 I am. I am indeed. Yeah. Okay. We're going to start recording then. All right. So we're very happy to have Nima deliver the last lecture. Okay. Okay. Everyone. Good morning again. Um, uh, and, um, actually I wanted to begin by, um, uh, by, uh, uh, by making a comment. Um, as I mentioned a couple of times, um, uh, uh, I'm, I'm going at the roughly, uh, half the pace that I was, um, planning on for these lectures, uh, which means that the, even though, um, I had a, I had a sort of complete story that I wanted to tell that would, uh, have by today gotten us to, uh, actually talking about, um, uh, you plus or minus colliders and proton proton colliders and muon colliders, maybe, uh, is instead putting me today in the middle of, you know, finishing a rather hopefully interesting, but sort of, uh, formal set of, uh, uh, topics about, um, uh, the inevitability of the laws of nature. Um, so, uh, I would, I want to make the following comment. If there is any of you who are actually, um, interested in, uh, um, the, the sort of last part of this, uh, uh, discussion, um, as well as just sort of more, more, more generally, uh, uh, more of a conversation about, uh, some of these questions about what's, uh, special about the Higgs, what the hierarchy problem really means, um, et cetera. Uh, what I'd like to do is the following. Um, uh, those of you who are interested in this, um, please feel free to send me, um, an email, uh, and, um, uh, uh, uh, uh, the following week, um, is a little bit, uh, uh, crazy for me, but the week after that, uh, is more or less, uh, open. And so I encourage all of you to just, um, who are interested, um, if there is a time in the, uh, in not next week, but the week after next, um, that, uh, that works for you to be able to sort of get together for some, um, informal continuation of these, uh, lectures. Just, um, email me. Um, and what I'll do is just whoever emails me, of course, zero pressure, but whoever emails me, I'll sort of collect, um, uh, just, just email me with the, with the, uh, some dates in that week that, uh, don't work, um, uh, or maybe better, some dates in that week that do, then I'll try to take an overlap of whatever the group of people who emails me is, and I'll take the overlap of, uh, as many of the dates that work as possible. Um, and I'll make a proposal for, uh, for, uh, a day of, you know, maybe a couple hours or a more open-ended set of discussions after that so that we can properly finish what, um, at least somewhat properly finish what I had planned to, uh, uh, actually get to in these lectures. So anyway, uh, zero pressure. People who, um, don't want to do it, uh, obviously I'm not gonna, I'm not gonna force you, and it's all very anonymous anyway, so it's not like, uh, I'm taking notes about who is and isn't showing up, but if you do want to, uh, um, have this discussion, uh, uh, do send me that email, I'll go through everything, I'll overlap the list, and then we can, um, uh, we can, uh, we can do this, uh, uh, not next week again, but the week after next. Okay? Yeah. And so my, uh, email address is just, uh, you can also see it on the web, but it's just arcani at ias.edu. Okay? Um, okay, so I encourage you to, uh, uh, do that. Again, just send me an email with, uh, what times, not next week, but the week after next might work for you. Um, hopefully, uh, if you can say what, what, what those times would be, if you can translate them to Eastern time, my time, so that, uh, I don't have yet an opportunity, and yet another opportunity to screw up, uh, that, that, that would be great, and then we'll arrange this to happen. Okay? All right. Thank you very much, and okay, so let's, uh, let's get to the, uh, let's get to it today. Hmm. I remember what we're trying to do is, um, uh, understand something about, um, the possible consistent theories of, uh, massless particles. So what are the consistent theories of, uh, massless particles that we can have? And, um, uh, our sort of journey so far has been in lecture two, roughly speaking, we learned what are massless particles. How do we think about them? Um, in other words, um, they are irreps of Poincare, and they turn out to be labeled by Helicity in addition to Momentum. And in lecture three, last lecture, we, uh, talked about, uh, what you might have thought of as the sort of first non-trivial interactions could be, um, which are, uh, when we have massless particles with massless particles with Helicity's H1, H2 and H3, um, we worked out that these guys are determined, fully determined by, uh, the structure is fully determined up to strength, up to coupling constant by, uh, Poincare as well. So despite the fact that, so this begins to look like, um, it depends on, uh, you know, maybe you have to make choose a Lagrangian or, you know, talk about how you're gonna write down, uh, the field and contract all the indices and all that kind of stuff. In fact, um, it's also totally determined just up to the strength of the coupling constant by, uh, Poincare. And, uh, and, uh, we, we started doing this, um, by, uh, introducing also the spin or Helicity variables for massless particles. Okay. Um, and, uh, this representation on the one hand makes it manifest that P is null. This guarantees that P squared is equal to zero. Uh, but on the other hand, uh, uh, precisely because, um, under lambda goes to T lambda and lambda tilde goes to T inverse lambda tilde. P goes to P. We discovered that this is in fact the simple action of the little group and therefore amplitudes are really functions of lambda and lambda, lambda tilde. So, so just, uh, if you, if we have an amplitude for n particles, massless particles, uh, with Helicities, with momenta labeled by, uh, spin or Helicity variables, lambda, lambda tilde and Helicities HA, then, um, uh, then, uh, the group theoretic property of this is that, one, uh, it has to be a function of only these sort of brackets like lambda i, lambda j, uh, which we called ij or lambda tilde i, lambda tilde j, if we called the square brackets ij. So these are the only Lorentz invariants that we can have, but two, it has to have a particular so-called Helicity weight. So if I rescale the eighth particle by, uh, TA and TA inverse for the lambda and lambda tilde, uh, the whole amplitude has to pick up a weight given by TA to the negative 2 HA. Okay, so, so, and this now contains all the information, the kinematic information that we have about, um, massless amplitudes and the correct, uh, the fact that, uh, when you do a Lorentz transformation on the momenta, they have to pick up a little group phase, um, as we discussed, uh, those things are not, um, are only redundantly described by polarization vectors and momenta, for massless particles, because the polarization vectors don't exist, they only exist up to something shifted by the null momentum of the massless particle, but they're directly functions of the spin or Helicity variable. So, um, as I said, if you've taken a, uh, if you've taken a field theory course, you've probably never actually seen an amplitude before. You've seen some expression, um, with things dotted into polarization vectors, but you've never actually seen the actual amplitude, um, uh, do actually see the actual amplitude when we work with these, uh, spin or Helicity variables. Okay, so, um, and to remind you where we got to, uh, when, uh, we have this nice expression when the Helicities are h1, h2, and h3, this is equal to, um, um, some coupling constant g times 1, 2 to the, um, uh, h3 minus h1 minus h2, 2, 3 to the h1 minus h2 minus h3, 3, 1, uh, to the h2 minus h3 minus h1, and, uh, this expression is valid, uh, when the sum of the Helicities is negative and some other coupling constant, um, with the other kind of bracket and with the exponents reversed. Okay? So, when the sum of the Helicities is positive. Okay, so, um, now we're going to move on, uh, so this is, this really completes the essentially kinematical part of the story. So now we're really going to move on to interactions more, uh, interestingly, we're going to talk about the four-particle amplitude. Okay, so there's some amplitude now. Uh, we're four particles. And before doing anything fancy, um, let's just go back, let's, uh, remember things we, we know just from, uh, you know, our, our, our simple courses. So, um, so let's say, let's say that we, we had an underlying pi cube theory. Then, then you know that at tree level, the amplitude would just be the sum of these diagrams, right? So this we often call the S channel, the U channel, and the T channel. Um, and the corresponding expressions would be, um, that this would be, if the strength of this coupling constant there is mu, this would be mu squared one over S plus one over T, uh, plus one over U, or the way I wrote it there, plus one over U, plus one over T. Again, where as usual, S is P one plus P two squared, um, which for massless particles would just be two P one dot P two, um, and, uh, U would be P one plus P three squared to P one dot P three and T would be P two plus P three squared. And, uh, from now on in this lecture, I'm probably gonna, um, very, okay, very often be sloppy about these factors of two, okay? So, uh, okay, so, um, that's what the, uh, that's what the, uh, amplitude is in this case quite trivially we see from the, uh, Feynman diagrams. Um, but I want you to see from here, um, uh, that let's say that, um, I mean in this case it's so, so simple that we just compute, we computed the, uh, Feynman diagrams, we just did it, but, um, there's a way that you can check whether the result is right or wrong, um, even if you hadn't done the, uh, competition, okay? And the reason is that the, that this four particle amplitude in this approximation where the amplitude only is at