 So let's continue with our discussion of the divergences. You remember in the last class, we studied the volume between the actions, and we decided to compute the expectation value of pi1 x1 by pi of x1, pi of x4. In this theory, we moved to Fourier space to do the calculation. Fourier function of the subject. And one of the contributions in perturbation theory, the leading connected contribution in order lambda squared, we come to understand what these terms mean a bit, it turned out to be lambda squared by 2, 1 by, so what we did was, same as we have, we will remember this Feynman graph, if p was the momentum flowing through the Feynman graph, if p1 has been 2, it's equal to minus p3 plus p4, we got a point over lambda squared by 2 squared, and that is the loop momentum. This object was divergent for derivative of 4. We were wondering what to do with this. So this lecture will start trying to understand how to make sense of what it is. So quite as I explained last time, it's fine to be in theory, it's not the kind of, but it's defined by a cut-off path in here, where the parameters of the Lagrangian functions with that graph, and the cut-off is eventually taken to infinity with the parameters scaled with the curve of the infinity, get away. And we're going to try to understand that statement. As we're going to try to understand this, following the beautiful paper by Paul Gypsy, you know when I was a student, I was studying finance and quantum theory, this business of renormalization just didn't make any sense. You know, I read Bjorken and Drell, and I read Itzik and Zoba, you know there was a lot of books, and everything, in a sense, just didn't seem to make any sense. It seemed like completely disconnected from the rest of the books, you know, that way of thinking. It didn't make any sense to me. I first felt I understood what was going on when I read this book, so by the way, I'm introducing the subject in this one. Okay. Okay, excellent. So, let's go. So, as we've seen, a diagram like this could lead to divergence. This divergence comes from the algebra. By which I mean that some intermediate particle momentum in a Feynman's graph is this one. It's becoming very large. By the way, writing perturbation theory in terms of Feynman's graphs and so on, that's of course familiar to you people. I did that very fast last time. Were there any questions about that mechanical stuff? Were there any questions before we... So, we had this divergence where one of the intermediate momentum is going very large. That was the part that contributed to this. Just as an aside, suppose m was equal to 0. Then there are Feynman's graphs that will give you divergences from the other end. Namely, they are r equals 0. Okay. There are two kinds of divergences that can occur in Feynman's diagram. Actually, it's really hard to explain. Okay. There are divergences where... Okay. Yeah. The opinions based on two kinds. The divergences where... The divergences where a very large momentum, divergences a very small momentum. And because these can... It sometimes seems like there's a third kind of divergences where the momentum goes on to share. The momentum goes on to share. We'll talk about all these things as we go. Okay. But these are actually very different factors. Okay. Divergences at low momentum are infrared divergences. And are some real things. Okay. They're not to be dealt with by changing the definition of the theorem. They represent a real physical thing. And they tell you something about what's going on in the theorem. Now, we have one of these ultraviolet divergences. That means it's happening in very high energies. Okay. Divergences happening in very high energies need to be dealt with. Now, before we start actually technically dealing with this, let us see what the main worry is. You see, what this suggests is that the Lagrangian as we wrote it down does not define... does not give us a well-defined path integral. And that is true. What it also suggests is that in order to get well-defined path integral, we need to modify the Lagrangian at very high energies. Okay. Because, you know, you get an ultraviolet divergences and you need to do something in very high energies. And some of you might be thinking, well, okay, it's not unreasonable. After all, we work at some energy scale when we're doing physics. Writing down this Lagrangian and pretending that it works at every energy scale should be a fact to see when we come to... Say, we're writing down the Lagrangian for QED if we come to the scale of the NHC. It's more effective than the other interventions. So, we write down the Lagrangian, the standard model we go to, some other scale, perhaps a plan scale. Okay? It's like, okay, fine. It's an ultraviolet divergences, some ultraviolet effect comes and cures it. And that may be... that may well be the case. But the issue is then, one, practicality. Is it the case that in order to do physics at any given energy scale, like our energy scale, we need to know everything arbitrarily, accurately. We need to know the details of what's happening on the plan scale. Is it possible that the logic sequence is installing? But it will probably be in the end of physics. Because it's very unlikely that human beings will ever get the full theory all at once. And the details of what's happening here depend, you know, the spectrum of the hydrogen atom, depends on the detail. In the detailed way, what's happening at the plan scale. In order to get the hydrogen atom, we have to know the whole theory all at once. And, okay, you know, we're limited creatures. It's unlikely that we will do that. Okay? So the issue is not so much one of, okay, this makes ultraviolet modification, deal with it. The issue is, if that's the case, then can we still have predictability without knowing too much about the ultraviolet physics? And the correct thing about one of these theories is that the answer to that question is yes. Okay? But even though we have to modify this naive definition of the theory in a way that, in some senses, that's sensitive to the ultraviolet, what we're going to try to say in this lecture is that having done that, we can hide our ignorance of what happens in the ultraviolet in a few fudge barometers. Okay? So we get a predictive theory, because one or two barometers allow us to predict what's happening in the universe. Independent of the details of what's happening, only a little bit of information about what's happening in the ultraviolet makes a way. That's the end. Okay? So this is where we're trying to go. So now let's get out of this philosophical moment and start everything. Okay? So the first thing we want to do is to take our path integral and turn it into a well-defined path today. So we're ready to point step to that by going to the Euclidean space and then the integration. But it wasn't going to happen. We see the references. So let's, just as a mathematical device, we find a regular, roughly speaking, what we want to do is to cut off all the movement at some fixed scale amount. Okay? But we want to do that in a nice and systematic way. So what we do is the following. Let's take this Lagrangian, whatever it is. Suppose we write this as a momentum space, because we have a phi of p in square root of p squared. Phi of p is p. And we multiply this by the power of k of p squared by minus p. And then we add the interaction. I'm going to allow myself to be more general than what we did here. I'm going to arbitrary interact here. There's an interacting Lagrangian to be here. It doesn't have this quadratic piece that we don't want to do. Okay? But this is this way. Now this function k here. Okay? This function now is a non-action. That's the function of the pi and the lambda. This is lambda. Lambda is an object of dimension loss. And let me choose this function k. And to be of the following form, let me choose the function k to be of the following form. k is a function of x of i. It's a function of a talcimant. It's some functions that at 1 is 1. Sorry, at 0 is 1. And it goes strictly to 0 and it has some number of order 1. That's true function. It doesn't matter. That takes this general form. Okay? And we will also choose it to be everywhere differentiable. Even though it goes strictly to 0 after this argument. You know, this is possible. What? No. This is the graph of T. Okay? Now we see why. Let's say that this is 1, 2, doesn't matter. Some number of order 1. Now what is the effect of this modification? The effect is entirely changing the problem. The propagator is the inverse of the x squared by the x squared. And the new propagator is k of t squared by lambda squared. Now because of our choice, it's k rather than k inverse because of the propagator you have to invert. Okay? Notice that for this choice of k, what happens is that the propagator vanishes for p larger than lambda times some number of order 1. Okay? So, if we were to recompute this graph with this action with the same lambda phi to the form, or we would get as a factor of k of r squared by lambda squared k of r plus p of x squared by lambda squared. Now there's no divergence. So when the arc comes large enough, we have actions. Okay? This action here, this modified action here, is called a cut-off action. It defines an infinitely cut-off path integral. Because k is 1, when the energy is much, much less than the scale lambda, our cut-off action is the same as the regular. At p much, much less than lambda, this object is 1. And our cut-off action is much, much less than the regular. That's true. 2 to the c squared is most