 OK, good afternoon, everyone. Is there any question, some other thoughts on yesterday's lecture before we start? OK, incidentally, the slides should be online. In fact, they are online already, I verified. So if you go to the agenda, you can get the slides both of yesterday and of today available. So here is a reference that includes a very nice overview of QCD aspects of LHC physics. It's not up, you know, 2006. So it dates before the start of the LHC, but the contents are still of value today. So yesterday, we discussed the evolution of initial states. And we focused on Drellian, which is a process in which initial state evolution is the only thing that really matters. Now today, we focus on the evolution of a final state. And then to make our life easier, we start with the complementary process of Drellian, namely a process in which the initial state poses no problem from the point of view, at least, of QCD. And that is E plus C minus annihilation into hadrons. So one thing that we know from asymptotic freedom is that if we are at sufficiently high energies, the cross section for E plus C minus to go to hadrons is well described by the cross section that we can calculate at the pattern level for E plus C minus going to quarks and gluons. At leading order in perturbation theory, the equations that are needed are those that you saw this morning in the electro week lecture by Professor Peskin. In particular, the cross section going through just the virtual photons, so in a regime in which the resonance is not playing a key role, is just proportional to alpha square, of course, plus the sum of the square of the charges of the quarks. And that leads to the very famous prediction for the so-called R ratio, which is the ratio between E plus C minus going to hadron to E plus C minus going to mu plus mu minus, which is just proportional to the number of colors and the sum of the square charges of the quarks. If we move into a region where the Z pole is dominant, then, of course, we need to correct these by the relevant vector and axial couplings of the quark to the Z. And you get more complete expression again from this morning's standard model lecture. So if we add higher order corrections from QCD, namely the emission of an extra gluon or the virtual exchange of a gluon between the QQ bar, we get the correction, which is of the order of, well, which is equal to alpha S calculated at the center of mass of E plus C minus divided by pi. If we take, in particular, a collision at the Z peak, so around 100 GV, this correction is of the order of 3%. Alpha is of the order of 0.1, 0.12, 0.118. If you want to be more precise, divided by pi, that's about 3%. So it's a very small correction. Nevertheless, it leads to an evolution, of course, of the rates when we take the ratio with the E plus C minus goes to mu plus mu minus. E plus C minus goes to mu plus mu minus, the Q dependence due to the running of alpha retro-magnetic is very, very, very slow. So we don't really, well, it's in principle measurable, but it has more correction compared to the running of alpha S. And when we take this ratio as a function of energy and we plot the extraction of alpha S that we obtain from this formula, we clearly see the running, the evolution of alpha S. And the agreement with data is excellent, provided, in fact, that the number of color is equal to 3. So the effect of N equals to 3 comes from here when we take the ratio, because then if we go to one loop, this expression will be Nc times the sum of the charges times 1 plus alpha S divided by pi. So we certainly have to put the number of colors to be equal to 3 in order just to match things. And then, of course, the number of colors also enters in the running of alpha S. If you go back to yesterday's slides, I gave the expression for the alpha S running. It was 1 divided by B0 log of the scale. And that B0 contains the number of colors. It's 11 times the number of colors, which is 33, minus twice the number of flavor divided by 6 pi. Sorry, the question was, where does the number of colors enter here? Yes? I thought this is due to the addition of 5 and less. I thought that this term may be. So the question is, what is the scale that should enter here in alpha S? The scale that enters in alpha S is the scale it's square root of S hat. It's the energy in the center of mass of E plus C minus collision. This is because if we do, and it's dominated by the result of the virtual gluon exchanges, OK? There are logs that appear in these calculations, OK? Log of the kinematics, log of square root of S hat. And they can be reabsorbed into the running of alpha S. You could write this expression with alpha S at the fixed scale. And then what you would find is that there are additional contributions which are proportional to these logs. And those contribution proportional to logs of the scale at which you are taking the data divided by the reference scale can be absorbed into a redefinition of alpha S by using the running alpha S. So this is the most efficient way to incorporate the effect of these logs into the running of alpha S. That's the manifestation of that, OK? Now, the one thing that I want to point out here is that you see we have corrections coming from the emission of a third particle, which are of the order of 3%. So the bulk of the process, when we look at it from the point of view of perturbative QCD, is just 2, 2, 2. It's the production of two quarks in the final state. When we go and we actually look at these final states and we ask ourselves how many particles are produced in the plus and minus collisions, say, at the z pole? Well, and we just count, say, charge tracks. So pi pluses, pi minuses, protons, charge chaos. What we find is that this is the distribution. And the scale is, you know, the peak is at about 25. The average number of charged particles that are produced in each of these collisions is 21. The probability that we have only two or three particles, in fact, you see that there is no data here. It looks like it's practically 0. Certainly it's 0 on the scale of 10 to the minus 4, 10 to the minus 5. I don't think that it's ever been detected a decay of a z boson. And there is a statistics of about 10 million events from Lape and SLC. A z boson going to two and only two charge tracks. That is exactly 0. So that's a bit bizarre. Because on one side, we do perturbative calculation. It's a 2 to 2 process. And we get pretty much 100% of the cross-section. Then we go and look at what appears in reality. And it's never two particles. It's a gazillion particle. It's 20 on average, but it can become 50. And of course, if we add neutrals, there is 50% more. So that is apparently a puzzle. And proving that this is, of course, understood in the context of QCD will be the main focus of my lecture. And the starting point that guides us to the solution visa posteriori, of course, is to look more closely at the structure of these events. Events in which the Z decays to this multitude of 20, 30, 40, 50 hadrons. How do they look like? Are these addrons going all over the place? Or, well, this is the most typical final structure. Namely, there are two so-called jets, so streams of pions, chaos particles going back to back. There are also events where, instead of having two streams, we get three streams of particles going back to back. They tend to be very collimated. You see them as being spread because there is a magnetic field inside the detector. That's why some of these tracks curve, because they have low momentum. And then they just veer away in the magnetic field. But typically, these are really highly collimated objects. So these we interpret as associating the jets as being the manifestation of the evolution of the quark. The quark is not, of course, an observable particle. When we produce a quark, there will have to be