 That I will do should I Have just one quick announcement before we get started all the students Will have their certificate of participation in their mailboxes. I think some of you need it for you know reimbursement and things like that So everybody will get the certificate of participation in their mailbox Let's get started for the day Those lectures The structure of the recursion relation we derive from a contour integral and we found that you can write the n particle Scattering amplitude again. I'm using capital a here. This is the color ordered Partial amplitude as a sum of residues of this complex function in the complex plane and This is defined. What is this function a of z? This is defined as a one complex parameter family of Deformations of our scattering amplitude of interest where we deformed the kinematics or particles one and two now one of the questions that arose in last lecture at the very end when I didn't have time and I had to postpone it was How do I know there's no contribution from infinity so when I derived this formula? I took a contour and I pushed it off to infinity Picking up all the poles you get along the way Indeed the pole at infinity is something you have to worry about in Engage theories and gravity theories you can prove that there is no contribution at infinity I won't give that proof right now because it's a little bit orthogonal to the treat to the direction I want to go in and it doesn't really teach you very much Except the fact that there's no pole at infinity of course One thing that works is you can always you I mean you can always take any formula for an amplitude and check that it Has no pole at infinity. So I'll just write here It is oh, sorry Sorry, that shouldn't be there. So let's see Thank you here. We've shifted the lambda tildes Here we've shifted the lambdas and I just put the hat here the hat is a conventional notation to remind us that these are the two shifted particles Oh, and it's very important to emphasize. I picked particles one and two here But there's absolutely nothing special about that you could pick you could pick any two particles and do the shift equally well it is a fact that You can always choose two shifted particles so that a of z Goes to zero at least as fast as One over z As z goes to infinity and that implies that there's no contribution From a pole at infinity So the last thing I had done at the end of the lecture was to discuss where this object has poles There does Have poles so in our complex z plane It has poles anywhere where Internal propagator can go on shell. It's important to emphasize Although we in Internal although we we we try to avoid computing anything Sorry internal propagator On shell although we avoid computing anything by Feynman diagrams It's important for us to always keep in the back of our mind that Feynman diagrams exist So if we need to appeal to any knowledge or properties of Scattering amplitudes well We always know that they can be written as sums of Feynman diagrams and at tree level Sums of Feynman diagrams will always give you rational functions of kinematics And moreover, we know exactly where the possible poles are because poles in any Feynman diagram come from Internal particles going on shell so this gray blob means so any Feynman diagram of this form any Feynman diagram of this form as a pole and this is Just what I was writing at the end of last time Z equals one half P2 Hey squared over this funny Bracket object it's important to notice that Because oh and notice of course that the external particles are given their fixed external ordering Because I happen to choose particles one and two to do my shift It's important that one and two be separated in this diagram. So let me make a side comment here Diagrams one and two appear on the same side Have no pole. Okay, so if you had a diagram like this one two and Then something else three up to n This propagator here is one over P1 plus P2 squared but One hat plus P2 hat is exactly equal to P1 plus P2 So that's the important property of the shift that we discussed last time it preserves energy momentum conservation, so If you shift both P1 and P2 this sum remains constant So this propagator here doesn't depend on Z So this does not have a pole in Z Okay, propagator does not depend on Z. So there's no pole Okay, so poles only come from diagrams where the two particles that you have chosen to shift are separated from each other by some Propagator, okay, it remains now only to calculate the residue Loop diagrams open a whole new can of worms because in addition to poles you have branch cuts so the the discussion gets much more complicated there and People have looked at one loop successfully, but beyond that these methods aren't really applied directly to loop amplitudes There there there is a more sophisticated version of a recursion relation for integrands of loop amplitudes Let's compute the residue Let me call this zk Let's compute the residue of a of z over z at z equals zk well going to be equal to So whenever you're calculating a residue You can always pull out anything that's non-singular Okay, so it's going to be one over zk I can just pull that factor out times the residue of a of z and this is going to be one over zk times the residue of Let's call it a left Times one over p2 plus right Okay, what have I done here? this thing this thing is the piece of a of z containing all terms That have This propagator What I mean is Again, imagine you were to write down all possible Feynman diagrams that contribute to this amplitude We'll never do that, but imagine in your head you do that Well, they're separated into two categories. Some Feynman diagrams have this propaganda. Sorry this propagator You know, there may be many other propagators inside here and inside here Only these terms only these diagrams contribute to that residue Any other Feynman diagram that does not have this propagator Does not contribute to the to the residue it doesn't have that pull Okay, now When I look at the Feynman diagrams