String Theory and M-Theory | Lecture 6
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Susskind, Leonard
(October 25, 2010) Leonard Susskind focuses on the different dimensions of string theory and the effect it has on the theory. String theory (with its close relative, M-theory) is the basis for the most ambitious theories of the physical world. It has profoundly influenced our understanding of gravity, cosmology, and particle physics. In this course we will develop the basic theoretical and mathematical ideas, including the string-theoretic origin of gravity, the theory of extra dimensions of space, the connection between strings and black holes, the "landscape" of string theory, and the holographic principle.
 Stanford University. Okay, so, fermionic strings have been a part of the subject since almost the beginning, not quite the beginning. And as I said, they solve two important problems. I forgot what they are, though. So they give you fermions, and they get rid of the tachyons. They leave you with ten dimensions instead of twenty-six. So they don't solve that problem. That problem turned out to be less of a problem and more of a feature, feature means a good feature, than people had expected at the time. But we'll come to it, and it's a subject known as compactification. Using the extra dimensions and doing something with them to make them innocuous, or if not innocuous, an interesting feature of the theory. But we'll come to that not now, though. What I wanted to talk about a little bit was not really historical, but the scattering of strings. The subject really began with studying scattering of particles. Elementary particle physics was always about scattering of particles, not because it's the most interesting phenomena that can happen. It's not. It's rather dull. You send some particles together, and a bunch of junk comes out in all sorts of directions. But it's about all we can do in the way of experiment. And so we try to unravel from the scattering data what was going on inside the collision. And inside the collision, of course, means the properties of particles and so forth. So the natural tool of experiment, the scattering, and the natural thing that a theorist would ask is if I have a theory of particles, how do you compute the scattering amplitudes? The scattering amplitudes, what is a scattering amplitude? A scattering amplitude, you have some incoming particles that are part of your incoming information. I'm going to have time running this way tonight, horizontally instead of vertically. I don't know why. Variety. Particles come in, something happens inside a black box, and particles go out. Not necessarily the same number of particles. The particles come in, and they carry momentum. Of course, they carry other things. They carry spin. They carry charge. They carry labels, like, for example, is it a muon, or is it a whatever it happens to be? But let's simplify the story and ignore everything except their momentum. So particles come in, and they carry four momentum. Four momentum means energy and momentum. Let's write down what a four vector of energy momentum is. The energy and the three components of momentum. Of course, if we're working in 26 dimensions, we have 25 components here, but I'll just write down four of them. That's a four vector, a relativistic four vector. Each particle has a momentum, and we're going to call it k. k mu, where mu goes from one to four, from zero to three. This is usually called zero, one, two, three. Doesn't matter. Four components of momentum. Now, what do we know about the four components of momentum of a particle? They have something to do with the mass of the particle. Well, of course, they are the energy and the momentum. What's the relationship between energy and momentum and mass for a relativistic particle? Anybody remember? c is one. We will take c equals h bar equals one. What's the connection between the components of energy, momentum, and mass? e squared, e squared equals p squared plus m squared. Here, let's write it the following way. e squared minus p squared equals m squared. Just in order to keep my notation consistent with notations of physicists for many, many years, I am going to write this as p squared minus e squared equals minus m squared. In formula, I've just taken the negative of it. Now, that can also be written in terms of the components of k. In terms of the components of k, that's, what is it? It's k naught squared, sorry, it's k vector squared minus k naught squared, naught standing for the time component, e. Spatial component squared minus the time component squared. This is often just called k squared. Just call it k squared. The left-hand side here is the square by definition. It's a definition of the relativistic square of a vector. It's called p mu p mu. Or to simplify it, let's just call it k squared. It's the space component squared minus the time component squared. That's called k squared. And for every particle, k squared is equal to minus m squared. You can't vary k squared. Of course, k squared consists of the energy and the momentum. When you say you can't vary k squared, it doesn't mean you can't vary the momentum. It means when you vary the momentum, the energy varies in a certain way. And the way that it varies is that e squared minus p squared or k times k naught squared minus k space squared equals minus m squared. So that's the first thing about, it's not even about collisions. It's just about particles. You characterize them by their four momentum, three components of which are independent. The fourth component has to be subject to this constraint. Now we put in a bunch of momenta. Let's call this k1 for the first particle. Is that the way I label them? I like to keep my notation straight. Yeah, I think I call this one k1, this one k2. And then outgoing over here, we'll call the incoming momenta k. I'm going to call the outgoing momenta q. These are also four momenta. I'm going to call it q3, q4, dot, dot, dot. Let's take the very simple case in which two particles go to two particles. q4. These particles come in, those particles go out. How do you represent momentum conservation? Momentum and energy conservation. Moment energy conservation are simply that k1 plus k2, thought of as four vectors, is equal to q3 plus q4. All components, the space components define momentum conservation, the time components define energy conservation. OK, now because of the perversity of physicists, what physicists like to do is to redefine the outgoing momentum and think of them as incoming momentum. Now that's crazy. The outgoing momentum are outgoing, the incoming momentum are incoming, but to do it, all we really have to do, to make it symmetric with respect to incoming and outgoing momentum, is to take each q, q4, change its sign and call it k minus k3. That means changing the sign of its energy, means changing the sign of all of its momentum. It's a trick to be able to write this in a symmetric form. Instead of writing k1, well let's see what it is. k1 plus k2 minus q4 minus q3 equals zero, now becomes k1 plus k2 plus k3 plus k4 equals zero. You treat all particles as incoming, but you have to remember that the label for the outgoing particles is labeled with minus the actual momentum of the outgoing particle, but once you do so, momentum conservation is completely symmetric between the four particles. A useful trick. It's a useful trick that keeps labeling especially consistent. Notice that when you change the sign of the momentum to redefine the momentum with a minus sign, it does not change the fact that the square of the momentum is equal to minus m squared. So we have, in this particular process we have four momenta, think of them as all incoming, although the energies of two of them may be negative, the outgoing ones, each one of them subject to this constraint. The question that a physicist would ask about this collision is what is the amplitude? What is the probability? The thing, the amplitude is the thing that you square, the complex number that you square to find the probability for that collision. But the collision is a function of a number of variables. It's a function, or the probability for the collision is a function of the momenta of the incoming particles and a function of the momentum of the outgoing particles. So it's a function of the k's. Let's call that amplitude a. It's a thing that you square. And it's a function of all of the k's, k1, k2, k3, k4. But there's some redundant information here. First of all, the momenta have to be conserved. Second of all, the square of each k has to add up to minus the mass squared. So there's really too many variables here. There's too many independent variables. How many independent variables are they? Before we impose any constraints, each momentum has four variables. This is four plus four is eight, plus four is 12, plus four more is 16. That's a lot of variables for something to depend on. Fortunately, you really don't depend on that many variables. Let's think about physically now how many variables there's a scattering amplitude or a scattering process to depend on. Well, the first thing you can do is whatever the momenta are, you can use relativity to go to a frame of reference where the center of mass is at rest. In other words, where the two momenta, the two space components of the momentum are equal and opposite. You can always do that. If the particles are both moving down the z-axis, well, you just move fast enough to be halfway in between them. If they also happen to be moving in some other direction, just move in that direction, you can always go to a frame of reference where the particles are equal and opposite, the space components, in other words, the spatial momenta. Next, you can always rotate the system so that the momenta are coming in along the x-axis. What's left over in the initial state? What does it depend on if you know that the momenta are equal and opposite? What does the whole thing depend on? What does the initial state depend on? The initial state only depends on the magnitude of the momentum, not the magnitude of the square of the momentum, I mean, sorry, not the magnitude of the four vector. The four vector has to have magnitude m squared, but the magnitude of the space momentum or the energy, if you like. Once you go to the center of mass, the only thing left over is the total energy of the collision in the center of mass frame. So that's one thing, e center of mass. Now let's suppose that our particles are all of the same kind for simplicity. The particles collide, what can you say about the outgoing momentum? First thing is they have to be equal and opposite. Why? That's momentum conservation. Moment conservation says they have to be equal and opposite. What about the energy of the outgoing state? Has to be the same as the energy of the incoming state. What is the only thing that can differ between the incoming and the outgoing particles? The angle, that's it, the angle of scattering, theta. So the scattering amplitude, although it's written in terms of 16 variables, really only depends on two. Two independent variables. Those variables should be expressible in terms of relativistically invariant things. We can think of it in the center of mass, but we also ought to be able to think about it in any frame of reference. We should be able to construct two independent invariants, relativistic invariants, which are enough to completely characterize the scattering. So let's talk about the invariants describing a scattering process. Two particles come in, two particles go out. Let's think of them as all coming in, changing the sign of the momentum. What is this? K1, K2, K3, K4. What can you do with a four vector? What can you do with four vectors to construct invariants? There's only one thing you can do, really. You can square them. That's about all you can do. But you don't have to be talking about K1 squared or K2 squared. You could be talking about K1 plus K2 squared, right? That's a good invariant. K1 as a four vector. K1 plus K2 squared. Now remember what that means. That means the space components of K1 plus K2, which are labeled with arrows, squared, minus the time component, which means the energies. Let's label them. K0, K0 means energy. K0, 1 and K0, 2 squared. You sum the momentum, square it. You sum the energies. Square it and subtract. And that's called K1 plus K2 squared. Let's see if we can figure out what it is. It's an invariant quantity. Let's see if we can figure out what it is by going to the center of mass frame. In the center of mass frame, the momentum of this particle, what should we call it? Well, it's K1. It's the space component K1. What about the momentum of this particle? It's equal and opposite. Minus K1. Right? So, the total space component of the momentum in the center of mass frame is zero. We don't have to worry about this one. What about this one over here? That's the energies. We didn't talk about it, but what about the energies of the two particles in the center of mass frame? They're equal and opposite. They have the same mass. The energies are equal. The center of mass frame is one in which the particles are perceived as moving with exactly the same momentum, magnitude of the momentum. And so the two energies are the same. So this just becomes twice K naught quantity squared. Very simple. This is the center of mass energy here. In the center of mass frame, you add the two energies. It's the total energy, and the square of it is just the square of the center of mass energy. So this quantity here, which is called minus s, it's just called minus s, is nothing but the square of the center of mass energy. Just look at it. Or s is the center of mass energy. Why? In the center of mass frame, this is zero, and this is just the total energy. So minus s is just the square of the center of mass energy. Square of the center of mass energy we call s. Now can you think of any other invariant that you can build? How about K3 plus K4 squared? We took K1 plus K2 squared. How about K3 plus K4 squared? It's the same thing. Why? Because K2 is minus K3 plus K4. Energy conservation. So we don't get anything new there. K1 plus K2 squared is the same as K3 plus K4 squared. It's the energy, and it just says that the energy in the initial state is the same as the energy in the final state. What about K1 plus K3 squared? That could be something new. So let's see what it is. Let's see if we can figure out what K1 plus K3 squared is by working in the center of mass frame. In the center of mass frame, all the particles in and out have the same total energy, have the same energy. Each particle has the same