hadronization. Something will have to happen for it to turn into a visible object. And the result of that process is the buildup of this jet. So this is the plus and minus. If you wish, these are final states arising from a plus and minus going to QQ bar. These are final states arising from a plus and minus going to QQ bar plus a gluon. So this is that famous 3% additional contribution that comes from order alpha s. And indeed, if we go and we count how many free jet events there are relative to two jet events, it's a number of the order of few percent consistent with being an order alpha s process. So this is the physics as nature tells us. And now we have to understand where in the Lagrangian of QCD this behavior is hidden. Where does it come from? So if you want to understand, of course, we go from two quarks to many particles by radiating something. That's the way in which we build up a multiplicity. So we start by understanding how the multiplicity builds up. And to do that, we start from the simplest case, which is the emission of a single gluon from the QQ bar. So the first emission. Now, these are the amplitudes. Let's assume that the Z boson is on shell. We can consider these also being a virtual photon. It doesn't make any difference what we're about to say. We can write out explicitly the exact three-level matrix element for this process. What I call gamma mu here, this capital gamma mu, is just the vertex we put at this point. If it's a photon, capital gamma mu is just a standard Dirac gamma mu. If it's a Z boson, there will be a contribution from gamma mu gamma 5 and a purely vector gamma mu. Now we can just reorganize the fermion propagators. And we end up with an expression that has these and that propagators explicitly as to p dot k and p bar dot k. So these denominators, these propagators, become singular when p dot k goes to 0. p dot k, we can write as the product of the energies of the core k is the momentum of a gluon. p dot k times 1 minus cos theta. Theta is the emission angle relative to the p leg. So when cos theta goes to 1, collinear emission, this goes to 0. And the amplitude receives a growing contribution. And likewise, when the gluon energy goes to 0, we have a singular contribution, which means these are the two regions that are going to dominate the emission. That's where the gluons are going. The collinear emission is not going to alter the structure of our event. Collinear emission, in fact, goes in the direction of this fact that we observe, namely that as the quark evolves, it evolves into a stream line, a stream of particles. Collinear emission means that indeed if it emits something, that something tends to go in the same direction of a quark. That's exactly the evidence we have from the data. It's not altering the global structure of a final state. Soft emission, on the other hand, gives rise to large contributions no matter what the angle is. So a soft emission can take place even at 90 degree at a very large angle from the quark. More than anything, soft emission means taking away color from the quark because the gluon is colored and it takes away color. And if we take away color from the quark, we are distributing color in phase space. And it means that at the end of the day, when we have to neutralize color to construct color singlet hadrons, we will have to reorganize and rearrange all of these particles going all over. If I have a gluon going at a very large angle relative to the quark and taking away color from the quark, the two objects are going away from each other in phase space. As I take two charges away from each other in phase space, the strong interactions become large, become non-perturbative. And therefore, at some point, I develop very, very strong interactions between these two objects. If these two objects have gone far away in phase space from each other, and all of a sudden, I have this huge long range interaction, I will have to this as the potential to completely scramble the kinematics that's been generated at the part on level. In other words, when I have a collinear gluon, the gluon goes in the direction of the quark. They will be close to each other. So when I have to neutralize the system, they will be going together. And whatever happens there will maintain the pencil-like nature of a jet. But when I have a soft gluon, I could well end up at the end of the day that the kinematics is completely controlled by non-perturbative effects. Before jets were observed, in fact, in the experiments, there were people who thought that the final structure of the plus and minus going to Hadron's did not look like jets, but looked more like what was called at the time as a fireball. What they thought was the following. We created a couple of QQ bars. The QQ bars go away from each other. At some point, since the carry-opposite charge, the confinement force will take over. So they will bounce back. And then they will bounce back again in a different direction. The whole system just needs to have overall momentum equals to 0, which means that they can go any direction they want. And overall, out of this motion, that's entirely driven by long range, strong non-perturbative interactions, I would build up a final state that has pretty much a uniform structure. I would be completely losing these part on level naive intuition that the final state is well represented by the two quarks. So if I want to do physics, if I want to be able to do a calculation in perturbation theory of quantities such as kinematical distributions, angular correlations, I need to make sure that indeed the parton well represents, at an inclusive level, what will be the final observables in my lab. And this is the problem that we have to face. We have to prove, in other words, that the emission at a large angle of these soft gluons is not going to screw up completely the picture that we developed in perturbation theory. So let us study now in more detail these soft gluons. And if we want to study soft gluons, first thing to do is to put the momentum equals to 0 wherever we can. So soft means momentum goes to 0, so very long wavelength. So we take this expression. We put the momentum of the gluon going to 0, except, of course, in the denominators, because that's where it's singular. But here, for example, we can take k goes to 0, and we are just left with p slash. With a k goes to 0 here, we are just left with p slash. And as an exercise, you just play with the Dirac algebra here. And what you can find is that this amplitude with the gluon emission in the soft limit can be written as the borne amplitude, so the one for, say, the photon of the z going to q, q bar, times a kinematical factor. Now, this kinematical factor, which is called the iconal factor, is completely independent of what the borne process is. And that's why I kept this capital mu here. You can verify, and it's simple, that it doesn't matter what the nature of this coupling here is. This could have been the coupling with the photon, with the z, even with the scalar particles, such as a Higgs boson. It could be a tensor-like coupling, like a graviton decaying to a no-shell graviton decaying to q, q bar. This will always be true. So we do have a factorization between the hard process and the effect of emitting a soft gluon. The reason why we have this factorization is, again, very similar to the factorization we discussed yesterday in the case of the initial state evolution of the big lap evolution. Namely, dealing with a soft gluon means dealing with an object that has a long wavelength. If it has a long wavelength, it cannot be sensitive to what happens at very, very short distances. Yesterday, we were discussing about the decoupling of time scales. Today, we're discussing about the decoupling of length scales, but it's pretty much equivalent. So a gluon with a long wavelength