All right. So, Nima, I'll be sharing today. You let me know when you decide when to take the break. Right? Just to decide. Are you ready? Can we start? Yeah, absolutely. OK, great. So let's start with a second lecture by Nima Arcania-Hameda, please. All right, guys. Good morning, everyone. So last time, we left things off by going back to a very basic question. What is a particle? Remember, my goal here in these first couple of lectures is to give you a way of thinking about what particle physics is, really going back to even what a particle is from this point of view that emphasizes the dramatic difference between massive and massless particles, and is also associated more generally with the way of thinking about things where the properties, the invariant properties of the particles themselves are the star of the show, rather than the somewhat more conventional textbook point of view where we talk about quantum fields and so on first and particles come out later as some excitation of the quantum field. The point of view that I'm exposing here is one where it's really the sort of particles of the fundamental objects from the get-go. And so the things that we'll be talking about today are really going back to this very basic question, what is a particle? And the deep distinction that we talked about already between massive versus massless. And I will begin to talk about why the case of massless particles especially are incredibly constrained. They're incredibly constrained by both the Poincarean variance, so by the space-time symmetries, which goes in again to even what we mean by what a particle is, that's what we're going to start with, and unitarity and quantum mechanics. And the amazing fact that I alluded to last time is that really these two things in a very direct way, in a very direct way, these are the principles of space-time and quantum mechanics tell us that we have this minuscule menu that we're allowed to choose from in order to describe any possible universe compatible with the principles of quantum mechanics and even special relativity, that the structure of the spin 1 must be Yang-Mills, that the structure of the spin 2 must be GR, that the spin 3 have must be associated with Susie. And now these are all things that of course are standardly discovered from a more conventional point of view of quantum field theory as well, but in a more roundabout way, at least from my point of view in a more roundabout way, I want to describe things this way because I find them very vivid, you see very clearly how directly the principles of space-time and quantum mechanics are reflected in the possible theories that can describe nature. And all of this will give you a deeper appreciation for what is special about spin 0. So, spin 0 is the story of the Higgs spin 0, it's the, as I mentioned last time, Higgs is the first spin 0 elementary particle that we've ever seen. And I believe after this discussion, you'll have a better appreciation for what is really so special about, okay, so, okay, and then after this discussion, we'll move on to say more things about the Higgs specifically, but today we'll mostly be about talking in more detail, even though we understand it intuitively quite well, talking in more detail about the difference between massive and massless, and going through aspects of this argument for why the dynamics of massless particles, especially with higher spin is so ridiculously constrained. I think this story that I'm telling you is one of the really great stories of theoretical physics and gives you a sense for why it is that relativity and quantum mechanics put such an insanely tight straitjacket on our imaginations for what might actually be happening as we understand nature more and more deeply. So, all right, so, so let's go back to this question, what is the particle? And I, as I emphasized last time, sort of more abstract answer to this question that scores very closely connected to the intuitive way of thinking about it is it's the particles are irreducible representations of the Poincare group. And we diagonalize translations by working with momentum eigenstates. And so we have states that are labeled by a momentum P mu, which satisfies the on-shell constraint that P squared equals m squared. But the states are labeled by P mu and some other labels sigma. And all of our discussion to begin with is going to talk about what these other labels sigma can possibly be. As I mentioned before, this is really following Wigner and the notation and the argument of that you can find in Weinberg volume one. Okay. All right. So these are the states, but what we're interested in figuring out is how can we possibly, if we take a general Lorentz transformation, lambda goes, P goes to lambda, let me just write it explicitly once. If we take a general Lorentz transformation, lambda mu nu, so that we have P mu goes to lambda mu nu P nu. So I will abbreviate this often just as P goes to lambda P. Well, we want to find a unitary operator u of lambda associated with that Lorentz transformation. And we want to see how does u of lambda act on these momentum eigenstates. And the most naive answer would be that u of lambda acts on P sigma to give lambda P sigma, where the sigma labels are the same. So we can ask, is this the answer? And this is not the most general answer. And it's obvious that it can't be the most general answer because as we discussed, there are certain Lorentz transformations P that leave, so take those Lorentz transformations lambda such that lambda P equals P. So take these special Lorentz transformations such that lambda star P equals P. For example, if the particle is just at rest so that P is equal to M zero, then these lambdas would just be rotations. Okay. So just to take a very simple example for a massive particle at rest, clearly there are Lorentz transformations that leave the momentum invariant, but they have to rotate something else. We know that the particles have spin, so they have to rotate something else. They have to rotate the spin index. Okay. So this formula cannot be correct, at least intuitively, we know cannot be correct because there has to be something that reflects the action of rotations that don't of the Lorentz transformations that leave the momentum invariant, but which somehow mixed up the spin labels. Here we're cheating a little bit because we're jumping to the end that we happen to know that massive particles have spin. And so we know there has to be something missing. But we're now going to discover that spin is one of the quantum numbers that can label a space-time quantum number that labels massive particles and that analogously it will only be a helicity that labels the quantum numbers for massive particles. Okay. So in order to do this, we have to be careful about how we actually label all of the states. So this was really sort of Wigner's idea. Let's begin by carefully labeling all the states and defining all the states. Begin by defining what we mean by all these states. Defining all the states, P sigma. And Wigner's idea was to begin with a reference momentum. So let's say you have a reference momentum k mu. So for example, for a massive particle, it could be a particle at rest in some frame. For a massless particle, it could be a particle moving in the z direction. So you begin with a reference momentum and then you find some way of writing your general momentum P mu as a particular Lorentz transformation on that reference momentum. Okay. And notice that this choice we're making for the Lorentz transformations, I'm going to write P mu as some Lorentz transformation l mu nu that depends on P. It will depend on this reference momentum k times k nu. Okay. And here I'm making a choice. This guy is not unique. And again, we've seen why it isn't unique because once you make some choice to make the Lorentz transformation to take you from k to P, you can still do more Lorentz transformations that leave P invariant. All right. So this choice is not unique, but pick one.  And now what we're going to do is this is going to be a definition now. We're going to define P and sigma is defined as this unitary transformation now for this special Lorentz transformation l of P and k on k and sigma. And notice this is a definition. The important part of this formula is that the sigmas on these two sides are exactly the same. Okay. So in other words, I have to tell you how I'm comparing what I mean by these new labels of the particles for one momentum and another momentum. And I'm simply defining that label for a general momentum to be what I would get from the reference momentum by doing this specific Lorentz transformation that took me from one guy to the other. Okay. So this is a definition for what I mean by all the states. Once you make a choice for the reference momentum and a choice for the special Lorentz transformation, this is now a definition for the general state. With that definition now, now we can ask what happens when I take u of lambda and I act it on this state P and sigma. Okay. And so what is this? This is u of lambda and just the definition that I just gave you for P and sigma. Okay. And our previous naive feeling for what the answer would have been as u of l of lambda P and k on k and sigma. That was our sort of naive idea because if we did that, this would have been lambda P and sigma, right? With the same sigma. But obviously, this is not true. Okay. The constant we can make it to be true is to go back here and just shove in front of this expression one. Okay. I will write this as u of l of lambda P times the inverse of this guy, l inverse of lambda P. Okay.  And I should have said explicitly, of course, a very important property of these unitary transformations is that they should faithfully represent the Lorentz transformation so that u of lambda one, u of lambda two is u of lambda one, lambda two. Okay. Okay. Now, let's look at this expression. I'm going to define this guy to be a Lorentz transformation w that stands for Vigner, w that depends on lambda P and of course this reference k. But what's the important part of w? What does w do to k? Okay. What does w do to k? w takes k to l P of k is P to lambda P. That's after acting with lambda. And then l inverse of lambda P brings us back to k. So what that tells us is that this u of w, I'll call it from now on, on k and sigma, whatever it is on the right hand side, it has the same sigma, it has the same k, because wk is equal to k. And therefore, this is something of the form, the sum sum over sigma prime, d sigma sigma prime of w on k and sigma prime. Okay. So that's all I know about this guy. Some linear combination, both the same k of k and sigma prime. And therefore, if I slap on this u on the outside, what I finally get is that u of lambda on P sigma is the sum over sigma prime of this d sigma sigma prime of w. Since the k is the, since I got k back here now, when u of lambda P acts on the k, it's just going to give me lambda P. Okay. So times lambda P and sigma prime. Right. That's the final expression that we want. So we have learned how, we have learned how to represent the action of a general Lorentz transformation on our state. Okay. And as said, this Lorentz transformation now mixes up the other labels, whatever these other labels are. Okay. So we've learned the rule by which they mixed them up. But we also learned what the other labels are. The other labels are a representation of the so-called little group. Okay. So what is the little group? The little group is the set of all Lorentz transformation. All Lorentz transformations w such that w mu nu k nu is equal to k mu. Okay. And that's again, what we expected intuitively to begin with, that for example, if we have the massive particle, if we do the Lorentz transformation on it, it should mix up something. It should, we should see the spin, which is associated with the rotations, which are those Lorentz transformations that leave the massive particle at rest invariant. And now we've seen this in a slightly more formal way.  