tree level, when the amplitude is at tree level, um, it has a very simple structure, right? Uh, the only singularities that it can have are poles. By the way, it could have, of course, there could also be pieces, I mean, I might have, I might have had a four particle amplitude too, right? In other words, I might have had a five to the four coupling, so that would have just been some, uh, a constant, okay? Or even maybe I have a d five to the four coupling, so I might even have some, some correction that goes like s or t or u, okay, times some other coupling constant, or s squared, t squared, u squared, times some other coupling constant, um, and so on, okay? So these are all things that are polynomials in the momenta, and, okay, so whatever the couplings are in, in, in the theory, I'd have to have, uh, I can certainly have these polynomials, I can have as many of them as I like, but if there was a five cubed coupling, the very presence of the five cubed coupling means that the amplitude has to have something which has a pole in it. It can't just be smooth as s, t, and u goes to zero. It has to have a pole in it, and the pole corresponds, of course, just to the exchange, that we see in these, uh, uh, diagrams, okay? So if there was a five cubed coupling, in other words, the very fact that we have this coupling, um, means that the amplitude has to have a pole. And note that this is true at tree level. Uh, we know that we could have more complicated singularities than, than poles once we go to loop level. For example, uh, at loop level, I would have something, I could have something like this, and these diagrams can have logarithms in them. They can have things like logarithm of s, even more complicated functions of s. Um, uh, but we're going to be working at tree level, and you might ask, what do I mean by tree level? Uh, that, uh, you know, I keep talking about the fact that, um, we want to talk about things without talking about fields and, uh, and Lagrangians and Feynman diagrams and so on. And then all of a sudden I'm saying the word tree level. That sure sounds like I'm thinking about Feynman diagrams. But in fact, what tree level means is a little bit more invariant. Tree level is reflecting, uh, what kind of singularities the amplitude has. And so we can sort of reverse it. We can say that, that the amplitude could, could be completely smooth, okay, at, at zero energy, so without just having polynomial interactions, that's the simplest possible analytic structure. The simplest possible case is it has no singularities at all. At least no singularities of zero energies. It would have singularities at infinite energies, but no singularities at low energies. Okay, so no singularities here, um, uh, at, at low energy. The next simplest kind of, uh, uh, the next simplest kind of, um, structure it could have is to have poles. And then that does have a singularity. And then after that, uh, the next most complicated structure of that would be to have a logarithm which has a branch cut. Okay, so if I look, if I think in the complex S plane, I can have a pole at S equals zero, for example, or I can have a branch cut that would correspond to, uh, um, that would correspond to a, uh, log. Actually the branch cut is here. This is in the S plane. And then these could get more and more complicated. We go to higher and higher loops. This would get more and more complicated. But we're going to imagine that there is some kind of weak coupling between the particles. After all, we're talking about, uh, a scattering amplitude. So we imagine the particles are escaping to infinity. And that means that we're going to restrict ourselves to the case where there's only poles. Okay, so that's an approximation. But in the approximation of those only poles, when we get things from Feynman diagrams, that's tree level. But you can think of it more invariantly as we're, you know, we're, uh, organizing the amplitude in how complicated the singularities are. And, uh, so we're beginning with the simplest possible singularities where the only singularities are poles. Okay. So let's say that we've done that now. So now, so now, so now what we're doing is we're talking about a, a, uh, a four particle amplitude. Okay. Um, and, uh, where it can only have poles. So again, this is quote, unquote, tree level. And now there are two physical requirements. And these two physical requirements are inputting the physics of locality and unitarity in the answer. Okay. So these are the constraints that we're now going to talk about. So one locality says that the amplitude should have simple poles, only simple poles that go like one over S or one over T or one over U as S and T and U goes to zero. All right. So what are some things, examples of things that are not allowed? I cannot have one over S squared or one over S cubed and so on. Can't have that. I can't have one over ST minus U squared or more complicated things like that. I mean, if I'm just writing down random formulas with poles in them, these are all kinds of things that I could have. No, the only kind of poles that I'm allowed to have are just one over S, one over T and one over U. Okay. So these are the only allowed poles. And this has to do, again, this reflects something that we're intuitively familiar with, that where are these poles coming from? These poles are coming from the fact that normally the amplitude is smooth, but the poles can occur when you can produce some intermediate state and the intermediate state, this is in position space, the intermediate state propagates a long distance and decays out somewhere else. Okay. But in order for this to be what is happening, so where this is a long distance, all right, the image of this back in momentum space is that in momentum space, we have to have a one over, so if this momentum is capital P, we have to have one over P squared, which is literally the propagator of this particle, right? Okay. So that says that the pole corresponds to the propagation of a particle over a long distance. All right. So that's requirement number one. Let me say it again. So here I have my four point amplitude and requirement one is that it should only have poles and only simple poles as S, T and U go to zero. Now, requirement two, and so this is locality of interactions. Yeah, Max, go ahead. Yeah, hi. I can ask us some more thing about this relation between locality and simple poles, because if I think of a pole like one over P to the four, then I usually think of that this perhaps maybe two particles, one of them is a ghost. So can I use this intuition to understand why this would be a non-local theory in some sense? Yeah, I mean, so in fact, if you just hang on a second, you'll see that this is very closely related to the second point. And this is, by the way, it's kind of a, already we've seen this very simple example, kind of a fascinating relationship between these two notions of locality and the uniterity. They're not totally decoupled from each other, and it's very much related to what I'm about to say. You see, one way of phrasing what's wrong with the one over, it's obvious that the one over ST minus U squared is crap. Okay, that doesn't mean anything. But you're right. You could say, well, what about one over P squared squared or one over P squared to the fourth? From the perspective of the Lagrangian, as you say, you can think of them as a pair of particles with opposite signs in the appropriate. So what's going on there is that the residue on P squared equals zero is actually vanishing. And it's vanishing because there's something with a plus sign and something with a minus sign. So that's how you see that it's violating, that we can't interpret the produced particle as having produced something with positive probability. And that's actually just the second point I'm just about to come to, which is to give an interpretation of what the residues on the poles are as ST and U goes to a zero. So if you like, I could say that maybe I'll allow things that have higher powers of ST and U, but that'll now be killed by my second requirement that I'm just about to talk about. Now, so what is the second requirement is that let's say on the poles S goes to zero, say, I should be able to see that the amplitude there goes like one over S, that's my pole, but then I should be able to interpret in the neighborhood as S goes to zero, see the whole point is in the neighborhood as S goes to zero, this intermediate state is now on shell. That intermediate state is now on shell. So I have to be able to interpret the residue that I just produced a particle here. And then the particle produced it one to distance and it made the other guys. So that means that, and if here it has Helicity H and it's outgoing on this leg, it would be incoming with also the Helicity H, but since I'm labeling everything with incoming momenta, this would have to be Helicity negative H. So in other words, if I say it more technically, the residue of the pole as S goes to zero has to be interpreted as the product of three particle amplitudes summed over everybody that can propagate in the middle here, whatever the states are, give them some label A and with Helicity is H and negative H. Okay. And you see that's cool because we just learned that these three particle amplitudes are fully fixed by Poincare symmetry. So now we can begin to see that we have a constraint. We have an interesting constraint that the four particle amplitude is forced to have poles because there are three particle amplitudes, but on those poles it has to factorize into the product of three particle amplitudes in the appropriate way. Okay. So that's it. These are my two requirements. Let me summarize again. M4 should have just a simple poles as