important. Okay? Because the device is not going to show the problem. Well, whether 1 or 2 orders are the same. It's not going to show the problem. Yeah, we can make it the same order. Okay? It doesn't matter. That's true. 0 of lambda is completely out of order. It's so far completely out of order. Completely out of order. So we try to play some tricks. And then, at the end of this, you will see what? Yeah. So far, it's completely out of order. We could, as well, by the way, just, you know, put a sharp cut-off. Okay? But we're going to want to differentiate things. And it's hard to differentiate with a sharp cut-off. That's why I'm doing this. Okay? So this defines a cut-off path integral. Okay? As it was commented, the action is completely out of order. Just a definition at the moment. What, how are we going to use it? The thing to notice, of course, is that the action reduces to what the problem we actually want to solve at a low end of the meter. And we will, in the end, have a value. Define this action with some cut-off. With a slump. And you might wonder, you know, what value should we choose for this slump? As it was pointed out, there's an action. You know, maybe in some physical problem, there's an actual cut-off. Maybe in the client's case. But philosophically, our approach is that we don't only use information about what's going on at very high scales in the real physical problem. So what value of lambda will be convenient for our analysis? Basically, there's no distinguished value of lambda. Anything we do should work at all. At all. And so now, there comes an interesting question. The question that comes about is the following. Supposedly to find z to be, d phi exponential of e to the power minus s of pi lambda plus integral j phi. What is the source term here? This j's source term. So that if I want to compute not just the function, not just the integral, but correlation functions. I have all the information I need. Okay? I have all the information I need. So that I can differentiate. Maybe I'll choose actually this k function to be, let's choose it like this. Let's say that k from 0 to 1 is just 1. From 2 onwards, it's just 0. And that's smoothly interpolated in an infinitely differential way interpolated. Okay? And then everything I do, I do, I do, I do, I do, I do, I do, I do, I do, I do, I do. Okay? And then everything I do, I will choose my source function. So it has non-zero support in momentum space. Only below that. So my sources are put in the part of the problem, which is unaffected by this, by this curve. Okay? So this is my z of j and the question that you could ask is the following. Suppose I take the z of j and lambda Okay? Suppose I take this z of j and lambda and I differentiate it with respect to lambda. That's like change lambda. Okay? Will it change the z of, I mean, does the z depend on this capital lambda? And of course it does. For instance, if you change lambda, change lambda from a very high number to a smaller number, this graph, which is very high, very large, and lambda is very large, then it becomes smaller. So it's smaller. Okay? But now, when we ask the following question, you do Wilson. Wilson asked the following, or Kenneth Wilson asked the following question. Suppose I do the following thing. I can disguise and also make a function. The rest of the gradient and also make it a function of lambda in such a way that this partition function is independent. Is that possible to do? And if it is possible to do, how exactly is it? You understand the question? We've got a cut-off. We want to change the cut-off. But also change the interaction part of the function. Is it such a way that this partition function, more precisely, this partition function divided by a z of 0? What correlation functions? Okay? Are they independent of lambda? Okay? I'm going to try to solve the problem of finding the whether it's possible to change this kind of function of lambda to achieve that, and if so, how it has to change in order for that to be achieved. This question I've said. I've said to you how to ask myself. Okay? You may wonder why I'm asking this question. Hang on for a minute. The point of this will come eventually. Okay? So, let's see. Let's first do some simple algebra. So, what we want is that this thing is independent. So, we want something like a d by d lambda of 0, which is d by d lambda. That's outside. Is equal to d by d lambda to right-hand side. So, now, what's d by d lambda? Right-hand side. See, d by d lambda, where is lambda at the right-hand side? It enters in this action. And it enters in two places. Okay? The first place is that it enters because k is going to be lambda. The second place is that it enters because l is an object. So, this object here is the expectation value. Expectation value meaning path integral. This does the same action. Of what? phi of v m squared plus v squared by 2 phi minus v. Then, d by d lambda that's the first term. There's a minus sign because we're doing it into the power of minuses. And then because this minus sign has no minus sign here, we'll be right. Plus, of course, there is dL by d lambda. Which in this case, they cannot d by d lambda or lambda d by d lambda. Just to follow the difference. Just to follow the difference. So, look at it. And now I'm going to make the way that I'm modeling this object. We'll see how we can have thought of this significance. The first thing I'll make is the observations. And then we'll discuss it. The observation is this. Suppose we choose lambda d by d lambda. Suppose we choose lambda dL by d lambda which is different. We're working in four dimensions. This discussion is easy. Of what? We hope we multiply by dL by dL while we choose this L here to satisfy that equation. And if this L satisfies this equation, so this whole thing multiplies this inside the integral. If we choose this L to satisfy this equation, then we've achieved what we want. Namely, Z, the more precisely the J after over our constant, over our normalization, Z is independent. That's the kind of thing about this definition. It's the kind of thing. So we want to try to establish this. And we do this like we often do in these manipulations, we use the path of thickness, as we've already seen in the right-wing machine a nice equation. Or in the right-wing machine a nice equation. That is, we use the fact that the integral over total derivative. And the total derivative that we're going to choose to establish this fact is the volume. So what is the claim? The claim is that if we put this in here, we find that it's equal to zero. Then we state that what we want to show is that d phi of exponential of minus s plus d phi of this whole thing minus phi Yeah. The expectation of i to the phi minus p is just 1 upon exponential of this. The other term is... No, no, no, no. The expectation of phi to the phi minus p is 1 over the exponential of this. The expectation of phi to the phi minus p is true in the freezing. We get perturbations of all phi. So this is some complicated thing. And here, have I forgotten about some two pi's? The two pi's mean to us. So suppose we choose minus phi or d. I'm not sure that this is zero or more precisely that it's zero up to attempt to be absorbed to some degree. I see. The expectation for lambda can give me lambda to zero. Yes, this is the shortest way. I'm saying that this is necessary in order that this equation be true. Naturally, as I'm thinking and you're saying that you need to justify the problem. I'm going to derive this. I'm just giving you the answer. Then we're going to derive that. I'm saying that if you choose d by d lambda to obey this equation then I'm claiming that this will be true. I'm going to derive that. That's z of lambda to be independent of lambda. That's the coordination function. The computer from this path of time would be independent of lambda. You have to change your intellectual function according to this equation. That's it. It's a flow equation for this interaction equation. I think here, say that this is one of the possible changes that you can do or this is the only possible change that you can do. No, it will be only. The game is that this you might be able to do some other trivial things but it's essentially and you may be able to change the constant part of the energy because that isn't the equation. Yes. Is it even really the inverse of this? It was the problem. Yeah. The first term in this Okay. There is a reason. We'll see. So now before we get to interpreting and so on, let's derive this. So as I said, we're going to start with the fact that the integral of a total derivative matches. So the total derivative that will prove useful for us is the following. So that d4b by d4b lambda d by d1 d squared in this thing integrated d5 d5 of the inverse d squared by d1. These derivatives this whole thing will be applied by d to the power of s plus d5. Is what I mean by that? Where? This is d lambda by d dk by d1. Yeah. Wait. All right. Suppose I just look at this. If 1 is 0. Is it clear what I'm doing? This is a half integral. That path integral depends on momentum because we take the path integral with some distinguished momentum insertion. And then I integrate that path integral into all of this over momentum. Okay. This is obviously 0. This is obviously 0 because each of these path integrals is 0. Because it's a path integral of a total derivative. Now that we got this cutoff and we're in a nuclear space it is actually true that the path integral of the total derivative matches. Because any phi was large enough to get 0. It's probably going to