OK, well, thank you very much. It's a real pleasure to be back in Trieste. So I thought since, in fact, we realized that I'm actually the only lecturer, I think, who's speaking about early universe, the very early universe, that I thought it makes more sense to not just focus on alternatives, but also put in the context of inflation and then some alternative. So let's just say, in general, my lectures will be entitled The Very Early Universe Cosmology. And by way of motivation, I think, as we try to decipher what happened shortly after the Big Bang or perhaps before the Big Bang, really our unique probe to figure out that question, the unique probe that we have are cosmological perturbations, primordial cosmological perturbations. So really, this unique probe is encoded in temperature fluctuations of the microwave background, or if you want curvature perturbation, the various moments there of the two-point function, the three-point function, et cetera. And, well, the sort of general theme of these lectures will be that, in fact, early universe physics is encoded in the symmetries of these correlation functions. Specifically, we will see that symmetries constrained these correlators much like they do in particle physics. So can they constrain these correlators? And correspondingly, there are ward identities we can write down associated with these symmetries. And this is what we'll derive. So specifically, the outline that I have in mind for these lectures is as follows. So in the first lecture, we're going to focus on, in some sense, the simplest case, which is what I call multi-field or spectator field. By the way, can you read in the back? Is it big enough? Yeah. Multi-field inflation. And we will see in this case, this is the story of having spectator fields around that fluctuate but are not affecting the background expansion very much. And in this case, the symmetries that govern fluctuations are, is the full SO4 comma 1 group of isometries of the sitter space. Then in lecture two, we're going to consider what seems to be an even simpler case, but in fact, from the point of view of perturbations, it's more subtle. And that's the case of single-field inflation. And we will see that single-field inflation can be thought of, at least the scalar perturbations thereof, can be thought of as the symmetry breaking pattern of SO4 comma 1 being spontaneously broken to spatial translations and rotations. And finally, in the very last lecture, we'll discuss an alternative mechanism, which in general terms can be called the conformal mechanism or conformal scenario. And this mechanism relies on the full conformal invariance on 3 plus 1 dimensional spacetime. So at the end of the day, it describes a symmetry breaking pattern, SO4 comma 2 being spontaneously broken to SO4 comma 1. And part of the reason I find this alternative appealing is that on the one hand, it relies on symmetries. So it's not just some action that we cook up out of the blue. It relies on conformal symmetry. And there are various different realizations of the scenario, one of them, as we'll see, being Galilean Genesis of Alberto Paolo and Friends. And there's also a fight of the fourth example that was proposed by Rubakov. So there are different ways of realizing this particular alternative, as we'll see. OK, so that's the outline. And by the way, if you ask me what is my, some people have asked me what's my favorite alternative. What's my favorite scenario? It's not even on this list, OK? So it's probably so provocative I shouldn't start by talking about it, otherwise you won't listen for the rest of it. But indeed, or you'll skip the rest. But OK. I'm actually serious, but I'll wait till the end. I'll keep the suspense a little bit. OK, so let us start with a multi-field or spectator field, inflation. So for our purposes, inflation in these lectures is really a phase of approximate decider expansion. And by multi-field, we simply mean a field. So we really mean a spectator field, which is a field that is light. It's present during inflation. It fluctuates, but does not affect the background much. So one example of this is something like the Curveton, where you have some massive scalar, called its sigma, some quadratic potential, whose mass during inflation is much less than Hubble. So this field has a mass much less than Hubble. And the field, it generically will be displaced from its minimum. It will be perched somewhere on its potential. And by virtue of the fact that the mass is much less than Hubble, Hubble friction sort of keeps it there. It doesn't move at all. But nevertheless, it fluctuates quantum mechanically. It acquires a scalar and spectrum. And then later on in the evolution, it can decay and convert its spectrum onto the curvature perturbation, which we later observe. So for such