okay, so what about 2d now? So in 2d you can do exactly the same treatment. It's a bit more complicated from a purely technical point of view because in one d I was, since I had only one dimension, my atoms had to move along this direction. Now in 2d, the atom may move in this direction or that one, so I have to, when I look at the sound waves, I have to look at the so-called polarization of the sound waves, the orientation of the motion of the atom. So technically it's a bit more complicated but here I'm not so much interested in the technical details, I just want to look at the global structure of the result and the global structure of the result will be exactly the same as in one d. That is, I will still get, when I calculate the variance of the displacement of uj minus u0 here, I will still get something proportional to the aperture kBt. I will still get an integral of the Brouillon zone which is now not a segment but it is now a square which goes between minus pi over a and pi over a with qx and qy. In this integral as before, I have a sine square of q times the distance between the two atom I'm considering divided by q square because again I'm looking at sound wave and now integrate not over a single dimension of q but over dqx and dqy because I am in 2d. So the only difference at this stage is these two here telling me that I have to integrate over two direction and not one. But this is enough to already change a lot the physics with respect to 1d. Indeed, when you want to calculate this integral, so I can, in 1d I could calculate this exactly. 2d I have to make some approximation. So an approximation that I can do if I assume that I restricting myself to q on the order of pi over rj or larger is to replace a sine square by one half. Okay, I could do something more precise but at this stage I can do that to get a rapid result. So I replace a sine square by one half. So you can get rid of these two here and I'm left with one over q square times d2q. d2q is the element of surface in the green one zone. So if I take polar coordinates, this is 2piq dq. And 2piq, yes, you have a question? Of course it's, what is that? That. So I look at the displacement of the atom j with respect to the equilibrium position. I call it uj. I look at the displacement of u0. It's not rj, okay? It's a displacement with respect to the equilibrium position. Okay, thanks for clarifying this. And so here I get, because I'm in 2d, I get in polar coordinates, 2piq dq. And so I'm left with an integral over q which goes, I have the one over q square times q, so I have one over q and the integral of one over q is a log, okay? So I get log of the upper bound divided by the lower bound. The upper bound is pi over a. This is the edge of the green one zone written here. Pi over rj is the wave vector q for which the larger, the smallest wave vector q for which this approximation holds. And then I'm left with this log of rj over a. And this is a result I wasn't seeing at the beginning. This variance of the displacement with respect to the equilibrium of j minus zero here is proportional to temperature, t. And there is logarithmically with the distance rj, which means that again if I take rj very large, even if I know perfectly well the value of u0, the value of uj will not be defined within better than little a, so I will not know actually where the atom j sits within a unit cell. So again, I do not have long range order, that is I, knowing where an atom in the crystal is not enough to know where all the other atoms are if I look far enough. However, you see already a difference respect to 1d. In 1d, the divergence was linear in rj. Now it's only a logarithmic divergence. So it's not as strong as a divergence, okay? So the loss of long range order is not as big and people often speak because the only logarithmic speak of quasi-long range order instead of the total absence of long range order. And just to finish the argument, you may wonder what happens in 3d.  Here now you will have a volume in momentum space, d3q, and when you replace this by polar coordinate, d3q will become q square dq. The q square that will appear here will cancel with this one. And now I have no divergence at all for low momentum, okay? So I had linear divergence in 1d, I had a logarithmic divergence in 2d, and now in 3d there is no divergence at all. So I can calculate completeness integral and I get something which will be kbt over kappa independent of rj. And this tells us that indeed in 3d we can have true long range order. If this quantity that I have here is much smaller than a square, which is again the square of the crystalline parameter, then if I know where an atom is sitting in this crystal, I will know where all the other atoms are sitting within some fluctuations which are given here, but these fluctuations are small compared to the lattice spacing, okay? And there is a criterion that this should be actually less than 10% of the lattice spacing if we don't want the crystal to start melting. So in 3d, with this argument, I get true long range order, but in 1d and 2d I do not. And the physical reason for that, which is hidden in this mass about Jacobian, is the fact that when you are in 1d or 2d, you don't have enough neighbors actually. When you are in 3d, and if an atom wants to move out of its equilibrium position, it's going to frustrate many people, okay? People on the left and on the right and in front of me, behind me, people above and below. And so it's very unlikely that an atom can move a lot because there are many frustrations that are caused when the atom moves. When you are in 1d, you have only two neighbors, one on the left and one on the right. And if you move like this, if this neighbor accepts to move also on the left, it's easy to have a big