energy. Particles just get scattered through an angle. That's all that happens. So they come in and they go out with the same energy, but they scatter through an angle. So let's take K1 plus K3 squared. Here it is. K1 plus K3 squared, K3 squared, K1 K3 squared. How about this? This is in the center of mass frame. It is same as K2 and K4, but that doesn't help. I want to know what it is, though. How big is it? All the particles have the same energy, incoming and outgoing. All they do is scatter through an angle. Zero. Zero. Why is zero? Why not twice the energy? Yeah, because K3 and K4, we flip the sign on them. So this isn't there. And this is just K1 plus K3 squared. So let's see if we can see what that is. K1 comes in, particle 1 comes in, and of course also particle 2. Particle 3 goes out. Here's particle 3. Time goes in, 3 goes out. But if we label these particles with the Ks, K1, and let's call this one now Q, Q3. Let's make it an outgoing particle. Then really what this is, is it's K1 minus Q3 squared. It's really the difference of the momentum of the incident particle and the final particle. It's called the momentum transfer. It's the momentum transferred, if you were to think of particle 1 and particle 3 as the same species of particle, then it would just be in the collision, it's the momentum transferred from 1 to 3. Bing, bing, there's a momentum transfer, and that's the momentum transferred from 1 to 3. That's what this is, momentum transferred. It can also be expressed in terms of the angle of scattering. I will tell you what the formula is. The formula for K1 plus K3 squared is it's just twice the energy squared minus the mass squared. We already worked out what the energy is, the energy is the s variable. It's twice e squared minus m squared in the center of mass frame times 1 minus the cosine of the angle of scattering. This is the interesting thing here. The angle of scattering is the angle between 1 and 3. K1 comes in, Q3 goes out, the particle gets deflected through an angle, and it's that deflection angle, here it is, theta, the deflection angle of the particle between the incoming state and the outgoing state, and that's what's here. Why is that interesting? Well, of course, that's what's measured in an experiment. You put detectors in different places, you scatter particles, and you find out the probability for them to get deflected through an angle at different values of the energy. This variable, the K1 plus, let's see, what did we do here? I'm sorry, I, I, the definition of s, I forgot to erase the s over here, the definition of s was K1 plus K2 squared. That was the definition of it. One plus two coming in, and it was the square of the center of mass energy. So s equals energy center of mass squared. K1 plus K3 squared, that's called, anybody think of a name for it? T. T, good, T. It is called T. I knew you were going to get that. And that is some combination of the energy and the momentum transfer. It's the center of mass energy squared minus the mass squared, big deal, we already know what the center of mass energy is, times one minus cosine of the angle of scattering. All right, so we have K1 plus K3 squared. How about K2 plus K4 squared? Is that different? Well, if you go back up to here, K1 plus K3 is minus K2 plus K4. So K1 plus K2 squared is the same as K3 plus K4 squared. It's also T. Is there another combination? We have K1 plus K2, we have K1 plus K3, K1 plus K4, right? That's another one. All right, so one more quantity is K1 plus K4 squared. What can that be? Well, if you were going to give it a name, what would you call it? Minus T equals K1 plus K3 squared. U minus U equals that. These are the STU variables. But what's going on here? There are only two quantities that the scattering depends on, energy and angle of scattering. We seem to have three independent things. Well, the answer is there are not three independent things. If you use the momentum conservation and you use the fact that each K squared is M squared, what you'll find is that S plus T plus U is equal to minus 4M squared or something like that. There's a constraint among them. There are not three independent ones, only two. These are the two interesting quantities. The third one is dependent on the other two. Do you have to consider the vacuum at all in these collisions? What does that mean? Is the vacuum going to participate in the collisions? I'm not sure what that means. The vacuum is all over the place. Participates like all hell. What does it mean? What do you mean by participate? Maybe the vacuum could just slow down a particle going through free space. Well, we know that doesn't happen, don't we? It's momentum conservation. Okay, well, just a wild idea. Yeah, it could be. I mean, it's a possible world. The vacuum has friction. It's not our world. Moment conservation says that particles aren't slowed down by the vacuum. At some point, I think you said that 3 and Q4 have the same energy. In the center of mass frame. In the center of mass frame. Why can't there be a, I mean, K1 and K2, the center of mass frame is kind of defined that way. Yeah. But then the center of mass frame is defined as a frame where the momenta are equal and opposite. If the momenta are equal and opposite, the energies of the particles are the same. If their masses are the same. Now they scatter, but they have to go out equal and opposite for momentum conservation reasons. Yeah. So that's, all right, these variables, ST and U, the other one which is there, those are called Mandelstam variables. And they're very symmetrically defined. Mandelstam, Mandelstam, Mandelstam. Mandelstam is a currently active physicist in Berkeley, but this notation comes from the early 60s, the very early 60s, Mandelstam variables. And notice what they have. They have this beautiful symmetric structure. That's what's interesting about them. Now let's talk about Feynman diagrams for a moment. And I'll just tell you what the answer is for certain Feynman diagrams. Let's suppose two particles collide. This is a Feynman diagram. Create a third particle of a different mass in here. Let's call it capital M. And then those particles go out like this. Those come in, Feynman diagrams, so they come in together. They coalesce to form a second particle or a third particle or whatever, and then that particle decays and it goes out. That's a Feynman diagram. That Feynman diagram has a value. I'm going to tell you what it is. It's the product of two coupling constants. There's always coupling constants and Feynman diagrams. And then there's the propagator of the particle in between. And that has a very simple form, one divided by s minus capital M squared. It has a s minus capital M squared. That's the characteristic structure of a scattering amplitude where two particles come together and merge and form another particle. And notice it's a function of s, the energy of the process. It doesn't depend on t. It doesn't depend on the other variable. You want one that depends on the other variable. You draw a similar diagram, except in which instead of one and two merging, you have, I'll draw it over here, one and three merging. One, two, three, four. It looks like exactly the same diagram except turned on its side. But it represents a very different kind of physical process. One and two come in, exchange a particle between them, and then go out as three and four. What would you guess if you had to guess the amplitude for this process? It's going to have g squared again, the g's are the vertices. What else? Exactly, one over t minus M squared. Now you can imagine a third process which would be one over u minus M squared, and it would look like this. One, let's see, that's a little hard to draw, one. It looks like this. One and four merge, one and four merge, little, you have to switch the lines like that. Let's ignore it. It's there, it's there, and it's important. But it's too hard to draw. I don't like drawing cross. They're perpendicular to the other two? They're not perpendicular, it's just a question of which prior particles came together, and I hate drawing cross lines on the blackboard so we won't. But there is, in principle, there can be another one. This one depends only on the energy. Doesn't depend on the angle of scattering altogether. T contains the angle of scattering. S is only a function of the energy. That means that when this process happens, to say that it doesn't depend on the angle of scattering means that every angle of scattering is equally probable. When the particles go in, they have equal probability of getting deflected through any angle or whatever. That seems odd, but that is the property of this process. And the reason is very simple. The particles come in from some direction, they form this compound state, and then when they decay, they've forgotten which direction they came in from. They come in, they form the composite, and then when the composite decays, it decays at an arbitrary angle uncorrelated to the initial directions. That's this process here. This one is different. It depends on the energy, but also on the angle of scattering. This is the one that depends on the angle of scattering, and if you work it out, it's easy to see that it favors small angle scattering, this favors large angles. So this one depends on angle, and this one depends. But notice how similar the two expressions are. There's a real symmetry between them. In fact, if you interchange S and T, the scattering doesn't change. This is a property of relativistic scattering amplitudes that they have this kind of symmetry. They sort of forget which were the incoming and which were the outgoing particles, but only if you express it in terms of these kind of invariance. Good. So now we have some basic idea of what it is that a theoretical physicist wants to compute about a model of particles to compare, not because he's so interested in it, but it's the thing that he can give to the experimentalist and say measure the probabilities for scattering as a function of energy and so forth, and here's my prediction. Okay. Are you saying that the dependence of S, T, and U causes this amplitude to what, shift around if you, I mean you still have three terms, but. But the value shifts around if you. Depends on the angle of scattering. All right. So if all that was there was this, this would favor small angle scattering and be, not forbid, but make it less likely. Let's see if we can see why. I think here it is. I think that you have a third term. The third term is there. What about it? Oh, well, what you say that you didn't want to write it up there or did it, you didn't want to write the diagram. I didn't want to draw the diagram, but if there was a process in which one in four could come together and do the same thing, then we would want to put a G squared, one over T, one over U minus M squared. And because of three are not completely independent. Yeah, they're connected. Yeah. Then that adds back to the. One over U minus, okay, so what is U? It's interesting. U is equal to E center of mass squared minus M squared times one plus cosine theta. So both, so they're clearly related. If you know S, you know the energy. Therefore, if you also know T, you know the angle of scattering, and this is just a function of the energy and the angle of scattering. So they're all connected with each other. Another way to say it is if you add S and U, you'll cancel out the angle altogether and you'll get just a function of energy, which is clearly dependent only on S. But they're all, but the Cipri's finding diagrams and in general you have to add them all. Okay. But I want to focus on this here. This does not describe the scattering of mesons, for example, very well. It gives a poor description of the scattering of mesons. And the reason is simple. The reason is that there are many, many different particles that can be produced when two mesons collide. Whole stacks of them with different mass and different angular momentum. We haven't described what the angular momentum would do here. What the angular momentum would do would be to change the dependence on the angle of scattering. Let's not get into it. Formulas like this are too simple to describe the realistic scattering of mesons. It's too easy to excite higher vibrations of these particles here. And when two particles collide, they could make some ground state. They could make some first excited state. They could make some next excited state. They can make a whole raft of different particles that can go in there. And people in the 60s, basically beginning sometime around 1965, 66, 67, tried to concoct all amplitudes with interesting mathematical properties that would represent all of the possible particles which could go in here. This was trial by trial and error. It was just making up formulas to try to represent the different particles which could go in here and the different particles which could go in here. The first attempt tended to be to add particles. This is called an S-channel process. Why? Because it involves one over S minus M squared. Particles coming together, K1, K2. That defines S. This is called the T-channel process, K1 and K3 coming together. So they began by saying, let's just add more and more stuff into here, S-channel, and let's add more and more stuff into here, T-channel, for all the particles that could be scattered. Well, that lasted for a certain period of time, adding up the various particles which could go in there. And then people tried to find more comprehensive formulas just by, I don't even know what the right word is, sort of a kind of curve fitting but a very sophisticated kind of curve fitting and a rather dramatic formula was discovered which contained the physics of all of the particles in the S-channel and all of the particles in the T-channel. Just replaced this combination here with not just one particle but many of them, whole towers of them with a formula which I'm going to just, this is of historical interest mainly, with a formula which I will write down and you can explore its beauty, its great beauty. And this is very, very simple, it's called the Veneziano amplitude. Young Italian physicist, he was young at the time, he's not young anymore but he's still Italian. He just, I don't know how he made this guess, he just randomly wrote down some things which had some right properties and which really did look like adding up things like this. It was a function of S and T. Everybody know what the gamma function is? Now the gamma function, you don't need to know it but I'm just going to write down the formula for fun. The gamma function is a generalization of the factorial function. For the factorial function