is a classical object in relation to what happens at very short distances. And as a classical object, it doesn't really care about what happens at short distances. In particular, it doesn't care about the spin of the particles that are involved, because the spin is a purely quantum concept. And this is why the Dirac algebra somehow works out very easily to give out such a result. You see here, all of the Dirac matrices, all of the complexity of spins, et cetera, are reabsorbed into the born amplitude. But here, there is no gamma matrices. There is no Dirac. This is completely independent of the spins of the particles involved. I could have used the scalar quarks. Instead of fermion quarks, I would be getting exactly the same result. So it's a fully universal result, which I encourage you to prove. Furthermore, I've used here for simplicity, massless quarks. You should do the exercise, putting the mass to the quarks. So that changes slightly the nature of the propagators. But you will find that, again, this is the result that we get. So it's independent of the spin. It's independent of the mass. It's independent of the specific nature of a hard coupling. So it's really a factorization. I have a slide here, which unfortunately is not on the version on the web, because I added it later. But I will replace the version on the web. And this will be there. Here there is some useful results in the algebra of colors. When we take the amplitude and we square, we have to sum over colors. And here there are some conventions and definitions on how we deal with these lambda matrices, describing the coupling of a gluon to the quarks. The one thing that if we leave out the issues of the color algebra, this is our soft amplitude. This is the overall iconal factor. When we take the square of the amplitude, we get this expression times you see p and p bar play out in this way times the sum over polarizations of the polarization vectors of a gluon. This object is g mu nu. p square is equal to 0. p bar square is equal to 0, because we're assuming mass less quarks. So what we're left with is p dotted into p bar over pk, p bar k times the boron. Now, this current here is what you, if you go back to, say, your classical electrodynamics, say Jackson, for example, of course, you will recognize these as being the product of the current p over pk minus p bar over p bar k. It's the electromagnetic current generated by a charged particle that is moving with momentum p before and then is deflected to momentum p bar as a result of interaction with an external field, for example. So if you want to look at what is the amplitude for the emission of radiation, that is what we have. So this reflects, once again, the classical nature of the soft gluon radiation. There is another simple way to extract these soft gluon emission Feynman rules. Let's look at this process in which we create a quark, we annihilate a quark, and we emit a soft gluon. So we have a psi bar psi with different momentum, p plus k and p. In between, there is a gamma mu dotted into the polarization of the gluon. And let's now take the soft gluon limit k goes to 0. So if k goes to 0, this becomes psi bar gamma mu psi. But psi bar gamma mu psi is nothing but the usual charge current of a free fermion. So as a matter of convention, that is equal to 2p mu. So the result here is 2p dot epsilon. And that's the p dot epsilon that appears in the numerator of the soft gluon Feynman rules. If we do it in the context of the whole diagram where this is not created out of a vacuum, but it's just an off-shell quark created in a process, you repeat the exercise, we take k goes to 0, and we recover exactly p dot epsilon over p dot k times the wave function of the electron. So p dot epsilon over p dot k is the Feynman rule for the emission of the soft gluon. And if we attach a soft gluon to an internal leg, so to an off-shell propagator that remains off-shell, so this is a whole process, and we just come with a gluon and we put it inside, going to touch one of internal legs, you can find out that as the energy of a gluon goes to 0, as k goes to 0, this expression is finite. There is no singularity. So that means that when we have a soft gluon emitted, the attaching the gluon to the external legs always gives rise to a singular process. So it's the dominant contribution. If we put the gluon in one of internal legs, it's finite, so it's negligible. Which again is consistent with the fact that if we have an off-shell internal leg, it's off-shell, so it means that it has a lifetime which is very short and the long wavelength gluon cannot see it, cannot be sensitive to it, right? So long wavelength gluons, soft gluons only attach to the final states, and that's an immense simplification in the analysis of our problems. So before we looked at the case of a Z boson or a photon goes to QQ bar emitting a gluon, so the initial state is color neutral. If we look at the gluon going to QQ bar and emitted in gluon, things are a bit more complicated because now we have color also in the initial state, in particular, the gluon itself has to radiate, can radiate the gluon, so we have three diagrams to put together. And we have a rather complex network of possible color flows. If you take the picture in which the gluon between a QQ bar, the vertex between a QQ bar and a gluon is, from the point of view of color, a way of transferring the quark color to the gluon and the gluon anti-color back to the quark that comes out of the vertex, we can use this notation to look at the possible color flows that take place in these three diagrams. And when we look at the soft gluon limit, we find that there is not an exact factorization as we had before, an iconal factor times the born amplitude, but there is a factorization in terms of an iconal factor times the part of the diagrams that have that specific color structure related to this iconal factor. For example, you see we have the case in which the color of the initial state gluon gets transferred to the final state quark. And then it's the other leg, the anti-color leg of the gluon that emits a gluon, whether in the initial state or the anti-quark in the final state. So this is a very specific color structure. And here the red line, the color line as you see is never enters in the mission of the gluon. So it's really these other color currents that contributes and therefore the iconal factor is built out of these color line. In electrodynamics, the charge current and the flavor current are exactly the same. If we have an electron, it's the charge of the electron that emits the photon. So the current that emits radiation is obtained by following the electron line. In the case of QCD, current is due to color moving. And since color can be transferred from the quark to the gluon and vice versa, the current emitting radiation does not necessarily follow just the quark line, but it can move from the quark line to the gluon. And this is what is reflected in these expressions. But otherwise, the basic iconal structure remains there. So as further exercises, playing with the color algebra that I gave you in the previous slides, you can then try to work out these two separate contributions in which it's the color line that emits the gluon or the anti-color line that emits the gluon. And what you can find is that these two interfere if we do the square of the total amplitude and we look at the interference between these two, these interference is of order n while the leading order term square is of order n cubed. n is the number of colors. So in SU3, n would be equal to three, but we can do it in general for an arbitrary gauge group SUn. In other words, that's what we say that the interference is suppressed as one over n square. n is equal to three, one over n square is equal to 110. There are other aspects, there are