fields, the key point is that for such fields, to a good approximation, we can treat the background as exact the sitter. We'll see that we cannot do that for single field inflation, but for this case, it is a good approximation to do so as exact the sitter space. So for starters, let us explore the properties of the sitter space. So four-dimensional the sitter space we can think of, by definition, if you want, as a four-dimensional hyperboloid in five-dimensional Minkowski space. So here's my hyperboloid. And here's Minkowski, the embedding Minkowski. So I'll denote by capital X is the coordinates of this embedding space. X0 will be time. And then I have spatial coordinates. Let's call it X1 and X2. OK? So this is Minkowski, five-dimensional Minkowski. So the line element in the radius, I should say, will denote the radius of this hyperboloid as H inverse. And of course, this H will be nothing but the Hubble radius or the Hubble constant of the sitter space at the end of the day. So we have, as our embedding space, five-dimensional Minkowski space. So five-dimensional line element is just eta eb dxa dxb, where a and b run from 0 all the way to 4.  It's simply eta eb xaxb is equal to 1 over h squared. Now we can choose coordinates on the hyperboloid in various different ways. We will choose coordinates. So the coordinates relevant for inflation are the spatially flat slices, I should say. OK, so if you take your hyperboloid, you can slice it in different ways. Yes? The only motivation from my point of view is that the symmetries are easier to describe. Because here you have the full SO4 comma 1 group. So that's the logic I'm following. We'll see that in this case, correlation functions are quite constrained by virtue of the fact that they have the full SO4 1 group. While when we get the single field, then if you want the fact that the inflaton is both driving the background and generating the fluctuations, its fluctuations are directly sensitive to the fact that you're not in exact the sitter space, that you have some finite epsilon, some slow-wall parameter. And so as a result, the fluctuations of the inflaton, they're actually described by a spontaneous breaking pattern of this kind. So it's just for simplicity. We lack scalar fields. Yeah, we have 1 and 1. It depends. So we only know we only have one at the TV scale, but who knows at very high energy scales? Yeah, so in general, in string theory, string theory is infested with scalar fields. They have more than they need. So at high scale, why not? And the fact that you would have some that are light, well, in my opinion, if you have one that's light to drive single field inflation, why not have many? OK. Anyways, if you don't like multi-field, this is just a warm-up if you want. OK. So as I said, you can slice this guy in many different ways. In particular, if you slice it with space-like planes, you'll get the closed slicing of the sitter space, the one where you have positive spatial curvature. If on the other hand you slice it with time-like planes, then you will get the open slicing. So the one that corresponds to spatially flat slicing is the one where you slice it with null planes. So let's draw again. So here I will use X4 for concreteness. This will be X1, let's say, X0. So if you pick a null plane, so here's null surface, intersecting this guy.  They will intersect the hyperboloid into these kinds of surfaces, going on the other side, and so on and so forth. OK. Each of these slices obtained by intersection with a null plane corresponds to a surface of constant time in the inflationary universe. Does that make sense? Yes. So in fact, we can write this using equations. One way to do this is to write the constraint equation for the hyperboloid in this way. So we're going to introduce a tau parameter, introduce a tau parameter which splits the constraints, which splits the constraints into two in the following way. So that constraint that describes the hyperboloid, we can introduce a parameter to break it into two constraints as follows. So I'm going to write it as X0 squared minus X0 squared plus X4 squared, will be equal to 1 over H squared, 1 minus little x squared over tau squared. And then the other constraint is just the sum of the spatial Xs, X1 squared, X2 squared, plus X3 squared is equal to 1 over H squared, little x squared over tau squared. So clearly, if you sum these two, you recover the constraints that we had. And now we have this parameter tau, which is an arbitrary parameter. Where tau here, I'm going to take to run from minus infinity to 0. So one way you can implement these two constraints is as follows. We can make the full, exactly, you'll see that in a minute. So now I'm going to write it explicitly. That's exactly right. So the tau parameter is parameterizing which surface or in which null plane is intersecting the hyperboloid. I will make this clear now. Thank you. So indeed, a nice coordinate system consistent with this parameterization is the following one. So let's write the embedding coordinates as follows.  