Okay. They're a representation of those Lorentz transformations w that leave k invariant. Okay. So what we've learned is that the sigma labels are a representation of this little group. All right. Any questions about this so far? If not, let's proceed. So now let's look at this question of massive versus massless. And so first let's look at the case of massive. And we have to figure out what the little group is. So let's choose for a reference momentum, the most obvious choice m zero. Okay. And therefore the little group are just rotations is just the rotations. For in four dimensions, it would be s o three. In d dimensions, it would be s o d minus one. Okay. So that's the but concretely, let's look at infinitesimal Lorentz transformation and infinitesimal Lorentz transformation is lambda mu nu is just the identity plus something small. Okay.  The fact that these Lorentz transformations, lambda mu alpha, lambda nu beta, a to alpha beta has to equal a to mu nu. Right. That's the whole point of the Lorentz transformation is that leaves the metric invariant. This tells me that this omega mu nu, if I lower the indices is using the metric is anti symmetric. Okay. And so if I want to figure out if I have some some lambda k equals k, this means writing out the indices that omega mu nu, let's say lower indices, k nu has to equal zero. Okay. So if we want to figure out what are the generators of the little group, we have to find those Lorentz transformation generators that annihilate the corresponding momentum. All right. And therefore, if I now choose k, k nu equals m zero. What do I learn from this formula? Well, if I put mu equals zero, I have that omega zero i k i is equal to zero. Okay. So I don't learn anything from that because the spatial components of k were already zero. Okay. But if I put mu equals i, I learned that omega i zero m is equal to zero. Right. Because the other components are zero. So that just tells me that omega i zero is equal to zero. And therefore, since omega is anti symmetric, I also learned that omega zero i is equal to zero. Okay. So if I look at the matrix of omegas that annihilate k, if I write here zero one, two, three, it's an anti symmetric matrix zero one, two, three. So I just learned that this that these top components have to be zero. And the rest of it is just an anti symmetric matrix. And of course, this is what we're familiar with with rotations. Right. So this is just going to be some a negative a b negative b c negative c, but it's only non-trivial in this lower block. And these are rotations. Okay. So that's quite familiar. And so we see that the massive little group is just saying it again is is s o three s o d minus one in general, and are just the rotations. Okay. Now let's proceed to the massless case where things are going to be different. Sorry. And so if I look at the what's the what's the algebra? What's the algebra? It's the usual one that we that we know and love where we have j x and j y is i j z. And here, sorry, where these j's are implicitly negative i omega. That negative i is just the conventional. And if I say this more nicely in general dimensions, instead of j x, I would replace that with j two three. Okay. So so the usual so I could write, for example, j z and j y is j x. I'm not going to write the eyes, because really I'm talking about omegas that don't have the that don't have the eyes. Okay. So so this would be j z and j x is j y something like that. Okay.  And you can just read off these commutation relations, even if you didn't know about it, just by taking these matrices, the the piece that's proportional to a and the b and c are three generators, and you can just work out what the what's the algebra is. Okay. All right. So now let's proceed to the mathless case. And now I'm going to choose k mu to be e e zero zero. Okay. So now this is the this is a mathless particle. And now let's go through the the the same argument. So I need to have omega mu new k new is equal to zero. Okay. So if I put new equals zero now already I learned something that omega zero one k one plus the rest of zero is equal to zero. So I learned that omega zero one is equal to zero. And therefore omega one zero equals zero. That's actually what I learned by putting mu equals one is that omega one zero k zero plus dot dot equals zero which already also tells me that omega one zero equals zero. So that's good. And so now finally let's put new equals two. If I put new equals two, then I learned that omega two zero k zero plus omega two one k one and everything else is zero is equal to zero. And since k zero is equal to k one this tells me that omega two zero is equal to negative omega two one. Okay. And similarly I learned from you equals three that omega three zero is equal to negative omega three one. Okay. So so if I now look at the omega matrix given what I've learned if I write out the same anti-symmetric matrix. Okay. So so I've learned that I have to have zeros oops let me write this little bigger sorry. So I've learned that I have to have zeros here. And so if I put if I call this omega zero two let me call a and omega zero three let me call b then I learned that then I learned that omega two I learned that omega two zero is equal to negative omega two one. Okay. So that means that if I put it by anti-symmetry this would be a negative a here and a negative b. So here this is a and b. Okay. And similarly here I get negative a and negative b. And then I know nothing about omega two three so there's just a c there. Okay. All right. So notice that once again there are three generators three independent generators that annihilate the light like momentum but they look different. They look different than the ones that we had before. Okay. So obviously the number of symmetries doesn't jump. So there's three generators for rotations but when it's massless we have these new three generators. And so let me define I can define jz to be the one that I get just from the bottom component the familiar one. This is just the rotations but let me do the one associated with the x let me call it tx instead of jx I'll call it tx which is zero zero one zero zero zero minus one zero zero zero zero zero zero that's the one that's associated with a and I'll call ty the one that's associated with b is zero zero one zero zero zero minus one zero zero zero zero. Oops sorry too small minus one one zero zero. Okay. So these are my three generators and now you can see something interesting obviously the commutator of tx and jz is equal to ty you can you can check this easily but I'll give you the sort of intuition for it in a second the commutator of ty and jz is equal to negative tx but so that's just like looks just like you know ordinary angular momentum if x and y were tx and ty were jx and jy but clearly tx and ty commute with each other okay tx and ty just act in different spaces so tx and ty just commute with each other okay all right so what is this algebra okay are you familiar with this algebra have you seen this uh uh algebra before well this is in fact the the uh the algebra of translations and rotations of the plane just the two-dimensional plane okay so it's called the Euclidean group and uh so what what uh so why why why is that well if we have this if we have a if we have the two-dimensional plane x y uh I can have the translations in x I can have translations in y and I can have rotations in the x y plane so that would be uh uh jz and clearly these translations commute with each other uh and clearly they have the the the corresponding algebra uh with z right because uh because if we do if we if we rotate a little bit in z we mix tx with with ty and ty with negative tx right so so this algebra is just the symmetries of translations and rotations uh on the plane okay so if we summarize then we see that the little groups are different they're sort of qualitatively different for massive particles um the massive little group is rotations so three in d dimensions so d minus one but the massless little group is the uh is the Euclidean group um in uh uh in in two dimensions and in general it would be the Euclidean group in d minus two dimensions if we did this in two in d space dimension okay um now let me give you uh a little bit of uh intuition maybe more intuition about why this discontinuous the uh transition happened and um uh so to see the transition between these two things um a little more smoothly uh uh suppose that in the massive case uh suppose in massive case um we had taken the reference momentum not to be k mu a particle at rest but a particle boosted uh in a particular direction so let's say I had taken k mu to be uh uh m gamma uh m gamma v uh zero zero okay with the usual gamma is one of one minus v square okay so um that's the uh that's the that's the boosted particle and you can you can do the analysis for the little group I'll leave this it's a very very short exercise but I'll leave it an exercise for you to work out that uh that the um that the uh if I if I call this the uh the uh yeah what what what you find for the for the uh little group generators are j two three exactly as you'd expect the rotation's there but uh for the other two generators we're going to get a linear combination of j one two that's the uh rotation in one two uh plus something that's the boost in the two direction and j three one minus the uh this velocity that's the boost in the three direction okay so um uh that's what you'll find if you work out the a little group in in this case um so as when the velocity is zero we have the usual three uh uh generators of rotations but as the velocity goes to one um they they shift to this j one two minus k two and and j three one minus the k three and if you work out the algebra what you find uh if I call this guy t hat one two and this guy t hat um two three uh three one then you find that that the algebra is that t hat one two t hat two three sorry t hat three one is one minus v squared times the the rotation in the two three direction while the other ones are um what uh what what you would be the usual thing that we would have expected is this is t hat three one and uh t hat um three one j two three is minus t hat one two okay but the uh interesting part is this one minus v squared here okay so so long as v is not equal to one okay so for any velocity that's not equal to one you can always rescale the generators uh so that the symmetries are those of the rotation group okay um but if v is strictly equal to one you can't do that uh uh rescaling okay so this is very much the way in which the Galilean