is only defined for integers. The gamma function is defined interpolated between integers. Gamma function for the integers is equal to n minus one factorial, gamma of n. Now it's defined for non-integers too and I won't tell you, it's defined by an integral and it continuously interpolates between the integers. This is the Veneziano amplitude multiplied by the coupling constant squared. If you examine this amplitude, you'll find out, I'm not going to go into its mathematics, its mathematics is very simple. All you have to know is about gamma functions and if you want to explore it, it's actually quite a simple construction. But I'm going to tell you what it looks like. It looks like an amplitude that you would make by summing up large number of particles, this is representing some composite particle or some particle it could be in there, of different masses in the S channel. But it also looks like, oh, notice that it's symmetric. It's symmetric with respect to S and T, exactly like this is. The peculiar thing about it is you can represent it as a kind of sum of Feynman diagrams with all the particles in the S channel, diagrams like this. But because it's symmetric under S and T, you can also represent it, you don't add, it is also equal to the same kind of thing going this way. Something odd was afoot about this formula. It had all the important features that the scattering amplitude should have. It could be analyzed as if a whole bunch of particles were produced and then decayed, but it could also be analyzed as if a whole bunch of particles were exchanged. This was something new. This had not been seen before. Previous to this, everybody would have added contributions for S and T. And this thing replaced this. The question was, what is this? Where does this come from? What kind of physics gives rise to this? What kind of physics can you imagine would give rise to this? To a formula like this? The answer, of course, turned out to be string theory. The invention in the discovery of string theory was just looking for a physical model which would give this for its answers, for its answer for scattering of two mesons or two particles. I'm going to tell you, without a lot of drama, what physical model gives rise to this scattering amplitude. I'll show you a little bit. I'll show you what the logic that went into it was. It wasn't very hard to guess that this was a theory of strings. It wasn't too hard just because, well, it wasn't too hard. You said exchange particles versus… The S channel and T channel exchange, direct production and exchange, sometimes called the direct channel and the cross channel. The S channel is called the direct channel. Those are the direct particles come in, coalesce, and then go out. The cross channel is when the particle jumps across from one side to another. I'll show you what the physics of that formula is. You begin with two strings. I'm not going to do a calculation, but I'll show you everything that went into it. It's a little bit tedious, what you have to know about. You have to know a lot about harmonic oscillators, and that's all. Nothing much more than that. You start with two strings. Here's a string. Now remember, the string has a coordinate along it which we called sigma. Let's draw the sigma axis over here from sigma from zero to pi. Here's the string. But the string propagates in time. Let's draw time horizontally. Here's the end of a string at sigma equals zero, sigma equals pi. These are open strings. The string of course moves around in space time, but it's always located between zero and pi. That's not spatial position. That's just its parameter along the string. The material of the string, the particles that make it up, are in here. Each point in the history of the string, think of the string as sweeping along, here it is. It's sweeping along some space time sheet. It's called a world sheet instead of a world line. Each point in that world sheet is characterized by a point sigma and a time which has called tau. Tau goes this way. It's a time, and sigma is a coordinate along the string. Not a real spatial coordinate, just a label for labeling points along the string. Now is this time that we've used previously the infinite momentum time, but that's not what's important. What's important is that the idea of a world line becomes a world sheet. And instead of being parameterized by a single variable tau, it's now parameterized by two variables. Each point in here has a position in space time, x, x mu, or just x. And we've already worked out what the equations of motion of x are. They are the wave equations describing waves moving up and down the string. The wave equation. Let's write it. This is d second x by d tau squared minus d second x by d sigma squared equals zero. That's the wave equation that described the oscillations of the string. But we don't even have to think about this. We can just imagine that this string is a collection of a large number of particles. Place the world sheet by a bunch of world lines narrowly spaced with springs between them. With springs between them. That's the picture of the evolution of a particle. Now what we're going to do is begin with two particles. We're going to have two particles coming in from the past. And we're going to put them right next to each other. Let's see. I think we need another color for the second particle. We're not going to put them right next to each other in space necessarily, but just in the parameter space. Here this goes from zero to pi. We need to put another one in going from pi also from zero to pi. Here's the other string. There's no meaning to the fact that I've drawn them right next to each other. I've just drawn them. Their actual position in space-time might not be adjacent. So this just parameterizes, this half parameterizes the world sheet of the left particle. This half parameterizes the world sheet of the right particle. And they might be far away from each other. That means the X's over here may be very different than the X's over here. Now with any luck at all, with some probability, the end of the string might touch the end of that string. And when it does, they can coalesce. That's an assumption that they can coalesce. But that's the basic process of string theory, that they can coalesce and then form a single string. Now they really are connected. If you like a new spring developed, when these touched each other, a spring appeared connecting the last particle of this string with the first particle of that string, and the whole thing becomes one string. That condition persists for a while, it persists for a while until a quantum mechanical event happens and the strings separate again. Randomly, but you can transform around cleverly to make it be at the same place along the string. There's enough symmetry, enough symmetry of the equations that you can put this point at the same horizontal level as this point. So what is the nature of this process? Given this, it's possible to guess what the answer is for the O. The important quantity in here is the amount of time that the string spends coalesced. It's the amount of time that the compound state stays before it breaks up into its final constituents. And let's just call that tau now. Let's call that from this time here, tau. I'll tell you what we do with it. In the end, we integrate over it, but let me tell you what you put in. If you're interested in a quantum mechanical amplitude, you want to start with an initial state. The initial state, let's start with one particle. The initial state can be thought of as the state of a whole bunch of little points, namely the points that make up the string, x1 through xn. Let's start with the first string. x1 through xn. It's just a collection of mass points, x1 dot dot dot through xn. And the wave function of it is just a function of x1 through xn. That's its wave function to begin with in the start. What do we know about the wave function? Does everybody understand why a wave function is a function of the n positions of the particles? That's quantum mechanics. Quantum mechanics says wave functions or state vectors are functions of the position of the constituents. What's the probability to find the particles at position x1 through xn? Psi star psi. So that's what this is. This is the wave function of the starting assemblage of particles. Of course, I'm purposefully not taking the continuum limit to show you what goes on. At the end, you have to take the continuum limit, but let's not do that. What do we know about this? Well, the first thing we know about it is that this particle comes in with momentum k1. I'll tell you exactly what that says. That says that this wave function contains a factor e to the i k1 times the center of mass position. This is the wave function e to the i kx is the wave function of a particle with definite momentum. What momentum do you use? You use the center of mass momentum. So what is the center of mass position? Sorry, the center of mass position. What's the center of mass position of these points? It's the average position. X1, the sum of the x's divided by n. So you have x1 plus x2, up to xn divided by n. That factor, and that factor alone, tells you that this initial particle here had momentum k1. Now, what about the rest of the wave function, which depends on the relative coordinates, not the sum of them, but the distances between the neighboring particles? That's some wave function which characterizes the ground state. It depends on everything except the sums of the x's. It's some wave function, let's call it sine naught for ground state, and it depends on the x's, same x's, but it actually doesn't depend on the sum of them. The sum of them here, the differences between them here. Differences of x's here, neighboring x's and so forth, here, some of them here. And this wave function is computable. It's just the ground state, describing the ground state of all the harmonic oscillators making up the string. And it can really work out. With enough room on the blackboard, I could tell you exactly what this function is as a function of a collection of x's. It's not very hard. It's a bunch of exponentials. It's workable, you can do it. This is the wave function of the first particle. What about the wave function of the second particle? It's exactly the same kind of thing, except it doesn't depend on these coordinates, it depends on the coordinates of the red particles here. So let's write it down. It's e to the i k2. Now, what shall I write? Shall I write x1 through xn? No, that's the original particle, those are the original constituents of the first string. I want the constituents of the second string. So let's write xn plus 1, xn plus 2, all the way up to x2n. The second half of the particles is grouped together into the second string, also divided by n, also times psi naught of 1 through n. This is xn plus 1 to the end, to the end of the chain. Can that be read? It is read. But can you read it? OK, you can read it. Now, that's the initial state. But now you say with some probability, the two endpoints merge. To say that the two endpoints merge simply says that you set xn equal to xn plus 1. You look for that piece of the wave function where the two endpoints are at exactly the same point. So you begin with this. The next step is to say, let the nth particle on the first chain be at exactly the same place as the n being at the same place as xn plus 1. So we're going to put in here then xn. All the others left unchanged. That's now not, that's the wave function of the state right at the point after the two particles have coalesced. When they're coalesced, at the point where they coalesce, they come together. Nothing happens to the rest of the chain, but the two particles, so this is, if you like, it's the amplitude that the two particles, the chain touched. This is the amplitude that the two particles, the endpoints touched. Now we have a new starting point, which is a function of x1 through x2n. It's a state of that many particles. And what do we do with it? We have to evolve it. We have to evolve it using the Hamiltonian. Remember what a Hamiltonian is. A Hamiltonian is a thing which updates you from one instant to the next. So you take this initial state. It's a well-defined thing, and you propagate it forward in time using the Hamiltonian. Take it. What's the right rule for updating a state from one, from an instant to a later instant? You multiply the state by something. e to the iht. e to the i. So you take this wave function, and you evolve it. You solve the Schrodinger equation for it. But that's the same as multiplying it by e to the i, the total Hamiltonian times tau. But what is the total Hamiltonian? The total Hamiltonian is just the collection of springs and mass points, a collection of harmonic oscillators. It's just a collection of harmonic oscillators. We know how to do this, and that gives us the state of the system after time t. What's the last step? The last step is to project the final state, here it is, onto two separate particles again to project it onto a state with two separated particles of momentum k3 and k4. It's very, very straightforward if tedious, slightly tedious. You take the two initial states of the two particles, they're well-defined ground states of the particles. You insist that the two endpoints are at the same place, that constrains the wave function. You let it evolve as a single string for a ways, and then you let it break up again, and that simply means multiplying it again by some final state. I'm not going to write it all out. It's the same kind of thing. And that gives you the transition amplitude. To make a long story short, you start with the two particles, you constrain it so that they're at the same place. You evolve it, and then you project it onto the final state. A very, very well-defined thing to do. That gives you the amplitude that the strings coalesced for an amount of time tau. But how should you choose tau? Those things in black and red there are added together, and then you multiply. Those are multiplied together. Now, if you have two separate systems, each having its own wave function, you multiply to create the wave function of a composite. So this is just a wave function of a composite of two particles. To begin with, you don't set the first particle equal to the last particle here. Then you look for the amplitude, you look for the piece of the wave function where those two particles are at the same place. You say, aha, now that they're at the same place, fuse them together and evolve it as if it were a single string, a single collection of mass points and springs. For an amount of time tau, and then