Let's write capital XI as 1 over h little xi over minus tau. And finally, X4 is equal to over 2 minus tau 1 over h squared plus tau squared minus little x squared. And you can convince yourself in the privacy of your room tonight that indeed this coordinate system obeys these two constraints. These sort of exercises you don't do in public, but you can do it alone tonight. Very nice. Now, coming back to the question that we just asked, indeed, you can show by construction it's manifest from what I just wrote in particular that if I add x0 plus x4, and this will get right to the question I was asked, so if you take x0 plus x4, then clearly the tau squared x squared bits cancel. And I'm just left over with 1 over h squared tau. So this sum is 1 over h squared minus tau. And given that tau runs from minus infinity to 0, this quantity is positive and describes, indeed, the fact that this coordinate system only covers the upper half, only half of the hyperboloid, you see? Because this null coordinate, if you want, has to be positive. So this doesn't cover the entire hyperboloid, only half of it. You can also check tonight that using this coordination of the hyperboloid, that the induced metric under hyperboloid ds squared. So all I'm doing is I take ds squared, which was the embedding metric. And I evaluate this guy using these various coordinates, so x0 equals dada dam, xi dada dam. I do that, plug it in, use a chain rule, and at the end of the day, this gives, I should write this as ds4 squared, you get, at the end of the day, the familiar decytermetric, 1 over h squared tau squared minus d tau squared plus dx squared, written in terms of conformal time. So if you've seen this before in terms of exponential, that's just when you transform to proper time. This is written in conformal time. So in particular, the scale factor in this coordination is just 1 over h minus tau. OK, question so far? Crystal clear? OK, very nice. Now, I'm sure you've seen this before. The reason I take this review route is to get to the symmetries. So if you want this hyperboloid in Minkowski space, is exactly the analog of a sphere in 3D Euclidean space. So it's like a sphere. And as you know, the sphere gets induced on it the rotational symmetry of the ambient space, of Euclidean space. So similarly here, the ambient Lorentz transformations, the Lorentz transformations of the 5D space will map points of the hyperboloid to other points on the hyperboloid. So they correspond in our coordinate system on the hyperboloid to an isometry of the metric. So the hyperboloid, in other words, the hyperboloid is preserved by 5D Lorentz transformation, 5-dimensional Lorentz transformations, with algebra SO4 comma 1, since we're in 4 plus 1 dimension. And the generators of this algebra we're going to denote as jAB, as usual, xA dxB minus xB ddxA, just like the angular momentum generators in quantum mechanics. And these guys obey the SO4 comma 1 algebra. So the two commutators, jAB, so the commutator, jAB, jCD, is just equal to a combinations of eta's and j's, AD, jBC. It obeys this algebra. This is just, of course, a generalization to 4D of the usual angular momentum, commutation relation in quantum mechanics, the SO3 commutation relations. That's a generalization to 4 plus 1. Now if we restrict ourselves, so now we ask how do these act on the hyperboloid, we can just use the chain rule. So in these tau x coordinates, it's another nice exercise you can do tonight, in these tau x coordinates, you can derive fairly straightforwardly that these guys will induce particular symmetries on the hyperboloid. And they're given as follows. So let's write them down. So first of all, there's going to be jij, where I take, so i and j, of course, run from 1 to 3. So these are like spatial coordinates on my hyperboloid. You can show that these components simply reduce to xi d dxj minus xj d dxi. So they just generate good old spatial rotations on a hyperboloid. I can take a linear combination of a rotation involving the fourth spatial dimension and i with a boost, j0i. And that you can show, again, just using chain rule, that this is 1 over h times partial partial xi, which therefore generates a spatial translation. By the way, as a trivial point, but it's worth emphasizing, it's not translation in the ambient space that does this, because translation of the ambient space doesn't preserve the hyperboloid. It's a boost combined with a rotation that does it. And you can visualize it this way. So here is the hyperboloid. Remember, and let's take this is the x4, x0 coordinates. And here are my