symmetry emerges from the Lorentz symmetry the limit as you spend this uh and the limit as you spend the speed of light to uh infinity is a technically to contraction of the of the underlying uh uh group um and another intuitive picture is that uh when the particle is at rest we can think that uh we have rotations we have a symmetries of the sphere but what's going on when the when you boost the particle is it's like you're making the radius of the sphere it's like you're looking in the neighborhood of the of the uh north pole the z direction but it's like you're making that the radius of the sphere go like one over root one minus v squared and this radius goes to infinity as v goes to one okay so what's going on is that we always have the symmetries of the sphere but as you take v to one it's like your your this rescaling of generators it's forcing you to look closer and closer to the north pole so that in the limit it's the flat earth limit okay in the limit as the v goes to one all you see are the uh in the limit as v goes to one all you see are the symmetries associated with the flat version of the sphere um and so uh so we still have jz but what what used to be jx and gy are now just turning into translations in the x and the y directions okay all right any questions about that so now we've seen sort of more more technically this sort of intuitive difference between the uh little groups for massive and massless particles okay very good so um now uh this difference let me just do one more thing and we'll take our break um so uh this has uh this difference between the little groups has a profound difference for what these labels are what these labels can uh these labels sigma can possibly be so uh as we've said already if the particles are massive since the little group is uh so three these labels sigma are just spin sigma labels are just spin that's the definition of what spin is spin are things that are representations of the rotation group but now let's look at the uh in the massless case let's look at what the representations of the little group could be right now notice remember in this usual story of spin in order to talk about any representation to begin with you just diagonalize as many things as you can right so that's why uh jx gy and jz don't commute with each other and so we pick one of them we diagonalize jz that's spin in the z direction and then okay then uh then then the other two um the other two generators we can take linear combinations to make ladder operators uh and that's how we understand spin in uh undergraduate quantum mechanics the situation is different than the euclidean group we can diagonalize more to begin with because tx and ty already commute with each other okay so tx and ty commute and then we have tx and jz is ty and ty and jz is minus tx but tx and ty already commute with each other okay so these already commute so i i can label my states naively by the eigenvalues under tx and ty right so that's what i'm going to do i'm going to label my states by whatever their eigenvalues are for tx uh and uh ty i've just written those as little tx and little ty so capital tx y on the state tx and ty is just given by uh tx little tx or y on this state okay so it's just eigenstates of the translations they're mutually commute so i can uh i can do that however let's say that tx and ty even one of them is non-zero then i have a problem um if they're non-zero now let me act on the state with a rotation in the z direction okay so let me see what happens if i take the state and i rotate by like e to the i theta jz on tx and ty well you know what this is going to do like tx and ty are just a two-dimensional vector and so under rotation this gets to kind of rotate to another vector so this is going to be you know uh this is going to be tx cos theta plus ty sine theta uh ty cos theta minus uh tx sine theta right but the important point is that if i have any tx ty non-zero here then by acting with jz i'm going to produce all these other states on the circle right and i will have all these states i must have all of these states for all these different values of t uh here okay because why because starting from one of these states uh by applying um rotations i'm going to get other states that's the whole point of the i mean i start from any given state i apply the symmetry i get other states so this tells me that whatever these labels sigma are have to be continuously infinite all right so on that these are called continuous spin representations and that's what comes out of the box uh when you think about the representations of the little group it's a very peculiar thing okay that that the most obvious representations of the little group for massless particles actually have an infinite continuous number of spins it's not like the massive particles where we have you know you have to say spin one and you have three states spin a half you have two states here there's a continuous infinite number of states uh if you allow tx and ty to be non-zero okay well that's odd we haven't seen these in in nature we have not seen massless particles that have a continuous infinity of a number of uh degrees of freedom and this actually remains kind of an interesting area of research every now and then people come back to try to see if it's possible to make sense of these continuous spin particles um but uh we haven't seen them in nature that's not a very good argument but we haven't we haven't a we haven't seen them in nature b the attempts to make theoretical sense of them are confusing at best uh so far even though some progress has been made over the last five ten ten years or so um so what we're going to do at the moment is to sort of punt on this problem and we have to somehow avoid this infinity avoid continuous infinity how do we avoid this continuous infinity the only thing we can do is to choose choose tx and y to equal zero okay if we choose tx and y to equal zero then we're nailed to the origin all right but then we just have one state right i have one state and i know that e to the i theta jz acting on the state where tx and ty are zero is just going to give me something proportional to the state again and so it's just going to be some e to the i some value i'll call h theta uh zero zero right and so we've learned what is the label associated with the massless particle it only has a single label okay and the single label is this holicity h is that clear so if i say this uh if i say it again uh and i say it in d space time dimensions the statement is that massive particles are uh representations of s o d minus one massless particles will always have these funny continuous spin representations of the euclidean group of d minus two see that has the same number of dimensions as s o d minus one okay that's just the story we're talking about but these are the funny continuous one if we want to avoid the continuous ones so avoiding continuous nails us to this t to the origin and now it gives us representations of s o d minus two okay so that's the sort of famous drop in the uh in the number of degrees of freedom now in general for general space time dimension between massive and massless particles uh and to say it one last time the smooth thing that carries over smoothly with the number of the amount of symmetry doesn't change the number of generators doesn't change is rotations s o d minus one to the euclidean group the uh uh euclidean d minus two but the the representations with the euclidean group have this continuous infinity in them unless we further send things to the origin after we send those t to the origin the only thing that we're left with are rotations around the direction of motion of the massless particle which is s o d minus two and that's this drop in the number of degrees of freedom between massive and massless particles okay but again in four dimensions in the familiar case of b equals four dimensions um uh this is just a massive particle s o three is spin s uh that has two s plus one degrees of freedom but massless that's massive uh massless is just the helicity so h equals just plus or minus s um uh and as i alluded to last time nothing at this point in our discussion even forces us to have both plus and minus in theories that have parity symmetry uh uh at least approximately in parity symmetry we're forced to have both the holicities but at this point in our discussion all we've learned is that uh is that uh we can either have just plus or just uh minus okay perhaps this is a good time to um take a break and uh ask for uh questions questions yes this one by please go yeah go ahead uh yeah about this fact that we're kind of uh throwing away uh possible representation yeah in the massless case yeah this contradicts you know galman principle of you know the totalitarian principle absolutely yes you're right you want that well uh the uh that uh as i said it uh in um in theoretical physics it's never a good idea to throw things away that that appear to come out of the theory um just because we haven't seen them yet in nature because they might eventually show up somewhere in uh in the nature so so we get to be um uh chauvinists as theoretical physicists uh to just study what our theory tells us um is uh interesting and you know eventually the experimentalists will find it um uh but in this case um uh it's a more confusing situation because uh when you start studying whether it's possible to describe these continuous spin particles in a consistent way um uh it's not obvious that there is a consistently interacting theory for them right so um and in fact uh uh the whole story of continuous spin particles or the development of them is a good example of this alternate point of view of thinking about uh uh the interaction of particles from this particle centered on um on shell uh point of view because um certainly you don't run into them obviously with Lagrangians or anything like that right and when you play with Lagrangians you don't see these things come out but um but you can nonetheless sort of play with them to see uh from the perspective that i'll talk about um in the next lecture whether it's possible to write down consistently interacting amplitudes for uh for continuous spin particles um with ordinary particles with the gravity um and uh uh it's a complicated it's a confusing subject that there is no known working example yet um but there isn't a theorem yet either that it's impossible okay so uh so um uh yeah so that's so it's in a little bit of uh uh limbo it's also a little bit of a backwater i mean people aren't very actively working on this subject there are some nice papers in the 70s there are some very nice papers uh five or ten years ago by Philip Schuster and uh uh Italia Toro that returned to this uh subject so if you want to learn more about it i would encourage looking at these papers of Schuster and Toro it's it's an interesting subject um uh people had a few kind of throw away no-go theorems like oh if you have infinitely many particles then you will screw up thermodynamics because uh you know you would thermalize them and you'd have infinite specific heat and so on i mean uh you know you'd have some difficulties like that uh those are not good arguments because when you have uh you know uh there might be infinitely many particles but you