just manage it. Is this clear? So this thing here is 0. What we're trying to do is process that object and let's see what we get. Just to open it up. Well, there are various things. Firstly, this function of derivative acting on x. We can just do it. Because it's an integral of a total derivative total derivative of some function that matches it. I mean, it's just that Are we assuming that it matches it? No, the action at very large values of 5 matches. You know, with integral. Suppose we've got integral dx d by d x of f from 0 to f. Okay, so that's... Okay. When does this vanish? I work with f of infinity and f of minus infinity, actually. But this whole object here has this e to the power minus x. And look at the power minus x. It's very rapidly. That's not a good thing to do. But I don't know about whether some spurious divergence is of the sort if you look at it. Good. Now with the cut-off. There isn't... I mean, many, many more on-shadows. Yeah, you never know if there's no on-shadows. Yeah, there's no issue. Listen. In the previous phase, with or without, this is just... Did you have a question? Did you have a question? Yeah, I'm sorry. I'm sorry. Okay. Now let's open the statement up. Okay. Because there's two functional derivatives. So let's take them one by one. Firstly, when we take this functional derivative, they only place an axis here. Okay. So, I'll just... I'll take one of them like this one later. Let's run the meters and evaluate the function. That's functional. This is d phi. Del by del phi of... I'll take that derivative later. phi of p k inverse of p squared to the lambda squared plus half to phi of p squared to the lambda squared. And then what do we get from here? From here, we get, you see, remember what s was. Right? If minus s was minus phi p p squared to the m squared k inverse by 2. By 2. Because there are two phi's, the by 2 goes away. So this term here is... This problem is not a place. Because this camera... I'll just write it. Okay. Okay. So whatever. If we have this term multiplied something, I'll just write this bracket. This bracket is equal to phi of... minus phi of p k inverse of p squared by lambda squared. And... phi of p k inverse of... You'll forgive me if I don't put the 2 phi's. The 2 phi's will all work out again. It's too much to keep track of. Phi of p k inverse of p squared by lambda squared. That's minus half. Okay. Actually, that means... Because the 2 phi's are actually... Times... Right. Right. That's one term. And the second term is plus dL delta L by delta phi of minus... I get the factor of half. Because it was a half in here. So you have to use all the half in here. You're right. This is just minus... That half is just... Thank you. I think multiple... I also have J.T.C. okay, good. But... Going to do this derivative in p here. At some value of p where there's no J. And you remember that... We have this k function that looks like that. I'm going to make sure that my p is somewhere here. Not my J's ideal. So you don't have to worry. Suppose I combine together these 2 numbers. This guy... With this guy... Is the same as this term. Except that it has... It comes with a minus half. So it just changes the factor of this to plus. Because this guy is by itself. So let's write out what we have there. We've got d phi delta by delta phi of p. Times phi of p v squared right now. That's well... 2 phi to the power 4 by p squared to the same squared. Delta by delta phi of p minus p. Times into the power minus p. Is that clear? The second function there. Is this clear? Now I'll take the second function. So the second function derivative has 2 terms. One term here... Is this active of this. That's 3 kinds of terms. One term here is this acting of this. That term is an insertion inside the path integral to the independent of phi. So once we do that integral over the speeder, it's just proportional to the path integral itself. That is telling you that apart from everything else we do, we will also have to rescale the path integral. Rescaling for not counting as significant. So we're not going to forget about that. Because we're not going to forget about that. Only terms which have high dependence somewhere matter. Now the way I'm saying it is that what we've not kept track of, in addition to this rescaling of the Lagrangian, there's also a rescaling of the constant term in Lagrangian. Which rescales the whole path integral. We're not going to keep track of that. So this term, there are 2 other kinds of terms. There's a term where this d by d phi hits this L. Or there's a term where the d by d phi hits the minus s. Let's look at each of these two terms. Let's first look at the term where the d by d phi hits the minus s. That term is just of d phi times, sorry, sorry, yes. So, so, so, so, so. See, suppose you have this L phi. In addition to changing L by this way, if you have to change it by some constant, some constant that's indeterminate of y, what will that put to the partition function? It'll just give you an overall rescaling of the partition function. This is a triviality because it cancels between numerator and denominator. It cancels between numerator and denominator in computing correlation functions. Such a term here could be absorbed into such a shift. Any change, any change like this that is phi independent can be absorbed into such a shift. And it's not to be kept right. I could have kept track of that. Change the renormalization equation to include this constant piece. But I'm only giving phi independent pieces. Is this correct? Let's look at this. So, we have this times this guy acting on this. This guy acting on this we just worked out here. It was this. So, this is... Let's put an overall factor of this m squared plus b squared m squared plus b squared by 2 pi to the 4 root. So, that is m squared plus b squared by 2 pi to the power 4. So, let's take a count plus 2 pi to the power 4 b squared by 2 pi to the power 4. That's what we worked out, right? There are two terms. There's the term for this. There's this piece and this piece. Is this correct? Now, the important point... This is why we chose everything the way we did. The important point is that when this appears with this plus this this appears with this minus this plus this. Okay? So, what we got apart from the fact that there's a p goes to minus p which doesn't matter because we're doing an integral thing. Okay? The p's are coming back, you know? So, under the integral sign it doesn't matter. This is basically the sign for a plus b in an a minus b. Okay? So, you can check that by doing the appropriate change of variable the cross-terms cancel. So, this expression here becomes m squared plus b squared by 2, 2, 5, 4 times 3 by... First time we're neglecting because that's a fine argument. Yes. Why are we neglecting? We're not neglecting. We're just not going to regret. There are three things. There's this and this. There's this and this. We're not neglecting. There's this and this. Let us also write down the number that you wanted. This is the thing. Yes. I'm clearing this up. Yeah, my sign is all connected. Anyway. So, this is the plus 2, 5, 4. Okay, that was this. We've got this overall 2, 5. We've got an overall 2, half out here. We've got 2, 5, 4. So, this will be plus if we... We've got 2, 5, 4 by b squared plus m squared e to the power minus s of the integral part of that object. But what we actually wanted was to multiply this part of the integral part by dk by d lambda or lambda dk by d lambda and integrate over or go ahead. We lost somewhere. This dk by d lambda and integrate over or go ahead. Now, so the thing that is actually 0 is this multiplied by lambda e by d lambda and integrate over or go ahead. This we see is equal to 0. It's equal to 0. So, now let's process it. You see, lambda dk by d lambda times... Okay, so first let's process. That's simplified a bit. Let's simplify a bit. So, lambda dk by d lambda times k inverse is simply d by d lambda of k inverse up to a minus sign. So, this term here is... This term here is m squared plus b squared by 2 lambda d by d lambda and put in the expectation value sign to not have the integral part of it. So, you see that this is the term. If I look at this term, this minus I'm using the formula that 1 by a dot if I want dot of k inverse that's dot of k divided by k squared by minus sign. So, I kill this minus sign and I kill the k inverse. Okay. Yeah. It's k dg dg inverse. Yeah, it's k inverse. Right, yes. That's this term. What about the remaining terms? The remaining terms we had these two guys multiplying these two things. So, that kills one of these factors. Okay. So, plus half of 5 to the power 4 by 3 squared plus n squared 2l by del 5 into n. That is exactly what we wanted to prove. Remember what we wanted to prove? We wanted to prove that if we replace this del l by del lambda by this expression, we would get 0. So, you remember what we wanted to prove? I hope that was a sign. Apart from the sign. We wanted to prove that you see, we want to prove that d by d lambda lambda d by d lambda of z of lambda is 0. lambda d by d lambda of z of lambda is 2 times. One was minus d by d lambda at the k function of the k inverse. That was exactly this. So, we don't have minus. Do we have minus? No. We don't have minus. Okay. The second term was this minus sign is this. Okay. It's what you put aside. So, the second term was d l by d lambda. Okay. So, let me multiply this whole thing by x. Minus and minus. Because something is 0, then minus 0. Okay. We've shown that this is 0. This is precisely d by d lambda d by d