fluctuation propagating. And in 2d it's marginal and this is why we get a log in 2d, okay? So the idea that I want to convey here is this lack of neighbors in low dimension which forbids to have two long range order. Yes? Ah, good question. Yes, how does this apply to graphene? So there are two answers to that in the literature. And I don't think people have really completely say which one is correct. So one first reason is that when you take this in graphene and you take the value of kappa, the distance rj over which you lose the long range order is very, very long, much larger than the graphene sheet that can be produced. So it's not so relevant for graphene from a practical point of view, taking into account the fact that the graphene sheets are micrometer size or tens of micrometer size. Now there are some more elaborate arguments which are not actually, which could apply to even if we could produce infinite graphene sheet, which is that graphene is not really flat. Graphene can bend and because of bending this introduce kind of long range interaction that is if I take into account the fact that the graphene sheet can bend like this, then I have interaction between two atoms. Let me maybe draw my graphene sheet so you see what I mean. So I'm not sure I can draw graphene sheet easily, but do you see the graphene sheet that I'm trying to draw? Yes, so I have an interaction between these two atoms when it bends. So this introduce a small component of long range interaction in the system because of the possibility of the graphene sheet to bend and this is enough to restore, from a purely theoretical ground, the true long range order because of the bending. So the graphene being not something which is flat but having this possibility to bend, it's a crippled surface if you want. Yes, yes. So you... It's already a small bend. Yes, a small bend is enough because a log is very fragile. So it's easy to get rid of a log if you introduce some new physics in the system. It's not solved and I'm not sure that this is relevant for any practical graphene sheet, okay? Yes, yes, then it might be relevant. Yes, yes. Okay. So now that we have seen this example for the crystal, let's go to both ancient conversation which is what we are interested in in this set of lectures. And first of all, I would like to start with the case of an ideal gas, no interaction. So again, these are probably some things that many of you have already seen. So I take particular obeying both statistics. I confine them in a box at a non-zero temperature. And the first thing I would like to say is that irrespective of the dimension, when I take a box or actually any kind of potential, I always get something that one can call both ancient conversation. That is, the number of particles that can be placed in the excited state of the box is always bounded. And indeed, this comes from the Bose law that I will use a lot, so I might as well write it on the board. So the number of particles on a state of energy, E, as you know from the Bosenstein law, is the exponential of E minus mu divided by KBT minus one, okay? So this is something you have all seen. Let's suppose that the lowest energy of the system is called E zero, and I set it to zero, just a convention of energy. Then in such a formula, the one which is written on the board, mu has always to be smaller than E zero, which again is zero here. Otherwise, if I allow myself to take mu larger than E zero, this will become negative, and it will be nonsense, okay? So in a Bose-Einstein system, by contrast to Fermi system, the chemical potential is bounded. It cannot take any value, it should always be lower than the energy of the ground state. And therefore, it's just a mathematical exercise to see that if I take the sum of all the population bat the ground state here, so I exclude P equal zero from this sum, this sum here is always smaller than the same sum without the mu here. And this quantity, if I take a discrete sum, is finite, I can calculate it. And therefore, the number of particles I can put in the excited state for this temperature and this chemical potential mu is always bounded, which means that if I take the same temperature, and the same mu, and I constantly add particle, well, same temperature, excuse me, and I constantly add particles that mu is going to vary, because mu is always smaller than that, so the number of particles in all the excited state will never exceed this value, and therefore, all the extra particles have to accumulate in the ground state. So in this sense, Bose-Einstein condensation in a finite system with a discrete level is very simple, just an inequality like that, and you don't have to discuss anything more. And this will be true in 1D, 2D, or 3D, okay? Now, the question that was not trivial in Einstein's paper, and which has actually caused a lot of debates, and the notion of thermodynamic limit was actually introduced because of this problem raised by Einstein, is what happens now when the size of the box tends to infinity. If I keep constant the temperature, and I let the size of the box tend to infinity, keeping the same density, then what happens? And when you do that, so the discrete structure here of the level will disappear, I will transform my discrete set of states into a continuum. So I have to go to replace this discrete sum again by an integral, and this integral involves the density of state, d of e divided by the same denominator, the expansion of e over kt minus one, and all the question is, again, created to low wave vector, low momenta, or if you prefer low energy, all the question is, does this integral converge, and does it converge in equal zero? And as you probably have all have seen, it depends on the shape of this density of states. So if you