don't have the same but it's not obvious that the strength of the couplings to each one of the sectors would thermalize all of them anyway that that's that those arguments are not good arguments um the really strong question is whether um uh you can uh write down that the really invariant question is is there a formula you can write down which is the which is local and uh conserved probability is unitary um for the interactions of these particles again i'll talk about that in the much more down-to-earth context of uh simpler context of ordinary massless particles um uh just in a moment now um uh but people have not managed to uh uh to write down fully consistent interactions for these guys yet uh someone in the chat asked whether they could have a reference um uh i i don't have the reference off the top of my head but if you just go to inspire and look at the philip shuster um yeah shuster uh and toro and if you just look at the words continuous spin if you just google that um then you'll find the papers that i'm not talking about and they'll have references to the older literature as well any other questions yes yeah uh so you said that in recent years like some research that developments has been made in traditional particles have they been able to pinpoint like an energy scale in which you might be able to see them this at all uh yeah i mean what they're um uh the the things that have been done are i believe if i remember properly something which uh kind of uh something that does make sense is the interaction between uh ordinary matter so this kind of uh vertex ordinary ordinary continuous spin particle um people have figured out what this uh uh what that vertex should look like and um and from there i think there's a proposal for what this uh for what that process should be so i think um uh there there is probably some you know you might look for massless remember these are massless you'd look for some sort of long range interaction maybe with very weak coupling some long range interactions associated with these continuous spin particles i can't remember if um uh any phenomenology was done with this um because again it's at a very primitive stage to really even see if these kind of interactions are possible um but uh but philip and uh atalia who wrote these papers are are dyed in the wool people uh with connections to experiment so if it was possible to um uh if it was possible to talk about experimental signals i'm sure they did it in their paper and uh in terms of the things that are being done is it largely uh are they suggesting more bosonic particles or for the particles are necessarily it's not i i think they were imagining the uh they were imagining the or no yeah i think that they're they're imagining uh uh bosons yeah and you know that people have uh you know what is this let me just say say something about it uh quickly um what what you should do in talking about this of course is instead of talking about this continuous t um you can go to a Fourier basis right you can go to a Fourier basis and um in that Fourier basis you would just have an infinite number of spins okay so so i'm just usual thing instead of talking about the position eigenstate on the circle you go to the momentum eigenstates on the circle and so you can think of the continuous spin representation as being labeled by the radius so this is the number row you can think of it as labeled by the radius and just an infinite number of spins okay so n is just an infinite number of spin n equals minus infinity to infinity okay um so i'm just saying that that i can represent any point on on the circle here any t i'm going to write um as uh as uh uh or any t i'll write as a row either the i theta uh and this state is then labeled i can write as a sum of row and n e to the i n theta okay as n goes from minus infinity to infinity so really you should think about these continuous spin representations as a representation that's labeled by this number row um and an infinite number of spins so somehow it looks like a a particle with an infinite a massless particle with an infinite number of spins and there has been suggestions now and then that these might be related to uh strings in the high energy limit if you imagine looking at strings at energies vastly bigger than the string scale well that also looks like somehow a system with an infinite number of spins of higher and higher tower of spins but because you're going to infinite energies compared to the mass of the string states that kind of looks like a naive roughly looks like a system with infinitely many spins infinitely many mathless spins so that's why there's some suggestion now and then that this continuous spin limit might be related to the high energy limit of string theory and again now and then people were interested in that question because maybe there are some you know the magic of string theory is associated with a huge enhancement of symmetries in the high energy limit um so that's uh that might be one setting in which these things uh matter okay but anyway that those are just words that haven't been made particularly concrete but um uh but in any case it is sort of it is kind of fascinating that that it kind of sits there like a lump okay that this is the whole story of Wigner I mean I I I hope I did justice to it but I think it's kind of a spectacular story that you go very basic things you just look at what is the particle the representations boom boom boom and then wham this weird thing happens in the middle which forces you to go to the origin and gives us dramatic difference between massive and massless but it's just kind of sitting there that we didn't really you know address this other part of the story well yet totally well yet and it's lying around it hasn't gone away it has not been fully killed one way or the other so this is something that would be great to understand more at the but once we decide that we only care about finitely many particles um uh which appears to be the case in nature um then then then we then we go from these continuous infinite things to the opposite limit where we lose degrees of freedom right where somehow we where the number of degrees of freedom jumps discontinuously between masses and math thank you questions um yes so um like maybe see the question but like all um so also the multiple particles that we have seen so far are tied to this like um um zero representation right yeah okay so is this just like another possibility that there are particles who have like infinite numbers okay good i just want to make it another possibility as i said the and it has not been settled one way or the other but there's no known working example of an interacting theory with these things no known fully working example okay um but nor is it totally ruled out so that that's why it's an interesting subject of uh of uh research as i said it's not a super active subject of research because it's very confusing that's of course a good reason for it to be actively studied i mean you know we should always be attracted to the most confusing things that's uh um especially when they're sort of deep things um but uh it's not anyway it's it's not yeah that you you should read these the papers and form your own opinion of the state of affairs but um yeah but they were quickly thrown out as i said in the early days not for very good reasons a we haven't seen it in nature b you would instantly thermalize all of them and you would screw up but i don't know the specific heats everything would be screwed up i think those are not very good reasons either one of them um a better obstacle would be if it's somehow impossible to build consistent theories by the way it's not inconceivable that it's impossible to build consistent theories because as we'll discover um even for these ordinary representations uh it's impossible to have consistent theories of massless higher spin particles if we have gravity i will explain what why that is um uh that's also something that's simply impossible so just because things are allowed by the representation theory does not mean that you can build consistently interacting theories for them so it could just be that it's similarly impossible to build consistently interacting theories for these continuous spin particles i'm just saying that no sharp statement about this has been made in the year 2021 uh one way or the other yeah this is another question by maria hi hi um i wanted to ask you about something that we talked about yesterday um you said that the reason why i don't um is massless is not an argument of symmetry but you do the discontinuity between the numbers of degrees of freedom of a massless photon sorry which is as we just discussed which actually is an argument of symmetry but it's an argument of the representations of the of it's a what is the particle argument which boils down to representations of uh uh the punk array group just like we uh discussed i meant it's not a consequence of gauge symmetry that it's kind of backwards that gauge symmetry is something that we introduce in order to be able to describe massless particles in order to be able to like take into account this discontinuous difference between the number of degrees of freedom of massless and mass yeah but sorry i didn't mean to interrupt go ahead yeah um yeah my question was that um if it's not so much an argument of gauge symmetries um why do people um normally solve these kind of problems of huge corrections with symmetry arguments such as introducing supersymmetry oh beautiful beautiful because uh uh because uh there's a there's a very important distinction here between gauge symmetries and global symmetries okay um uh gauge symmetries well we'll talk about it more in just a moment but let me just say now it's a question it's these gauge symmetries that are a little bit in the mind of the theorists they're not really there in nature they're in the mind of the theorists as a very useful tool to describe the dynamics of massless particles with spin but the whole point of gauge symmetries uh and really the more correct word is gauge redundancies is that the physical states are gauge invariant okay so that's uh again we'll talk about this point more in a moment but uh but global symmetries are different global symmetries relate different states to each other right so um and so global symmetries have consequences global symmetries are not in the mind of theorists global symmetries are real things they have consequences for example you know uh in the uh you know they imply the different states of the system are degenerate uh with each other uh when you spontaneously break a global symmetry we got goldstone goes on so global symmetries are perfectly fine to rotate things uh into um uh uh each other then you can ask uh and so and so what what what people often use in trying to solve the hierarchy problem is to find a reason why uh the Higgs is related by a global symmetry to uh to a something else who's um uh that that we that we can understand but you can also see this from the point of view that I'm talking about um which is that the real mystery of the Higgs is that the the uh the real mystery of the Higgs is that um it has spin zero and and that doesn't have a jump in the number a number of degrees of freedom between massive and