other things that make these interferences less singular. So it's an excellent approximation at the percent level to indeed neglect these interferences between different color structures. And what is true is that therefore in the large n limit and in the soft gluon limit, one could consider independently and separately these color currents as being emitters of a soft gluon. That simplifies all of our calculations because we don't have to take care of the interferences. Now each of these objects, however, itself contains two diagrams. And when we take the square, there is in principle the interference between these two diagrams. And now we are going to analyze what happens when we take the interference of these two diagrams. Now, before I do that, let me explain to you why it's interesting, why am I wasting time of interferences because it's not like we cannot do these calculations, we could do them exactly and everybody can take care of the interferences. There is one place where we are heading to now is we are developing an understanding of what happens when a single gluon is emitted. But as we saw in the final state, there is 20, 30, 40 particles. So we have to assume that at the end of the day there will be a huge number of gluons being emitted. And one way in which we can describe the emission of these multiple gluons is by going through an iterative process in which the gluons are emitted one after the other. So I attach a probability to emit the first gluon then there will be a probability to emit a second gluon from the final state and then a third gluon, a fourth gluon, et cetera. Now, in the context of quantum mechanics, I cannot go through such an iterative process because I cannot just go forward in time without checking the impact that next emission had on the previous history of the system. Another way of putting it is I have to take into account all of the possible diagrams and all of the possible interferences. In real quantum mechanics, I cannot make a statement about the gluon being emitted before another gluon because there is no time ordering. There will always be a probability that it's emitted before and a probability that it's emitted after. So if I can find a way in which within a well-understood and defined approximation, I can introduce a time ordering in the emission of the gluons so that interferences are suppressed and small. I can implement a numerical algorithm which is what we call a Markov chain. It's like a tree, an evolution tree, to describe as subsequent emission probabilities the evolution of the quark. This is what will give rise at the end to what we call shower Monte Carlos, the way in which in practice today we are able to describe quantitatively the evolution of a jet. So I just proven to you, well, you will have to do it yourself by proving this statement, that the emission from independent color lines does not interfere at the leading order in one of the NC square. So that is a definite statement that can be used. And the second statement is that if I look now at the two diagrams that contribute to the individual amplitude which is leading order in NC, I can describe that as a sum, you see there are two diagrams, a gluon emitted from one quark, a gluon emitted from the other quark. So there is the square of the two and then there is an interference. But there is a, let's call it a theorem, angular ordering that tells us that in the soft gluon limit, this square is equal to the incoherent sum of the square of the iconal amplitudes. So positive objects with a constraint, all of the interference boils down to the so-called angular ordering criterion. In other words, I can treat the system as if the gluon were emitted from this quark line, but with a constraint that the emission angle phi one has to be smaller than the angle between the quark and the anti-quark. Plus a contribution which is again a square, a positive number, a probability, times, well, with emission constrained with an angle phi two, which is smaller than again the separation between the two. Now the overall positive contribution here is the iconal amplitude square I gave you before, so p dot p bar divided by p dot k p bar dot k. And then again, I just have to separately and incoherently add a probability, a positive probability of emitting from one leg plus a positive probability of emitting from the other leg with a constraint that the emission angle has to be smaller than the angle separating these two. Now, I will give you first an intuitive explanation of this angular ordering concept. This is a fundamental input for the construction of the shower Monte Carlos. So let's take an off-shell, for example, a situation in which we have an off-shell photon that splits into a QQ bar pair. For simplicity, we're looking at this into a frame in which the photon is moving very fast in one direction. So this splits into a QQ bar pair with an angle phi between the two. And then in the final state, a gluon will be emitted. If the gluon is emitted here, this intermediate state quark is off-shell. Because of the uncertainty principle, it can only live for a finite amount of time. We cannot keep a state off-shell for very long. In particular, it can only stay off-shell for a time which is proportional to one divided by its virtual mass, multiplied, of course, by a gamma Lorentz factor. So the lifetime of this excitation before the photon is emitted has to be less than 1 over mu, where mu is the mass, the virtual mass of this intermediate state, times gamma. Gamma is the energy of a quark divided by mu. So tau is less than e over mu square. Mu square, we can calculate, the algebra is here, by, as a function of the energy, the momentum of the gluon emitted and the emission angle relative to p. And at the end of the day, you do some trivial algebra and we get that tau has to be less than 1 divided by the separation angle, the emission angle, times k-perp, where k-perp is the transverse momentum of a gluon, emitted gluon, relative to the quark. So if k-perp is equal to 0 or theta is equal to 0, in other words, if the gluon is emitted collinear to the quark, then the virtuality here is very small and the lifetime can be very, very long. If the virtuality is small, it can stay there for a long time. It's only when it's off-shell by a lot that it has to decay right away. Now, during this time in which we can maintain these before it radiates, during this time, the quark and the anti-quark will separate from each other because there is the photon that then decays and they go away from each other. And they go away from each other, we assume that they're moving at the speed of light by an amount which is proportional to how long we let them go. Times the angular separation. So the distance they travel from each other in the transverse plane is equal to the separation angle times the time. And we put in this formula and we get the expression between the distance and k-perp and the emission angles. Now, after this time, the photon is emitted. They're separated by a given amount. And if the gluon, sorry, the gluon is emitted, the gluon that will be emitted at this time must have a wavelength which is shorter than the separation between the q and the q-bar. Because if the wavelengths were much bigger, well, the q-q-bar system on the scale of this wavelength is seen as a color singlet object because the quark and the anti-quark, the color charges cancel each other. And of course, if I have a color singlet object, it cannot emit radiation. So that means that there will be a suppression of radiation of gluons with the transverse wavelength, which is longer than the separation between the quark and anti-quark. The quark and the anti-quark act as a dipole. And as a dipole, the field goes like 1 over r squared as opposed to going like 1 over r. So it doesn't radiate at long