go, sorry, we've written this integral here, if you go in 3D, the density of state in a cubic box, when the size of the box tends to infinity, varies like the square root of the energy, and the denominator here for low energy is typically linear in energy, so you get an integral which is square root of e divided by e, so one over square root of e, and this integral converges in zero, and therefore the number of states you can put in the excited states remains bounded when you take the number of particles, excuse me, or more precisely the densities that you can put in the excited state when you take the thermodynamic limit, remains bounded, and therefore this Bose-Einstein condensation that we had for a finite size system survive when you take the thermodynamic limit, because this integral converges, okay? So again, for a discrete system, you always have BC, and when you take the thermodynamic limit, if this integral converges, it means that the density that you can put excited state is bounded, and therefore you will survive, your BC will survive, and in 3D, this is the case. When you go to 2D, the density of state d of e is constant, and therefore the integral that you have to calculate now is one over e, because the denominator is partial to e, and this integral diverges, which tells you that the Bose-Einstein condensation phenomenon does not survive when you take the thermodynamic limit. The density that you can put in the excited state now can be infinite, and therefore there is no macroscopic accumulation of particles in the ground state, okay? So I wanted to make this clear for everyone. Is it okay? No question? Okay. So still it's interesting to see what happens when you take a Bose gas in the thermodynamic limit into dimension and see where does the particle go. I told you that they do not accumulate in the ground state, but still there is a kind of saturation which occurs, and in order to show that, I have taken this this Bose-Einstein law, which is written here. So I take, I'm in a box, so I take e equal p square over m, over two m, excuse me, and I plot here the density distribution in this box. For values, phase space density rho times lambda square. So this is the quantity that I will use a lot. So the phase space density is a density, special density rho times lambda square, lambda t square, where this lambda t is equal. I just rewrite it here, square root of m kBt. So varying the phase space density is a way to vary the degeneracy of the gas, to go from the classical regime to more quantum regime. So here I have plotted in values colors, the momentum distribution, N of p in the box, as a function of momentum p, for values phase space density. So let's take phase space density of 0.1, so gas which is very weakly degenerate, kind of Maxwell-Boltzmann gas. So N of p is just a Maxwell distribution, a Gaussian distribution. So here I have taken some logarithmic coordinate here, so a Gaussian with a logarithmic semi-logic coordinate correspond to a parabola like that, and this is a blue curve here. The unit that I put here, so it's the momentum p, but I rescaled it to have a dimensionless variable, so I use p lambda t over h bar. When this is equal to one, if you square it and you replace, you take the value of lambda t which is written on the board here, you get p square over 2m equal 1 over 4 pi kBt, so this corresponds to the situation, one here is when the kinetic energy is on the order of kBt. So at low phase space density, no surprise, this is Maxwell-Boltzmann. When I increase the phase space density, I see that I have a bump of particle accumulating in slow momentum, momentum which are such that p lambda t over h bar is smaller than one, or if you prefer, the kinetic energy is smaller than kBt. So you see that when I put more and more particle, I don't get a Bosnian condensate, because I'm in 2D, a Bosnian condensate would be an accumulation of particle in p equal zero here, which is not what I see. But I see that still that the particle accumulates at low momenta here, the high momenta here are nearly saturated, so putting more particle in the system does not put more particle at large momenta. All the particles go to low momenta, but not to a single one like in 3D, but to a continuum of low momenta which is here. And this is what happens for this ideal gas. And if I look at what happens at phase space density of 10, I actually, this is not obvious because I'm using logarithmic coordinate, but most of the particle are now sitting in this low energy tail here, and very few particles are here by contrast. And if I now ask myself what is the shape of this momentum distribution close to p equal zero, then I can take again this equation here, and I say that E minus mu over kT is a small variable, so I can expand the exponential, and I get now that this N of p when I expand will be kBT divided by p square over 2M plus mu, which is a Lorentz distribution. So in 2D, when you cool a gas, a Bross gas, what you get is that most of the particle when you have strongly degenerative regime, most of the particle accumulate in the low momentum class and they have a momentum distribution which is a Lorentz distribution like this. So now you may ask what does it mean with respect to the absence of long range order, and the long range order in a Bross gas, this is characterized by quantity I have introduced in the colloquium yesterday, this is the G1 function, this is how the phase is correlated between one point in R, one point in R prime. So the thing I had yesterday was, I mean here is the two differential equivalents, so G1 of R, yesterday I think I wrote that it was psi dagger of R, psi of R prime, where psi is the field operator, but this can also be written, it's equivalent to R, the one-bodied estimatrix, row one, R