basically you take your scissor and just cut the spring in the middle and let it evolve after that. That calculation is quite doable, not even very hard. It's just too much for the blackboard for one morning, but I'm going to tell you what the answer is. I'm going to write down the answer for you. Don't try to remember. This is the answer. It's an integral over the tau. Did we say that we integrate over tau? I think I said we integrate. What do we integrate over tau, incidentally? Why do we add up the amplitudes for all possible times that this composite could exist? This is the Feynman rule of summing over all possibilities. The only parameter here is the time that it takes for this to break up again. Feynman's rule is sum over all paths, which in this case just means integrate over the time that they spent evolving together. That gives you the amplitude for the two particles in the initial state to become the two particles in the final state. The result is an integral. It's the integral over the time that they spend together. That's it. I'll tell you what the integrand is after a certain amount of calculation, which is actually not very hard. It becomes the integral of e to the tau. That's the time. Times s plus 1. This s plus 1 is exactly this s. It's the s variable or the center of mass squared, the center of mass energy of the two particles. You have a factor like that. You have another factor, which is 1 minus e to the tau to the minus t minus 1. t is the momentum change between the initial particle here and the final particle here. Remember, those momenta are coded in this wave function, which I've erased. They were coded in the wave function in terms of those exponentials. Sorry, the center of mass energy. Here is the momentum transfer. It appears in the formula. Then you have to integrate it, d tau. There happens to be another factor, I think, of e to the minus tau in the integrand. You're right. I have d tau twice. e to the minus tau, yeah. The first term is tau times s plus 1. Yeah. No, yes. Yeah, that's what you get. Is that plus e to the minus tau? What's that? When you combine the e to the minus tau hanging out the end, is that multiplied or how do you multiply this? I've left it out here on purpose. I've left it out here on purpose, d tau. Now that doesn't look particularly symmetric between tau and between t and s, but after it was computed, it took about a half hour to realize that you should change variables in this integral. Change variables from e to the tau, from tau to something called z. Let e to the tau equals z. Let e to the tau equals z. Now we write this. e to the tau is z, so this becomes integral of z. I'm missing some minus sign. No, e to the minus tau must be, this must be minus here. It is e to the minus tau there. This thing just becomes z to the minus s plus 1. e to the minus tau, so this is minus, minus, e to the minus tau z, by definition, just to change your variables. What about 1 minus e to the minus tau? That becomes 1 minus z to the minus tau minus 1. Oh, incidentally, this integral goes, I think, from 0 to infinity. Is that right? From 0 to infinity. Yes, it does. This becomes 1 minus z to the minus tau minus 1. What is d tau times e to the minus tau? It's just d z. It's just d z. I think I got it. It's just d z. Did I get that right? I've lost track of whether this is plus or minus here. I don't remember. But it's just d z. This is the formula. And where does the integral go? The integral goes from tau equals 0, where z is equal to 1 to, this is minus. It is minus. I'm sorry. I'm making a mess out of it. e to the minus tau is equal to z. To tau equals infinity. In other words, you have tau equals 0. That's when it very suddenly merges and then falls apart instantly. The tau equals infinity, when the separation is infinite here. And what happens when tau goes to infinity? z goes to 0. So it's an integral from 1 to 0, or 0 to 1. There's probably some sign in here of an integral that looks like this. The amazing thing about this integral, the whole upshot of it, is that this integral is completely symmetric between s and t. Can you see that? How do you see that it's symmetric between s and t? Right. You just substitute for z, z minus 1. You make a change of variables between z and z minus 1. And you see that this integral is completely symmetric. So although a completely unsymmetric starting point between energy and momentum transfer, somehow it wound up giving a completely symmetric answer between the two of them. I lost track of the minus sign. The minus sign comes from e to the minus tau d tau going to dz. Yeah. I've lost track of the euro. So you get a minus sign, that's why you switched to order. There's a minus sign somewhere. I'm saying e to the minus tau d tau. Yeah. And then you're going to make that just be dz. Yeah. So you get a minus sign. Yeah, there's got to be a minus sign out here, right? Right. Then you switch the order of integration and it becomes from 0 to 1 like this. That's right. That's correct. It's symmetric whether or not you switch the order of integration. Right. And that's the answer. It's symmetric between s and t. And what is it? It's a process in which two particles join, form a composite which wiggles around for a while and then breaks up. In other words, it's the analog of the Feynman diagram in which a composite is formed and then decays. Composite is formed and then decays. But it winds up being completely symmetric between s and t. In fact, this function is called the Euler-Beta function. It's a function of two variables, s and t. It's called beta of minus s and minus t. It's the Euler-Beta function. Same as function of mathematical physics. And guess what it's equal to? It's equal to the Viniciano amplitude. It's exactly equal to the Viniciano amplitude, namely this product of gamma functions over another gamma function. How did it get to be that it was symmetric between the two? Now this is not obvious at all. It's symmetric between the s channel and the t channel. It's symmetric between the s channel and the t channel. I'm going to come to that next time. That has to do with a fundamental, extremely deep symmetry of string theory called conformal symmetry. It's a symmetry which allows you to take these world sheets and deform them in crazy ways as if they were Turkish taffy and stretch them out in different directions. And for example, turn this picture into a picture which looks much more like two particles coming together and exchanging something. It's a character. Yeah? So the gamma gamma function is reminiscent of the probability function of I think it's factorial A, factorial B over factorial AB, the permutation. Right. Comparatory coefficients. Yeah. So what's the time in between the two? It is true that the beta function evaluated at integers is the combinatoric coefficients, the inverse of the combinatoric coefficients. You know what he's talking about. He's talking about n factorial, m factorial over n plus m factorial, which occurs all over the place in combinatorics. This just happens to be the same function. No simple connection. First of all, it's the inverse of it. The inverse of it is not a combinatoric coefficient, but it just happens. It's an integral which defines the same combination of gamma functions. There is no simple relationship. It's not that something combinatoric went on here, at least not to my knowledge. And it's certainly not the way Viniciano found it. I don't know what magic he pulled up to find it. So this was, if you like, partly historical, but part of the important logic of the theory is that number one, you can calculate with it. It's all it is, is harmonic oscillators. Everything can be done with harmonic oscillators. It's a bunch of harmonic oscillators. You break the process up into pieces, and then you integrate over the time in between. You can calculate, you calculate, and you find an integral. The integral, at this point, by magic, has the property that it's symmetric between s and t and somehow looks like not the sum, but has the features of having processes where particles coalesce. That's the direct process I drew here, but somehow buried in it is also somehow processes where particles are exchanged. And that was the magic of it. What was the surprise that a whole new logic of putting processes together to make new processes? Okay. Any questions about this? Yeah. I think I'm missing something. In the beginning, we described two strings called less than you know, the string of n oscillators and n oscillators, they come together in two n, as it goes on. But remember, when you're taking the limit n goes to infinity, two n is the same as n. Sure. But it still has the property of a boson or fermion. Oh yes, so far, yes. The question is, when you wrote down the wave equation, you had two points coalescing, which is two n minus one, so it changes from positive to, it seems inconsistent, I'm sure. You lose two, not one. Well, n and n plus one becomes one. No, actually n and n plus one disappear basically. Okay. If you like, I mean, I think that's the right way to think about it. Yeah. Yeah, they eat each other. They eat each other like that. And you're left with, right? So the new spring, the new spring that forms connects these two. Yeah, otherwise we'd have big trouble. We would lose one fermion. Right. So that's a good one. One way to think about it is that a particle and an antiparticle color less than the, if you thought of these as quarks on ends of strings, you could think of the last one being a quark which annihilated with the first quark of the other string. So yeah, you lose two of them. But who cares about two when there's an infinite number? Yeah. Right. Yeah? Right. But you're right. You have to be careful not to lose an odd number of fermions. That's something that shouldn't happen. Yes. The next quarter you continue on with string theory and talk about superstring. Well, we're not going to spend a lot of time on superstrings, although the things that I will tell you, strictly speaking, apply to superstring. And not to the bosonic string. No, I'm not going to go into the heavy mathematics of superstrings. What I think we're going to do next time is I'm going to tell you, well, I'll tell you a little bit more about string theory and about the properties of world sheets and the symmetries that allow you to stretch this thing like Turkish taffy and make different kinds of Feynman diagrams out of it. But then we're going to move on to another subject called M theory, which is another way of deriving string theory from a totally different van- well, I think we have to talk a little bit about compactification, about what you do with these extra dimensions, and we will maybe next time. But then we will move on to a totally different origin of string theory that evolved much, much later in time, sometime around 1995, 96, which started with a completely different picture, and in which a great many of the features, the more complicated features of string theory are completely transparent, and it's called M theory, and we'll start a different starting point, which leads to exactly the same physics. So this process you just described was based on the clues from two Masons colliding? Yes, but it's also the way to photons collide in string theory. It's the way any two open strings collide in string theory. Incidentally, the analog for closed strings is very similar. You have two closed strings, and you pick a particle from here and one from here, and you require that they be at the same point so that they do this kind of thing. And then you evolve it as a single string, same rules, same kind of rules, and that would correspond to the scattering of two closed strings. If you scatter two closed strings, the intermediate thing that you make will again be a closed string. Does this only apply to strings where the particles are fermions? No, no, no, no, no, no, no, no. This calculation was originally done for the Bosonic string, and of course it's more complicated when you have to keep track of the fermions also. Extremely similar, extremely similar, some slight differences that are not important. What I should say about this is that the scattering amplitudes, we said that there was a photon and a graviton in the system. All right, it looked like there was a spin two particle and a spin one particle, and you could force them to be massless if you wanted. You didn't have much choice. There weren't enough components to make a massive particle. So you did that. Well, then you take these particles and you collide them, and you work out scattering amplitudes. Scattering amplitudes are very distinctive and characteristic for emission of photons and gravitons. They are not just any old scattering amplitudes. They have very, very definite properties which make them extremely special. The emission and absorption of photons cannot be mistaken for the emission and absorption of scalar particles or other things. They satisfy some very, very important rules. Those rules originate from the conservation of electric charge. Conserved electric charges emitting photons is almost very, very little ambiguity in what the emission and absorption of photons look like or what a scattering of photons by charge particles look like or even the scattering of photons by photons. And it was at this point calculating these diagrams where it became completely clear that these things which we were calling photons were behaving exactly like photons and the things that we were calling gravitons were behaving exactly like gravitons. That they satisfied all the rules for graviton-gravaton scattering. Just saying there was a particle that looked like a graviton was a very weak thing. When all of the scatterings were constructed and the rigorous tests of whether it was satisfying the rules for scattering of gravitons by massive things, by massless things and so forth fit perfectly, exactly, then people realized that they really were dealing with something that looked like the scattering of gravitons and photons. So the scattering amplitudes played a big role in establishing with precision that we were dealing with objects that did behave like photons, gravitons, and it. Okay, I think we're finished. Yes. Where does charge come from? Where is what? Charge. Oh, yeah. We haven't talked about that, but okay, we'll talk about it next time. Remind me. Remind me. For more, please visit us at stanford.edu.