STU goes to zero at most. And two, on those poles, we need to have factorization. That the residue, let's say as S goes to zero, so we get one over S, but then we should be able to interpret it as the sum of anything that's propagating between them, but now these are on shell, okay, with Helicity H and negative H. And again, one is loosely, somewhat loosely, locality, and two is uniterity. All right. So now that is our task. So now we have a, now we're starting to have a well-defined task. So for instance, let's go back to our example where the amplitude was one over S plus one over T plus one over U. This is an extremely simple example. Okay. And let's see what happens as S goes to zero. Well, we have one over S times mu squared. And indeed, this three particle amplitude was mu. That other three particle amplitude was mu. So mu squared is mu times mu. And so indeed, this amplitude works. Great. Okay. Let me just show you one more example. Let's say that, let's look at another, so that's a pi cubed theory. Let's look at a Yukawa theory. Okay. So here's a, let's write what a, what a Yukawa coupling looks like. We didn't do that example last time, so let's do it, let's do it now. So let's say I have a massless fermion, now a velocity minus a half, minus a half, and a spin zero particle. So this is some pi. So this is some fermion side, some spin zero particle with velocity zero. So if this is particle one, two, and three, then by our rules, from last time, what is the amplitude? Well, you can shove it into that general formula, or we could just, I can just write it down. And you know there's a unique expression with the correct weights. So all you have got to do is write the expression with the correct weights, even if we don't remember the formula. And okay, that's the expression. Okay. Very good. Okay. So, okay. So, and notice that remember this was supposed to have units of, this was supposed to have units of mass. So from here, we see this already has units of mass. So we see that y is dimensionless. Okay. So that's just like what you expect for a yukawa couple. All right. And so let's say we want to have a four particle amplitude now, where let's say all of them are minus for fun. Okay. So particle one is minus, two is minus, three is minus, four is minus. Okay. So what's the, what's the amplitude? Well, I'm just going to write down the answer right away. It's y squared times one, two times three, four over s. Okay. And let's see what, why it's correct. And so this is a slightly more interesting example. Now I only have one channel here as s goes to zero. But as s goes to zero, this looks like one over s times this three particle amplitude, one minus two minus five. And then there's a five, three minus four minus on the other side. Okay. Good. So this was y times one, two. This was y times three, four. And so liquidy sweat, we've written down the amplitude for scattering of fermions through a yukawa interaction. And I hope you're already maybe slightly impressed with this, because we didn't have any gamma matrices, we didn't have any, you know, U bar, V, all that crap, but I certainly never remember how to think about it. Okay. Without looking it up, the conventions in some book and gamma five and all that junk, we never see any four component anything here. It would just wrote down the answer. All right. Okay. In fact, I encourage you to do and the exercise, try to write down the four point amplitude for like a Compton scattering in this example, like where it would look like five. So let's say something like this. So one minus two for the phi and then three for the phi. And let me do like four plus. Okay. Let's actually see if we can do this example ourselves. Let's see if we can do it now, because this is gives us a good, this should also give us a good, it'll give us a template for what we're about to start playing with. There is a question of the chat whether we have the, whether we have the other, whether we have the other channels as well. I don't necessarily have to have the other channels. It could be that this was the only, in other words, these particles one, two, three, four, they might have, they might be different species of particle. So that this is the only, so the only coupling I have is one, two to five and three, four to five. But I don't necessarily have to have a coupling of two and three to five. If I did, I'd have to include the other channels, too, and then you would do it. Okay. So let's say that, in fact, let's say that all of these guys were identical. Okay. If all the fermions were identical, then indeed I would have to, I would have to add the other channels. So I'd have to write something like two, three, one, four over T and add the other ones as well. Okay. So you can also finish doing that exercise. But let's look at this one. And we can start seeing a strategy for writing down the answer, because it looks simple, right? It looks like as S goes to zero, what should the residue be? Okay. What should the residue be? Well, as S goes to zero, let's just, let's just write it down. It would have to be this product of three particle amplitudes. Okay. So this would be one minus and some intermediate minus here. And this would have to be the same intermediate plus. And that's why I put a plus here. Okay. So if I put a plus here, then here I can have four plus, let's say. And so this would be a different coupling. Okay. So if I write down the kind of couplings that I'm allowed to have, we could have the one that we wrote down where it's one minus two minus five. And that would be some y12. I could also have some one plus two plus five. And that could be even some different y prime times the square bracket for 1, 2. And notice that y prime doesn't have to equal y, which means that we don't have to respect parity. Okay. So, okay. So, okay. So let's go back to this example now. Let's work out what the residue is over here. And let's try to do it a little bit carefully, because on this side, now we are encountering something slightly new. We've got to worry about this intermediate guy. Okay. So on this side, on the left, I have something like, so let me write a big again as f goes to zero. I have one minus two intermediate minus intermediate plus three, four. So on this side, on the left, I have a residue, which is y1i. On the right, I have y prime i4. Okay. And so this is now slightly cool, because what is it, how do I figure out what this is? I don't, how do I figure out what this, what this product is? One i, i4. By the way, notice something nice here that there is no little group weight under i, right? And that's good, because the answer doesn't, on the outside, doesn't know anything about i. It should only have little group weight under the, under the external particles. And that's good that the i occurs here with an angle bracket and a square bracket. So the little groups cancel on the inside. And that's the reason that's another way we can see that we have to add a minus here and a plus there. That's the only way that the little group weights can cancel between one leg and the other leg of the i. But how can we figure out what this is? Well, we just have to remember that pi is actually equal to lambda i, lambda i tilde, right? So this is a massless particle. i is a massless particle. The whole point is that it's on-shell. So pi alpha, alpha dot, lambda alpha i, lambda tilde, alpha dot i. So what is this expression? This expression is lambda one, alpha, p intermediate, alpha, alpha dot, lambda tilde four, alpha dot. If the, for once I, I write it out explicitly like this, we can also write it as one p intermediate four, okay? So I hope that notation is clear. All right, but now we're going to write p intermediate is equal to p one plus p two, okay? I could also write it as negative p three, negative p four. These are two different ways that I could write this p intermediate. And so, so therefore, p intermediate, lambda one, let's say I chose this expression. This is equal to one p one plus p two. But the point is that the p one part cancels. Why do the p one part cancel because p one is massless, right? So p one, alpha, alpha dot, lambda one, alpha, this is equal to zero because p one is lambda one, lambda one tilde, right? So the lambda one, lambda one is giving me zero, okay? So from here, I see that one p intermediate four is equal to, in fact, one, two, four. So I could write this as one, two, two, four. So that's a nice expression, right? Because p two itself is massless and is lambda two, lambda two tilde, right? So I can could have also written it as one, negative p three, negative p four, four. And now the p four part goes away and this would be negative one, three, three, four. So this is something we're going to see over and over again in the four particle kinematics is expressions like this. So that when we have this guy, this is the same as one, p two, four, which is the same as negative one, p three, four. And so I could write this many ways. I could, I could write it in a more symmetric way between particles two and three with a one half, one p two minus p three, four, for example, okay? So these are all the equivalent ways of writing it all the support of momentum conservation, all right? But anyway, one way or the other, we've now learned what is residue. We've now learned that this residue is pi one minus two, intermediate minus intermediate plus three, four. Let's even just choose one. Let me just write just in terms of two, just the is like one, two, two, four, okay? So it looks like life is easy. That's what the residue has to be when s goes to zero. So the answer has to be the amplitude has to be one, two, two, four over s. Done. Okay, with a y, y prime, okay? So there's a y and a y