prime. These two definitions are equivalent, this can be shown easily. And so this one-body correlation function, I can let you show this as an exercise, it's relatively easy to show, can always be calculated from the momentum distribution because these two quantities are just rated by a Fourier transform, that is if you want to calculate G1 of R, you can take the momentum distribution N of P, and N of P is the Fourier transform of G1 of R, and vice versa. So now we know that the momentum distribution is, this is what was here, is a Lorentzian at least at low momenta. So we have to know what is the Fourier transform of a Lorentzian, and the answer is, you know what it is? An exponential, an exponential minus R over L. And so what we get for this ideal both gas is that indeed pi R's again is correct, there is no long range order, the G1 function decays exponentially with the distance between two points I'm considering, and with a length L, which increases when I increase the phase space density, but still it always decreases exponentially, exponentially fast. Okay, so how am I doing in time? Yes, I still have 15 minutes. So are there questions at this stage? Yes, yes, yes. So the question is, do you need a Boschian condensation to get ITC co-superconductivity or to get superfluidity? And the answer is no. You can get, this is precisely the work of Berezky-Cosser, St. Aulets. You can get superfluidity without Boschian condensation, and this is what I want to show later. Yes, no, no, you should not be sorry, this is a very important point, thanks for pointing it, yeah, that's, so. Okay, so probably what Emmanuel had in mind was a quasi condensation, so quasi long range order, which was enough to ensure everything he was saying. Also he has a pretty finite size system, so this was the answer for the graph in sheet. If you're looking at a short distance where the log is not time to rise, then everything happens as if you had long range order because you are looking between points which are relatively close to each other. Okay, other questions? No? So let me now go to the case of an interacting Bosch gas which is more interesting, but also more challenging, and you will see that interactions still do not violate Mermin-Vagnott theorem, but still they change a little bit the picture that we have for the ideal gas of this exponentially decreasing G1 function. So in order to discuss the interacting Bosch gas, one has to do some approximation, and you can never treat exactly an interacting many body system in at least in two-dimension or three-dimension, and the approximation I'm going to use is a so-called classical field approach. That is, this classical field approach amounts to say that we have a true many body wave function which is a complicated object, but we are going to replace it by an ansatz which is often called the Hartree ansatz, which is to say that all particles occupy the same state psi, so I write this either as n psi like this, or if you prefer n many body wave function which is the particle one inside psi r1, particle two psi r2, particle n psi rn, but the psi itself is a fluctuating object. If psi was the ground state of the box, this would be a Bosch ansatz ansatz, okay? But here I take for this psi a fluctuating object, a fluctuating field, but I assume that all the particles are occupying this field. So it's not an exact result, but it's an approximation, and the rule of the game is to find the correct probability distribution for this classical field psi. So this may sound a bit strange as an approximation, but for those of you who are familiar with electromagnetism, this is exactly what you do when you do classical electromagnetism. When you, if you want to describe a field using quantum electrodynamics, you have to use some very complicated state which where you have to specify the number of photons with momentum k1 and polarization epsilon one, the number of photons with k2 epsilon two, this is very hard to manipulate, but we all have learned classical electromagnetism where we describe our electromagnetic field by a classical field, actually two classical fields, the electric field and the magnetic field, okay? So when you are doing classical electromagnetism, you are doing exactly what I'm doing here. Here you are saying that all the photons, so to speak, share the same wave function, it's always dangerous to speak of the wave function of a photon, but are described by the same field characterized by these six real numbers, EXEYZ, BX, BYBZ, okay? So this is what we are going to do for our matter wave field. We are going to assume that all the atoms share this same psi, but this psi is a fluctuating object. What are the limits of such an approach? So again, maybe a good way to see that is to come back to the electromagnetic field. When we do that, when we take a classical field approach, we neglect the fact that the light is composed of elemental individual photons, so we neglect the corpuscular nature of light, and if you want to understand what a photon is, you have to look at, say, a spontaneous emission process where you have photons emitted one by one. So when are these process relevant, this spontaneous emission process? When will a classical field approach be bad? Which will be bad when actually you do an experiment which is sensitive to the commutator between A and A dagger, which will tell you that it's not the same to take A A dagger or A dagger A. And if you look at what is the matrix element of A dagger A and A A dagger, since this commutator is equal to one, this one will be equal to N, where N is the number of photons in the mode K epsilon, and this one will be equal to N plus one. So if I have a mode which is heavily populated, if I have a number of photons N per mode, which