 Stanford University.  Okay, so, fermionic strings have been a part of the subject since almost the beginning,  not quite the beginning.  And as I said, they solve two important problems.  I forgot what they are, though.  So they give you fermions, and they get rid of the tachyons.  They leave you with ten dimensions instead of twenty-six.  So they don't solve that problem.  That problem turned out to be less of a problem and more of a feature, feature means a good  feature, than people had expected at the time.  But we'll come to it, and it's a subject known as compactification.  Using the extra dimensions and doing something with them to make them innocuous, or if not  innocuous, an interesting feature of the theory.  But we'll come to that not now, though.  What I wanted to talk about a little bit was not really historical, but the scattering  of strings.  The subject really began with studying scattering of particles.  Elementary particle physics was always about scattering of particles, not because it's  the most interesting phenomena that can happen.  It's not.  It's rather dull.  You send some particles together, and a bunch of junk comes out in all sorts of directions.  But it's about all we can do in the way of experiment.  And so we try to unravel from the scattering data what was going on inside the collision.  And inside the collision, of course, means the properties of particles and so forth.  So the natural tool of experiment, the scattering, and the natural thing that a theorist would  ask is if I have a theory of particles, how do you compute the scattering amplitudes?  The scattering amplitudes, what is a scattering amplitude?  A scattering amplitude, you have some incoming particles that are part of your incoming information.  I'm going to have time running this way tonight, horizontally instead of vertically.  I don't know why.  Variety.  Particles come in, something happens inside a black box, and particles go out.  Not necessarily the same number of particles.  The particles come in, and they carry momentum.  Of course, they carry other things.  They carry spin.  They carry charge.  They carry labels, like, for example, is it a muon, or is it a whatever it happens to  be?  But let's simplify the story and ignore everything except their momentum.  So particles come in, and they carry four momentum.  Four momentum means energy and momentum.  Let's write down what a four vector of energy momentum is.  The energy and the three components of momentum.  Of course, if we're working in 26 dimensions, we have 25 components here, but I'll just write  down four of them.  That's a four vector, a relativistic four vector.  Each particle has a momentum, and we're going to call it k.  k mu, where mu goes from one to four, from zero to three.  This is usually called zero, one, two, three.  Doesn't matter.  Four components of momentum.  Now, what do we know about the four components of momentum of a particle?  They have something to do with the mass of the particle.  Well, of course, they are the energy and the momentum.  What's the relationship between energy and momentum and mass for a relativistic particle?  Anybody remember?  c is one.  We will take c equals h bar equals one.  What's the connection between the components of energy, momentum, and mass?  e squared, e squared equals p squared plus m squared.  Here, let's write it the following way.  e squared minus p squared equals m squared.  Just in order to keep my notation consistent with notations of physicists for many, many  years, I am going to write this as p squared minus e squared equals minus m squared.  In formula, I've just taken the negative of it.  Now, that can also be written in terms of the components of k.  In terms of the components of k, that's, what is it?  It's k naught squared, sorry, it's k vector squared minus k naught squared, naught standing  for the time component, e.  Spatial component squared minus the time component squared.  This is often just called k squared.  Just call it k squared.  The left-hand side here is the square by definition.  It's a definition of the relativistic square of a vector.  It's called p mu p mu.  Or to simplify it, let's just call it k squared.  It's the space component squared minus the time component squared.  That's called k squared.  And for every particle, k squared is equal to minus m squared.  You can't vary k squared.  Of course, k squared consists of the energy and the momentum.  When you say you can't vary k squared, it doesn't mean you can't vary the momentum.  It means when you vary the momentum, the energy varies in a certain way.  And the way that it varies is that e squared minus p squared or k times k naught squared  minus k space squared equals minus m squared.  So that's the first thing about, it's not even about collisions.  It's just about particles.  You characterize them by their four momentum, three components of which are independent.  The fourth component has to be subject to this constraint.  Now we put in a bunch of momenta.  Let's call this k1 for the first particle.  Is that the way I label them?  I like to keep my notation straight.  Yeah, I think I call this one k1, this one k2.  And then outgoing over here, we'll call the incoming momenta k.  I'm going to call the outgoing momenta q.  These are also four momenta.  I'm going to call it q3, q4, dot, dot, dot.  Let's take the very simple case in which two particles go to two particles.  q4.  These particles come in, those particles go out.  How do you represent momentum conservation?  Momentum and energy conservation.  Moment energy conservation are simply that k1 plus k2, thought of as four vectors, is  equal to q3 plus q4.  All components, the space components define momentum conservation, the time components  define energy conservation.  OK, now because of the perversity of physicists, what physicists like to do is to redefine  the outgoing momentum and think of them as incoming momentum.  Now that's crazy.  The outgoing momentum are outgoing, the incoming momentum are incoming, but to do it, all we  really have to do, to make it symmetric with respect to incoming and outgoing momentum,  is to take each q, q4, change its sign and call it k minus k3.  That means changing the sign of its energy, means changing the sign of all of its momentum.  It's a trick to be able to write this in a symmetric form.  Instead of writing k1, well let's see what it is.  k1 plus k2 minus q4 minus q3 equals zero, now becomes k1 plus k2 plus k3 plus k4 equals  zero.  You treat all particles as incoming, but you have to remember that the label for the outgoing  particles is labeled with minus the actual momentum of the outgoing particle, but once  you do so, momentum conservation is completely symmetric between the four particles.  A useful trick.  It's a useful trick that keeps labeling especially consistent.  Notice that when you change the sign of the momentum to redefine the momentum with a minus  sign, it does not change the fact that the square of the momentum is equal to minus m  squared.  So we have, in this particular process we have four momenta, think of them as all incoming,  although the energies of two of them may be negative, the outgoing ones, each one of them  subject to this constraint.  The question that a physicist would ask about this collision is what is the amplitude?  What is the probability?  The thing, the amplitude is the thing that you square, the complex number that you square  to find the probability for that collision.  But the collision is a function of a number of variables.  It's a function, or the probability for the collision is a function of the momenta of  the incoming particles and a function of the momentum of the outgoing particles.  So it's a function of the k's.  Let's call that amplitude a.  It's a thing that you square.  And it's a function of all of the k's, k1, k2, k3, k4.  But there's some redundant information here.  First of all, the momenta have to be conserved.  Second of all, the square of each k has to add up to minus the mass squared.  So there's really too many variables here.  There's too many independent variables.  How many independent variables are they?  Before we impose any constraints, each momentum has four variables.  This is four plus four is eight, plus four is 12, plus four more is 16.  That's a lot of variables for something to depend on.  Fortunately, you really don't depend on that many variables.  Let's think about physically now how many variables there's a scattering amplitude or  a scattering process to depend on.  Well, the first thing you can do is whatever the momenta are, you can use relativity to  go to a frame of reference where the center of mass is at rest.  In other words, where the two momenta, the two space components of the momentum are equal  and opposite.  You can always do that.  If the particles are both moving down the z-axis, well, you just move fast enough to  be halfway in between them.  If they also happen to be moving in some other direction, just move in that direction, you  can always go to a frame of reference where the particles are equal and opposite, the  space components, in other words, the spatial momenta.  Next, you can always rotate the system so that the momenta are coming in along the x-axis.  What's left over in the initial state?  What does it depend on if you know that the momenta are equal and opposite?  What does the whole thing depend on?  What does the initial state depend on?  The initial state only depends on the magnitude of the momentum, not the magnitude of the  square of the momentum, I mean, sorry, not the magnitude of the four vector.  The four vector has to have magnitude m squared, but the magnitude of the space momentum or  the energy, if you like.  Once you go to the center of mass, the only thing left over is the total energy of the  collision in the center of mass frame.  So that's one thing, e center of mass.  Now let's suppose that our particles are all of the same kind for simplicity.  The particles collide, what can you say about the outgoing momentum?  First thing is they have to be equal and opposite.  Why?  That's momentum conservation.  Moment conservation says they have to be equal and opposite.  What about the energy of the outgoing state?  Has to be the same as the energy of the incoming state.  What is the only thing that can differ between the incoming and the outgoing particles?  The angle, that's it, the angle of scattering, theta.  So the scattering amplitude, although it's written in terms of 16 variables, really  only depends on two.  Two independent variables.  Those variables should be expressible in terms of relativistically invariant things.  We can think of it in the center of mass, but we also ought to be able to think about  it in any frame of reference.  We should be able to construct two independent invariants, relativistic invariants, which  are enough to completely characterize the scattering.  So let's talk about the invariants describing a scattering process.  Two particles come in, two particles go out.  Let's think of them as all coming in, changing the sign of the momentum.  What is this?  K1, K2, K3, K4.  What can you do with a four vector?  What can you do with four vectors to construct invariants?  There's only one thing you can do, really.  You can square them.  That's about all you can do.  But you don't have to be talking about K1 squared or K2 squared.  You could be talking about K1 plus K2 squared, right?  That's a good invariant.  K1 as a four vector.  K1 plus K2 squared.  Now remember what that means.  That means the space components of K1 plus K2, which are labeled with arrows, squared,  minus the time component, which means the energies.  Let's label them.  K0, K0 means energy.  K0, 1 and K0, 2 squared.  You sum the momentum, square it.  You sum the energies.  Square it and subtract.  And that's called K1 plus K2 squared.  Let's see if we can figure out what it is.  It's an invariant quantity.  Let's see if we can figure out what it is by going to the center of mass frame.  In the center of mass frame, the momentum of this particle, what should we call it?  Well, it's K1.  It's the space component K1.  What about the momentum of this particle?  It's equal and opposite.  Minus K1.  Right?  So, the total space component of the momentum in the center of mass frame is zero.  We don't have to worry about this one.  What about this one over here?  That's the energies.  We didn't talk about it, but what about the energies of the two particles in the center  of mass frame?  They're equal and opposite.  They have the same mass.  The energies are equal.  The center of mass frame is one in which the particles are perceived as moving with exactly  the same momentum, magnitude of the momentum.  And so the two energies are the same.  So this just becomes twice K naught quantity squared.  Very simple.  This is the center of mass energy here.  In the center of mass frame, you add the two energies.  It's the total energy, and the square of it is just the square of the center of mass  energy.  So this quantity here, which is called minus s, it's just called minus s, is nothing but  the square of the center of mass energy.  Just look at it.  Or s is the center of mass energy.  Why?  In the center of mass frame, this is zero, and this is just the total energy.  So minus s is just the square of the center of mass energy.  Square of the center of mass energy we call s.  Now can you think of any other invariant that you can build?  How about K3 plus K4 squared?  We took K1 plus K2 squared.  How about K3 plus K4 squared?  It's the same thing.  Why?  Because K2 is minus K3 plus K4.  Energy conservation.  So we don't get anything new there.  K1 plus K2 squared is the same as K3 plus K4 squared.  It's the energy, and it just says that the energy in the initial state is the same as  the energy in the final state.  What about K1 plus K3 squared?  That could be something new.  So let's see what it is.  Let's see if we can figure out what K1 plus K3 squared is by working in the center of  mass frame.  In the center of mass frame, all the particles in and out have the same total energy, have  the same energy.  Each particle has the same energy.  Particles just get scattered through an angle.  That's all that happens.  So they come in and they go out with the same energy, but they scatter through an angle.  So let's take K1 plus K3 squared.  Here it is.  K1 plus K3 squared, K3 squared, K1 K3 squared.  How about this?  This is in the center of mass frame.  It is same as K2 and K4, but that doesn't help.  I want to know what it is, though.  How big is it?  All the particles have the same energy, incoming and outgoing.  All they do is scatter through an angle.  Zero.  Zero.  Why is zero?  Why not twice the energy?  Yeah, because K3 and K4, we flip the sign on them.  So this isn't there.  And this is just K1 plus K3 squared.  