massless every other spin even spin a half has a jump in the number of degrees of freedom between massive and massless um when for even for spin a half when you take into account that really we only have one holicity either the plus or minus holicity so when we have something massive we have both of okay um so uh so from that point of view in order to the the approaches to the hierarchy problem can be thought of as one way or another finding a symmetry to relate the spin zero particle to a particle of a different spin so and in supersymmetry there's a global symmetry that'll mix up the spin zero particle with the spin a half particle um of course that's close to the conventional way of talking about things too where we say chiral symmetries can protect the uh spin a half masses and supersymmetry lets you inherit the chiral symmetry for the spin a half partners for the uh for the uh scalars okay thank you all right so all right more questions can i ask a question nima uh no i was wondering about the continuous spin um so you mentioned that uh maybe this paper of shuster and toro provided the the the two two interactions of higher of this continuous spin with ordinary masses in particles so i was wondering in principle one could um take these amplitudes and and then construct side as you're running for for the interactions and has someone checked that no no i think and and it and you you'll see that the that the state of affairs is um uh uh um i mean that they've written down they've they've written down sort of on-shell uh three particle amplitudes um and even before getting to the rg running and things like that trying to understand the kind of tree level exchange is already um uh interesting i think that they got to the level of understanding the the tree level exchange but um i think really that the sort of question that's remaining is to establish the consistency of this beyond the very simplest things that they've looked at before going to the questions of the running and stuff stuff stuff like that that um i mean that the state of affairs is much more primitive than uh than than even what you have in have in mind i think but i i don't actually remember the details of this and there may have been more development since i've paid attention so i would just encourage you to look at the uh the paper it's very clearly written written set of papers okay thanks i'll check okay little break all right great five minutes oh sure yeah sorry i thought we i thought we had the break already okay that's fine we'll come back in five minutes great you in five minutes if there any students on the line maybe i can um if you guys can raise your hand just just out of curiosity um uh how many people uh i'm just curious uh the the uh the stuff of this lecture today how many of you have seen this sort of standardly in your uh graduate courses and so on okay thank you thank you very much so it seems to be a relatively small fraction of of uh of you guys so uh it seems worth so i hope you guys found it found it useful um but we're we're going to be turning now just from uh the basics of this representation theory to some physics so um uh should we start uh andrea or wait a little bit please please okay there we go so let's go back here all right so um now that we've learned uh uh what particles are uh let's ask about interactions so we're going to be interested um in scattering processes involving uh involving particles um and um so uh just talking about so let's imagine that we're just going to scatter massless particles for now so we're going to focus on massless particles for the rest of this discussion um because that's where something really new is happening um and so uh what is the data then what is the data for for a scattering process you have to give me the momentum um where all these p-squareds will be zero so so all these p-a-squareds i'll usually label the n particles by the by a so a equals one through n so all the p-squareds are on shell um but i also have to give you the helicities of the particles okay so so the labels for the scattering process are p1 h1 up to pn hn and i'm working in the convention that all the momenta are incoming okay so that momentum conservation is just the statement uh um that the sum of the momentum is zero okay of course when we do that some of the energies have to be positive and some of them have to be negative and we think about the energies that are positive as incoming states and the energies that are negative as outgoing states okay so that's the if we want to connect to the usual in-out scattering process but this is the more convenient way of uh labeling the states and so um if we just think for a moment uh what is the sort of group theory what's the representation properties of the amplitude if i have this state that's labeled by momenta and helicities and i have an amplitude that depends on uh these labels okay then what does this object have to do um under Lorentz transformations well what it has to do is to follow it that if i Lorentz transform the momenta i have to pick up a little group phase okay so this is what an actual amplitude does right and that's just uh uh and i hope that's uh that's obvious um we imagine that the actual that that the that the scattering matrix is actually punkarain variant so it has to have the delta function for momentum conservation okay um uh that's translational invariance but under Lorentz transformations we just saw that if i do a Lorentz transformation i need to pick up uh the little group phase on the state okay and so so that means that this amplitude is not Lorentz invariant but if i do a Lorentz transformation every particle has to pick up its uh corresponding uh little group phase okay right so these are the transformation properties of the amplitude now i hope you notice that this is not exactly what we talk about in courses when we uh talk about the amplitudes let's say for the scattering of photons okay just to give uh an example so the in the usual formalism in the usual formalism um of field theory uh and Feynman diagrams uh let's say we have some amplitude for photons well we immediately talk about not an amplitude but uh what's sometimes called the Feynman amplitude which is an amplitude in quotation marks okay the Feynman amplitude which now has some Lorentz indices on these m's right m mu 1 up to m mu n you know we see this like all the time when you draw some diagram like this you just see it nakedly because there's a mu index here and a new uh uh index there okay so these Feynman amplitudes are are just Lorentz tensors right they have these they have these mu indices and the actual amplitudes are not Lorentz tensors notice there's no muses here right there's no muses here at all there's no muses here okay there's just something which has the helicity label there's not any muses okay so now how do we actually deal with this well the way we deal with this is something which in courses is often sort of said quickly or at the end or you meant you don't really care about it um uh that much is you say that at the end of the day you first calculate this guy with Feynman diagrams and then you dock into polarization vectors right so in other words the the imagined connection is that the amplitude for p and h is this m mu let's say just had spin one particles is this guy that now just depends on p but now you multiply by some polarization vector epsilon mu one for helicity h1 uh and momentum p1 up to epsilon mu n hn uh for pn okay so here we multiply by uh polarization vectors okay okay now and this is really not a coincidence okay because what is the polarization vector doing okay um the fact that in the usual formalism we have even this notion of a polarization vector um a polarization vector uh the presence of polarization vectors is an explicit is an explicit reminder that uh that we are using fields to describe particles right so why is it that we have polarization vectors because really we imagine that that a mu business that m mu business is about some field a mu of x and that in the free approximation well we just have free propagation we write a mu of x is some polarization vector epsilon mu e to the i p dot x right but what what is this doing this is solving the classical equation you know something like box a mu equals zero right so this is putting the field on shell field on shell is giving me particles right but the actual m mu one up to mu n thing that we're computing is really about the underlying fields and then we restrict the fields on the outside to be on shell in order to uh describe the scattering of the particles okay so i'm so i'm just emphasizing to say that this very notion that you have so i'm just to emphasize again what we talk to late with finding diagrams is not an amplitude this part is not an amplitude okay that's uh there's no relation to an amplitude we convert it to an amplitude by dotting into uh polarization vectors okay now you might this might already bother you a little bit you know why are we talking about something that has nothing to do with the actual uh scattering process that we care about and involves these other things and we dot into polarization vectors okay that that's uh this is the a kind of a zero order point in our in the development of the subject we've learned what particles are uh we learned the difference between massive and massless we start thinking about what amplitudes are and here there's a bifurcation point where where you either sort of go down the direction of fields and if you can read more in Weinberg volume one about why fields are inevitable and nice and everything is great in order to describe the interaction of particles that discussion actually begins with spin zero particles where everything is very simple and then there's more and more complication as you go to higher and higher spin okay um and we're going to take that attitude now okay so don't worry we're not going to jump to something uh very different but i want to stress that already at this point there is a second strategy that you could take which is to never talk about the fields only talk about particles never see polarization vectors absolutely directly talk about what the amplitudes are and and and we might talk about that too depending on how i decide to run things we may or may not talk about that tomorrow we won't have time to talk about it today but anyway we're going to proceed with this the more usual way of thinking about things with polarization okay now for a massive party we're talking about massless ones let me just remind you for massive particles there's no problem with polarization vectors okay there's no problem with polarization vectors and let's think about what i mean so for example for spin one i would introduce an epsilon mu of p and now epsilon mu how many degrees of freedom are there an epsilon mu there are four degrees of freedom right epsilon zero one two three um but we know that that uh that uh that massive spin one has only three degrees of freedom so what is this difference between three and four well we know that the polarization vector satisfies the constraint right what is the constraint that it satisfies it satisfies that epsilon dot p equals zero okay so that's one constraint on a polarization vector and again if we go to the rest frame