that do have this propagator It's clear that all of them that the entire sum over those sets that set of Feynman diagrams Can just be written as the sum of all possible Feynman diagrams in this gray blob Times that propagator times the sum of all possible Feynman diagrams in the right side in the right gray blob So that's what I mean here in this formula a left means This amplitude Literally the sum over all Feynman diagrams of this topology and a right means the sum of all those Feynman diagrams But those things are non-singular So I can also pull them out of the residue the only thing that's oh And I guess there should be a hat here because I want to have a z dependence Okay, so we have residue of this thing um Plus squared minus 2 z 1 p 2 plus a 2 We had this formula last time if you expand If you expand out this denominator and you recall so I'll put it up here recall Equals p 2 minus z lambda 1 lambda 2 tilde if you plug that in here and then expand it out you get these two terms and This residue it's easy to compute is nothing other than minus z k over Us the residue at z equals okay, so in conclusion the amplitude we're interested in on through n and Be written as a sum over k And let's think for a moment about what's the range of k? Well, you need to have at least three particles on both sides Okay, other of both sides of this cut otherwise you get zero So the minimum value of k We're oh I'm sorry. Here's k. So the minimum value of k is three and the maximum value of k well, you need k plus one You need K plus one the biggest value can be is n so the maximum value is n minus one if a left One hat sorry sorry sorry a left starts with Uh-huh like this k plus one k plus two up to n up to one hat See I'm reading it cyclically you always read these things cyclically so the left amplitude reading it cyclically is k plus one through one The right amplitude then I've got this explicit propagator down here and these hats mean Valuated z There's only one more slight subtlety, which I glossed over in that I Said oh when you look at any kind of Feynman diagram of this form any kind of Feynman diagram of that form the slight subtlety is There are two such classes of Feynman diagrams Because gluons being massless vector particles come in two polarization states or helicities So you can have Feynman diagrams where the helicity coming out of this leg is plus or minus one But then since it connects into that diagram going inward the corresponding helicity there has to be the opposite Okay, so in addition here. There's a sum over the helicity of the exchange particle. Oh and oh my gosh and Yeah, I forgot all this I forgot about the I Forgot about this extra leg So let's use capital P to denote the momentum carried by that leg By momentum cons by energy momentum conservation. This capital P is going to be the sum of all of these So I apologize for that the capital P here with helicity h and I Need a minus capital P here with helicity minus h and that means that that just means so let me be explicit capital P Well, you can take it to be P2 It's the same thing that's squared in the denominator. Yeah Okay, so the important form point about this formula is that These two amplitudes that you need to know here. They're always smaller than the one you started with Okay, so it's a genuine recursion if you want to know the eight particle amplitude you can compute it via three particle amplitude times seven particle amplitude plus Four particle amplitude times six particle amplitude etc So you can always break everything down recursively until you run into your basic building blocks Which are your three particle amplitudes? So this recursion is Seated you always whenever you have a recursion you need to specify the initial conditions is seeded by the Three particle amplitudes and I I discussed their construction last time are there any questions about this? formula so this formula applies in any In any it applies for gluon scattering amplitude in any gauge theory at tree level all right, let me tell you about the simplest set of solutions to this recursion relation Now when I make that comment, I'm being anti-historical because I'm going to present results that were discovered in the mid-80s Long before this recursion was known the simplest Amplitudes those with negative Felicity and And minus two positive Felicity These are called MHV. We use that word last time, but I I never told you what it meant and I promised I Would tell you what it meant and that's where I'm getting to now Which are given by If you if you tell me that you have n particles all of which have positive Felicity Except for i and j that have negative Felicity then the answer is really simple It's ij to the power 4 over the cyclic product 1 2 2 3 Up to n 1 No, this should be n-plus These are all Of them are positive Felicity except for i and j the three particle amplitude. We talked about last time is A special case of this It's really a useful exercise The school is almost over so I don't have time to assign homework But it's a very useful exercise to check that these are a solution of this recursion relation and It turns out to be very simple If suppose you wanted to check that that formula satisfies this recursion relation You'll very quickly find that in this sum In fact in this double sum only one term Contributes so no matter how no matter how big n is There's only a single term in the sum that contributes all the rest are zero and you basically get a trivial relation between this thing for n particles and This thing for three particles and this thing for n minus one particles. It just pops out algebraically So I encourage you to give that a try if you're curious This is called the park Taylor formula or park Taylor Barron Schiele. It was discovered in the 1986 1987 and it was really sort of the underlying motivation behind a lot of Amplitudes developments because think about it this simple formula encapsulates a huge I mean an orbit since this is for arbitrary and it incorporates an Imprinciple arbitrary