lambda of z. The first term was here. The reaction of the k inverse. And the second term was d by d l l. But we assume that d by d l is this. Is this clear? I'm sorry. The algebra is metairy. It's the kind of thing that's harder to explain in class than to work out in paper. Just because you can't write everything in one mind. Okay. But look. What we've done. You see, we managed. Not bad. We managed in 20 minutes. We got the signs right. We got the two parts right. Better than expected. Okay. Okay. And you say that. So, now. Any questions or comments over there? It seems to me that the full background d by d on the side doesn't play a role. No, no. It did play a role. It played a role. Do you remember that this a plus b played minus b? Yes. That was really a plus b played minus b. It was a of b plus b of b. Into a of minus b times b of minus b. Okay. Okay. The vanishing of the cross terms and that expression require the integral. Yes. Any other questions? The initial answers that we started out with like when we do that over there, it will be zero. Was that at all? Or was there any difference with the promotion? No, no. You see, what we want is something like this. We want to find that this term plus something will give us zero. Whatever the plus something is, we will identify with at del b l by del l. Okay. Since we have proved this, if I was to present it logically, what I would have done is start with the total. Never tell you what it is. I would have started with the total element. Okay. Did you use this formula? Okay. And then notice that therefore, if we chose d l by d lambda to be this quantity, z of lambda remains in the hand. The equation is like that. You can form the derivative in the derivative of this derivative. Yes. How did we get it? That's right. Yes. We wanted something that would include this piece plus something else. That was what we wanted. I mean, fooled around, you know, that was a creative guess. Of course, actually, we know the answer. I explained to you in two minutes how diagrammatically you do one thing. I'll explain that in two minutes. But, okay. Very accurately. See, this change of the derivative is very intuitive. You see, because by changing lambda, what are we doing? What we've got is a path integral where instead of integrating over a wall momentum, we're integrating over a smaller interval. We're doing this in a smooth way. But if for a moment you allow me to make a cut-off, this k function, shout, that the change in lambda essentially means that the original is a grand chain, had an integration over a small momentum chain between lambda and lambda plus d lambda. That the new path integral does not have. So, what should the new Lagrangian be? Suppose I did this infinitely. So, for a moment, suppose that we were using a cut-off of the momentum piece. Just to see this clearly. If I change lambda, I wonder if the path integral not to be changed, not to be affected. What I have to do is to do the integral over phi of p from lambda minus d lambda to lambda. That integral is going to change the action as a function of phi of p for phi of p less than lambda. Doing the integrals can be done using phi. It's just performing a Gaussian, some sort of Gaussian kind of integration because of Lagrangian. We want to do this integral. There are essentially two kinds of diagrams that we can contribute. When we remember that we're doing an integral we are very interested in going to the next one. The two kinds of diagrams that we can contribute is one. Let's suppose I've got some legs in an interaction that I would see running. One thing that can happen is that you can close the legs. What do I mean by that? You see, I had the action of a Lagrangian in which phi went over all momentum ranges. It went over the momentum ranges here which I call a low momentum. And it went over the infinitesimal momentum range here which I call high momentum. Now we want to do the integral over phi over high momentum. Insertion of two phi high momentum. I do this integral that's given by these five numbers. The phi is the answer it produces is an object with two phi is less. With two phi is less. Okay? Times whatever you get by doing the integral here over this momentum. Over the high momentum. What is this term that's how we identify this term in that? It's this term. You knock out two of the phi's that we have. We get this instead of just integral lambda plus d lambda. It's the fact that we call a small graph. This term here is simply such that what's going to happen is this. We've got two lines that are rounded back then and you do a tree exchange with high momentum that we have. Which term is that here? This this and and the this here. You can ask why not diagrams like this? Why not for instance you know how complicated that is. Why not diagrams like this times this? Can anyone give me an answer? Remember this is an example. Each of these graphs where you do a look and it's proportional to the phase space over which you're doing that. There's nothing. Over which you're doing that. Look. If you do two such integrals you've got two factors of this. Well, infinitesimally because that's how we do it. We can say that the infinitesimally these are the only two types of graphs that come together. And negative these two types come together. So what you are doing is integrating out the fields at high moment and changes the actions to the field at low moment. And the great thing is that you can do this or rather you can do this very simply. Because in fact you're only doing an infinitesimally moment to be taken by that. At a time. So we've got a rigorous derivation, rigorous non-petterative derivation of this thing. But this equation if you ask any reason of the physicists he would have told you what the answer was before finding the derivation. There clearly is a question that got the answer first is by integrating out the fields. And then you have to find some sophisticated argument to make it something. It's a very intuitive answer. Is this clear? So the question is why you chose that? You know you knew what you wanted to get. There was a new method that is also like higher over time. Yes. There's both. Both proportional to you see in this you see in this in this sharp sort of way this tree I will guide you a bit about. Because it only contributes when these momenta are poised so that the intermediate momentum will lie inside that shape. That's one of the advantages of this movement. That contributes over a major momenta but its contribution is proportional to data. So these two I Okay. Any other questions about this? Okay great. So we've derived this equation. This equation has a name it's called the exact renormalization combination. It's a beautiful equation. Unfortunately here there's been no envelope of this this equation for let's say there is not a non-million gauge would be sort of nice to have an exact renormalization of gauge theories and we discuss as we go along one of the obstacles and findings. But not a theory of a pure scale of feeling leaves it because you can generalize many scale of feelings so on. This is this is a very beautiful clean exact equation. Okay. The next step of analysis what we're going to try to do is to try to understand okay we've got this nice equation. So far of course for both of you this is totally unmotivated. Who cares that we have this equation? What does it have to do with me or you? I sympathize with that question. It's coming. Okay. Now I'm going to draw everything on the body except the equation. Okay. So we forget about the equation. We look at the answer. The structure of this answer. This answer is the first this object is the first order differential equation. Let's let's call lambda is e to the power e. So e by d lambda lambda e by d lambda. So in renormalization group flow sort of quote unquote time this is nothing to do with time in space. Just observing it. This is the first order differential equation. What's the structure? It's a structure that tells you that if I know what L is at any given time I will I will predict what L is at every future time. Is this clear? It's like an evolution equation. Notice that the right hand side depends only on L at that time. It's explicitly on time. There's no explicit factors in lambda. That's true. There's a k Okay. It's true. It's true. It's true. It's true. It's true. I said that wrong. There is an explicit factor of lambda and k which as we will see is basically true. In the non-trivial models No. Yeah. Yeah. There is an explicit factor there in that k. In this part there isn't and that will be enough for what we need to speak out. Okay. This is an equation that tells you if you know at some time what your Lagrangian is you can predict it in the future. Now suppose you take in your future meaning lower scales lower lambda. Suppose you take your Lagrangian Okay. Maybe I should have so that increasing times decreasing scale. Okay. So suppose you know your Lagrangian suppose you know your Lagrangian at some time and it's very simple Lagrangian. That's just 5 to the 4. At later times we're going to because this term here will generate a 5 to the 6. Then once we generate the 5 to the 6 term we'll also generate 5 to the 8 5 to the 10 terms and so on as we teach them. Moreover this term will feedback on the 5 to the 4 terms. Also the Lagrangian we started with could have very simple momentum events. It was just local in 10 to the 5 to the 4. Well because all these factors of momentum floating around we will generate very complicated momentum events which in position space will involve