where p mu is equal to m zero this just says that epsilon mu is equal to zero epsilon right and that's perfectly good so these are my three spin states okay and if you think about where the if you think about uh where this comes from for the you know massive spin one field equation d mu a mu equals zero which is in momentum space p dot epsilon equals zero is a consequence of the equations of motion right so so as you would expect when you solve the equations of motion um i'm just relating all the pictures to each other you solve the equation of motion for massive spin one you discover the solutions are epsilon mu e to the ipx where epsilon dot p equals zero okay so no problem the degrees of freedom match between massive and massless of course you have to put a constraint on the polarization vector okay and and uh and the underlying field picture that comes from the equation of motion but even if you didn't know about the underlying field um if you just wanted to somehow convert these m mu one up to mu n you wanted to convert them to something that only transformed under spin by inventing these polarization vectors uh it would be fine there's a simple constraint on the polarization vectors that makes the degrees of freedom match the difficulty for massless particles is the following so what we want to imagine is that there is a polarization vector epsilon mu maybe it depends on the helicity and p right and we want to uh uh assume that this polarization vector exists such that uh when i do a lorenz transformation epsilon mu h of of lambda p well but this is just lambda you know mu new epsilon new h of p and maybe it picks up a little group phase okay so if i'm talking about spin one uh maybe this is what i uh what i uh wanted to do okay now notice if we have such an object if these polarization vectors exist then if i use them to dot into the uh the Feynman amplitude m mu one through mu n um this object would transform correctly so that's the kind of point is that these polarization vectors are kind of uh they they the the the polarization vectors have are supposed to have one index that's lorenz and one index the helicity that's little group and so they're supposed to transform on one side with lorenz and on the other side with little group and in this way when you contract them into a lorenz tensor they give you something that just transforms under under the little group okay so that's our desire for what polarization vectors should be the analog of that definitely works in the massive case okay of course it's not a phase it's but but uh it's but but the uh but the the polarization vectors transform like lorenz transformations on the mu index and like uh like spin one under their other spin index okay so this is what we might hope to want for massless spin one but there's a fundamental difficulty no such polarization vectors exist and it's easy to see why first let's just see it from a degree of freedom argument okay so we know that massless spin one has let's say two degrees of freedom um epsilon mu uh equals zero one two three as usual has four degrees of freedom so we see that already in the massive case well as in the massive case let's say we impose that epsilon p equals zero okay if we do that we go from four to three degrees of freedom but then we're done there's no other lorenz invariant constraints that I can put on this polarization vector okay there's nothing I can do there's no other uh constraint uh I can put on it so it just seems that I have three degrees of freedom so three is not equal to two and what is the what's what's the problem uh uh the the problem is that if you give me any epsilon mu that satisfies these constraints then epsilon mu plus anything times p mu will also satisfy these constraints um uh because if epsilon dot p equals zero then epsilon plus alpha p dot p equals zero as well uh since p squared equals zero so that's our first hint that there's something wrong there's no way that we can uh pick out two degrees of freedom from a polarization vector um we can get from four to three but then we're stuck okay there's no way of uh uniquely picking out uh two uh uh there's no way of canonically picking out two degrees of freedom out of the three uh for the polarization vectors that satisfy epsilon dot p equals zero okay and um uh so so what does that mean in practice so I'm telling you polarization vectors don't exist right or better yet the polarization vector the word the for the polarization vector for a given holicity particle doesn't exist nice nima what are you talking about um I've learned in school ever since I was a child that the polarization vectors and I'm even very fancy for plus or minus holicity I know there's a one over root two and a one plus or minus i because I'm so cool I know about eyes okay so these are the polarization vectors for um for uh massless spin one particle what are you talking about screw you know they're they're there they are uh what do you mean there's no polarization vector well what I mean is that these zeros here don't mean any okay so this is a polarization vector for a particle uh that's supposed to be moving in this uh uh that's supposed to be moving uh like like like this right um sorry let me uh let me let me do this so if we have a particle moving in the z direction let let let me let me do it right if we have a particle I wrote it wrong if we have a particle p mu uh let's say moving in the z direction then we say that the polarization vectors plus or minus are like uh zero one over root two uh plus or minus i over root two zero right this you've seen in uh books all the time um what I'm saying is that these zeros don't mean anything okay that you could have equally well written this as any alpha one over root two plus or minus i over root two alpha okay and there's no Lorentz invariant way of of uh declaring those entries to be zero and in fact you can just check this is a this is a check that you can do check okay there are Lorentz transformations lambda p that leave p invariants they're part of the little group okay so you can just check if you take a Lorentz transformation that leaves p invariant when p is equal to you know e zero zero e and that that uh that lambda epsilon um will not look like zero uh uh one over root two uh uh plus minus i over root two zero it won't have the zeros there anymore in general the zeros will get turned into some nonzero alpha which depends on the Lorentz transformation okay so this should really bug you that uh that you so you know we've done all this work with with the Feynman diagrams we calculate all of these Lorentz invariant tensors everything is very beautiful so everything is naively Lorentz invariants and it is Lorentz invariant at the level of the Feynman amplitudes but at the final step when we dot into polarization vectors epsilon mu of p one blah blah epsilon mu n h n of p n these things don't actually exist they don't exist in other words uh they're not they don't exist in a Lorentz invariant sense okay if i make a choice for them then uh if i do a Lorentz transformation i'll get another choice those zeros will be turned into uh will be turned into something else and therefore we have to introduce a new idea here so we have to introduce a new idea that we don't associate a state a particle state is not associated not associated with a single polarization vector epsilon mu but instead we have to associate it with a whole equivalence class of polarization vector okay so we have to associate it with the equivalence class of polarization vectors epsilon mu where epsilon mu and epsilon mu plus anything times p mu are identified in other words we have to declare that if i make if i choose any polarization vector from the set they all describe the same physical state okay because while an individual polarization vector is not Lorentz invariant obviously the whole equivalence class of them is Lorentz invariant if you do a Lorentz transformation on one polarization vector you'll land on another polarization vector in this set the the new polarization vector will only differ by the old one by something proportional to p mu okay so if you if you so if you just say that that you're not going to label a state with a unique epsilon but only by a member of this class uh where two epsilons are identified if they differ from p mu then that is a Lorentz invariant way of talking about the estate okay is that clear any questions about that and notice that this alpha can depend on p in an arbitrary way okay so just some general alpha of p so just to be a little bit loose now we've seen our first instance of gauge redundancy and you see why the word is redundant right because we are using we are saying that epsilon nu and epsilon nu plus any alpha p mu are the same state the whole point is that we have to enlarge what we meant by uh by labeling a state to allow all of these labels to describe exactly the same thing in order for now and by the way why are we doing all this because we made the decision early on that we're going to first compute Feynman amplitudes and then dot into polarization vectors if we do things in another way where we never talked about Feynman amplitudes and directly talked about the particles to begin with as I alluded to earlier we never have to see this and again I might come back depending on how I'm doing for time I might just discuss this more fully later but having made this choice to describe physics in this language we're forced into this gauge redundancy to talk about massless particles with spin one in this example and any any other spin as well okay but I hope it's clear why the word is redundancy unlike ordinary symmetries that relate to different states to each other here this is something we are inventing from the outside in order to allow many different labels to describe the same state okay and why is the word gauge here um well uh because this is the momentum of space analog of what we're used to seeing in position space so in position space we have a mu and a mu plus d mu alpha okay and we say that these are meant to describe the same configuration okay so that's the that's the gauge transformation and in momentum space for the polarization vectors that's just this right so uh the derivative here turns into a p mu so this is epsilon mu plus epsilon mu plus p mu alpha okay all right but there's a very big consequence of uh of uh what what there's a very big restriction on uh on on the Feynman amplitude there's a restriction on uh this amplitude mu one through mu n um if indeed uh the the the physics only depends on this equivalence class of epsilon mu with the whoops with the epsilon identified with epsilon plus alpha p this just says practically that if you dot epsilon mu one through epsilon mu n m mu one mu n to get uh to get the actual amplitude um we should get the same answer if we replace any given epsilon mu with epsilon mu plus any alpha p times p mu okay and that tells us that whatever these m whatever these m mu one through mu n are they have to satisfy that if I take any one of the indices m mu p mu and whatever the other indices are that if I dot that into p mu I should get zero right because again that's just what I get if I take any one of these epsilon's and I shift it to epsilon mu plus alpha p mu if I'm going to get