So let's see if we can see what that is.  K1 comes in, particle 1 comes in, and of course also particle 2.  Particle 3 goes out.  Here's particle 3.  Time goes in, 3 goes out.  But if we label these particles with the Ks, K1, and let's call this one now Q, Q3.  Let's make it an outgoing particle.  Then really what this is, is it's K1 minus Q3 squared.  It's really the difference of the momentum of the incident particle and the final particle.  It's called the momentum transfer.  It's the momentum transferred, if you were to think of particle 1 and particle 3 as the  same species of particle, then it would just be in the collision, it's the momentum transferred  from 1 to 3.  Bing, bing, there's a momentum transfer, and that's the momentum transferred from 1 to  3.  That's what this is, momentum transferred.  It can also be expressed in terms of the angle of scattering.  I will tell you what the formula is.  The formula for K1 plus K3 squared is it's just twice the energy squared minus the mass  squared.  We already worked out what the energy is, the energy is the s variable.  It's twice e squared minus m squared in the center of mass frame times 1 minus the cosine  of the angle of scattering.  This is the interesting thing here.  The angle of scattering is the angle between 1 and 3.  K1 comes in, Q3 goes out, the particle gets deflected through an angle, and it's that  deflection angle, here it is, theta, the deflection angle of the particle between the incoming  state and the outgoing state, and that's what's here.  Why is that interesting?  Well, of course, that's what's measured in an experiment.  You put detectors in different places, you scatter particles, and you find out the probability  for them to get deflected through an angle at different values of the energy.  This variable, the K1 plus, let's see, what did we do here?  I'm sorry, I, I, the definition of s, I forgot to erase the s over here, the definition  of s was K1 plus K2 squared.  That was the definition of it.  One plus two coming in, and it was the square of the center of mass energy.  So s equals energy center of mass squared.  K1 plus K3 squared, that's called, anybody think of a name for it?  T.  T, good, T.  It is called T.  I knew you were going to get that.  And that is some combination of the energy and the momentum transfer.  It's the center of mass energy squared minus the mass squared, big deal, we already know  what the center of mass energy is, times one minus cosine of the angle of scattering.  All right, so we have K1 plus K3 squared.  How about K2 plus K4 squared?  Is that different?  Well, if you go back up to here, K1 plus K3 is minus K2 plus K4.  So K1 plus K2 squared is the same as K3 plus K4 squared.  It's also T.  Is there another combination?  We have K1 plus K2, we have K1 plus K3, K1 plus K4, right?  That's another one.  All right, so one more quantity is K1 plus K4 squared.  What can that be?  Well, if you were going to give it a name, what would you call it?  Minus T equals K1 plus K3 squared.  U minus U equals that.  These are the STU variables.  But what's going on here?  There are only two quantities that the scattering depends on, energy and angle of scattering.  We seem to have three independent things.  Well, the answer is there are not three independent things.  If you use the momentum conservation and you use the fact that each K squared is M squared,  what you'll find is that S plus T plus U is equal to minus 4M squared or something like  that.  There's a constraint among them.  There are not three independent ones, only two.  These are the two interesting quantities.  The third one is dependent on the other two.  Do you have to consider the vacuum at all in these collisions?  What does that mean?  Is the vacuum going to participate in the collisions?  I'm not sure what that means.  The vacuum is all over the place.  Participates like all hell.  What does it mean?  What do you mean by participate?  Maybe the vacuum could just slow down a particle going through free space.  Well, we know that doesn't happen, don't we?  It's momentum conservation.  Okay, well, just a wild idea.  Yeah, it could be.  I mean, it's a possible world.  The vacuum has friction.  It's not our world.  Moment conservation says that particles aren't slowed down by the vacuum.  At some point, I think you said that 3 and Q4 have the same energy.  In the center of mass frame.  In the center of mass frame.  Why can't there be a, I mean, K1 and K2, the center of mass frame is kind of defined that  way.  Yeah.  But then the center of mass frame is defined as a frame where the momenta are equal and  opposite.  If the momenta are equal and opposite, the energies of the particles are the same.  If their masses are the same.  Now they scatter, but they have to go out equal and opposite for momentum conservation  reasons.  Yeah.  So that's, all right, these variables, ST and U, the other one which is there, those  are called Mandelstam variables.  And they're very symmetrically defined.  Mandelstam, Mandelstam, Mandelstam.  Mandelstam is a currently active physicist in Berkeley, but this notation comes from  the early 60s, the very early 60s, Mandelstam variables.  And notice what they have.  They have this beautiful symmetric structure.  That's what's interesting about them.  Now let's talk about Feynman diagrams for a moment.  And I'll just tell you what the answer is for certain Feynman diagrams.  Let's suppose two particles collide.  This is a Feynman diagram.  Create a third particle of a different mass in here.  Let's call it capital M. And then those particles go out like this.  Those come in, Feynman diagrams, so they come in together.  They coalesce to form a second particle or a third particle or whatever, and then that  particle decays and it goes out.  That's a Feynman diagram.  That Feynman diagram has a value.  I'm going to tell you what it is.  It's the product of two coupling constants.  There's always coupling constants and Feynman diagrams.  And then there's the propagator of the particle in between.  And that has a very simple form, one divided by s minus capital M squared.  It has a s minus capital M squared.  That's the characteristic structure of a scattering amplitude where two particles come together  and merge and form another particle.  And notice it's a function of s, the energy of the process.  It doesn't depend on t.  It doesn't depend on the other variable.  You want one that depends on the other variable.  You draw a similar diagram, except in which instead of one and two merging, you have,  I'll draw it over here, one and three merging.  One, two, three, four.  It looks like exactly the same diagram except turned on its side.  But it represents a very different kind of physical process.  One and two come in, exchange a particle between them, and then go out as three and four.  What would you guess if you had to guess the amplitude for this process?  It's going to have g squared again, the g's are the vertices.  What else?  Exactly, one over t minus M squared.  Now you can imagine a third process which would be one over u minus M squared, and it  would look like this.  One, let's see, that's a little hard to draw, one.  It looks like this.  One and four merge, one and four merge, little, you have to switch the lines like that.  Let's ignore it.  It's there, it's there, and it's important.  But it's too hard to draw.  I don't like drawing cross.  They're perpendicular to the other two?  They're not perpendicular, it's just a question of which prior particles came together, and  I hate drawing cross lines on the blackboard so we won't.  But there is, in principle, there can be another one.  This one depends only on the energy.  Doesn't depend on the angle of scattering altogether.  T contains the angle of scattering.  S is only a function of the energy.  That means that when this process happens, to say that it doesn't depend on the angle  of scattering means that every angle of scattering is equally probable.  When the particles go in, they have equal probability of getting deflected through any angle or  whatever.  That seems odd, but that is the property of this process.  And the reason is very simple.  The particles come in from some direction, they form this compound state, and then when  they decay, they've forgotten which direction they came in from.  They come in, they form the composite, and then when the composite decays, it decays  at an arbitrary angle uncorrelated to the initial directions.  That's this process here.  This one is different.  It depends on the energy, but also on the angle of scattering.  This is the one that depends on the angle of scattering, and if you work it out, it's  easy to see that it favors small angle scattering, this favors large angles.  So this one depends on angle, and this one depends.  But notice how similar the two expressions are.  There's a real symmetry between them.  In fact, if you interchange S and T, the scattering doesn't change.  This is a property of relativistic scattering amplitudes that they have this kind of symmetry.  They sort of forget which were the incoming and which were the outgoing particles, but  only if you express it in terms of these kind of invariance.  Good.  So now we have some basic idea of what it is that a theoretical physicist wants to compute  about a model of particles to compare, not because he's so interested in it, but it's  the thing that he can give to the experimentalist and say measure the probabilities for scattering  as a function of energy and so forth, and here's my prediction.  Okay.  Are you saying that the dependence of S, T, and U causes this amplitude to what, shift  around if you, I mean you still have three terms, but.  But the value shifts around if you.  Depends on the angle of scattering.  All right.  So if all that was there was this, this would favor small angle scattering and be, not forbid,  but make it less likely.  Let's see if we can see why.  I think here it is.  I think that you have a third term.  The third term is there.  What about it?  Oh, well, what you say that you didn't want to write it up there or did it, you didn't  want to write the diagram.  I didn't want to draw the diagram, but if there was a process in which one in four could come  together and do the same thing, then we would want to put a G squared, one over T, one over  U minus M squared.  And because of three are not completely independent.  Yeah, they're connected.  Yeah.  Then that adds back to the.  One over U minus, okay, so what is U?  It's interesting.  U is equal to E center of mass squared minus M squared times one plus cosine theta.  So both, so they're clearly related.  If you know S, you know the energy.  Therefore, if you also know T, you know the angle of scattering, and this is just a function  of the energy and the angle of scattering.  So they're all connected with each other.  Another way to say it is if you add S and U, you'll cancel out the angle altogether and  you'll get just a function of energy, which is clearly dependent only on S.  But they're all, but the Cipri's finding diagrams and in general you have to add them all.  Okay.  But I want to focus on this here.  This does not describe the scattering of mesons, for example, very well.  It gives a poor description of the scattering of mesons.  And the reason is simple.  The reason is that there are many, many different particles that can be produced when two mesons  collide.  Whole stacks of them with different mass and different angular momentum.  We haven't described what the angular momentum would do here.  What the angular momentum would do would be to change the dependence on the angle of scattering.  Let's not get into it.  Formulas like this are too simple to describe the realistic scattering of mesons.  It's too easy to excite higher vibrations of these particles here.  And when two particles collide, they could make some ground state.  They could make some first excited state.  They could make some next excited state.  They can make a whole raft of different particles that can go in there.  And people in the 60s, basically beginning sometime around 1965, 66, 67, tried to concoct  all amplitudes with interesting mathematical properties that would represent all of the  possible particles which could go in here.  This was trial by trial and error.  It was just making up formulas to try to represent the different particles which could go in  here and the different particles which could go in here.  The first attempt tended to be to add particles.  This is called an S-channel process.  Why?  Because it involves one over S minus M squared.  Particles coming together, K1, K2.  That defines S. This is called the T-channel process, K1 and K3 coming together.  So they began by saying, let's just add more and more stuff into here, S-channel, and let's  add more and more stuff into here, T-channel, for all the particles that could be scattered.  Well, that lasted for a certain period of time, adding up the various particles which  could go in there.  And then people tried to find more comprehensive formulas just by, I don't even know what  the right word is, sort of a kind of curve fitting but a very sophisticated kind of curve  fitting and a rather dramatic formula was discovered which contained the physics of  all of the particles in the S-channel and all of the particles in the T-channel.  Just replaced this combination here with not just one particle but many of them, whole  towers of them with a formula which I'm going to just, this is of historical interest mainly,  with a formula which I will write down and you can explore its beauty, its great beauty.  And this is very, very simple, it's called the Veneziano amplitude.  Young Italian physicist, he was young at the time, he's not young anymore but he's still  Italian.  He just, I don't know how he made this guess, he just randomly wrote down some things which  had some right properties and which really did look like adding up things like this.  It was a function of S and T. Everybody know what the gamma function is?  Now the gamma function, you don't need to know it but I'm just going to write down the formula  for fun.  The gamma function is a generalization of the factorial function.  For the factorial function is only defined for integers.  The gamma function is defined interpolated between integers.  Gamma function for the integers is equal to n minus one factorial, gamma of n.  Now it's defined for non-integers too and I won't tell you, it's defined by an integral  and it continuously interpolates between the integers.  This is the Veneziano amplitude multiplied by the coupling constant squared.  