prime here. So times y, y prime, which again, if I wanted to write it more symmetrically, I could write as one p two minus p three, four, okay? Any questions about this? Go ahead, Fabio. Hi, just to be sure, all contractions are done with the Epsilon tensor, right? All contractions are always done with the Epsilon symbol. That's the only answers we have here. That's right. Yeah, okay, right, thanks. So what would happen if we're introducing a pole as one over s if we had like the new channel how would that look like? Oh, what do you just repeat exactly the exercise? You want to do it? Okay, so we can do, yeah, so let's say it looked like this instead, right? I mean, just the u channel in this case would just literally be reversing particles two and three, right? Right, so it'd be p three minus p two, I'm assuming. Yes, that's right, exactly, exactly. Okay, and so would the final amplitude be some linear combination of s and u? Yeah, we'd have to have a pole in each one, exactly. So if these scalars if these scalars were identical, then I would have to have the other channel as well, and then I would add those two channels, exactly. Okay, great, thank you. Okay, any other questions? Yeah, I'm just doing one of the residues at a time. In fact, shortly we're going to be talking about more than one residue at a time for a reason that will become obvious in a second. Maria, you had a question. I have a question that goes back to the reasoning about the poles of the four-particle amplitude. Yes. Like, we know that it has poles when you perform the canonical calculation with the Lagrangian and so on. Right. But using your argumentation line, I think I got lost. Why do you deduce that it must have poles? Well, I think there is a longer answer to this question, but the shorter answer and maybe the sort of best answer right now is that on quite general grounds, we don't know a priori what the analytic structure of the amplitudes actually has to be. Okay, so part of the little bit of the art in this subject is to just from what we know, you see, what is the basic problem? The basic problem is that the amplitudes are something that you measure from far away. Okay, so the sort of data that labels the process is some things that come in, something scatters and they go out. Right. And all you have access to is what happened at infinity, the infinite past and the infinite future. What we're trying to figure out is what property of this answer somehow encodes that it could come from a process that could be given a local description if you wanted to. Okay. Now, we know that if you do start with a Lagrangian and so on, then that will certainly be the case. You're giving it a local description. So what you try to do is look at what the answer looks like, but not in detail. Look at what just some of the gross features of the answer looks like that you get when you start from a Lagrangian, and try to use that to sort of guess what the more abstract properties are that you could impose on the answer that carries a fingerprint of that property of locality that you wanted to. Okay. So you could say that this zeroth order is kind of cheating a little bit. You look at the answer and you go back and say this is what you want. In this case, there's slightly better answers to your question, but not that much better. And it would take a lot longer to explain those very slightly better answers. So just right at the moment, just think of it this way. It's like we don't want to compute the Feynman diagrams. We're lazy. We don't want to do all these horrible things and some polarization factors and all that stuff. We're enjoying seeing how directly we can determine all these things, but somehow we can write down a million formulas that are not what we get from when we play with the Lagrangian and Feynman rules. So we just want to sniff a little bit, just a little bit of information from what we get from Lagrangian, which is just the kind of poll you have, and then say, okay, that's all I need from you. Go away now. And then we try to impose those con additions and see how far we can get. Your question is a very good one. And while at tree level, I can give you a slightly better answer. When we get to loop level, we really can't. Okay. And so one kind of looks a little bit at what the kind of functions are, what the kind of analytic structure is, and then you go back and pretend that you didn't look and just try to impose that and see how far you can get. In the end, the most adventurous answer to this question is that ultimately what we're doing now is trying to impose consistency conditions to determine the answer. But those consistency conditions partially are coming from thinking about maybe these things came from Lagrangian's after all. The most sort of intellectually exciting part of this whole program is to try to sever that tie totally, right? And just try to think about the amplitudes and the answer to some completely different question. And if we eventually find the amplitudes, even for general scattering problems in the real world, if we can find them as the answers to radically different questions, then the nature of that question would tell us what kind of poll it's supposed to have, what kind of singularity it's supposed to have, and so on. And then we would read backwards from that very likely quite abstract seeming question, why we are forced to have all these properties that we then ascribe to locality and uniterity, having space time and quantum mechanics and all of that. We're not talking about that at all, right? That's a much more speculative part of this subject. I think an exciting part of the subject, a much more speculative part of the subject. Here we're just trying to impose consistency conditions. So as I said, we have to sniff a little bit what's going on from the answer and then say, okay, that's all I need from you. Go away and we continue. Okay, thank you. Okay. All right. Any other questions? Julian? Yes, so, sorry, I may have got lost somewhere, but is there an easy way to understand when, like, which or an intuitive way to understand, like when to take which vector parallel, which not so, when to use the, which bracket. Yeah, we sorry, we had that rule, that the rule is when the sum of the helicities is, when the sum of the helicities is negative, you have to take the lambda tilde parallel. And then the amplitude is only written in terms of the angle brackets. And when the sum of the helicities is positive, it's the other way around. Okay, sorry, maybe I have missed that. Thank you very much for clearing it up. And let me just remind you why that is. Okay, so just with an example. So if it was like one minus two minus three plus, there are two expressions that we can write down. One two cubed over one three two three, or the reverse one three two three with the square brackets over one two cubed. Okay, these are two expressions with the identical helicity weights. But as we go to Minkowski space. So remember in Minkowski space, everybody becomes parallel, because if the, this is the configuration of the lambda tildes are parallel, but in Minkowski space that forces the lambdas to be parallel as well. Okay, so as we go towards Minkowski space, everyone is forced to be parallel. But then this guy goes to zero, but that guy goes to infinity. Okay, because as the lambdas become parallel, there's the upstairs is vanishing like the cube, the downstairs is vanishing like the square. Okay, so it's that requirement which tells us that when the sum of the helicities is negative, we have to write things in terms of these brackets and when it's positive, we have to write in terms of those. Okay, did that make sense? Yes, thanks a lot. Uh, Nima, one question. Yes, you mentioned that when the sum is equal to zero, yeah, one can show it inconsistent. Can you give a, can you give a gist of how, how I, oh, you'll see just of it. In fact, I'll leave that as an exercise for you after you see what we do in the other examples. Okay, there's a question in the chat by Oscar. Yeah, who says he was thinking about the sum over intermediate states. Yeah, no, it's not always the most forgettable factor. I mean, here we've just done, it's very simple examples with a few states. You know, you have to sum, you know, in total generality, I mean, if I want to say it stupidly, abstractly, we have, you know, a bunch of labels, helicities, and also labels a three, and you just have to sum over all of them, right? So it's a sum over all of these intermediate labels that you could possibly have and some coupling constant here g a one, a two, a three. Okay, so, so you really would have to have, if you have one, two, you'd have to have the sum over g a one, a two, b, g b, a three, a four. Okay, so you really have to sum over everyone. Yes, and there's another question. This is, this is just looking at the, at tree results for the amplitude so far. We can now start pursuing this philosophy further to get to a loop level as well, but here we're just doing trees. Maybe you're not so impressed yet. I mean, I'm sufficiently technically incompetent that I'm already impressed that I could write down these Yukawa Compton amplitudes in 30 seconds without worrying about gamma matrices and stuff like that. But if you're not impressed now, you'll be impressed momentarily. I promise. Or you'll never be impressed. So, you know. Okay, shall we take a little break? Sounds good. So let's come back in five minutes. Five minutes, perfect. Looks like a really nice day out there, Giovanni. Yes, no bad. I literally hear birds singing in the background. Oh, it's great. It's very idyllic, son. You know that we've had been under attack here? There was a 17-year cycle of the of these insects. Decades. It's unbelievable. I'd never experienced it. It's literally biblical. I mean, literally, you know, it's like it wasn't enough that we had the plague. Now we have the locusts. What's the period again? 