wavelength in a dominant way. So that means that d, the distance, has to be larger than the wavelength. The wavelength being 1 over k-perp. So if we put this expression larger than 1 over k-perp, 1 over k-perp cancels 1 over k-perp, and we get the phi over theta has to be larger than 1. And therefore, we get that theta has to be smaller than phi. In other words, the emission angle has to be smaller than the angle separating the quark from the anti-quark, which is exactly the color ordering constraint that we obtain. So this is just a statement about the coherence, taking into account the fact that if we have a quark and an anti-quark, they certainly can emit as individual radiators. On the other hand, if we're dealing with long wavelength longer than the separation between the two all of a sudden, they're not monopoles any longer, but it becomes a dipole, and it cannot radiate. So that's the statement of angular ordering. There is a formal proof. These are words. You can actually work out all of the expressions doing all of the integrals, averaging in azimuth, et cetera, et cetera, and you get exactly the result I gave you in a formal way. When we say that the radiation outside these cones is equal to 0, it's actually only in the case, again, of course, of a soft gluon emission, because we can radiate a hard gluon wherever we want. And it's true under these azimuthal averaging, but this is a technical slide which I will not go through. It's there for you if you want to look at it. So that's the emission of a gluon. And now we go to the next gluon. We have q, q bar with a gluon emitted, but now we understand how gluons are emitted. So we can take this final state of q, q bar plus a gluon and we can apply again our soft gluon emission for the next gluon being emitted. And we can calculate the iconal factor for the gluon being emitted by every, each one of the new color currents that we created, still with a constraint of the angular ordering. So if this was the first gluon being emitted, the second, now, the new color current will go from the gluon to the quark. So this is the new angle that we have to use for our angular ordering. And the next gluon, we have to be at a smaller angle. And the next gluon that we emit, we have to be at a yet smaller angle. So we'll have a subsequent emission of gluons that have to be emitted at angles smaller and smaller and smaller because they're always confined by the angular ordering that keeps the angles getting smaller and smaller. And this is the key to maintain, you remember the beginning I said collinear emitted gluons that preserve the pencil-like nature of the jet, but soft gluons can go everywhere. No, it's not true, the soft gluons can go everywhere because color ordering coherence will force the jet to self-collimate. Even the soft gluons will be forced to go in the direction of a primary quark. So even then we'll contribute to developing this pencil-like structure. Question? Yeah, but the gluon is when the quark is, so he's asking why do I have to repeat exactly the same color angular order in now that I have a gluon instead of having a QQ bar pair? When the quark, this is the quark, but then it's a gluon. If I look at it from the point of view of the way the color flows, say that this is the quark and this is the anti-quark, what happens is the following. This is the way the color now flows. So the color of the quark gets transferred to the gluon and the anti-color of the gluon is transferred back to the quark. So after emitting the gluon, the quark does not have the same color it had before. It has a different color. So now this is the color current. We've completely decoupled the quark from the anti-quark, okay? And therefore now the next gluon will have to use these as a reference current. The iconal factor for the emission of the next gluon will be built out of the momentum of this gluon and the momentum of this quark. If this is momentum K and the next gluon has momentum K prime, then the iconal factor probability will be Q dot K divided by Q K prime times K K prime. The anti-quark plays no role. Of course, on this side, the next gluon being emitted then will be emitted by the current formed by the anti-quark and again the gluon. So the gluon has two possibilities to emit through its color line and through the anti-color line. Okay? And each of them will have its own iconal factor and each of them will have its own angular ordering. So if this one decided to come closer to the quark, at that point on this side, everything is going to be defined by the evolution of this jet. So the next gluon now will be at a smaller angle and the next one at an even smaller angle and so forth and so on, okay? So the thing just collimates. Sorry? Yes, the gluon carries color and anti-color. Because it's like a matrix, right? It acts on the color of the quark and transforms the color of the quark to a different color. So it's a three by three or n by n matrix and the gluon is the field that carries the quantum numbers of these matrix, of the color matrix in the joint representation. Now one thing which is important to one of the key outcome of this discussion is the following. Down here you see we're following just evolution of the quark. It emits many, many, many gluons and the question is what is the total color of this system? You know, as we saw, the gluon has color anti-color. I emit many gluons, it's a mess, right? But what is the total color of this system here? The total color of this system is equal to the color of the quark that I started from because of color conservation, right? So it doesn't matter how many gluons I have, still this system only carries the color charge of the quark. And you see, and this is where it's absolutely crucial to keep track of interferences, for example. Because if I didn't take care of interferences, in particular this angular ordering, and I were to assume that each of these final state gluons now acts as an independent emitter, and I were to calculate the probability of the next gluon being emitted, I would have a fixed number, which is the probability of this gluon emitted the next one, plus this, plus that, plus that, plus that, plus that. And this would be the incoherent sum of probabilities, but that would amount to assuming that this state has a large color charge. So I would be making the same mistake that I could do in electromagnetism. If I were to look at a system like this, in which I have an electron that emits photons that split again into plus and minus pairs, so the system here you see contains several electrons and positrons, and if I go and I calculate the potential far away from here, if I were to add incoherently, say to the force, the force coming from all of them, if I do take into account the fact that they have opposite charges, of course, and I write the potential far away from this line, I will have a leading term, which is just the charge of the electron divided by the distance, plus something that goes like one over R squared times the sum of the charges weighted by this distance relative to the electron, and this is exactly the electric dipole moment of my system, right? But the electric dipole moment has a field that goes like one over R squared, not that goes like one over R, and that is absolutely crucial, right? It makes a big difference in electrostatics or anything if I put N times the electric charge, or if I put the total charge plus one over R squared times the dipole moment, okay? So in QCD, taking into account the interference is what allows us to properly account for these quantum correlations and dealing with the interference effects that otherwise would be neglected and would lead to an artificial growth in blue on emission probability. So the picture that we have now of the evolution of a quark is the following. Let's take, for example, a