If you examine this amplitude, you'll find out, I'm not going to go into its mathematics,  its mathematics is very simple.  All you have to know is about gamma functions and if you want to explore it, it's actually  quite a simple construction.  But I'm going to tell you what it looks like.  It looks like an amplitude that you would make by summing up large number of particles,  this is representing some composite particle or some particle it could be in there, of  different masses in the S channel.  But it also looks like, oh, notice that it's symmetric.  It's symmetric with respect to S and T, exactly like this is.  The peculiar thing about it is you can represent it as a kind of sum of Feynman diagrams with  all the particles in the S channel, diagrams like this.  But because it's symmetric under S and T, you can also represent it, you don't add, it  is also equal to the same kind of thing going this way.  Something odd was afoot about this formula.  It had all the important features that the scattering amplitude should have.  It could be analyzed as if a whole bunch of particles were produced and then decayed,  but it could also be analyzed as if a whole bunch of particles were exchanged.  This was something new.  This had not been seen before.  Previous to this, everybody would have added contributions for S and T. And this thing  replaced this.  The question was, what is this?  Where does this come from?  What kind of physics gives rise to this?  What kind of physics can you imagine would give rise to this?  To a formula like this?  The answer, of course, turned out to be string theory.  The invention in the discovery of string theory was just looking for a physical model which  would give this for its answers, for its answer for scattering of two mesons or two particles.  I'm going to tell you, without a lot of drama, what physical model gives rise to this scattering  amplitude.  I'll show you a little bit.  I'll show you what the logic that went into it was.  It wasn't very hard to guess that this was a theory of strings.  It wasn't too hard just because, well, it wasn't too hard.  You said exchange particles versus…  The S channel and T channel exchange, direct production and exchange, sometimes called the  direct channel and the cross channel.  The S channel is called the direct channel.  Those are the direct particles come in, coalesce, and then go out.  The cross channel is when the particle jumps across from one side to another.  I'll show you what the physics of that formula is.  You begin with two strings.  I'm not going to do a calculation, but I'll show you everything that went into it.  It's a little bit tedious, what you have to know about.  You have to know a lot about harmonic oscillators, and that's all.  Nothing much more than that.  You start with two strings.  Here's a string.  Now remember, the string has a coordinate along it which we called sigma.  Let's draw the sigma axis over here from sigma from zero to pi.  Here's the string.  But the string propagates in time.  Let's draw time horizontally.  Here's the end of a string at sigma equals zero, sigma equals pi.  These are open strings.  The string of course moves around in space time, but it's always located between zero  and pi.  That's not spatial position.  That's just its parameter along the string.  The material of the string, the particles that make it up, are in here.  Each point in the history of the string, think of the string as sweeping along, here it is.  It's sweeping along some space time sheet.  It's called a world sheet instead of a world line.  Each point in that world sheet is characterized by a point sigma and a time which has called  tau.  Tau goes this way.  It's a time, and sigma is a coordinate along the string.  Not a real spatial coordinate, just a label for labeling points along the string.  Now is this time that we've used previously the infinite momentum time, but that's not  what's important.  What's important is that the idea of a world line becomes a world sheet.  And instead of being parameterized by a single variable tau, it's now parameterized by two  variables.  Each point in here has a position in space time, x, x mu, or just x.  And we've already worked out what the equations of motion of x are.  They are the wave equations describing waves moving up and down the string.  The wave equation.  Let's write it.  This is d second x by d tau squared minus d second x by d sigma squared equals zero.  That's the wave equation that described the oscillations of the string.  But we don't even have to think about this.  We can just imagine that this string is a collection of a large number of particles.  Place the world sheet by a bunch of world lines narrowly spaced with springs between  them.  With springs between them.  That's the picture of the evolution of a particle.  Now what we're going to do is begin with two particles.  We're going to have two particles coming in from the past.  And we're going to put them right next to each other.  Let's see.  I think we need another color for the second particle.  We're not going to put them right next to each other in space necessarily, but just  in the parameter space.  Here this goes from zero to pi.  We need to put another one in going from pi also from zero to pi.  Here's the other string.  There's no meaning to the fact that I've drawn them right next to each other.  I've just drawn them.  Their actual position in space-time might not be adjacent.  So this just parameterizes, this half parameterizes the world sheet of the left particle.  This half parameterizes the world sheet of the right particle.  And they might be far away from each other.  That means the X's over here may be very different than the X's over here.  Now with any luck at all, with some probability, the end of the string might touch the end  of that string.  And when it does, they can coalesce.  That's an assumption that they can coalesce.  But that's the basic process of string theory, that they can coalesce and then form a single  string.  Now they really are connected.  If you like a new spring developed, when these touched each other, a spring appeared connecting  the last particle of this string with the first particle of that string, and the whole  thing becomes one string.  That condition persists for a while, it persists for a while until a quantum mechanical event  happens and the strings separate again.  Randomly, but you can transform around cleverly to make it be at the same place along the  string.  There's enough symmetry, enough symmetry of the equations that you can put this point  at the same horizontal level as this point.  So what is the nature of this process?  Given this, it's possible to guess what the answer is for the O.  The important quantity in here is the amount of time that the string spends coalesced.  It's the amount of time that the compound state stays before it breaks up into its final  constituents.  And let's just call that tau now.  Let's call that from this time here, tau.  I'll tell you what we do with it.  In the end, we integrate over it, but let me tell you what you put in.  If you're interested in a quantum mechanical amplitude, you want to start with an initial  state.  The initial state, let's start with one particle.  The initial state can be thought of as the state of a whole bunch of little points, namely  the points that make up the string, x1 through xn.  Let's start with the first string.  x1 through xn.  It's just a collection of mass points, x1 dot dot dot through xn.  And the wave function of it is just a function of x1 through xn.  That's its wave function to begin with in the start.  What do we know about the wave function?  Does everybody understand why a wave function is a function of the n positions of the particles?  That's quantum mechanics.  Quantum mechanics says wave functions or state vectors are functions of the position of the  constituents.  What's the probability to find the particles at position x1 through xn?  Psi star psi.  So that's what this is.  This is the wave function of the starting assemblage of particles.  Of course, I'm purposefully not taking the continuum limit to show you what goes on.  At the end, you have to take the continuum limit, but let's not do that.  What do we know about this?  Well, the first thing we know about it is that this particle comes in with momentum  k1.  I'll tell you exactly what that says.  That says that this wave function contains a factor e to the i k1 times the center of  mass position.  This is the wave function e to the i kx is the wave function of a particle with definite  momentum.  What momentum do you use?  You use the center of mass momentum.  So what is the center of mass position?  Sorry, the center of mass position.  What's the center of mass position of these points?  It's the average position.  X1, the sum of the x's divided by n.  So you have x1 plus x2, up to xn divided by n.  That factor, and that factor alone, tells you that this initial particle here had momentum  k1.  Now, what about the rest of the wave function, which depends on the relative coordinates,  not the sum of them, but the distances between the neighboring particles?  That's some wave function which characterizes the ground state.  It depends on everything except the sums of the x's.  It's some wave function, let's call it sine naught for ground state, and it depends on  the x's, same x's, but it actually doesn't depend on the sum of them.  The sum of them here, the differences between them here.  Differences of x's here, neighboring x's and so forth, here, some of them here.  And this wave function is computable.  It's just the ground state, describing the ground state of all the harmonic oscillators  making up the string.  And it can really work out.  With enough room on the blackboard, I could tell you exactly what this function is as  a function of a collection of x's.  It's not very hard.  It's a bunch of exponentials.  It's workable, you can do it.  This is the wave function of the first particle.  What about the wave function of the second particle?  It's exactly the same kind of thing, except it doesn't depend on these coordinates, it  depends on the coordinates of the red particles here.  So let's write it down.  It's e to the i k2.  Now, what shall I write?  Shall I write x1 through xn?  No, that's the original particle, those are the original constituents of the first string.  I want the constituents of the second string.  So let's write xn plus 1, xn plus 2, all the way up to x2n.  The second half of the particles is grouped together into the second string, also divided  by n, also times psi naught of 1 through n.  This is xn plus 1 to the end, to the end of the chain.  Can that be read?  It is read.  But can you read it?  OK, you can read it.  Now, that's the initial state.  But now you say with some probability, the two endpoints merge.  To say that the two endpoints merge simply says that you set xn equal to xn plus 1.  You look for that piece of the wave function where the two endpoints are at exactly the  same point.  So you begin with this.  The next step is to say, let the nth particle on the first chain be at exactly the same  place as the n being at the same place as xn plus 1.  So we're going to put in here then xn.  All the others left unchanged.  That's now not, that's the wave function of the state right at the point after the two  particles have coalesced.  When they're coalesced, at the point where they coalesce, they come together.  Nothing happens to the rest of the chain, but the two particles, so this is, if you like,  it's the amplitude that the two particles, the chain touched.  This is the amplitude that the two particles, the endpoints touched.  Now we have a new starting point, which is a function of x1 through x2n.  It's a state of that many particles.  And what do we do with it?  We have to evolve it.  We have to evolve it using the Hamiltonian.  Remember what a Hamiltonian is.  A Hamiltonian is a thing which updates you from one instant to the next.  So you take this initial state.  It's a well-defined thing, and you propagate it forward in time using the Hamiltonian.  Take it.  What's the right rule for updating a state from one, from an instant to a later instant?  You multiply the state by something.  e to the iht.  e to the i.  So you take this wave function, and you evolve it.  You solve the Schrodinger equation for it.  But that's the same as multiplying it by e to the i, the total Hamiltonian times tau.  But what is the total Hamiltonian?  The total Hamiltonian is just the collection of springs and mass points, a collection of  harmonic oscillators.  It's just a collection of harmonic oscillators.  We know how to do this, and that gives us the state of the system after time t.  What's the last step?  The last step is to project the final state, here it is, onto two separate particles again  to project it onto a state with two separated particles of momentum k3 and k4.  It's very, very straightforward if tedious, slightly tedious.  You take the two initial states of the two particles, they're well-defined ground states  of the particles.  You insist that the two endpoints are at the same place, that constrains the wave function.  You let it evolve as a single string for a ways, and then you let it break up again,  and that simply means multiplying it again by some final state.  I'm not going to write it all out.  It's the same kind of thing.  And that gives you the transition amplitude.  To make a long story short, you start with the two particles, you constrain it so that  they're at the same place.  You evolve it, and then you project it onto the final state.  A very, very well-defined thing to do.  That gives you the amplitude that the strings coalesced for an amount of time tau.  But how should you choose tau?  Those things in black and red there are added together, and then you multiply.  Those are multiplied together.  Now, if you have two separate systems, each having its own wave function, you multiply  to create the wave function of a composite.  So this is just a wave function of a composite of two particles.  To begin with, you don't set the first particle equal to the last particle here.  Then you look for the amplitude, you look for the piece of the wave function where those  two particles are at the same place.  You say, aha, now that they're at the same place, fuse them together and evolve it as  if it were a single string, a single collection of mass points and springs.  For an amount of time tau, and then basically you take your scissor and just cut the spring  in the middle and let it evolve after that.  