17 years. Every 17 years. Yeah. And, you know, I like animals. I like creepypollies. They have no problem in general, but these things are a disaster. They're like, yeah, they really describe the birds in the background. You have birds in the background. We're now at the peak of it. It's going to last another, but at the peak, when you go outside during the day, it's deafening. They're so loud. It's literally deafening. So you have to close all the windows. Even in the middle of the night, you close all the windows, everything. You can still hear them through the window. So this sort of sound of just one lone bird tweeting in the distance is very it's like paradise compared to what we're going through here. All right. All righty-do. Shall we start back up? Oh, you want to hear someone ask me to open the window so they can hear the cicadas? First, it's not quite as loud now, but secondly, much as I would love to entertain you, there is a really highly non-trivial chance that in the 20 seconds that would leave the window open, there would be 20 of them in here. And my wonderful cat, who's the best cat in the world, I apologize to anyone else who has cats that they think are great. They're not as great as my cat. I promise you. My cat has spent the past four weeks dispatching them, but she's exhausted downstairs right now from dealing with them. So I apologize. I'm not going to, just for the sake of your entertainment, I'm not going to put further hardship on my beloved Gato. Okay. I hope that doesn't strain our friendship permanently, but okay. Very good. So now the excitement is going to begin. And in order to see what the excitement is, there's actually something interesting in, there's an interesting pattern we can observe in the three particle amplitudes that we didn't, that, you know, we just wrote down the answer, but we haven't stared at quite enough of them. So I want you to notice some, an interesting difference between the three particle amplitudes for different kinds of theory. So let's look at what I'll call boring theories to begin with. So boring theories are boring from the Lagrangian point of view. There are things like Phi cubed or Yukawa, like Phi, Psi, Psi. So they're boring in that you don't need any fancy things like gate symmetry and stuff like that to talk about them. Okay. Um, in fact, we just did these two examples. What do their Lagrangians look like? Okay. So, sorry, what do their amplitudes look like? Well, this one was easy. This one was one, two, three, was just given by mu. And this one, as we just said, is like one minus a half, two minus a half, and zero was given by y times one, two. But now let's look at exciting theories. So for example, let's look at the, a coupling of a, even the simplest thing of a, of a charge scalar to a photon. Okay. So there's some spin zero guy here. So this is spin zero, spin zero, one, two, and massless spin one, let's say with negative velocity three. Okay. Okay. Well, once again, we can, we can write down the same, we can write down the general formula, or we can just write down an answer here that has the correct weights. So what, so the answer that has the correct weights, it has to have weights of two in the particle three. So it's one, three, two, three divided by one, two. Okay. And there's some E here. And you can see that, that this thing already has units of mass so that E is dimensionless. Okay. So there's already something kind of cool here, right, which is this three particle amplitude already has a pole in it. Isn't that funny? Right. This is a three particle amplitude that has a pole. What the hell does that mean? Okay. Well, that's fine. You can ask, you know, I can't reach this pole in Minkowski space in real, in the real moment I don't see because the whole amplitude is zero. Okay. There is more things upstairs than, than downstairs. But it's kind of interesting that, that this amplitude actually already has a pole in it. Let's look at the same thing with gravity. Okay. So what, what if this was now coupling to a graviton, so something that a holicity minus two, then what would that, what would the coupling be would just be this one, three, two, three over one, two squared, right? Because that, that now has the correct weight in three. And now what are the units of this? This would have to have units of one over mass. So that's, that's like one over m-punk. So that, indeed, that's the coupling to a graviton. But you see that these exciting theories, they have the property. And by exciting, I mean, these are the theories that in the usual Lagrangian way of talking about things, we need gauge symmetry, gauge redundancy, general covariance, difumorphism invariance, and so on. And the difference between the boring and exciting theories is exactly the, that the, that the boring theories, there are three particle amplitudes are totally only numerators. There's no poles in them. But the exciting theories, the three particle amplitudes have poles in them. Okay. So that's just, just observe that interesting difference. So so far, we've only talked about cases like this. And now why does it matter? Because remember, what did we do? We thought it was super easy to build four particle amplitudes that were correct. All we have to do is look at, I mean, this was our strategy, right? We say look at the residue in the S channel, whatever it is. So there's some residue, RS in the S channel. And similarly, RT in the T channel and RU in the U channel, maybe some of these are zero. But anyway, we have these residues. And then we say, good, the amplitude is just RS over S plus RT over T plus RU over U. Okay, that was basically our strategy so far, right? Just look at the residue is get the answer and just slap on a one of RS in front of it. But now you see what the problem is. The problem is that when the three particle amplitude itself has a pole, when the three particle amplitude itself has a pole, then when you compute the residue in the S channel, the answer could well have a pole in another channel, right? So that if you write the expression RS over S, you might not be free and clear because that expression that you wrote down might now actually have a singularity somewhere else that you have to go check if that works out correctly or not. Okay? And we're going to see precisely that this is what actually happened. Now we can do many, many examples. I'm not going to do all of them, but I'm going to do some sort of canonical ones. So I'm going to begin with imagining can we have a single particle of spin S? So I'm going to, since there's also the S channel, I'll write the spin as curly S. So, but there's just one particle, I'm going to, and I'm going to imagine that it has this three particle interaction, one, two, three, so that this amplitude is some G, some coupling constant, one, two cubed over one, three, two, three to the power of this S. And similarly, there's some plus, plus, minus, which is G prime. Maybe I'll just, I'll just keep calling it G because this part is not going to matter so much. The other kind of bracket, one, two cubed over one, three, two, three to the power of S. Okay? All right. And so now what I'm going to do is look at the four point scattering for this configuration, one minus two plus, one minus two plus three minus four plus. Okay? Now, just ahead of time, by the way, guys, just so we know what we're looking for, I know ahead of time, whatever this is, is going to look like one, three to the power of two S, two, four to the power of two S. And then what's left is going to be some function of the usual Mandelstam variables. Right? And why do I know that? Because if I pull these factors out, I've already soaked up all of the weights, right? I've soaked up the weight on particles one and three. I've soaked up the weights on particles two and four. And so whatever is left cannot have any weights and that will be a function of the Mandelstam invariant, S, T, and U. Okay? So I know that ahead of time, we're going to see it explicitly anyway, but just so we're not, just so we're, we knew that it had to come out in that way out of time. All right. So let's try our strategy and look at the residue in the S channel. Okay? So, so there I'm going to get, so I have one minus two plus some, let's say, intermediate minus, then intermediate plus three minus and four plus. Okay? So the expression here is one i cubed over one two two i. And then on the other side, I'm going to get i four cubed over i three three four. So this is to the S and this is to the S. Okay? I'm not going to keep writing the coupling constants over and over again. And so we use that one i i four. So this is the same thing that we saw in our previous exercises. One i i four is one p intermediate four. And so this is either one two two four or equivalently is negative one three three four. Then we can simplify these, we can simplify this expression. So for example, from the one i i four, I can replace that with the with the one three three four upstairs. Okay? So if I just write it out like one i cubed i four cubed. So this is what I'm staring at right is one two i two i three three four all to the power of s. But this I can write as like one i i four. I can write for example as one two two four cubed. And down here, I can write it as one two. And this i two i three here I can write for instance as one two two three. Or let me write it. Yeah, let me write it as one four three four. And so you can simplify this is just one three squared two four squared over you, which is one three