is very large compared to one, the difference between N and N plus one is not significant, whereas if I'm doing something like spontaneous emission, a mode which is initially in the vacuum state and I emit a photon in it, there, of course, I cannot neglect the fact that the photons are quantized. So this classical field approach, a priori will be valid if I have a large number of particles per mode of this matter-wave field that I'm going to introduce. Another issue that you have when you use a classical field approach is the so-called black-bodier addition problem, a famous problem of the 19th century, that is when you are doing classical field physics and you are looking at the thermal equilibrium of a classical field, then you may have some ultraviolet divergencies, and therefore you need a cutoff. That is, you need not to consider frequencies which are very, very large. You have to put a cutoff in your frequencies, otherwise you have divergencies. So here we will do that. It's not so important for us because as we have seen, all the physics that Payor's introduced and what will come with BKT physics is a physics of, as I have explained, low-momentar, low-frequency. This is large distances, low-momentar. So what happens at very large frequencies, this ultraviolet cutoff is not so relevant for our physics, so we can put this cutoff. It may mean some uncertainty on the prediction, but at least the main physics will still be present because these physics belong to, I would say, infrared wavelengths and not ultraviolet wavelengths. So how does one under a classical field at, in thermal equilibrium, at finite temperature T? So here I'm doing it for the simplest field I can think of, a field like the electromagnetic field where the energy, I share it with a field that I have written by this e-caligraphic. This, the energy is promote to e-square, integrated of our volume L-cube. Here I'm doing it in 3D, but I could do the same in 2D or 1D. So again, the way to treat that is to expand in eigenmodes which are all independent from each other for this classic, very simple energy model. And so the state of the field is characterized by the amplitude e of q, the Fourier component of this expansion. And e of q-square is the proportion of the mode q. And the energy that was written here as integral, when I expand in Fourier space, I get something which is alpha L-cube, the integral of the whole volume, times eq, the eq-square, so the Fourier component square here. And now I just have to say that the probability to get a given realization for the field, that is a given set of amplitudes of eq, is just given by the Boltzmann weight, exponential minus the energy over kBt. And because the energy, excuse me, e of e is the sum of eq-square, I get that every eq-square has a thermal average which is kBt. This is very similar to what I said before, but now it's at the level of the field and not the displacement of the atom. And this linear variation of the position of each mode is something that is very characteristic of a classical field approach, something you have probably seen when you have studied black body radiation. So we'll do the same for our matter weight field. So let's see how it works with the matter weight field. So I take the energy of my gas. So this is my Hamiltonian I start with. So I have some kinetic energy. I have some trapping potential to hold the atoms. And I have some interaction between the atom, u of r, i minus rj with the one half here to avoid double counting. And in my classical field approach, I'm going to say that the wave function, macroscopic wave function of the system, psi of r1, r2, rn, is simply a product state, little phi over r1, little phi over r2, little phi over rn. And the normalization of the phi is such that the integral of phi square of the volume is equal to the number of particles n that I put. So saying that I get an energy which depends on the field phi, which I will have to treat again as a fluctuating object. And this energy corresponds to the sum of all components that are here. So the pj square, which are over here, will give me, you know, that pj in quantum mechanics, well, the momentum p is the gradient of the wave function. So the pj square, which are here, will give me h bar square over m, the gradient of the field phi.  And the interaction term here, u of r i minus rj, will be treated by a contact term. So it's an approximation, but it's a very good approximation for low energy physics for cold atoms.  And then when I plug my ansatz here with this potential, I get g over 2 phi square, phi 4. So a, here is a scattering length that Emmanuel already mentioned. So are there questions on that? No? You all know that? OK. So let's go now to 2d. And then probably I will stop at this stage. So let's go to 2d. So in two-dimension, I have to go to two-dimension. I have to freeze one direction of space. So I introduce a strong potential along the vertical direction. And I assume that my particle only occupies the ground state of the vertical direction. So this ground state, if I take a harmonic potential, it's just a Gaussian distribution with a size given by the size of the harmonic oscillator depending on the oscillation frequency on the potential along z. So the classical field that I was introducing before, this phi of x, y, z, is actually a factorized. I have this chi 0 of z, which tells me that I have frozen the vertical direction. So I know exactly the state of the particle for the vertical motion. And psi of x, y is still a random field that I have to understand. And so when I plug the previous gross potato ansatz that I had before into this with this ansatz, I get again a kinetic energy, which corresponds to a kinetic energy in the x, y plane. So this is a modulus square of the gradient of psi of x, y. I assume that I have no more potential in the x, y plane. So I have no potential energy. And the kinetic energy, which was partial to phi modulus to the 4, now is partial to psi modulus to the 4. And I have introduced here some parameter g tilde. And this g tilde involves a previous scattering length that I had before. And when I integrate over this chi 0, which is here, there is a little algebra which is not complicated, I get little a divided by aOH.  