the same answer it would better be that uh that the p mu dotted into uh uh m mu one gives me zero so this is a restriction then we thought when we're playing with Feynman uh when we're playing with Feynman amplitudes we thought that anything we did was Lorentz and Berrien everything is fine but it isn't in order for it to be physically Lorentz and Berrien for uh uh for describing the amplitude of the scattering particles whatever that the rules are that give us this m mu have to uh be such that they satisfy this uh important constraint okay and only then do you get something that's actually physically Lorentz and Berrien now in textbooks this is often called the on-shell word identity and in the usual way of thinking about things you begin by thinking about a Lagrangian and then you love gauge symmetry because it's so beautiful and lovely uh and then uh then you have to gauge fix because it's confusing uh you get something infinite if you don't gauge fix so you gauge fix the thing you have to gauge symmetry but then you gauge fix it and then uh consequence the later consequence of the underlying gauge symmetry of the Lagrangian is the on-shell word identity from the point of view uh where we're just thinking about writing down consistent interactions for the scattering particles in this form was in the on-shell word identity is the most important thing it's the it's the starting statement okay so you have to ensure that you have the on-shell word identity otherwise the scattering uh isn't uh despite appearances is not Lorentz and Berrien okay and I now want to show um uh in a way that's as closely related to the usual language as possible um uh uh how this implies that if we're going to describe the interactions of massless particles we need to have something like uh uh uh something like gauge invert okay so let's do a little example by the way uh uh Andrea how am I doing for a time uh in principle you were 10 minutes over time uh in principle I'm 10 minutes over time okay um let's see what I can do um uh can I have around uh 10 more minutes sure okay great all right um I just want to make two more points uh along along these uh uh these lines and then then then uh we can call it a day for it today okay so I want to do do do an example uh let's say I want to describe um I want to describe massless spin one um interacting uh with spin zero okay so just um but forget about all the stuff you learned in books uh uh for for a second other than the fact that we're going to write down Lagrangians and Feynman rules and and and stuff stuff like that okay um but uh don't worry about gauge invariance is it pretty not you know let's just let's just uh we're physicists we're not philosophers so we're just practically we want to go and build build a Lagrangian here so what what would I do so let's say for the spin one part um what Lagrangian would you write down um and and if you've never heard of anything and if you're honest with yourself you would never write down negative a quarter f mu nu f mu nu what a weird complicated thing f mu nu d mu a nu minus why the hell would you do that um what what you'd write down if you're just a dumb guy and you should be a dumb guy whenever you possibly can be is um is uh just d mu a new square okay so that's the that's the first thing that that I would do oh no we're scared it's not gauge invariant fuck that okay that's uh um if this gives you a perfectly decent propagator the propagator is you know minus eta mu nu over p squared everything is fine so we can start right nothing nothing is going wrong we're just we're just gonna gonna start by the way if you're nervous uh when we start with negative a quarter f mu nu squared and then we gauge fix this is what we end up getting after we gauge fix right if we do the usual Feynman gauge with like the you know alpha equals one or something so it can't be such a stupid thing to do because that's actually what what we get except instead of starting by saying we love gauge invariance and gauge fixing we just write down a propagator that works to begin with okay so so uh you should never be scared you should just do the most obvious thing first but remember what the important physical constraints are all right so so then what else could I do I have my scalar so the scalar would be some d phi squared okay and now let's just write what the dimensionless interactions are just by dimensional analysis between phi and this uh between phi and a so I can have one coupling which is the the the usual one five star d mu phi uh a mu okay and uh and this is nothing to do with gauge invariance it's just some coupling I can write down so I'll call that coupling g and there's another coupling that I could write down I'll call it g prime I'll just put the squared there for fun uh five star phi a mu a mu okay so if I know absolutely nothing else so these are the interactions that I would write down that are dimensionless uh in this in this model okay so just using dimensional analysis to write down some coupling concepts now because you're so smart you know that I'm supposed to choose g prime equals g so that's the covariant derivative blah blah blah I'm not so so smart okay I'm just writing down the the most obvious interactions there are and off we go we're going to go computing Feynman amplitudes in this theory and there's nothing wrong so let's let's compute the Compton scattering okay so I would take this the diagram and this guy okay so here this would have a g and a g g and a g and there would also be this diagram with the g prime squared again I just put the squared there for uh convenience and notice that no matter what g and g prime are these are going to give me Lorentz invariant Feynman amplitudes okay so nothing wrong they gave me Lorentz invariant Feynman amplitudes so um what's wrong here what is wrong is that when I attempt to convert these Feynman amplitudes to actual amplitudes by dotting with polarization vectors I have to make sure that they satisfy that that identity that that says that it doesn't matter which representative of the polarization vectors I use in order to describe the photons so I have to do a check that p mu acting on m mu nu actually gives me zero right and I was actually going to do this little exercise for you it's really a three line exercise so I won't do it for you I'll leave it for you to do it yourself literally just compute this diagram this one it's a three line computation and for example you'll find that if you just have these two that p mu m mu nu is not equal to zero and you're forced to have this other guy in order for p mu m mu nu to be zero and in fact uh having that is going to force that g prime is equal to g and so uh so beginning with this more primitive condition which to say for the fifth time is encoding the fact that that the Lorentz invariant Feynman amplitudes turn into Lorentz invariant scattering amplitudes for the massless particles um which forces this gauge redundancy in the description using polarization vectors um that tells you that that the Lagrangian you have to use has to have some special property has to relate different couplings to uh each other in order to make this on shall we're done in the truth and the fact that the underlying redundancy looks so similar to look so similar to the gauge redundancy in position space is a clue that one way of doing this is to write down a Lagrangian that has the gauge symmetry but remember the in the end of the day when you're done getting something you can calculate what the Lagrangian never has gauge symmetry it's always been gauge fixed and so the claim is that the Lagrangians that give you uh uh Feynman amplitudes that satisfy the on shall work at end these are just those Lagrangians that come from a classical Lagrangian where you add a gauge fixing term to uh to fix the gauge okay so that's the that that's that's an interesting non non-trivial claim going backwards so i'm not saying uh start with gauge symmetry the path integral uh it's infinite so you gauge fixed after a lot of work you get to the on shall work I'm showing go backwards begin with the practical question of trying to build Lagrangians that satisfy this rule and you'll discover the only way of doing it is to is to get your Lagrangian from a gauge invariant one by adding gauge fixing terms in an appropriate way all right so so I encourage you to do this exercise yourself and in in the last five minutes I actually want to talk about a more general thing that that follows from this kind of argument so I'm telling you how this is done in uh um uh in uh I just made a claim for you that any Lagrangian that you get uh any Lagrangian that's going to make this work has got to come from an underlying gauge invariant Lagrangian with gauge fixing um the same thing would be true in gravity if you had a massless spin two it has to arise from so let's say for a massless spin two just to stress the point massless spin two uh now I have a polarization vector epsilon mu new and after I identify epsilon mu new with epsilon mu new plus alpha mu p new plus alpha new p mu okay so that's the analog of uh of epsilon goes to epsilon plus p um uh if you have massless spin two and notice that this is the same in position space as h mu new is identified with h mu new plus d mu alpha new plus d new alpha mu and this is nothing other than linearized general covariance general linearized uh uh difumorphisms and uh and but it's exactly the uh the same claim is that if I have uh now uh in gravity an amplitude mu one new one mu two new two and so on I have to ensure that if I dot this into p mu one I get zero okay so this is the unshell worded and the for gravity and once again the claim is that the only way to have a Lagrangian for massless spin two particles that that respects this is if that Lagrangian comes from a generally covariance Lagrangian the full non-linear difumorphisms not just the linearized ones okay what we see on shell is just this linearized uh difumorphism but that's all that's needed okay so just the linearized uh difumorphisms um this unshell word entity forces the underlying Lagrangian to be generally covariant up to gauge fixing terms of course okay which then we we use to actually uh do the uh computation but instead of going through a sort of a proof of that uh statement I want to give you I want to give you a sense for where where it comes from and at the same time uh uh tell you two celebrated facts um or celebrated fact about the Weinberg software okay it's really fast so uh you'll see how how easy it is it's not um and um but from here we're going to we're gonna learn something quite uh dramatic so Weinberg asked this question in the 60s imagine you have some scattering process to begin with for some charged particles so there's some amplitude here uh um uh and he asked what is the amplitude for emitting uh a photon one more photon okay so maybe there's so there's a bunch of charged particles what is the amplitude for emitting a photon in the limit as the momentum of this photon becomes soft and the limit is the momentum of the photon goes to zero and if you think about it for a second um the dominant diagrams uh that contribute as the momentum goes to zero are exactly ones where the photon is attached to the external legs of the diagram rather than