number of Feynman diagrams into one very simple compact formula So the fact that Feynman diagram calculations are extremely messy yet. They add up to something so simple should be some kind of indication to you that there's some deeper mathematical structure and it's You know to your benefit to figure out what that structure is Because if you understand that structure better, it'll help you in your future calculations or just help you understand better the mathematical Foundations of the theory. Okay, so um, oh I still haven't explained the word MHV. Sorry MHV stands for maximally Felicity violating so you can check that using the recursions that if you have all of the same if you have all the same Felicity or Only one negative Felicity glue on The amplitudes have to be zero So the first case where you have a non-trivial answer is when two Particles have negative Felicity. Oh and just a quick comment Of course parody invariant says that you can you can conjugate everything So if you take this and do a parody transformation on it, you'll get n gluons all with negative Felicity if you parody conjugate this you'll get n minus 1 with negative Felicity and 1 with plus So there's a parody conjugate formula analogous to this that involves the square brackets instead of angle brackets. So just to be Absolutely clear. I'll write the formula for MHV bar, which says that 1 minus 2 minus i plus j plus n minus is equal to i j to the fourth over 1 2 2 3 2 n 1 There's a little bit of a puzzle here because if you look at the four particle amplitude With two positive and two negative This will give you a formula that will give you a formula. They look nothing like each other, but secretly they're equal to each other Okay If you do like 1 plus 2 minus 3 plus 4 minus on the one hand, this is equal to 2 4 the fourth power over 1 2 2 3 3 4 1 But on the other hand if you use this formula, it's equal to 1 3 to the power 4 over 1 2 2 3 So secretly these two are equal to each other though. It's not obvious and I bring this up now because remember some of my motivation in the first lecture was saying that we want to adopt a set of variables that are more naturally suited to To Providing exactly the right amount of physical information to specify a scattering amplitude without any redundancy Here we're starting to see that even these spinner helicity variables while they are useful. They don't yet completely achieve our task Because I mean this is a relatively simple example if you start going to an 8 or 12 or 15 particle amplitude It's easy to start writing Various different formulas that look nothing like each other you know that go on for pages and pages and pages and Mathematica and They don't look anything like each other, but they're secretly equal to each other That's a hint that we're still not using Absolutely the best variables because if you're using the best possible variables that Exactly cover your kinematics space with no redundancy And if you're checking two rational functions whether or not they're equal to each other It should be obvious that they're equal to each other You just subtract them in Mathematica and and do expand and you get zero Okay, that doesn't happen here because these variables are still constrained by overall momentum conservation so The next thing that we're going to do is to alleviate that problem But in order to Yeah, actually, let's just jump right in so this just takes us to the Variables called momentum twisters. So let's just recall the lambda lambda tear lambda tilde variables Satisfy p squared equals zero automatically and nicely encode Nicely encode Gauge invariance of the epsilon of the polarization tensors So that's we've just I've discussed the last two lectures. However They are still constrained By a quadratic relation. Okay, if you have any n-particle scattering amplitude Oh, and I didn't mention this this was implicit and what I was mentioning before but we always adopt the convention where In any in any scattering amplitude all moments are pointing outward. So there are no minus signs here in our energy momentum conservation relation So the point is the lambda lambda tilde variables are still constrained and they satisfy a quadratic Constraint and that's a little bit annoying. So again going back to this example It certainly is not true that this is identically equal to that No, that's not what we're saying. What is true is that this is equal to this on the locus Where this momentum conservation is satisfied? so what we would like to do now is to adopt a new set of variables which Trivializes this momentum conservation relation. So the first step in that construction is to define what we call dual variables or region momenta and Oops There's unfortunately some ambiguity in the literature different references use Some people but I minus one some people have an opposite sign there So the point is the following that if you parametrize your if you introduce variables Xi That are related to your momenta by this formula then they will automatically satisfy momentum conservation So then some I equals 1 to N PI is automatic and All indices are our cyclic Pn by this formula means xn minus x1 now You you gain the fact that momentum conservation is automatically satisfied There's there is still a tiny bit of redundancy left in the X variables because if I hand you a certain collection of momenta They don't determine the overall they don't determine where the origin is in X space You always have the freedom to do an arbitrary translation in X space, but That's fine. That's no problem. It just means that anything we calculate will have necessarily translation and variants in X space These of course You know these are everything here is a four vector So there should be some indices so the Yeah, so it's very important in everything that we do now It's going to be very very important that there is a cyclic ordering. So The first lecture where I said that we can decompose any amplitude Even at loop level if we focus on large n we can decompose any