three one all to the s. Okay, so I'll write it again. This is one three squared two four squared to the s over you to the s. Alright, so the summary is that the residue in the s channel is equal to, well remember, I promise that we'd have this factor one three to the two s two four to the two s out in front, ubiquitously, and it's now divided by you to the s. And similarly, we have the residues in the other channel. Okay, so so the residue in the t channel would be the same one three always the same. Now let's say over the s to the power of the span and the residue in the u channel would be over you over sorry t to the s. Alright, so you see what what the problem is right if obviously if s equals zero, then everything is okay all these residues are just you know all these residues have no poles. And so our strategy that m equals r s over s plus r t over t plus r u over you just gives me you know g squared times one over s plus one over t plus one over you and that's correct. That's what we did our very first example. But let's say that we now look at s equals one. Okay, so if we look at s equals one, then and then then the residue in the s channel is one three squared two four squared multiplied by one over you. All right. And similarly the residue in t is this thing multiplied by one over s and the residue in you is this thing multiplied by one over t. All right, well that's pretty interesting then because if I'm going to attempt to make an amplitude using the residue now you see the problem if I just take r s over s well this is the stuff here times one over s u and now this is going to have a residue in the u channel and that residue in the u channel has got to be given by this thing with the t and so on. Okay, so now I have a I have a subconsistency problem that I have to deal with. Okay, so let's write down what the most general ansatz is that I could have that's compatible with this kind of residue that I'm seeing here. So what's the most general ansatz I could have? The amplitude could be one three squared two four squared. Okay always factored out this is always what's been equals one but I can have some a over s t plus b over t u plus c over us. See this is already kind of cool because normally we think all the amplitudes are just one over s plus one over t plus one over u and this can't look like that here. It's going to look more interesting to even have a chance of reflecting this fact that we just talked about. All right and let's see how it works. Let's see what happens as s goes to zero. Okay, so we want to see the residue as s goes to zero and now here we have to remember that s plus t plus u equals zero. Okay, so as s goes to zero t is equal to negative u. Okay, so as s goes to zero I get a residue from this term and I get a residue from this term. So what is the residue as s goes to zero is equal to a over t plus c over u. Okay, which I can write as because t is equal to minus u I can write this as c minus a over u. Remember this is only supposed to be valid as s goes to zero. Right, so we're only interested as s goes to zero. So this residue is equal to c minus a over u. Okay, so beginning from the sansats we've worked out what the residues are. This is the residue as s goes to zero, sorry, and I'm factoring out this thing out in front. I'm not going to keep writing that over and over again. The residue as t goes to zero similarly is b over u plus a over s, which I can write as a minus b over s, and finally the residue as u goes to zero is b over t plus c over s, which is b minus c over t. And I'm writing it this way because remember what we worked out these residues have to be. This residue had to equal one over u, this had to equal one over s, and this has to equal one over t. So I have a constraint on these a, b, c. I have to satisfy that c minus a equals one, that a minus b equals one, and that b minus c equals one. Okay, that's what we have to have in order to satisfy our, our rules. And you see this is impossible because simply add these three equations, add this plus this plus this on the left hand side, we get zero on the right hand side, we get three. Okay, so very cool. We've already, we've discovered our first thing which is impossible. I cannot have, I cannot have single self-interacting spin one. And this is good because as I mentioned at some point in one of the other lectures, if we looked at this amplitude, this one two cubed over one three two three, this is here one and two would be fermions because if you interchange one to two, the amplitude goes to negative itself and yet the particles have even spin. Okay, so it's a good thing that we didn't find a consistent theory here because that would have violated spin statistics. Okay, so good. It's good that the, that our rules are telling us that this is not actually allowed. Okay. Okay, so what can we do? Well, let's now imagine we have multiple spin one and let's go back to this picture where this was one minus two minus three plus and I'll decorate this with an A, B, and a C, or maybe I'll call it A1, A2, A3, be more professional. So a, so I'm going to now introduce something. Oops. Now a new object here, F, A1, A2, A3, and still the one two cubed over one three two three. Remember, this is now just for spin one, all for spin one. Now we don't have to, we don't have to do anything, any extra work other than just decorating some indices here. So when we have this object, now we're going to say that the most general ansatz is one three squared, two four squared, and now we're going to have some A which has these four indices on it. So A1, A2, A3, A4 over ST plus B, just exactly what we had before just putting four indices on it, whatever we had, whatever we called it before, SU, all right. And when we go through the residues, it's exactly the same formula as before. Remember before we had, before we said we had to have C, oh, but sorry, but now what does the residue have to equal in the S channel? It's exactly all the angle bracket stuff is the same as before, but what is the dependence now on these indices? So here if there's some intermediate E, this would be F, A1, A2, E, F, E, A3, A4, okay. So going through exactly our same argument as before, from the residue in the F channel, we would discover that C, beforehand we had that C minus A equals one, and then we had that A minus B equals one, and then we had that B minus C equals one, but now instead of one, we're going to have C, A1, A2, A3, A4 minus A, A1, A2, A3, A4 has to equal some F, A1, A2, E, F, E, A3, A4, and then we're going to have another F, F, and we're going to have another F, F here with the appropriate things here and some E, okay. I'll leave for you to fill in that, that just fill in those indices, but I think it's clear to you what it'll look like, okay. And so now how do we avoid a contradiction? Before we got a contradiction, because when we added these three equations, we got zero on the left-hand side, but on the right-hand side, we had one plus one plus one, which was not zero, but now we're not going to have one plus one plus one anymore, are we? What are we going to get on the right-hand side? So in order for these equations to be consistent, as someone said already in the chat, we have to have that F, F plus F, F plus F, F with something, something, E, E, something, something, something, something, something, E, E, something, something, something, something, E, E, something, something equals zero. And where have you seen this before? This is the Jacobi identity for a Lie algebra. It tells you that the F, A1, A2, A3 must be structure constants, which furthermore tells you, from general things, you know that F, A, B, C have to be anti-symmetric, amongst other things, which then solves the, bows the symmetry problem of the amplitude, okay. So as I said, I hope you're impressed with this. Again, in some of the usual ways of talking about the Yang-Nouze theory, it's all about the beauty and how lovely it is that we can rotate the phase far away. We don't, well, we can rotate the phase here and there independently because it's so elegant and pretty and blah, blah, blah, blah, blah, but you see nothing to do with elegance or pretty or beauty, zilch, it's nailed on you. You can do nothing about it, okay. The only way to have a consistent theory of massless spin one particles is for the interaction to be governed by the structure constants of a Lie group, okay. I think that's an absolutely amazing fact, all right. Let's keep going with our exercise. Let's keep going and go to spin two, spin s equals two. Now, if we look at the residue in the f-channel, then we're going to get one, three, two, four to the fourth, but remember our formula was one over u to the s, so now this would be one over u squared, so this would be one, three, two, four to the fourth one over u squared. And now naively you think we're dead, right, because this residue is one over u squared and therefore if I write down the, if I write down anything I write down would have to be the amplitude would have to be, you know, one over s times one over u squared times this one, three, two, four to the fourth. And this has a double pole, which we're not allowed to have. So it looks like we're dead except you have to remember this residue, when we say it's one over, when we say the residue as s goes to zero is this thing to the fourth times one over u squared, this is only valid when s equals zero and remember when s equals zero u is equal to negative t. So we could equally well have written this as one, three, two, four to the fourth times with a minus sign minus one over tu and that few has simple poles and just t and u, but now if that's what we're going to do there's a unique thing the answer can be. The answer has