proton interacting with an external probe, a photon in this case, so this could be a deep and elastic experiment. As one of the quarks approaches the interaction with the photon, we start radiating, that's the D-glap evolution that we discussed yesterday. The color of the quark is taken away by the gluon. The gluon gives the anti-color to the quark that will actually go and hit the photon. As soon as the quark is hit, it will radiate a gluon because it's being accelerated. The gluon will take away the color and then will give a different color to the new quark. This new quark will emit another gluon, giving yet another color to the quark until the evolution stops because following the acceleration, the emission of radiation, at some point we get to a point where there is no phase space to radiate any longer. And after this evolution, and you see in this process, the color always, we never get into a situation in which in order to group together to form a color singlet, we need to go and look at very different places, regions in phase space because every time the gluon is emitted, it's the gluon that transfers the color to the new quark coming out. So if I go and look for color singlet pairs which are these light blue blobs, as you see, I always find them locally. There is only one exception which is the blob that connects the color coming from the initial state and the final state. These two objects, since one is being emitted from the initial state, the other is being emitted from the final state, they could be going to rather different points. So there will be some possible long range effect, but it's really localized and confined to a very small number of emissions. This is what is called the pre-confinement. In other words, before the actual confinement, adornation due to strong interaction state place, already the perturbative phase of evolution confines, in other words, allows the jet to develop in such a way that color singlet clusters are localized and close to each other in phase space. The other way to look at this is the following. Let's take the quark coming out and as you see, nevermind that here I put the same color red than I had here, right? Because I could have chosen completely. I only wanted to use a finite number of colors, so I repeat it, but it's obvious that this red does not speak to that red because they're very far away from each other in phase space and therefore, and this anti-red will be much closer to it. But you see, as the quark evolves, it always leaves the color behind. So we have this color, at the beginning I said, people before they saw jets, they thought that I take the quark and the anti-quark, they go away from each other and at the end of the day, there was this huge force between them that forces them to bounce back and God knows what happens, right? No, that's not what really happens because as soon as the quark emits a gluon, it's giving out its color. The quark that comes out has a different color and then it meets another gluon and the quark that goes out has yet a different color. So the color is always left behind. We see it here, right, in this evolution. The color is always left behind and that's why at the end of the day, after a quark has emitted 15 gluons, that quark does not have to go and look for its anti-color on the other hemisphere, on the other side because you will find the anti-color locally. So in a sense, the picture is similar to what happens in this little drawing where I'm looking at a dielectric and I'm coming close to it with, say, a charge and of course, if I get a charge close to a dielectric, I develop a charge on the opposite side. But what is happening is not that the charge is moving from this surface of a dielectric all to the other side. What happens is that the dielectric will get polarized. In other words, this charge will attract the closest positive charge by polarizing, say, an atom. So I develop a negative charge just in the first layer inside that will attract the proton from the next atom, polarizing it and it goes on. So I do see in practice a charge moving but this charge that goes from one phase of the dielectric to the other is not due to an actual particle moving but it's really the effect of polarization. And this is exactly what happens as the jet develops. Okay, I do have a charge that's moving but it's really a subsequent polarization. Now, I have like 20 minutes left. Let me jump across this next set of slides in which I would describe in more detail how shower Monte Carlo's work. And let me go to discussing jets at the end and if I have time I come back to this later. Let me just make one perhaps show you one result because it takes little time and it's important. So the picture we developed is one in which the quark evolves amidst gluon. At the end of the evolution, I have a system of quarks and gluons. That by itself still it's not hadrons. On the other hand, we run out of acceleration so there is no reason for these quarks to emit further gluons. So what happens is that the last particles that have been emitted, they are going in the directions they are going and they start separating from each other. So I have my last, let's say the last step of evolution, so here I have a sequence of gluon radiation, I have a gluon being emitted here. And this gluon is a quark, a calor, anti-calor pair. As these quark and gluon evolve, at some point, they get far enough from each other the strong interactions become very strong and perturbative. How do I proceed next? From this point on I cannot use perturbative QCD any longer so I use a thermological model to describe the next steps of the evolution. If we know that in this regime the force between, the potential between colors is linear in the separating distance. So as the quark, as the two color lines separate from each other, the distance grows linearly, the potential grows linearly. The potential grows, it means that the kinetic energy is reduced. So I have some energy available, that's the kinetic energy of these objects and as they get away from each other, they slow down and that gets turned into potential energy. It's like if I had a spring, right? Now when the potential energy becomes larger in value, then say twice the quark mass, in principle I have stored in my system enough potential energy that I can use E equals to MC squared and create a QQ bar pair. So when I go across this line, when I go beyond this distance, I can generate out of the vacuum, out of the energy that was built into the potential field a quark anti-quark pair. So why is that convenient? Because if I get a quark anti-quark pair out, then what I can do, I can take the anti-quark and match it up with the quark, the quark here and match it up with the anti-quark on the other side and now this is a color singlet object and this is a color singlet object. It's a bit like I take two particles connected by a spring, I give them a very, very strong momentum to separate them. The spring is not an infinitely solid string so at some point it breaks and if I give enough of a kick to the balls, at some point they will slow down but the moment I build enough energy that the string has to break, the string will break. And when the string breaks, of course I have two little strings which are not bound to continue extending and those two little strings can travel independently from each other. So I convert the energy that I built up into the potential energy, the field of the string into energy to break it but at that point the system is free, it's not interacting any longer. So this is the way and these objects will then turn into hadrons and these will turn into hadrons and these two systems are completely free. So now I don't have to worry about color confinement any longer. Of course it's a quantum mechanical process so it's not like rigorously, the