And this is an important point I want to make. In two dimension, at least within this classical field approach, the interactions are characterized by a dimensionless parameter. By contrast to 3D, where interactions are characterized by a scattering length. So the number that characterizes them is a dimension. Here, it has no dimension. In practice, it is a ratio between the scattering lengths and the size of the ground set of the Hamiltonian. But the fact that it is dimensionless, you will see tomorrow plays a very important role because this is what gives the scaling variance of the 2D formalism by contrast to 3D, where the interactions were introducing a length scale or an energy scale. Here, there is no length scale, no energy scale, which comes with the interaction. The only thing I need to know is the value of g tilde. And in standard experiments, it's between, say, 0.1 to 1. And this is the only thing that has to be known about the 2D Bose gas. This is this dimensionless parameter. OK, so what do you think? Should I stop here, or I can take questions if you want? I can continue, but maybe I can just show this slide just so that you can have lunch peacefully. You may wonder when I'm doing this classical field approximation whether there is any quantum feature left in my problem. The fact that I've replaced all the quantum commutation relations by c numbers, are there any quantum features left in the system? And the answer is yes. There is still something which is quantized in the problem, which is related precisely to the fact that we are using a classical field. So when I take an assembly of particles described by this classical field psi of r, I can interpret this classical field psi of r, which is a complex object, by a square root of a quantity which is the density of particles. The use of psi squared tells me the density of particles in r. And I have some phase factor into the i theta of r. And as I said yesterday, there is a velocity field, which is associated to this phase theta, which is just a gradient of theta times h bar over m. And the quantum feature which remains valid here is when you have such a classical field like that, if I take the integral of gradient of theta over any closed contour, I know that since this field is single valued, I have to get the same value of the field when I make my closed contour, which means that theta has to be the same within 2 pi. And so when I take the integral of gradient theta dr over any closed contour, then it means that this integral is n times 2 pi. And therefore, in terms of velocity, the motion that I can describe by this classical field approach are such that the velocity field has an integral over any closed contour, which is again quantized. It will be the same thing as this n 2 pi times this h bar over m, which is here. So there is a quantization of the circulation of the velocity, which is present in this classical field approach. So even though I'm doing some classical physics with this classical field, I still have something which is quantized. This is a circulation of the velocity. And this is the only ingredient I need to get super fluidity. So I've lost completely quantum physics by doing this classical field approximation. So probably it's a good place to stop here. So I thank you for your attention. I am ready to take any question. Thank you.


Hi everyone, my name is TJ and welcome to the ACMI Artist Painting Challenge. Each artist will take a blue bin of paint colors of their choice to paint this radiant sunset. Artists use your creativity. Mixing of paint colors is encouraged. An exact likeness is not expected, so after 15 minutes our special panel of judges will announce the winning artist. For our contestants we have JB, ACMI's news director. He's also a professional artist. Karen, an ACMI member who is writing and producing a short film. And Sol, an ACMI intern who loves editing. And now to introduce our judges, Katie and Joanne. You will have 15 minutes from the time of the bell. Grab your paint colors and aprons and begin. On your mark, get set, paint. Can it be all white? It's just my interpretation. Well, we have three things we're judging for. Creativity. Wow, this is great. Look at that. That's like a Rothko. Vibrancy and originality. Yes. Sky's the limit. You can be as creative as you want. Was that a ruler? He's getting his horizon lined in. They got rulers over there. Is that allowed? It must be rulers there. I see a ruler. Karen, you have a ruler. That's Sol's technique. You see that, right, Karen? J.B., oh, what? Karen, she has a ruler. Oh, yeah. A triangle there. Yes. Do you see your ruler, Karen? Yes. Right behind your camera. On the right hand side. There you go. J.B., you have your ruler? Nothing wrong with that. Maybe he doesn't want his ruler. He might be going abstract. No ruler allowed. Oh, he doesn't mean it. It's freehand. All right. This is professional. He's also, J.B., decided to not get his apron. Interesting choice. Interesting choice. No apron for J.B. Deep in concentration. It's going to be covered in paint after this. That's fine. Just challenge. That's the point. I think so. Oh, there's the first color. Look at that technique. Jeff Varn with that swish and swish of the paintbrush. I don't know. I don't know about that. So can they