to be with a minus sign, one, three to the fourth, two, four to the fourth over s times t times u. Exactly, Oscar. That's it and you see now we're done because now by symmetry when I checked the other channels it's also going to work. So we're done. There was a consistent theory of the mass list been two particle and not only have we confirmed that the theory exists, but we just computed the amplitude for it. Okay, so this is the four particle amplitude for gravitons and the coupling constant out in front is one of ram punk square. So this is the four gravity amplitude at tree level and as I said in an early lecture, good luck if you try to do this calculation using Feynman diagrams. Already the three particle of vertex is like a hundred terms in it and this is the most nightmarish calculation that you, I mean you could do it with computer but you can't do it by hand and indeed as Oscar says this was done in a tour de force amazing calculation by Bryce Duet in the late 1960s early 70s. I don't remember exactly what, but around that. All right, so we discovered that we are allowed to have mass self-interacting massless spin two and in fact you can go on from here and I'll leave this as an exercise for you that you can now imagine can we do this when we have multiple species. Can I do this if I have like a one, a two, a three and a multiple species of self-interacting spin two. So I would have this one two cubed over one three two three squared with maybe some g a one a two a three and what you can do is show that these g's have to satisfy similar kind of identity to that Jacobi but with some kind of signs messed around and you can show it's a nice exercise to try and show that there is no solution for those g's other than the ones indeed where they just decouple into sectors that don't talk to each other. Okay so in other words like g's are just decoupled into a bunch of totally non non-interacting things. So from there we also learned it is impossible to have a consistent theory of multiple interacting massless spin two particles. If you're going to have a massless spin two particle it's that they basically decouple into different into different universes okay they can't couple to each other in particular. All right so I hope you're starting to be more impressed. Let me just give you oh sorry and so and I forgot to say before too that we also wrote down the answer for Yang-Mills right because we saw for these a b and c okay so so let me summarize what we've seen so far. So we've discovered that for spin one there is a consistent amplitude but only when the f satisfied Jacobi and then we also wrote down the answer the amplitude is equal to ff over st plus a different ff over tu plus a different ff over us times the uh one three squared to four squared and for spin equals two we discovered that the amplitude is one three to the fourth two four to the fourth and even had to have a minus sign out in front one over stu okay with an over some implant square. By the way I'll leave it for you to puzzle out this minus sign that we have to have because we have to flip the one over u squared that's that gravity is attractive that's the opposite sign that it has relative to over here that's the fact that the gravity is attractive but I want you to notice something amazing about these these formulas it's sort of most amazing in the gravity case the gravity case the answer does not look like the sum over three channels it's not something over f plus something over t plus something over u it's the whole product one over stu comes together right that's just a fact about what the actual amplitude looks like the actual amplitude there's no way of expressing this formula as something over s plus something over t plus something over u so um so what's going on then when we compute these things from Feynman diagrams when we compute these things from Feynman diagrams clearly we get something like something in one channel with something in the other channel or something in uh in the other channel and even some contact terms okay so when we compute with Feynman diagrams we get something like that so this sure looks like it's four polarization vectors over s plus four polarization vectors over t plus four polarization vectors over u and maybe some contact pieces okay so we are learning something each one of these pieces cannot be Lorentz invariant by itself okay and in fact in the conventional language each one of these pieces is not gauge invariant by itself so when you break things up in this way there's no way to interpret the pieces as something Lorentz invariant we've written them in terms of these faco polarization vectors that are not really transforming like four vectors right remember in order for it to really transform like a four vector it had better be that if you replace epsilon goes to epsilon plus alpha p that the answer stays the same if i took just this term it couldn't if i took just this term it couldn't only the sum of all of them does okay so that's uh so that's a way of seeing why ahead of time any way of describing this physics in a way that breaks it up as if it comes over the sum of channels has to have something funny associated with it because the numerators cannot be Lorentz invariant by themselves and we see why we see why because the actual amplitude has this product structure not a sum structure and the same thing is there in yang mills it's not quite as dramatic because you don't see all the three poles but it's the product of sd the product of tu and the product of us rather than something which was sum over the channels at the same time and uh and so this was all foreshadowed in the very beginning of our discussion uh from from what we uh remarked on this difference between the boring and the exciting theory okay so all of this happened because right at the beginning we had uh this distinction these are the theories that have simple Lagrangians with no need for any gauge redundancy or anything like that and we see that that the three particle amplitudes don't have any poles but these theories the three particle amplitudes have poles and that means that by the time we get to the four particle amplitudes there's no way of expressing them in a way that looks like some of a simple pole they have to have these interesting product structures and that means that any way of computing them using polarization vectors has to have some kind of redundancy or some problem in it because the actual amplitude does not look like this okay so i hope you have a deeper feeling for why why physics is forced to be described uh in this way there's a question uh by Gautam yes um yeah so i was wondering if we could sort of use the same argument for spin equal to three yes let's just let's just do that right now let's just do that right right now so if we do spin equals three now we're just dead because the residue in the s channel is equal to something cubed now one over u cubed now we're just dead now there's nothing we can do okay we can't write this as one over u squared t t u squared whatever you do it has to have uh at least a double pole so we learn no self-interacting massless spin greater than or equal to three done okay so but then what happens when you do s equals three by two uh when we do s equals three have it works okay and um so uh and in fact so let me uh maybe i can just say um i know i've basically run out of time can i have maybe just another 10 minutes or so so i can finish this thread of very good all right so all right so um as i said uh uh you can look at a um a discussion along uh these lines is in section four of my paper with the uh with the the huang squared okay so if you look at this 2017 paper i think section four we discussed many of these things and there's earlier uh nice paper by uh migheidi and rodina um which discusses many examples like this but i think now you've gotten the hang of it you should try you should just do it yourself i mean there's a lot of fun and uh you'll you'll learn something but i want to just give you um uh without going through all the details just give you a sense for where the constraints come from it's all the same thing it's all checking that you look at the residue in one channel of forces pulls another channel and you have to look at their con consistency so if we look for example at the interaction between other spins let's say spin zero a half one general spin here imagine we have a general spin s here but interacting with spin one and we have um uh you know we have uh uh some plus minus and let's say uh uh plus here then um there's going to be this has some index a this has some index i and j so this is like some matter okay so i'm just going to call this t a i j i don't know anything yet about it being a generator or anything like that but when i look at the residue here when i look at the residue of this kind of thing and in the for this sort of uh amplitude i a b j when i look at the residue in the s channel then i find that it's given by one over you um two uh two one one three to the two minus two this spin the spin of the guy on the outside okay um uh times one three two four to the two s uh times some t a t b i j okay so that's just what you get when you uh and sorry this one i could write here this more a little more symmetrically um so two p one minus p four three but it's raised to this uh funny funny power uh two minus two s okay okay and so evidently okay and so evidently you see that that this is easy to see right because this is a t a here and a t b here so this is like t a i k if this guy's k t b k j so that's how you see that you get the structure t a t b i j but you get this uh interesting uh factor just from gluing together the three point amplitude okay so from