the internal ones and why is that that's because it's in this diagram here that I would have a propagator like one over p plus q squared minus m squared if this particle is mass m and this momentum is p okay maybe this is p p plus q all right so and uh because p is on shell this is equal to one over two p dot q okay q squared is equal to zero p squared equals m squared so this is one over p dot q and so this blows up as q goes to zero this uh uh blows up whereas um if you did this on any internal line uh the particles are off-shell in the in the internal line and so we don't get this divergence as q goes to zero okay so uh so therefore if we want to see what the amplitude is for emitting a soft photon it suffices to just look at the external legs now let's just think for a second what could this vertex be okay of course we know what we expected to be from the Lagrangian but um but even if we didn't know anything about the Lagrangian just this is particle j maybe there's some coupling constant e j that we could call the charge but the biggest interaction that it could have with the photon would would involve the polarization vector of the photon dot the momentum of this particle after all it could either dot the momentum of the particle or could dot the the q um of the photon and of course epsilon dot q is equal to zero okay so so um this is the largest interaction that I could I could have involving the hard momentum of the uh of the photon this is also the interaction that would give rise to long-range forces inverse square law forces and so on so that's the assumption that we're making we're looking at the strongest interactions that the biggest interactions that the photon could have okay and therefore uh with every leg we would have this factor associated with vertex and then we would have this propagator that we just talked about pj dot q coming from that so whatever else is going on here from the jth leg we'd have something like that and therefore Weinberg says that the amplitude for emitting a photon an extra photon is equal to the amplitude beforehand multiplied by this factor the sum over all legs ej pj dot epsilon divided by two pj dot q where q is the uh where where q is the momentum of the photon all right okay beautiful so this is what uh this is what dominates as as q goes to zero so that's the soft photon now let's apply this rule that the answer had better not depend on whether we choose epsilon or I should be able to shift epsilon to epsilon plus anything times q and so the answer should should be the same so what happens if I do that in other words what happens if I dot q into this into the corresponding uh Feynman amplitude or I do this shift here um well I find that I have to have the zero is equal to the sum over j uh ej and I'm going to replace epsilon with two so it's pj dot q divided by two pj dot q so that so the pj dot q is cancel and I learned that this has to equal the sum over j ej so I learned something beautiful that the only way to consistently couple the photon to these j particles is if the sum over the charges is equal to zero I have to have charge conservation I didn't put that in I didn't put anything with the underlying Lagrangian gauge invariance anything like that um but the only way to consistently couple a photon with the this leading dominant coupling um the only way to consistently couple a photon um is such that because uh because of the behavior of the amplitude and the stock limit is especially simple um we learned that we have to have that that that we have to have charge charge conservation all right let's continue this argument and now imagine that we are emitting a spin two particle okay so this would be some mu nu now exactly the same argument for the jth leg for the jth leg I would emit a spin two so I would write down the leading interaction I could write down would be epsilon mu nu pj mu pj nu okay so and some coupling constant I'll call it kappa j and uh once again from this propagator I'd have an over two pj dot q and so Weinberg would tell you that the amplitude for emitting uh anything plus uh a soft graviton with momentum q is equal to the amplitude without it multiplied by the sum over all j this coupling constant kappa j pj mu pj nu epsilon mu nu over two pj dot q once again if I uh replace epsilon mu nu with epsilon mu nu plus alpha mu p nu q nu plus alpha nu q mu I have to get the same answer and so that means that if I if I strip off the epsilon mu nu and just dot this into just q I should get zero okay so uh so that that that implies that the sum over j kappa j pj mu uh pj dot q over two pj dot q is equal to zero and so I learned that the sum over j of kappa j pj mu is equal to zero now that's really interesting that's the analog that we found before is the sum of ej has to equal zero this is again the sum of the charges has to equal zero but the charges and gravity have an extra factor of the of the momentum in them okay well if you think about this at first this looks impossible because I already know by momentum conservation that all the momenta have to add up to zero okay so this is putting yet another constraint on the momenta that's some different linear combination of the momenta have got to add up to zero and that would mean that the amplitude vanishes except for the maybe very special angles where this is possible right so in order for this not to force the amplitude uh to vanish what has to be the case all the cap of j's have to be equal to each other okay so we must have all the cap of j is equal they're all universal and equal to something that we can call one over m flunk so we've discovered the principle of equivalence the universal coupling of gravity but we've discovered it not by thinking about falling elevators and curved spacetime or anything like that we've discovered it as a consequence of the consistency of special relativity and quantum mechanics right okay and uh and as i said this is a special case of the more general fact that i was telling you that uh that these uh that the on-shell worded entities that are the uh that from this language are just the way of enforcing that we have a real Lorentz invariance of the scattering of these particles force you into the structure that we know and love of yang-mills and uh general covariance and so on is the love of the underlying Lagrangian but the Weinberg soft theorems are a sort of a quick way of seeing this in a certain limit um and if we go one step further and try to do this with a particle of spin three okay so some mu new gamma okay so then we would write down some epsilon mu new gamma pj mu pj new pj gamma and then we get it over to pj dot q well by the same logic and some coupling kappa j by the same logic we'd have to have that the sum over j of kappa j pj mu pj new is equal to zero and now this is just impossible there's nothing i can do this is now a quadratic constraint on the momenta and it's just impossible okay i can't there's no choice i can make for the kappa j's that does not uh that doesn't put so many constraints on the external data that that the amplitude would just have to vanish so from these amazingly simple arguments uh and what i hope you saw is that we just began with Lorentz invariance and and uh and just the just the uh and quantum mechanics just at the level of unitary representations of the punk array group we see this dramatic difference between massless and massive and when we describe physics with our standard formalism of polarization vectors and so on this forces us into gauge redundancy and it has these dramatic consequences that massless spin one has to have a gauge structure massless spin two has to have a generally covariance structure and you can't have massless spin three and and above okay so um that's uh i will um uh so someone asked for references yes this business about the Weinberg shop theorem is also very nicely explained in Weinberg's uh i believe volume one uh field theory book um so uh it's uh yeah that's the best discussion of it that i uh i uh that i know okay so um maybe i will uh stop here uh again that so uh what i hope you got from this lecture in in a much more maybe direct way than what we usually see with the books where we go through a sort of roundabout process of thinking about a Lagrangian and we like gauge symmetry because it's pretty but it makes the path on a girl divergent so we gauge fix etc etc and then right very very late in the day we come to thinking about the actual connection to the direct particles what i hope you see is how absolutely direct the connection is between these deep facts about the way the world in in the wind that we look that we see when we look outside the window um how the properties of that world sort of uh gross properties follow very directly from the from the deep principles that that that we uh uh uh know about uh as i said what i showed you in this lecture in the second lecture is the the the sort of best way of showing it in the usual formalism um uh there's a there's an even deeper and more transparent way of a more powerful way of seeing it in an alternate formalism where you never see these polarization vectors to begin with and um i will think a little today whether i'll come back and actually describe this tomorrow or move on to uh other things because i don't have i'm going as usual more slowly than i thought i would but uh i'll stop here thank you very much all right thank you very much a very nice uh enema maybe we can take a few questions you should don't mind this one from uh fabio thank you for your patience


Russia's nuclear-capable Tu-95, Tu-160 bombers deployed near NATO border. With Russia's continuous use of its strategic bombers, including the Tu-95 and the Tu-160, for launching conventional standoff attacks on Ukraine, satellite images have revealed that there has been an unexpected surge of bombers in Russia's north, close to the NATO border, according to Eurasian Times' media outlet. A fresh set of satellite images of the Olenya airbase in Russia's Kola Peninsula revealed that about 16 strategic bombers had been deployed to this airbase close to the NATO countries, Finland and Norway, the Barents Observer reported. The airbase is located in the closed town Olenegosk-2, which is an hour's drive south of The base currently houses 14 Tu-5 bombers, two Tu-160 bombers and two Tu-22M bombers besides other heavy transport aircraft as could be shown in the satellite images. Two Tu-160 are positioned at the southern end while at all other aprons large enough for heavier aircraft, 14 Tu-95 could be seen positioned side by side. The Tu-22M aircraft are relocated to the parking grounds in the airbase's northwest. The revelations came when several reports have indicated that Russia was launching attacks on Ukraine with its bombers from the Murmansk region of the Kola Peninsula. Bombers have typically been deployed to carry standoff missile strikes on Ukraine from within Russian airspace. The sudden movement of nuclear capable bombers close to the NATO border was alarming as Russia's frequent nuclear-saber-rattling accompanied it. The concern was exacerbated as Russia's central storage for nuclear warheads, Bolshoi Ramazero, is reportedly just 10 kilometers from Olenya airbase.


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