higher derivative terms in the action. So that's clear from the structure of this equation that no matter what you try to do initially if you care about the structure of the action at all times there's no particular simplification. It's not consistent as you in the action take some particular truncated form okay so you so our action as we flow in this space it's going to be arbitrarily complicated. Let's have a way of parameterizing it. Let's work in momentum space okay. So let's say the 5 to the 0 to the 5 so if we click this action here this definition of the Fourier transform this is what we're using all along okay okay now L this interaction 5 can be expanded first to data series in a number of 5 sessions okay so sum over N ln of 5 where ln by 3 will do some coefficient n p1 p2 pn delta function 2pi 4 and delta function p1 that's pn 5 p1 5 pn so the most general form it can take and there's no reason the selection won't take this most general form for arbitrarily complicated functions of the c okay the one thing though that we know is that because if we start with anything crazy the thing is that we're generated we always have always working with the counter and we work you see everything here can be expanded in in a power series moment these expressions these terms where they are will admit Taylor series expansion in piece so one way of parameterizing these is to set one of these momentum to be determined by momentum conservation and Taylor series a n1 n2 nn minus 1 p1 to the power n1 p2 to the power actually probably p1 n2 actually we should have just tried to the mu in this here because p is our momentum so n1 mu1 to the power n1 mu1 and so on I mean are you saying it's are you saying that you've been complicated Taylor series no I'm just saying I'm not saying you've been complicated I'm just saying that there is no obstruction to performing there is no non-analytic because all of it means momentum dependence from these propagators okay and they all have support only in the range in the integral which is powerful that's basically so p1 mu1 to the power n1 mu2 writing down some parametrization of this what this is so suppose you've got p1 mu1 to the power n2 p2 mu2 to the power n2 the coefficient of that term I call a n1 mu1 it's not a function of pn because what are the means of the n's mu's are the n's are the what are the n's n is just how many mu's are the regular n's p1 p2 p3 this has to be this has to be your n's it's true it has to be except that these are all very intense okay so we could try to parametrize that more seriously doesn't matter okay all I wanted to do was to notice that the data and this function is in terms of an infinite number of numbers infinite number of numbers are the various Taylor coefficients or variance so there's some high discrete infinity number of numbers the various Taylor expansion coefficients that completely characterize this this end end now one thing that's of interest to me is to find the scaling dimension to find the engineering dimension the dimension of each of these numbers okay I thought the dimension goes the only thing that matters is how many different n's are p's you have okay so I have this I have not been careful in parametrizing should put variance in there none of that's what matters let's suppose that the coefficient that we know a coefficient which is a n which means number of i's and d which means total number of n's total number of d factors okay I want to know the dimension of that coefficients so let's do some counting the dimension of phi of x as you all know you see this is not the kinetic d phi the whole thing squared times d4x has to be dimensionless so phi is dimensionless that's the dimension of phi that means the dimension of phi of p is equal to minus p because phi of p d4 p is phi of x so the dimension of phi of p plus 4 must be the dimension of phi of x okay now this term here enters this Lagrangian so the term with n of i's appears in a Lagrangian with firstly n minus 1 momentum momentum integrations n minus 1 because it's not okay so let's some of you have dimension of a n comma d whatever it is so all the dimensions of the momentum integrations the dimensions of the momentum integrations are n minus 1 times 4 plus the dimensions of the phi's so that's n times minus 3 plus the dimensions of the explicit factors of momentum plus v is equal to 0 because the whole Lagrangian is dimensionless in any term of dimensions that is the dimension of its coefficient okay is this equation clear? dimension of coefficient plus dimension of the integration factors plus dimension of the explicit dimension associated with h phi of p plus dimension associated with n times is equal to 0 so this tells us this tells us that a of n d is equal to minus 4 minus a plus 4 but as a sanity check let's suppose n was equal to 4 and d was equal to 0 we had 1 million phytophonic dimensions that should be dimensionless that is or the kinetic term n equals to d equals to is dimensionless it is the mass term n equals to d equals to 0 should have dimension 2 because this formula works okay now the thing about this the thing to know about this formula is that almost all of these quantities have a negative all of these coefficients have a negative dimension which ones don't the only ones that don't are the kinetic term the mass term and phytophonic let's suppose we heal to the theory where we got 5 was to minus 5 so we can never generally ordered number 5 so the only ones that don't have are kinetic terms so so this n is equal to d equals to 0 that's mass n is equal to so d is equal to so 2 because d equals to 1 is equal to that's kinetic term the only terms that you can write down is the dimension that has where full coefficients can appear and can appear in this in this order of this equation is one of these equations what if we take time to find q then we would have got we could have had mod we would have had 5 q would have been a most of the range let's know yeah that would have been okay now we see we do the following and I'm going to ask another question okay let me first ask the question there the question that you might ask is the following a question you might ask is suppose I come something in the grand show with all these terms and I already have some process I want to measure some dimension this quantity probability that you scatter between angles 0 and 45 in some scattering experiment at momentum at two parties at some other momentum p I want to know what is the chance that you will go and scatter between 0 and 45 0 and 10 something dimensionless question takes dimensionless question okay takes dimensionless question and you ask okay and you ask suppose I've got some Lagrangian with these parameters and you are very naive you haven't done the kind of sophisticated analysis we've already done and you are naive and you ask how does the answer depend on these various parameters how does the answer depend on these various parameters you might have guessed that the answer would depend on these parameters by dimensional analysis okay that if you have an object here which has this negative dimension okay now the dependence on a n p would come with your momentum scale in a manner the momentum scale to about 4 minus n minus a p that this would be the effective parameter the effective parameter determined how important that a and d was in the answer answering the question that was happening in momentum scale p okay and then you might have concluded that since this sorry since this number now is positive most of these guys if you take p goes to 0 the limit of low energies most of these characters will have no effect on any energy compared to you know the scale set by these by these parameters the point about the analysis that we we've done is that the the the the the analysis that we've we've we've done so far is that this this is 2 9 this is 2 9 because there is one more dimension scale in the problem which is the cutoff we're interested if we're in a theory momentum scale cutoff down down some scale we're going to measure p and there's a big gap between p and not okay then the dimensionless combination that appears here A and D that's lambda to the star that's a problem A and D times p to the star so it's lambda as much bigger than p these quantities could make an anomalously anomalously large contribution to any physical effect that is not seen by dimension analysis okay because there's a new dimension scale in the problem now Wilson's main idea Wilson's main idea was suppose we take this reorganization group seriously and we actually if we're interested in some process of momentum scale p let's actually do the integral run your Lagrangian so we lower the cutoff down to momentum scale of order p then whether you're not dimensionless these A's with lambda p it's more or less the same thing because then the same order so then A times p is a fair measure of the effect of this term on physics okay there is no hidden dimension large number namely the number of lambda divided by p in your problem if you're interested in doing a calculation at scale p it is useful to work with the Lagrangian where the cutoff is approximately could be twice p but approximately momentum scale once you've done that this is a fair estimate of the effect of such parameters on your physics okay since we are going to since we are going to since this is the attitude we're going to take in dealing with these parameters okay five minutes in dealing with these parameters in dealing with these various parameters we may choose except in dealing with these