interfere with each other? Is that allowed? Hey, did you have a challenge? A bit. Yeah. Have you heard that describing the rules? Exactly. Yeah, it was a lot of fun. Did we not count a whole gallery? Yeah, right downstairs at the Cutter Gallery right now. Very fine quotes at the Dallam Museum and Cutter Gallery. Oh. Will your finished products be on display in the Cutter Gallery? I don't know. I don't think so. I'll let you know. I'll see how this turns out. We'll send them the video and then they'll be convinced. Yes. Am I in my snuffier camera shots? I don't think she's in here. I'm feeling a little left out. I would like to paint as well. Wait, we can draw. Can we do a painting but we won't judge it? Of course we'll win. That's probably one. Are there any more canvases? Oh, look at you with the drip effect. What are you, Paula? What? Well, certainly for speed, I'd say JB's doing pretty good. But, you know, the story of the tortoise and the hare. Keep that in mind. Karen is carefully selecting her colors. Slow and steady. You know what they say. Isn't there a saying that he wins the race? That's it. I never understood why the fox is always antagonized. Or the antagonizer. Nobody likes foxes. They don't pay their taxes. Oh my goodness. Foxes are tax exempt. Wow, the time is going. It's taken by. I do like what I see from Soul right now. I can't explain it. But Karen's like color matching is pretty on point though, I must say. For the original. Compared to the original. Contestants, I think you have ten minutes left. The third of the way through. But really, I'm not impressed by any of these. This is all. This is all we got. I'm waiting to see something impressive. I'm not impressed. I could do better. Come on. You could do better than this. What are they doing? I don't know. Whatever they do. Simon Cowell. I have to wait for the reveal at the end. That's what falls the studio. Karen, how are you doing? Tough to say. It at least resembles something. Where's Cheryl? Who's Cheryl? The producer. The brainchild. She's getting some behind the scenes videos, I believe. Oh, I see, okay. I think she's very impressed.  Which is more than I can say. Get those colors going. Look at the camera. We can see something. Yeah, yeah. It's kind of an algae looking. Water there, JB. I'm sorry? It's kind of algae looking. It's his impression of a child's river, I guess. Or the mystic. That overflow. I'm inspired. I'm calling this first grader does sunset. I'm inspired. I think I'm going to take up some painting. There you go. You and me both. Do we know what kind of paint this is, guys? acrylic. It was oil, we'd be here for three months. I think souls blend in colors really well. From what I can see. We're down six minutes. Over halfway. Six minutes. Have you done any painting at all in your life? Yes. Joanne, yeah. Does the bedroom count? The bedroom counts? Sure. I painted the bedroom, I painted the kitchen. And toenails once in a while. Oh, sure. I did these. They're coming off, but it counts as painting. Yes, it counts. I painted my bathroom once a couple years ago. What color? A gray. A nice light. Gray, I believe. But I've since moved, so now I want to have to paint again. The only thing I don't like about the subject is I can't paint a happy little bush. Why can't you? That's your interpretation. I can't add anything. Oh, well, I see green, so who says the green isn't bushes? What does he say? Happy little accidents. Happy little accidents. That's what life is. I like that. It's a toss. You got three minutes. Three minutes. Three minutes, 50 seconds. Da Vinci had 20, at least.  Oh, you do gray? Oh, that's what you're saying. Yeah, back and forth. In cap issues? I couldn't get the paper out. It's been rigged, Karen. It's been rigged against you. It's really good. You went Pollock. Very cool. Oh, one minute. Yeah, less than a minute. 30 seconds. Seven, six, five, four, three, two, one. Paintbrush is down. Good job, everyone. Gosh, these judges were abusive. Oh, that tore my paintbrush. Man. I'm telling. Is there any way, can you guys like, lift him? Because there's like a glare. I don't want them to drip or get. Okay, so maybe she's moved to here. Yeah. Yeah, I think so. I don't know if I know. They're looking serious. What about this one? Yeah, okay. I think. I think we got it. I don't know. So I think we've just decided you guys can put your paintings back down. And so, thank you. At this point, do we give out? I think first, do we want to hear a little a story about tell us about your painting. Let's start with JB. Tell us about your painting. What inspired you? What happened throughout the course of these 15 minutes? Yes, the sun inspired me. What I wanted to do was I wanted to, since we only had a time constraint. Time constraint. Yes. I don't want to repeat myself.  You're trying to get on your good side. It's going to be tough. Keep talking. No, since we had a time constraint, I figured I'd eliminate the rate and I didn't have a chance to blend the sky in here. I wanted to kind of do like a pink sky at night sunset. And instead of the race, I figured I'll do the clouds. Throw in some birds, you know, class AVs. And then kind of do a a Van Gogh blend like sun with a very Homer-like rough sea. So I figured what I would do is I would, instead of the lines and everything else of the stratification, if you will. Yes. And I will. I'll tell you what I do as a rough sea. Very creative. Very creative choices. Very creative. We've heard you. My general inspiration actually came from a little experience I had driving back from Framingham. It was pouring out rain. It was near dusk twilight even. And because of where the sun was and how just the general positions and stuff, almost dark, and yet you could see this striking rainbow that