amplitude into a sum of of of Primitive objects each of which has a well-defined permutation or cyclic structure Okay, so Let's contract. So the next step in our construction is we're going to cut well, let me say Let me write the words before I do the calculations we'll appeal to the correspondence mean point in Minkowski space lines in P3 and Null lines minkowski space and points in P3 is complex projective three space Okay, so those are the words. Let's do the formulas now so that you see how the construction works first we take this formula and We contract it with Lambda So when I when I use the word contract I mean multiply by Epsilon a b lambda b on both sides This of course is lambda i a lambda i tilde a dot So on the left hand side we're going to get zero Because oh there should be an eye there I apologize on the left hand side. We're going to get zero because the Epsilon is going to kill we have lambda i Antisymmetric inner product with itself so we get zero equals And now I'm going to use a somewhat shorthand notation because I'm a little lazy to write all the indices You'll see what I'm doing in just a second So what I'm going to do and I'm just too lazy to write all the indices this thing has an a and an a dot This has a dotted index and an undotted index the dotted index I'm displaying explicitly there the undotted index of this and the undotted index of that are Contracted with the epsilon symbol so the general rule here is anytime anytime two common indices are contracted They're always contracted with the epsilon symbol, but I'm going to leave that implicit rather than having to write it every time Okay, so I'll just write it here once X i lambda i a dot means Epsilon a b X i a a dot lambda b Okay, so we get this formula and This formula tells us that these two things are equal to each other Okay, so let's give them a name Let's define You I a dot to be equal to that thing let's collect the four components for each particle I collect the four Components lambda one lambda two Mu one dot if okay This is not an eye. This is a one with a dot over it. Oh and each of these has a subscript i Let's collect these four components into a single object e I Now ranges over one two three four and I of course ranges from one to n now this at this stage We're doing violence to Lorentz symmetry Okay, because we're we're combining two objects that we don't really have a right to combine Something that transforms under the left-handed s u2 of the of the Lorentz group and something that can Conforms transforms under the the right-handed s u2 of the Lorentz group We're combining them into a single four component object But we're going to see it's to our advantage because we're going to get something that exposes an even much bigger symmetry Okay, we're going to get to see a symmetry that relate that rotates Makes arbitrary rotations in that four dimensional space um Now an important point is that z these z's are naturally variables in complex projective space so note easy are Indistinguishable for any non-zero complex number t and that's because in this defining relation if I Simultaneously multiply all of my mus and all of my lambdas If I simultaneously multiply all of my lambdas and all of my mus by a factor of t I mean this equation is still satisfied. So in other words There's an ambiguity by an overall rescaling and That is the very definition of p3 complex projective space p3 Have a formula here That I should show No, no, you don't need to scale the x's but I you On on this blackboard. I have not written the lambda tildes Of course, you would also have to rescale the lambda tildes by a factor of 1 over t And in fact the formula I'm looking for which I have right here. Here we go From that from the definition from these equations you can work out what the lambda tilda has to be I'll just write it lambda till the I Has to be equal to I minus 1 I I plus 1 plus I I plus 1 mu I minus 1 plus I plus 1 I Minus 1 mu I over I Minus 1 I I I plus 1 Everything here has an a dot Hey, where did this formula come from this formula comes from the fact that we know Well, you have to reverse engineer a little bit You know that the x's are related to lambda lambda tilda by this way and mu is defined by this So it gives you a system of equations and when you solve them You'll find that the lambda tildes are given by this formula and let's check let's check that this satisfies the correct Scaling property. So if we simultaneously rescale all the lambdas and all the mus by a factor of t The numerator here will scale by a factor of t cubed and the denominator Will scale by a factor of t to the fourth. So indeed this gives us the correct scaling note Lambda tilde scales like 1 over t lambda tilde as required Keep p equals lambda lambda tilde unchanged Are there any other questions? So where we are so far? Let's just review if if you give me a The the the If you give me the kinematic information of a scattering problem, I can encode that I can encode that in Endpoints in p3 Now I want to explain the reverse I want to I want to show that if you give me endpoints in p3 that I can reverse engineer the kinematic information So Okay, so suppose you give me endpoints in p3 Then I'll define the lambdas and the mus by defining the lambdas to be the first two components of each z and The mus to be the second two components of each c then consider the following equations a Dot is x i lambda i a dot and Mu i minus 1 a dot is X i lambda i minus 1 a dot Okay, these are almost the same equations I have over here, but not quite because It's important that I have the whole collection of equations at my disposal. So the point is here Here I've written the two equations in which x i appears Okay, so all I did was I took this relation here and Shifted i by one. Okay, so the the equations hold separately for each i So I can just take i and shift it by one and I get this So so the point is given a collection of lambdas and mus define X i To be the solutions to these equations. Okay, these are four Each x