parameters we will choose a n d times lambda to the power minus 4 plus n let's call these guys x n these are dimensionless numbers these are the guys that determine the there are fair measure fair measure of the impact of these parameters a and d on physics and a particular scale namely on scale lambda okay somehow one program might be the following one program might be the following take this equation here and rewrite it as a set of flow equations for the numbers x n d okay what do these flow equations look like as we have seen almost all of these x so what are these flow equations so suppose we write d by d lambda lambda d by d lambda of this value because of this explicit factor of lambda we get lambda d by d lambda of x n d is equal to n plus let me do let me do what I said before because lambda is equal to the power minus and look at the flow towards the insurance flow of positive so d by d d of x in the is equal to minus of n plus d minus 4 okay that's just a trivial part because we made this this readout and then plus whatever we get plus whatever we get from here I'm sure the answer is 4 because otherwise there is a large number in the problem on which things can be done meaning suppose you had a Lagrangian at scale lambda and you knew all the parameters in the Lagrangian at scale lambda okay that gives you an immediate estimate of their effect of business on a process at scale lambda x n d's are a measure of how important they are at scale lambda however suppose you had x n d's at some scale lambda and you already know what the effect of effect of that was on a process at scale at scale lambda you are not sure because there is a dimension in this parameter and the lambda divided by lambda which could appear in your process to any power okay there is a hidden scale in the problem that could completely ruin all estimates of how things should be done once however you brought your top scale down to your momentum scale there's no meaning there's a large ratio of scales left in the problem that could affect estimates you know nine estimates of them so I'm saying x n d at scale lambda is a fair estimate of the effect of that term on a process that happens at scale lambda x n d at some scale lambda is not necessarily a fair estimate in fact many examples that's very important it's like a QC the coupling constant can be very small that's a very large scale there's not a ratio of how important it is for scattering and scale down the scale of what the scale of the momentum can be scale down what? beta yeah beta here it's just some arbitrary function which doesn't have n or v yes which doesn't have n or v some arbitrary function that has no yeah now you see let's get back to this question you see this lambda here entered only p squared by lambda so now we make and that p squared is integrated so we make the change of variables we make the change of variables p goes non-dimensionalized momentum lambda just this this okay so this equation here when you write it in terms of these these x x d's what is the following structural form the equation has the structural form v by v d is equal to minus n plus b minus 4 x and b plus p down n d of all beautiful kind of equation every quantity here is dimensionless the information of dimension is all going to this one this one object this equation the first term I can understand because as you see the first term comes more and more irrelevant to x comes more and more irrelevant but the second term is not that simple because it's whatever it is it comes out of here right you just expand this one yes because I'm not familiar you get something yes but that as you see that might actually negate x it might it might it might no no no we don't analyze it the structure of these equations okay you guys okay are you exhausted you have something to do can I hit them on sure I don't I don't want you to be exhausted if someone want to cast a casting quote that's going to be slightly good like unless somebody says okay I'm not I'm not I can't absorb anymore we just go on again because I actually want to rest to discuss the subject as fast and interesting and intriguing this quantity so let's give us a little extra time we've got these equations like this and what we want to do is to understand 400 what we want to do is to understand how these what these these flow equations okay so now we're going to follow Pultz's view this is of course an infinite dimensional set of equations this is an infinite number of variables but many of the important conceptual features of this equation can be seen in a toy model in part of two such variables as we've already already remarked and as we will see in more detail as we go along variables come in two sorts there are the marginal and the relevant ones and then there are the irrelevant ones so we will make a toy model of these flow equations where we keep only one module of variation and one invariant okay and analyze that toy model and you'll see many of the features that are of interest to us let me do this analysis because we won't get to the punchline on this first see x and d is A and B lambda or something where lambda has been taken with that d d x and d gives I can see the first and I can see the second second term is whatever you get from here what you're supposed to do is to plug in what this Lagrangian is in terms of the A and B not that introducing that you'll see this perfection plug that in and you guys what I can say we're not even going to explicitly work it out explicitly that's a horrible mess this is an experiment to write that equation but you can see the structural form is this that's all we're saying actually the way I've written is here structurally you get by linears and linear terms so structurally you look very simple that will that hides the fact that there are infinite number of variables if you try to integrate something yeah it's not it's a complicated non-linear system okay so now in order to get a feel for the flows of these these quantities in order to get a feel for the flows of these quantities we're going to follow Gielski and analyze a simple drawing of course we get just one object of dimension 4 without an x4 one object of dimension 6 without an x6 for instance it would be the final 4 and the final 6 okay then the flow equations now d by d above x4 is equal to beta 4 of x4 and x6 and d by d of x6 is equal to minus 2x6 plus beta 6 of x4 these two flow equations do is is is the following what we're going to do is the following you see suppose we have this x4 and x6 base let's say x4 is the x6 we stand somewhere in the space these flow equations separate some of the lines and all of these separate we start here flow equations generate some lines we stand somewhere else and the flow equations separate some of the lines or somewhere else we get another line what we're going to demonstrate is that under appropriate certain circumstances these flows are convergent tools that is these reorganization flow lines tend to a single work work on a single if we wait this whether this happens or not whether the details are not we will find explicit conditions under which I choose some x4 and some x6 and some line and then we use these equations today because I don't care about this is something this is some other thing it's like a trajectory like when you plot an orbit it's true that orbit is a function of time you could just plot the orbit and we could keep trying to find out what time we're just using the orbit this could be the main point the main point is that these flows are going to be basically like convergent some consequence of that would be the orbit suppose we start here with flows which is zero but at different values we've done many different things I'm choosing this thing to do because that's what we do when we study reorganization so the invariant statement is how it's merged into each other but I'm going to actually demonstrate this statement by looking at a particular fact now as we start with flows whose all start off as x6 equal to zero but at different values of x4 this is the point that was related to to your comment if we looked at where we went at a given time this flow may have at some time this can be taken as here and this can be taken as here we may be quite far away there's no time we need to compensate by flowing a little extra there is a point on this flow where the points are varying that's what we're going to demonstrate now how do we say that invariantly what we're going to demonstrate is that what we're going to demonstrate is that if we choose one way of saying this in the book if we choose the flow lines we choose the flow time so that the two points are the same value of x4 I've wrote this guy to some time then I've wrote this guy to whatever time it takes to make the value of x4 if it's in x6 I'd like time to turn out to be reasonable this demonstrates the merger of the flow lines do you understand that so in order to understand this phenomenon what I want to do what I want to do is above I was trying to study this phenomenon but for infinitesimal changes of the initial initial additions it means that this flow line is so much so x4 is what x4 is the variation it indicates the presence of high-core binoculars in x6 because it's a high-six because there's very little very little and then when you start with the x6 which is zero or to begin with yes which means that there is no x6 multiplication no no it doesn't even there is no x6 we start with the x6 0 and x6 doesn't it's not 0 no no that is generated with the overtime the contributions right and then