moment I go across the distance, this QQ bar pair opens up, there is a probability that it happens, it can wait a bit more and if it waits a bit more, of course I can get to higher thresholds instead of breaking at the point where I have twice the quark mass, it can break at the point where I have four times the quark mass and then instead of creating a quark-anti-quark pair, I can create a pair of di-quark, anti-di-quark and then I can take these two quarks with a quark and I have a free quark system which is a barrier and here on this side they have an anti-barrier and this is the way in which I create protons for example in the evolution of a quark. And likewise I used here MQ but of course if it's a u-quark, it has a mass, if it is an s-quark, it has a different mass, if it's charm, it has a different mass. So in principle I can generate all possible different mesons or variance from this process. I cannot calculate, I cannot control these from first principles with exact calculations. It's just a phenomenological model so I touch parameters which is the probability that here I split into u-u-bar and d-d-bar that will be equal because of isospin. The probability that it goes into s-s-bar is a parameter. The probability that it goes to c-c-bar is a different parameter. The probability that I have a diverion is another parameter and I go, I compare with data from E plus E minus going to hadrons. I look at what is the multiplicity of strange hadrons, charm hadrons, et cetera. I tune the parameters in the calculation and once this is done, I have a phenomenological description of what happens in the course of hadronization. In this description, however, now I can use in any other prediction for any other process because it's universal. We got to this stage, we started from a z-decay but by the time we got to this stage there is no memory of where we came from because of the factorization. You remember, we were driven by the illusion of a soft gluon and the soft gluon emission probability doesn't depend on what the hard process was. So the development of a jet is independent of whether we were doing E plus E minus going to z, going to quark or a virtual photon, going to quark or a Higgs, going to quarks or anything else, okay? So that means that the description of hadronization is also independent of the hard process. So once we fix those parameters once using some set of data, we can use it forever. So in this respect, it's very similar to what we do with PDFs. We do a measurement, we extract the PDFs and then we use the PDFs to calculate different processes, okay? Yeah, the question is if in these two jets, one jet goes to say a baryon, should the other jet contain an anti-baryon? No, typically if a jet contains a baryon, the anti-baryon will also belong to that jet. Yeah, because typically the creation of a baryon is a result of something that's happening locally. I have the quark and the gluon down here at the end of the evolution and all of a sudden when they get further away from each other, I create a di-quark anti-di-quark. So this is the baryon, but you see the anti-baryon has to be nearby, very close. So typically I will find baryon and anti-baryon on the same side. It's an interesting prediction of a model which then one can go out and verify. How often does the anti-baryon belong to the jet where I measure the presence of the baryon as opposed to being on the other side? It's a possible interesting observable. So I skip these, let me do it this way because otherwise tomorrow we're still here. So this is just one example of what I said. These are tables of production rates of photons, pions, rows, charge rows, ethos, kaons, protons, delta resonances, you name it, all of the possible particles in the PDG you go and you look for in ZDKs and this is what is measured in terms of multiplicities and this is what different Monte Carlo's different modeling of this hydronization process and of the shower evolution we give rise to. These asterisks that you find occasionally correspond to places where there is a two or three sigma discrepancy between the expectation and the Monte Carlo that typically means that the Monte Carlo does not have enough free tunable parameters to accommodate the data, but as you see most of them in most cases the agreement is really exceptional, right? We get the right number of lambdas of delta plus pluses of cascades, omegas, omega minuses which is a very rare process because it's a free strange hydron, okay? The Monte Carlo predicts correctly how often that happens. Yes, which one? So the question is why do we get different results when we use different programs because different programs may have slightly different implementation of the algorithm for the perturbative shower. That's something that was in the slides I didn't show and they have a slightly different algorithm for dealing with the hydronization at the end. I just gave a physical picture of what are the ingredients but in practice you can do different things. You can use, for example, theoretical assumptions to reduce the number of parameters. You can ask, you know, the probability to produce, well, a rho, a rho meson is exactly the same quark composition than a pions. It's a u d bar. So positive rho is u d bar, the pion is u d bar, but the rho is a vector and has a heavier mass. So instead of using a separate parameter you could decide that you use, you multiply the probability to produce a rho by a factor of three because it's a vector, a spin one, so it has three degrees of freedom while the pion is a scalar, it has only one, but then it has a heavier mass. So you can put in like a Boltzmann factor, e to the minus the mass of the resonance divided by some constant. And in this way with just one parameter you can describe both the pion and the rho. Of course, if the assumption that you're making is not correct, you will find a slight discrepancy. So this is where, you see the point is that if you put too many parameters in a fit, you're not gonna be able to fit, right? Because if you have 150 parameters you really don't know what is happening. So there is a benefit in trying to reduce the number of parameters as much as you can and you have to find the balance between the optimal number of parameters and the accuracy of the predictions. I did not understand the question. The first part of the question because I was looking at the slides. What did you say? Let me say, Zajik, is only the radiated inside the Boltzmann going u, u, y? Yes. What about the Boltzmann? Oh, okay, so good question. I just dealt with the plus and minus going to Quarks to make it simpler. The question is if I go back now to Hadronico lesions the gluon could be emitted from the initial state, okay? It's a bit more complicated but the bottom line is exactly the same. One can build up exactly the same formalism also for gluons emitted from the initial state, okay? So it's just a bit more complicated but there is no qualitatively new result that comes out. So in the last five or 10 minutes I will say something about jets in Hadronico lesions. We discussed so far jets in E plus C minus but we're interested in Hadronico lesions. This is just a picture of jets in a standard presentation, experimental representation. So jets are interesting because it's the process among the hard processes, it's the one that has by far the largest cross-section. If we look at small q-square interactions in proton-proton collisions the vast majority of those are indeed quark quark or gluon-gluon collisions leading to jets. The predictions from QCD are known up to next to leading order accuracy. Next to next to leading order results are slowly coming out. Next to leading order was completed in the mid 80s, in 86, so it's about 30 years now and after 30 years we still don't have the complete next to next to leading