is four independent variables And this is a set of four linear equations But we can derive an important Consistency condition Having found those solutions They satisfy equals x i lambda i a dot mu i equals x i plus 1 lambda on a dot Okay, so here again, I've reverted to the original Set of equations by shifting the indices so it's I think I've said this too many times But we have the whole collection of equations. So you give me a whole collection of z's I Write for each i this set of equations on x i I then solve all those equations Simultaneously to find all of the x i Well, they that collection of x i satisfies this these equations and now if you subtract them You get zero equals and I'm going to use some slightly abusive notation here Again, this means contracted with the epsilon symbol. So this would be epsilon a b a a dot a a dot Anyway, the point is We conclude or this this formula implies that the determinant of x i minus x i plus 1 is 0 You see here the only way the only way you can find a Non-trivial solution think about this like a two by two matrix equation right pull the epsilon into there This is a two by two matrix acting on a two component column vector The only way you can have a non-trivial Solution to that is if the determinant of this two by two matrix is zero This implies that the determinant of p i is zero which implies that p i mu is null So we've proven that a collection of n ordered points b i in p 3 Contains exactly the same amount same data as a collection of n cyclically ordered I'll just put ordered null momenta in minkowski space satisfying overall momentum conservation But a picture is worth a thousand words So if you haven't appreciated this derivation, let's just draw the picture that goes along with this Yes Yes, I Will that's an important point and I'll discuss that in just a moment so You well, I'll discuss it now you can do this construction in any quantum field theory you want that has massless particles That have some natural ordering so engage theory in the planar limit. We have a natural ordering because of the trace structure okay But there's something very special about particular Quantum field theories, which is that they exhibit the full symmetry of that p3 space that you're not talking to But not all not all quantum field theories will exhibit that symmetry So, yeah, we'll discuss that Yeah, yeah, it's not quite symmetric product because you can't allow Yeah, I know yeah, yeah well Yeah, that's right. It's a We're gonna get exactly to that point. It's a space mathematicians called configuration of endpoints in p3 and that's exactly the kinematic space That we're going to be talking about so here we have Minkowski space And here we have p3 if you talk about a scattering process in Minkowski space you have a collection of null momenta Like this is p1 So this is a four vector in Minkowski space and it's null Okay, I can't draw it null on the blackboard, but of course it's null Now because of overall momentum conservation, I can put all of my p's back to back and they'll form a closed polygon so this is a polygon with Null edges So each edge represents a null vector in Minkowski space and it's a closed polygon because of energy momentum conservation the vertices here are the locations of the X's okay, so this would be the point x1 the point x2 the point x3 the point xn and As I mentioned earlier, there's an overall ambiguity. There's a translation invariance in this space Because it doesn't matter where You know the the p's are defined by differences of X's Oh, and I'm sorry I This is inconsistent with my earlier labeling this should be x2 Because p1 Would yeah, it's a little confusing because different references use different conventions Over here in Minkowski space the point is similar we have a collection of points z1 before and then we have a collection of lines That you can draw between successive points So this is the line One two line two three line Four etc. And eventually this will close as well over here. You have zn. Oh my gosh. I didn't mention it Well, if you look I didn't mention it before I if you look back at the equations here these incidence relations I didn't use that word yet, but these are called incidence relations If you look at the set of all if I give you a collection of X's okay Like if I give you a single X and you look at the set of all lambdas and muses that satisfy The incidence relation for that given X It's a set of linear equations The solution is a line. So each line in twister space corresponds to one of these points X So the line here to three Corresponds the point X to line one two corresponds Point X1 sorry. Yeah, that's right Yeah, P1 a complex line. Yeah, okay, so I'm out of time now X1 X3 Okay, I'm out of time now. So I just drew the picture So the point is the following The only important point you need to take away from this is that the Z variables provide a free Parametrization of exactly the kinematic space for mass of scattering if you if you draw n random points in Minkowski space It's not guaranteed that they will be null separated from each other Okay, like if if you pick if you're looking for a 10 particle scattering amplitude and you pick nine random null momenta and You let the 10th momenta be given by momentum conservation is minus the sum of the first nine It's almost guaranteed. It won't be null unless you made a spectacularly lucky choice So this geometry is very rigidly constrained by requiring each of these to be null and requiring it to be a closed polygon Here the geometry is completely unconstrained you draw n random points Boom boom boom and random points in p3 then you connect one point to the next by a line Okay, and then you follow the algebra which is on the blackboards here to reverse engineer and you'll find that each line here automatically Corresponds to some point here and those points are automatically null separated from each other Okay, so in my next and last lecture will we'll be using these Momentum twister variables Extensively, so thank you