That calculation is quite doable, not even very hard.  It's just too much for the blackboard for one morning, but I'm going to tell you what  the answer is.  I'm going to write down the answer for you.  Don't try to remember.  This is the answer.  It's an integral over the tau.  Did we say that we integrate over tau?  I think I said we integrate.  What do we integrate over tau, incidentally?  Why do we add up the amplitudes for all possible times that this composite could exist?  This is the Feynman rule of summing over all possibilities.  The only parameter here is the time that it takes for this to break up again.  Feynman's rule is sum over all paths, which in this case just means integrate over the  time that they spent evolving together.  That gives you the amplitude for the two particles in the initial state to become the two particles  in the final state.  The result is an integral.  It's the integral over the time that they spend together.  That's it.  I'll tell you what the integrand is after a certain amount of calculation, which is  actually not very hard.  It becomes the integral of e to the tau.  That's the time.  Times s plus 1.  This s plus 1 is exactly this s.  It's the s variable or the center of mass squared, the center of mass energy of the two particles.  You have a factor like that.  You have another factor, which is 1 minus e to the tau to the minus t minus 1.  t is the momentum change between the initial particle here and the final particle here.  Remember, those momenta are coded in this wave function, which I've erased.  They were coded in the wave function in terms of those exponentials.  Sorry, the center of mass energy.  Here is the momentum transfer.  It appears in the formula.  Then you have to integrate it, d tau.  There happens to be another factor, I think, of e to the minus tau in the integrand.  You're right.  I have d tau twice.  e to the minus tau, yeah.  The first term is tau times s plus 1.  Yeah.  No, yes.  Yeah, that's what you get.  Is that plus e to the minus tau?  What's that?  When you combine the e to the minus tau hanging out the end, is that multiplied or how do  you multiply this?  I've left it out here on purpose.  I've left it out here on purpose, d tau.  Now that doesn't look particularly symmetric between tau and between t and s, but after  it was computed, it took about a half hour to realize that you should change variables  in this integral.  Change variables from e to the tau, from tau to something called z.  Let e to the tau equals z.  Let e to the tau equals z.  Now we write this.  e to the tau is z, so this becomes integral of z.  I'm missing some minus sign.  No, e to the minus tau must be, this must be minus here.  It is e to the minus tau there.  This thing just becomes z to the minus s plus 1.  e to the minus tau, so this is minus, minus, e to the minus tau z, by definition, just  to change your variables.  What about 1 minus e to the minus tau?  That becomes 1 minus z to the minus tau minus 1.  Oh, incidentally, this integral goes, I think, from 0 to infinity.  Is that right?  From 0 to infinity.  Yes, it does.  This becomes 1 minus z to the minus tau minus 1.  What is d tau times e to the minus tau?  It's just d z.  It's just d z.  I think I got it.  It's just d z.  Did I get that right?  I've lost track of whether this is plus or minus here.  I don't remember.  But it's just d z.  This is the formula.  And where does the integral go?  The integral goes from tau equals 0, where z is equal to 1 to, this is minus.  It is minus.  I'm sorry.  I'm making a mess out of it.  e to the minus tau is equal to z.  To tau equals infinity.  In other words, you have tau equals 0.  That's when it very suddenly merges and then falls apart instantly.  The tau equals infinity, when the separation is infinite here.  And what happens when tau goes to infinity?  z goes to 0.  So it's an integral from 1 to 0, or 0 to 1.  There's probably some sign in here of an integral that looks like this.  The amazing thing about this integral, the whole upshot of it, is that this integral  is completely symmetric between s and t.  Can you see that?  How do you see that it's symmetric between s and t?  Right.  You just substitute for z, z minus 1.  You make a change of variables between z and z minus 1.  And you see that this integral is completely symmetric.  So although a completely unsymmetric starting point between energy and momentum transfer,  somehow it wound up giving a completely symmetric answer between the two of them.  I lost track of the minus sign.  The minus sign comes from e to the minus tau d tau going to dz.  Yeah.  I've lost track of the euro.  So you get a minus sign, that's why you switched to order.  There's a minus sign somewhere.  I'm saying e to the minus tau d tau.  Yeah.  And then you're going to make that just be dz.  Yeah.  So you get a minus sign.  Yeah, there's got to be a minus sign out here, right?  Right.  Then you switch the order of integration and it becomes from 0 to 1 like this.  That's right.  That's correct.  It's symmetric whether or not you switch the order of integration.  Right.  And that's the answer.  It's symmetric between s and t.  And what is it?  It's a process in which two particles join, form a composite which wiggles around for  a while and then breaks up.  In other words, it's the analog of the Feynman diagram in which a composite is formed and  then decays.  Composite is formed and then decays.  But it winds up being completely symmetric between s and t.  In fact, this function is called the Euler-Beta function.  It's a function of two variables, s and t.  It's called beta of minus s and minus t.  It's the Euler-Beta function.  Same as function of mathematical physics.  And guess what it's equal to?  It's equal to the Viniciano amplitude.  It's exactly equal to the Viniciano amplitude, namely this product of gamma functions over  another gamma function.  How did it get to be that it was symmetric between the two?  Now this is not obvious at all.  It's symmetric between the s channel and the t channel.  It's symmetric between the s channel and the t channel.  I'm going to come to that next time.  That has to do with a fundamental, extremely deep symmetry of string theory called conformal  symmetry.  It's a symmetry which allows you to take these world sheets and deform them in crazy ways  as if they were Turkish taffy and stretch them out in different directions.  And for example, turn this picture into a picture which looks much more like two particles  coming together and exchanging something.  It's a character.  Yeah?  So the gamma gamma function is reminiscent of the probability function of I think it's  factorial A, factorial B over factorial AB, the permutation.  Right.  Comparatory coefficients.  Yeah.  So what's the time in between the two?  It is true that the beta function evaluated at integers is the combinatoric coefficients,  the inverse of the combinatoric coefficients.  You know what he's talking about.  He's talking about n factorial, m factorial over n plus m factorial, which occurs all  over the place in combinatorics.  This just happens to be the same function.  No simple connection.  First of all, it's the inverse of it.  The inverse of it is not a combinatoric coefficient, but it just happens.  It's an integral which defines the same combination of gamma functions.  There is no simple relationship.  It's not that something combinatoric went on here, at least not to my knowledge.  And it's certainly not the way Viniciano found it.  I don't know what magic he pulled up to find it.  So this was, if you like, partly historical, but part of the important logic of the theory  is that number one, you can calculate with it.  It's all it is, is harmonic oscillators.  Everything can be done with harmonic oscillators.  It's a bunch of harmonic oscillators.  You break the process up into pieces, and then you integrate over the time in between.  You can calculate, you calculate, and you find an integral.  The integral, at this point, by magic, has the property that it's symmetric between  s and t and somehow looks like not the sum, but has the features of having processes where  particles coalesce.  That's the direct process I drew here, but somehow buried in it is also somehow processes  where particles are exchanged.  And that was the magic of it.  What was the surprise that a whole new logic of putting processes together to make new  processes?  Okay.  Any questions about this?  Yeah.  I think I'm missing something.  In the beginning, we described two strings called less than you know, the string of n  oscillators and n oscillators, they come together in two n, as it goes on.  But remember, when you're taking the limit n goes to infinity, two n is the same as n.  Sure.  But it still has the property of a boson or fermion.  Oh yes, so far, yes.  The question is, when you wrote down the wave equation, you had two points coalescing,  which is two n minus one, so it changes from positive to, it seems inconsistent, I'm  sure.  You lose two, not one.  Well, n and n plus one becomes one.  No, actually n and n plus one disappear basically.  Okay.  If you like, I mean, I think that's the right way to think about it.  Yeah.  Yeah, they eat each other.  They eat each other like that.  And you're left with, right?  So the new spring, the new spring that forms connects these two.  Yeah, otherwise we'd have big trouble.  We would lose one fermion.  Right.  So that's a good one.  One way to think about it is that a particle and an antiparticle color less than the, if  you thought of these as quarks on ends of strings, you could think of the last one being  a quark which annihilated with the first quark of the other string.  So yeah, you lose two of them.  But who cares about two when there's an infinite number?  Yeah.  Right.  Yeah?  Right.  But you're right.  You have to be careful not to lose an odd number of fermions.  That's something that shouldn't happen.  Yes.  The next quarter you continue on with string theory and talk about superstring.  Well, we're not going to spend a lot of time on superstrings, although the things that I  will tell you, strictly speaking, apply to superstring.  And not to the bosonic string.  No, I'm not going to go into the heavy mathematics of superstrings.  What I think we're going to do next time is I'm going to tell you, well, I'll tell you  a little bit more about string theory and about the properties of world sheets and the  symmetries that allow you to stretch this thing like Turkish taffy and make different  kinds of Feynman diagrams out of it.  But then we're going to move on to another subject called M theory, which is another  way of deriving string theory from a totally different van- well, I think we have to talk  a little bit about compactification, about what you do with these extra dimensions, and  we will maybe next time.  But then we will move on to a totally different origin of string theory that evolved much,  much later in time, sometime around 1995, 96, which started with a completely different  picture, and in which a great many of the features, the more complicated features of  string theory are completely transparent, and it's called M theory, and we'll start  a different starting point, which leads to exactly the same physics.  So this process you just described was based on the clues from two Masons colliding?  Yes, but it's also the way to photons collide in string theory.  It's the way any two open strings collide in string theory.  Incidentally, the analog for closed strings is very similar.  You have two closed strings, and you pick a particle from here and one from here, and  you require that they be at the same point so that they do this kind of thing.  And then you evolve it as a single string, same rules, same kind of rules, and that would  correspond to the scattering of two closed strings.  If you scatter two closed strings, the intermediate thing that you make will again be a closed  string.  Does this only apply to strings where the particles are fermions?  No, no, no, no, no, no, no, no.  This calculation was originally done for the Bosonic string, and of course it's more complicated  when you have to keep track of the fermions also.  Extremely similar, extremely similar, some slight differences that are not important.  What I should say about this is that the scattering amplitudes, we said that there was a photon  and a graviton in the system.  All right, it looked like there was a spin two particle and a spin one particle, and you  could force them to be massless if you wanted.  You didn't have much choice.  There weren't enough components to make a massive particle.  So you did that.  Well, then you take these particles and you collide them, and you work out scattering  amplitudes.  Scattering amplitudes are very distinctive and characteristic for emission of photons  and gravitons.  They are not just any old scattering amplitudes.  They have very, very definite properties which make them extremely special.  The emission and absorption of photons cannot be mistaken for the emission and absorption  of scalar particles or other things.  They satisfy some very, very important rules.  Those rules originate from the conservation of electric charge.  Conserved electric charges emitting photons is almost very, very little ambiguity in what  the emission and absorption of photons look like or what a scattering of photons by charge  particles look like or even the scattering of photons by photons.  And it was at this point calculating these diagrams where it became completely clear  that these things which we were calling photons were behaving exactly like photons and the  things that we were calling gravitons were behaving exactly like gravitons.  That they satisfied all the rules for graviton-gravaton scattering.  Just saying there was a particle that looked like a graviton was a very weak thing.  When all of the scatterings were constructed and the rigorous tests of whether it was satisfying  the rules for scattering of gravitons by massive things, by massless things and so forth fit  perfectly, exactly, then people realized that they really were dealing with something that  looked like the scattering of gravitons and photons.  So the scattering amplitudes played a big role in establishing with precision that we  were dealing with objects that did behave like photons, gravitons, and it.  Okay, I think we're finished.  Yes.  Where does charge come from?  Where is what?  Charge.  Oh, yeah.  We haven't talked about that, but okay, we'll talk about it next time.  Remind me.  Remind me.  For more, please visit us at stanford.edu.
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