it's decaying then no but it doesn't mean it's a zero it just means it just means that we will explain this in detail let's first understand the phenomenon and look at the simplification that's the main point of it but it means that that it means that we have a one parameter inside of this let's understand this let's wait wait for this first let's just understand the properties of these flows and let me understand what it means so is what I'm going to try to do today I'm going to try to study different flows parameterized by initial values which are x6 equal to zero and x4 equal to zero this is an arbitrary choice okay so one parameter initial conditions chosen these initial conditions because it's what we usually do when we study to normalize okay and then what I'm going to do is to vary those initial conditions infinitely and once I vary those initial conditions also varies the time of flow so as to ensure that both flows sit in the same x4 and then once I do that I will then study the change in x6 as a function of time and we write down an equation for that change in x6 that is our game any questions or comments we'll see we'll see that's quite a theory so let's go ahead and do this what we're going to do so suppose we've got some lambda for which is a function of lambda 4 and it becomes x x4 which is a function of x4 the initial value of x4 that's this guy of D and of D and we've got some x6 which is some x which is also a function of lambda 4 or not and D that's one of these curves okay so that's how we do it so we are setting D is equal to 0, 5 and then it reacts where x is going to that's it yes okay yes yes let's say that we choose a convention that P is equal to 0 and then we choose this curve here for instance is the curve drawn out like that I've written on the board what I want to do is to make a variation I want to vary both T and x4 in a way that does not change x4 at T okay so I want to move to a different curve that's this one but also to a different time so that x4 stays the same so I want to choose some so some delta x4 0 and delta to ensure that this guy x4 does not change okay now of course there is a particular if we choose delta x4 by delta x4 0 delta x4 0 delta x4 by delta T delta T okay this variation I want to change starting the change of the variation of x6 what is the variation of x6 which is because the zeta 6 delta x6 which is delta x6 by delta x4 0 delta x4 0 plus delta x6 by delta T delta T I want to study there is a function of changing x1 so I eliminate delta T from this one okay so zeta 6 is equal to delta x4 0 into delta x6 by delta x4 0 minus delta x6 by delta T times delta x4 by delta delta x4 0 over delta x4 delta T minus delta T this is the definition of zeta zeta 6 particular flow line I want to try delta x4 0 about it I will find the small variation zeta 6 what it means is given in this diagram the change in x6 at 90 where the where the p is the flow along the initial line you flow a little more on the on the perturb line source to ensure that you have the same x okay that's what that zeta 6 is now what I am going to try to do is to compute an equation for zeta 6 that is I am going to compute zeta 6 e by d zeta I have got this little vector I am going to see how it evolves let's say this is what I am going to try I am going to try to find the conditions under which that goes to 0 so the flow lines approach it I am going to find the differential equation for zeta 6 the great thing is that you are going to find the homogenous equation the right hand side to find the zeta 6 so let's do it just a matter of fact okay delta 4 x0 is a constant it's a small number because that is not a function of time so what are the functions of time in this equation so this quantity delta number 6 delta 6 delta x6 by delta 2 delta e of course is given by our basic formulas it's a beta okay so we have to x4 dot is equal to beta 4 and x6 dot is equal to minus 2x6 plus beta 6 in this algebra it's very inconvenient to differentiate this because of the beta 6 combination and the later substitute beta 6 okay so zeta 6 let's first try the formula zeta 6 is equal to delta x4 not into del x 6 by del x4 not plus beta 6 over beta 4 and del x4 definition beta of zeta and now I want to compute zeta okay so in order to do that I have to understand two things I have to understand how to compute beta lots and after that I understand how to compute these things beta lots have a history of flow because beta some function x1, x6 which is j more confusing initially is this object it's delta x6 by delta x4 not how do I think of that a way of thinking of places and of places all those suppose I've got any infinite I've got some flow and I've got any infinite variation around that flow you can see how those infinitesimal variations change in time by differentiating by just very precision any del any infinitesimal change which delta x4 not is equal to del beta 4 by del del x4 delta x4 plus del beta 4 by del x6 delta x6 and similarly delta x6 not if you put del beta 6 by del x4 delta x4 plus del beta 6 by del x6 delta x6 obviously any small variation around the given around the given in particular it goes for the variation that you get by doing the small change in the initial condition x4 so in particular it applies when delta x4 is equal to delta x4 not and it is delta x by delta x4 not by 6 these are just constants ok so applying these formulas to these let us tell us what d by dp of these quantities now suppose people we have you see what was this x4 what was this x4 x4 right because even the x4 as a function of d and x4 this zeta here is just this stuff evaluated at x4 not and d to differentiate that object as a function of d not as variable by varying is keeping g4 constant is keeping x4 all the definition of zeta 6 then zeta 6 zeta 6 this is a function of d true but then zeta 6 the definition of zeta 6 is not right no no as this derivative a change of x4 as a function of d we have we have we have changing the trajectories what are we doing we have one trajectory and we have another trajectory this statement is true differences in coordinates between two different trajectories as a function of d now I am looking at a a particular example of this let me take true neighboring trajectories to be the trajectory labeled by x4 x4 not and the trajectory labeled by x4 not plus delta x4 I am just applying the formula of this variation of trajectories is this clear so if we want to find the time derivative of this without any other derivatives we just use the formula is that clear I know it's confusing okay these things are always clear not clear tell me what sir sir this is like a Taylor expansion where we did the first product already let's take this actually that's right but there confusion is not of this formula this is applying what is between this which is okay this delta x4 is the formula this delta x4 is x4 of x4 not plus delta x4 not and d this one this one minus minus x4 x4 not and d that is this formula only holds for such a variation two trajectories okay this therefore applies to this object because this delta x4 not is a constant so we conclude del x4 by del x4 not dot is equal to l beta 4 by del x4 l x4 by del x4 not plus del beta 4 by del x6 del x6 by del x4 not and we conclude the del x6 dot by del x4 not this is equal to the same formula we will replace beta 4 by del x6 del beta 6 by del x4 del x4 by del x4 not not del beta 4 by del 6 by del x6 del x6 by del this formula for this dot dot means a constant x1 the x6 and the beta every 6 beta will you publicly not to write dptas also ok ok now that we got all that we need let's write down an equation beta 6 dot is equal to what? well it's equal to del x4 not now wonder if we just get directly for you del beta 6 by del x4 not plus del beta 6 by del x del x6 by del x4 not that's wonder the second we get directly from here plus beta 6 by beta 4 into del beta 4 by del x4 del x4 by del x4 plus del beta 4 by del x6 what difference do you see in this f matter of course we just get from the j so one term is let's say the beta 6 so del beta 6 by del x4 by beta 4 plus del beta 6 by del x6 by del x4 by del x4 not but the last term we get by difference in the beta which is minus beta 6 times beta 4 squared and del beta 4 by del x4 beta 4 plus del beta 4 by del x6 6 times del x4 by del x so the first term is beta 6 by del x4 not del x4 the first term yeah that's del beta 6 by del x4 thank you now we've done the algebra right it should turn out to be the case that the terms with it's not going to be the case it should turn out to be the case that the terms with del beta 6 by del x4 and del beta 4 by del x4 let's check that let's take look at all the terms here proportional to del beta 6 by del x4 del beta 6 by del x4 here it's del beta 6 by del x4 okay can anyone see another term here the part I'm thinking ah yeah where is my minus where is my minus that beta 6 by del x4 was minus thank you thank you okay so now we need to fix this let's see what what have we got wrong this one was just correct next one was minus so now this at this cancel each other because beta 4 by del beta now should also be true that del beta 4 by del x4 yeah let's check that del beta 4 by del x4 one term is here and one term is here beta 6 by beta 4 square beta 4 ah del x4 by del x0 cancels right yes so this ah and this and let's see what they evaluate cancel let's see what they evaluate you see so let's take the term that is del beta 6 and del x6 what is number it multiplies well from here what we get it was of course there's an overall then from here we