here already you see something uh interesting if the spin is greater than if the spin here is uh is uh uh is greater than three halves then two minus two s is negative and we're dead because this expression has a pole it has a pole in this crazy quantity that's not s t or u okay so we're just completely dead so that already tells you you cannot have um that already tells you that you cannot have and this is for this is for um s sorry greater than or equal to three halves okay that you cannot have charged particles uh massless charged particles of spin you have for higher that's impossible okay so for lower spins so for spin uh one and lower you can in principle have them okay but something has to happen with this t a t b business and now if you work out the consistency of the rest of the residues then you find you have to have that t a t b is equal to f a b c t c okay and the reason is that is that uh uh you get poles in s and u from this compton amplitude but they but those residues have a pole and t so that can only work if you use this interaction that three point interaction that we already talked about before which had to be f a b c okay so you learn that if the commutator is non-zero it has to actually equal f a b c p c right so we see just from these uh conditions that the only particles that can interact with massless spin one massless spin one can interact with itself with strength f a b c and also with spin equals zero and a half with some t a i j which satisfies that t a t b is f a b c t c so the entire structure of the yang mills theories is forced and i can't have no spin greater than or equal to three halves that are charged and if you continue to do this argument now the same way to spin two then you discover exactly the same argument you discover one um sorry uh this guy here is uh uh these guys are spin s and this is spin equals two then you discover that one the analog before that we couldn't have it for spin greater than or equal to three have here it just goes up one power this is impossible for spin greater than or equal to five halves okay basically the like power upstairs uh the in that residue calculation the power upstairs end up being uh two p one minus p four three to the power of four minus two s okay um and uh so if s is greater than or equal to five halves this thing would have a pole and this is dead dead right so we learned something amazing that if there is a graviton at all and if we're going to interact with it it's impossible to have massless particles of spin greater than two and we already discovered that if we do have multiple massless particles of spin two they have to be in decoupled universes okay so essentially tells us we cannot have massless particles with spin higher than two and the massless spin two one has got to be unique first of all and the second thing we learned is that the strength of all the other couplings for any spin here all of these have to have the identical strength all these have to have the strength one over amp one all equal and this is the equivalent principle but uh any everything else uh uh that i uh but much more than the uh equivalence principle of course okay just just nailed all of this right but this includes the spin three halves is okay massless spin three half is okay all right so massless spin three half is not ruled out massless spin five halves is uh is ruled out by these arguments uh and above uh so now we are almost done with our exercise this is the list of spins we're allowed to have okay and uh it's worth doing a little bit more investigation on this guy right so here's the uh the the final residue argument is uh it's perfectly okay to have okay to have okay to have one massless spin three halves and one graviton massless spin two this is okay but can i take the system and just add let's say i want to add one scalar to it one scalar five okay so let's now look at the amplitude for the scattering of the spin three halves guy so here's the uh spin spin spin three halves guy three halves three halves scattering against five five five all right and of course there's going to be some residue here in the t channel coming from the graviton well what do you think happened surprise surprise the residue in the t channel has a pole in the s channel that's we've seen that all all over the place in these arguments that means therefore that we have to have a particle in the s channel but what is that particle what could it be if this is uh spin three halves what could propagate here to give me a residue in the uh in the uh s channel well whatever it is has to be a fermion and we begin to see that this is only consistent if there are super partners okay so this i can't just take a theory of the massless spins we have with gravity and randomly add a scalar i have to add the scalar together with a fermionic friend such that the coupling constants here are also one of ramp walk so the coupling constants have to be related to each other okay and you can really continue along this argument and show that massless spin three halves with just one of them one massless spin three halves forces n equals one susie and you can really continue with this argument to show that you can have up to n equals eight massless spin three halves but not nine and that really completes the uh the elit so we can have particles of spin zero one half one three halves and two and uh already we saw this one is unique we can have at most eight of these n less than or equal to eight of these and then have to go along with susie and actually super gravity now we can have as many of these as we want so you get less constrained as we go to smaller spin as many of these as we want but it's associated with the yang mill structure and these are the more boring guys okay um they can interact with each other however they want but the interactions they have with the other guys are totally nailed okay well anyway as i said in the beginning of this lecture this is really what i wanted to cover in the first sort of lecture and a half of these lectures and it took me four lectures to do it but i hope what you've gotten out of this is a deeper appreciation for how absolutely directly the world we see when we look out the window its broad properties are totally like nailed by these uh broad principles of um uh of quantum mechanics and uh relativity it's really extraordinary again we didn't have to be einstein to come up with this literally you didn't have to be an einstein you have to know quantum mechanics and you know smaller einstein special relativity but you don't need all of the fancy stuff curved spacetime falling elevators all those thought experiments that's that's the beautiful it's the way einstein did it but it's somehow even more striking that it's just there's no other way it could be you know it's not like the universe could have been some ugly way and it could have been some pretty way and it decided to mean this pretty geometric way no it just couldn't be any other way it's an i think a totally extraordinary fact and it also shows however this huge difference between and all of this stuff where did all of this come from all of this came from this dramatic shift in the number of degrees of freedom between massive and massless particles this entire story right from the beginning uh uh came from that uh sort of uh innocuous seeming but rather dramatic thing this business about not being able to be a little massless a little massive being like a little pregnant okay so you can't do that there's something dramatically different between masses and massive and it goes right to the heart of all these things that we understand so incredibly constrained from our even conventional point of view and what the reason I wanted to do all of this even though it took me uh price was long to do two and a half times as long to do as I wanted to is that highlights even more how damn weird it is that we got this guy okay um this is the one thing which whose degrees of freedom does not jump as we go from massless to massive and that is the heart of what makes the Higgs so incredibly strange to begin with that's the heart of the hierarchy problem which is not an aesthetic problem is not a problem of people think something is pretty or not pretty it's a very sort of hardcore problem um uh um and um uh but even without talking about in the detail it should be clear that there is something weird about this guy and there's a reason why why we've seen these other things before we've seen many of these many of these okay this one there could only be one of them to begin with this one we haven't seen any yet but there's a reason why we haven't seen 10 thousand of these floating around of these guys okay there is something unusual about them and uh and that's the you know without really saying much else that the the zero-thorner justification for putting the damn thing under a microscope and studying it as the uh experimentally studying it as intensively as we can but anyway um those of you who are interested in hearing more about the kind of second half of these lectures and uh it's those of you who want to participate also um I mean I'll prepare some lectures but we can also really make it a long question and answer session as well so we could just chat about it that might be um also a fun way to do these things um but anyway I hope to talk about that as well as in a little bit more detail um uh how we can um further this program and possible future colliders and even what we can do today um uh in order to try to make these uh colliders uh have more of a chance of actually coming becoming a reality all right and so I guess with that I will stop thank you very much everyone right so let's start now the q and a don't do it