was just in the sky. And it was genuinely fascinating to me that just stark contrast. So I wanted to emulate that through the blues and the purples and such. I also noticed since they gave us gloves, I wanted to try to make a fun effect with them. With the water and stuff. It was a 50-50 on the texture but I think it gave a lot of depth to the water. It doesn't look much like water but it certainly gave a lot of color and texture to that and I'm actually quite happy with it. The rays sort of came in as a nice little contrast to that blend and that was mostly what I was going for and what I was hoping to see. I also like the intense hot light you have in the center there as you slowly diminish the light and go from yellow to orange. Very nice touch. From light to like a salmon color. Very impressive. For mine the first thing that inspired me was the sky. I wanted to create a galaxy feel where it's just the sky and somehow interpret the sun in there but something got lost a long way. I ended up making sort of a green sun but that didn't look right so then I just made it kind of a fiery red sun. The bottom is supposed to represent the sun somehow to get the top as well. Yes. Thank you for sharing Karen. Thank you very much for sharing. And thank you artists. You took us through a whirlwind of 15 minutes. Highs and lows, up and downs. I was anticipating the outcomes seeing how you all started up and how it ended. The colors that you chose and the techniques that you used, I found it very interesting. It's very creative. Your story is very inspiring. Yes. We're lining up our awards now and we have conferred with each other. It wasn't easy because as we were watching it unfold we thought we had it figured out well this is this and this is this then it wouldn't finish it's like no, this one is more this and that one is more that and that one is more the other. So the long and the short of it is for creativity creativity creativity. Last you'd come up to receive your award for this is what I was trying to focus on for creativity creativity creativity creativity take a shot at the time for creativity the sun's going down now we have Jeffrey Barn yes creativity you may approach the desk thank you thank you judges thank you I want to thank my mother I want to thank my dad I get to do the next one yes and for for a virus we present this award for a virus to Karen so vibrant so vibrant and last but not least for originality so and now the overall award for best story did you know there was a best story award? best story award and the explanation or presentation of their painting goes to so you talk about the drive to I believe framing him and the thank you all this has been an amazing experience and you have been watching the ACMI artist painting challenge thank you for joining us today artists judges and crew I'm TJ and we will see you next time take care see you next time bye


A group of vigilantes and hunters in Tarava State have rescued 40 hostages who had been languishing in a kidnapper's den in Aganduil. The rescued hostages were kidnapped from rural communities in Yorua and Zing local government areas of the state. Our correspondent Jaut or Jude Nuhugumdali's report is presented from our studio. This would be a major breakthrough by David Iloyanumon who was recently posted to Tarava State as a commissioner of police by the Inspector General. The battle to make Tarava State crime-free has been taken to the kidnapper's den located at Tab Gongomaliki Mountain by the hunters and vigilantes who were on a rescue mission. Briefing these men under development, the State Commissioner of Police David Iloyanumon said the rescued hostages were taken to the police clinic for examination and treatment after they regained freedom. He disclosed that one of the rescued hostages had one eye blocked out while another had gunshot injury and is still in the hospital. And accordingly we engaged the local hunters and the vigilante. So they all went there and the ones I am going to relate to you now is the one that happened at the Gomaliki Mountains and beyond. We have 40 kidnapped victims we are rescued from the kidnapper's den. The victims have various degrees of injuries and ailments including one Musa Mashi who has two gunshot injuries while the right eye of one Sunday was propped out by the kidnappers. May I conclude by applauding the unflinching support of his Excellency, the Executive Governor of Tarava State, Dr. Ago Kephas. The CB also paraded 10 suspected kidnappers who were apprehended at different locations in Arodo Kona local government area of the state. He informs that the state police command will continue to collaborate with local security apparatus to ensure the state is safe. The police has also arrested 10 suspects those ones you are seeing behind me these are suspected kidnappers. These were arrested at Mulum and Kaswanlai area of Arodo Kona local government while they were escaping from hot zone. Remember I told you the hunters were on the hills chasing them while they were escaping from the hot zone they were coming down the mountain and going to the communities. Meanwhile some rescued victims and community members share their views. The kidnappers are taking many people from our community are about almost 71 people we are very happy with the intervention of the vigilante people that when they are and rescue our people in the bush. Some hunters vigilantes beat the supreme price as their members lost their lives in the raids to the kidnappers then in Tarava state and it is expected that government will immortalize them in the future encouraging others. Hello hope you enjoyed the news please do subscribe to our youtube channel and don't forget to hit the notification button so you get notified about fresh news updates.