get del x6 by del x4 not okay so we get the term and del beta 6 by del x6 yeah minus beta 6 by beta 4 del x4 by del x what about the term that is that is multiplies what these two terms ah so this term here multiplies and I just say multiplies multiplies we come to the conclusion let's not tell this this zeta 6 dot is equal to del beta still done all the things del x6 ah minus beta 6 by beta 4 del beta 4 by del x it's quite beautiful equation now plugging it forward beta 6 still done this plugging it forward beta 6 still done this the terms which is minus 2 and then there was beta 6 so zeta 6 dot is equal to minus 2 beta 6 plus del beta ah ah this was del beta 6 by del x6 by the beta 6 by beta 4 del beta 4 by 4 now we can solve this this ah this problem by integration zeta so beta 6 still done ah beta 6 so what do we get? do you see how do we solve a differential equation like this? ah how do you solve it by by by by by by by by by by integration so zeta 6 is equal to this quantity zeta. Just by taking zeta 6 down here suppose that I want to find zeta 6 times zeta 2 to the power 2 t zeta is called as x it is what you are going to do x. So, here d x by d t is equal to this bracket like this. So, log x is equal to the integral of that bracket x is equal to e to the power that integral of bracket zeta say x is equal to e to the power minus 2 times e to the power integral 3 plus x to the power 3. Now, I will give this formula what this term tells us is that zeta 6 just decays. This term tells us that there are corrections. Unless these corrections give you add up to some time. So, this term tells us that zeta 6 is equal to e to the power 3 plus e to the power 3 plus e to the power 3 plus e to the power 3 plus e to the power 3 plus e to the power 3. This is not true what is true. However, is that a damping term makes the flow lines merge to a particular so what that means is let me take a allocation of initial addition one specialty the static points. And we tue the flow times to bring x 4 equal then x 7 Then the x-axis whatever they are function of time flow you know decay to 0 the difference in x-axis as we take a length time that is the effect on these flow equations of this damping term. It is not always true it can be over by it is not linear but it is true these things of it are small enough basic point clear what we are going to I probably will go through the inequality. So, the people are actually goes through the proof in the quantum field in case I will give you the example like clearly let me take one this is the irritating inequality but what we will now argue next time is that the same thing happens in quantum field that basically in this infinite dimensional parameter space all flow lines focus in on a at least in particular on a very small sub map since they focus in on a very small sub map since they focus in on a very small sub map ok we can it does not matter how our definition of the theory was tales of UV physics do not matter because as we have explained in order to understand physics at scale lambda it is sufficient to know the effective action at scale lambda it is not just sufficient it is like efficient it is both sufficient and efficient. So, if very widely differing starting points all flow to the same effective action up to some two parameter set then in order to understand what quantum field theory again we just need a labeling of this of where we are on this two parameter set of attractor flows. And one way to generate that is to start on an arbitrary two dimensional set two parameter set in a unit take to this two parameter ok and that is what we do when we start to think ok this is the thing I will try to explain to you in much more detail in next class. We have got all the algebra of this so the next class will be more fun. So, the question is can you flow to the UV rather than the IR? That is a very confusing question but from this point of view of course you might think why not. Actually let us the first thing to say about that is that even at the level of differential equations it is a much better posed problem to flow to the IR rather than ok. Because as we have seen changes in initial conditions damp out towards the UV but blow up towards sorry towards the IR but blow up towards the UV ok. So, it is a you know it is a little like trying to solve the equations of hydrodynamics discuss hydrodynamics in reverse time you will quickly get blowing up solutions. And that is basically the problem the problem is that if you flow towards the UV ok you might well find that in finite time you reach that you actually get the UV. Let me give you an example of an equation that we did. It is an example that comes when it comes about when we try to study that the renormalization would flow between. Of course the equation is such that B by 15 of 1 by x is equal to 1 by x is equal to ok this of course you can write equal to 0 as x dot is equal to A x square. Just by breaking up the negative x right. Now suppose you are going to flow like this by a couple constant x there is in finite time suppose you start with some particular x dot. Then in finite time namely t equals from namely 8 t and t equal to 1 by x dot this quantity 1 by x is equal to 0 and therefore x would have blown up ok. And that is of course some unreliable something. So the danger if you flow towards the UV is that you will hit singularities that these equations though well defined will give you some singular solutions at some point. This not only can happen but does happen. Sir. Sir. And it is not only can happen but does happen and usually indicates the following. You see the difference between the UV and the IR is there. It could well be that in order to get you know a sensible theory of the form that we have got you need some new degrees of freedom in the UV. You can let us say a new massive field of mass m that is larger than your auto that you integrated out to generate your rectum engine. Now when you are flowing up how procedure will never allow you to integrate in this massive field. It will just try to produce an effective action with the degrees of freedom that we have and not be able to do that because you need that additional matter and hence give you some sort of signature which is some sort of signature that something is going on. Okay. And flows towards the UV while formally defined could often suffer from such processes. Flows towards the IR and usually much better be it for the reasons that it is engaged with the flow towards quickly towards some universal flow lines. Okay. And typically you do not have a hit singularities except for the deep in the IR. Sir. Sir. There may be more to say. Yeah. Sir. Sir. Sir. Sir. So a single theory which is to find a lower which is some detective theory who can actually be object from a number of theories in an unique max so there is no unique theory which would flow. Okay. That is one way to say. However you know you might ask how do we see this from our differential equations. Yeah. You might ask how do we see this from here so procedure we used to generate is differential equations. But the differential equations themselves saved it to allow for evolution both forward as well back. The signature of trouble from the point of view of the differential equations. But I believe we see the realities. We should try to solve this. We can also come to a large group of operators in the lower-meditary which are irrelevant. Which are irrelevant? Irrelevant. Yes. Stop. So that could signal that you don't have any theory with me without singularities. I mean, you can't think of both ways. I didn't understand. I mean, you can have a large group. You can have different theories. Different operators of irrelevant operators are low-energy. And you have different course of moving. And you can basically... Irrelevant operators just drop out. Irrelevant operators, we have to look more and more important. Irrelevant operators, operators that naively are very important to high energies but are not important to low energies. Yeah. We will let's talk more about that. Sir, in terms of the equilibrium in terms of a reflection. Because in that case, as we go to the other place, the trajectory is sort of... Reflection of what is happening over here. Reflection being what? I mean, early... See, just see the reflection of the trajectory that we have plotted reflection. In an uncertain direction. So, what we have done is, like, the trajectory sort of merges as DT. Like, at least... Yeah. So, in the equilibrium it's like... It's an opposite thing happens. The trajectory is divergent. It's more symmetry depending on... So, we plotted on the x for x. So, if we just look at the... Reflection is what? They're twice as much as before the other... As x decreases. No, I'm sorry. Sir, I don't know this. We have to say something about this. This diagram will go. What reflection are you going to make? Yeah. Consider the... If you just imagine a mirror... Wait, let's... If you imagine a mirror, like, to the left of the entire axis. And you consider, like, the reflection of this thing. So, the merging will occur. But you're saying the beauty will take... No, no, let's go. Let's... Any other questions? Okay, so, in the next class, which is next, is Monday a holiday? No, no. In the next class, which is on Monday, because now that we've done the hard work, we will try to understand the implications. We've done... We're defining what will appear there. Basically, we're almost there. We've more or less understood how to define what will appear there. But we're not set with... Okay.