order calculation of jet production, that gives you an idea of how difficult it is, right? So overall the uncertainties are the level of about 10%. So it's a mild precision, it's not the precision we can achieve in a later week physics but it's a decent accuracy. The uncertainties due to the knowledge of a part on density vary from few percent at low energy to even 100% if you go to the highest possible energies and we'll be discussing that tomorrow. And of course jets are everything that interacts via strong interactions or in fact even by weak interactions but couples to quarks will end up in quarks with some branching ratio with some probability. The Higgs boson decays to quarks, mostly BB bar, W and Z decay to quarks, so and quarks become jets. So doing physics with jets is absolutely essential to do pretty much all of the physics of interest in a Hadron Collider. So as far as inclusive production of jets is concerned the dominant process is two jet production. So it's just scattering of gluons for example, there is four diagrams, there is ST and U channel, there is glue-glue and acylation into QQ bars. This is for example the process that's relevant when we want to produce new heavy quarks. The production of top quarks, we don't have tops to first approximation inside the proton already but we can get them by looking at glue-glue fusion going to top, anti-top. There is Compton scattering quark against the gluon. The QQ bar themselves can annihilate into gluons so it's the opposite process and glue-glue goes to QQ bar and then there is elastic scattering of quarks as opposed to quark anti-quarks, et cetera. So the process is two to two, okay? Now the initial state has transverse momentum zero so we only have two degrees of freedom which are the two energies of the initial partons. The final state has a QQ bar pair. The total energy is fixed because it's equal to the energy of the initial state and we have one degree of freedom which is the angle with which the jets can be emitted relative to the beam line, okay? So we have these two degrees of freedom, two degrees of freedom from the initial state and that fixes the system. Of course, our system can move because if the initial state partons don't have exactly the same momentum, the overall system will have a boost. So in the laboratory frame, I will not be seeing the quark and anti-quark going exactly back to back but they can also go at some angle, okay? So the individual emission angles relative to the beam line of the two quarks are two separate quantities. So the phase space at the end of the day can be written as you can do here, there is all of the steps, you can look at them as an exercise but at the end of the day, the phase space is proportional as three variables, the rapidity of one jet, the rapidity of the second jet and the PT. And the PT is equal for the two because in the transverse plane, they go back to back. The longitudinal momentum of the initial state is equal, sorry, the transverse momentum of the initial state is equal to zero. So they have to balance each other in transverse momentum although they can have overall some longitudinal momentum. And if you put together the phase space with a cross-section, this is the expression we have. It's a triple differential cross-section versus PT, eta one, eta two and it's proportional to the partonic densities of the appropriate values of X times the matrix element square. You can do the calculation of a matrix element. Some of them are relatively simple because they are similar to the calculations in electrodynamics, C plus, C minus going to mu plus, mu minus is very similar to q, q bar goes to, say u, u bar goes to TT bar. It's a very similar calculation. Some others are a bit more complicated. For example, those involving gluons. There are ways in which we can get to the bottom line with simpler, with minor efforts. And that is by using again our soft gluon emission rules. For example, if we look at the elastic scattering of two quarks, it's gluon exchanged in the T channel. This is the dominant process. And if we apply, instead of applying the full Feynman rule, we apply here the soft gluon emission. You see these current acts as a P mu. It's just a single current. So we have P mu on one side, q mu on the other side, one over T from the propagator and easily we build up the matrix element. If we take the square of these with some of the colors, we get this result. And this can be properly symmetrized for S over U for the proper exchange. And we get this result for the final amplitude square, which happens to be, in spite of the fact that we use the soft gluon approximation, this actually happens to be the exact matrix element. And you can do as an exercise similar cases like the Compton scattering with the gluon and the glu glu and two within 10, 20%. We get the right results also in this case. So these are just the numerical values for the scattering amplitude square. So this is a pattern level cross section for emission at 90 degrees. The matrix element of a two to two process has zero dimensions because the cross sections, the phase space goes like one over a mass square and the cross section goes like one over a mass square. So the matrix element itself is dimensionless and that's why there are just numbers here. This is independent of the energy. You see it's ratios of S, U and T in variance. So by looking at these numbers at 90 degrees, we get a sense of what processes are most important. And one thing that comes immediately to eyes is that glu glu goes to glu glu as a numerical value of 30 and it's by far the largest process of all. Following that is Compton scattering core gluon going to core gluon, which is still, it's only 20% of that. But glu glu goes to glu glu is by far the dominant process. To the extent that there are gluons. If we go to small x, you remember from the slide yesterday of the PDFs, gluons were dominating. So not only there is more gluons in the proton than anything else, but also the cross section for gluons to interact with each other and go to gluons is larger than anything else. And that's why at the LHC, for example, up to transverse momenta of the order of a few hundred GED it's practically 100% glu glu goes to glu glu. Most of the jets we see are indeed gluons. A process like glu glu goes to QQ bar. This is the process I told you gives rise to heavy quark production if I want to produce top quarks, for instance. And here there is no mass. This is done just for mass less quarks. But if I'm at high enough energy, the mass is not really important. But you see it's only 0.15. So if I go to high energy and I ask myself what's the probability of producing a TT bar pair relative to everything else? You would say, you know, I go to high energy. The top mass is not important. So I will be producing top quarks like I produced anything else because the mass is not important. That's not the case because glu glu, when they come together, they're much more likely by a factor of 30 divided by 0.15 which is like what 200, right? They are 200 times more likely to go to glu glu again than to go to QQ bar. And this is independent of the mass. It's just due to color factors, to the fact that we have a glu on exchange in the T-channel as opposed to having a quark exchange. So it's several factors that make it glu glu goes to glu glu. 200 times larger than glu glu goes to QQ bar. And that's why typically, if you go to large mass, the probability of finding a QQ bar pair in the final state is of the order of a fraction of a percent of a percent. That's true for PP bar, TT bar for BB bar for everything else, okay? All right, so with these, I stop here. And as I said, tomorrow we take all of this knowledge we have and we just projected onto few selected studies done on actual data from the Tevatron or from VLHC. Thank you.