null slices. So to generate a translation from one point to another, sorry, let's make it dramatic to this guy, what I'm doing is I'm doing first a boost along this guy and then a rotation. So this is the xi direction. So a boost along xi x0 followed by a rotation from in the i4 direction. And that's equivalent to translating infinitesimally. Now these are manifest symmetries of our 4D decyterometric, because clearly the spatial part is both rotationally and translationally invariant. But of course, there's more. So I can take the boost along the fourth dimension, j04. And that corresponds, that will define as d, capital D. And it corresponds to the following transformation. It's tau partial partial tau plus xi partial partial xi. And you recognize this as a dilation, spacetime dilation in which, so this is a spacetime dilation in which tau gets rescaled by lambda tau and xi gets rescaled by lambda xi. That's clearly a symmetry also of our decyterometric. Simply because, yeah, you just see it explicitly. If you rescale tau and x, similarly, it gets absorbed by the 1 over tau at front. So that dilation is an isometry. And finally, the one that's really non-trivial is this one. So it's when I take the other linear combination, so j4i plus j0i. And that guy will define as k sub i. There are three of these. And this, you can show, is given by the following expression. So it's h time 2xi tau d tau plus 2xi xj. Sorry, should be bottom partial xj minus minus tau squared plus x squared partial partial xi. Now that is less trivial. And that's something you can do tonight in the privacy of your room to check that this transformation is indeed an isometry of this metric. So that's going to be an exercise for you. Check that Ki is an isometry. The other ones are clearly isometries. So now I want to consider, so now we're going to move on to perturbation. So let's consider a massive scalar field in the sitter for simplicity. Let's consider a massive scalar on this sitter background.  So now when I specialize to this fixed metric, which is this sitter metric over there, you get the following. So you're going to get, so you just get powers of the scale factor. So you get 8 of the fourth from this, 1 over a squared from the inverse metric here. And so this guy together gives me a, so you get a squared, sorry, 8 of the fourth from here, 1 over a squared from there. So I'm left over with 1 over h squared tau squared. And then from this guy, I get 1 half phi prime squared minus a half grad phi squared. And now for the master, I get an extra power of a squared with this guy. So I have 1 half m squared h squared tau squared phi squared. Now this theory, by construction, better be invariant under our symmetries. So we have the following. So since phi is just a scalar, it must transform as a scalar under this coordinate transformation. So this s is invariant under the following field transformation. Must be invariant under the following. So if I change phi via dilation, so delta phi, so delta sub d, is lambda, where lambda is a small parameter, tau dd tau plus xi dd xi on phi. This is a symmetry of the action. And similarly, if I transform phi by this other transformation, there are three of them. So I have three parameters, which is a vector bi. And here is the transformation. It's just what we wrote down. So 2 xi tau dd tau plus 2 xi xj. And finally, minus tau squared plus x squared partial partial xi on phi. So by construction, these are symmetries of the action. And this will be relevant for us in a minute. Absolutely. That's right. Yeah, that's right.  They're just global transformations, as you said, isometries. This action is a scalar by construction. So it's invariant under any diff. In particular, it's going to be invariant under a Killing symmetry. So the only thing I've used here is the fact that phi is a scalar under diffs. And so I can ascertain how it transforms. So all I'm using, if you want to get to this point, I'm just using the fact that phi in the new coordinate system is just phi in the old coordinate system. And then I expand in small parameter away from that. So for small diffs, and then you get. So now in other words, yeah, so thanks for your question. So in other words, in this language, I don't think of it. This is maybe the passive or active form. I always forget. But this is the form in which, if you want, coordinates are staying fixed. But instead, I'm thinking of just transforming my fields. Yeah, thank you. OK, so now we can use as well this action to study the equation of motion. I want to emphasize, by the way, that although this looks classical, I'm going to think of this completely quantum mechanically, meaning I'm going to think of phi as a Heisenberg picture operator, in which case it obeys its classical equation of motion. So if you want, thinking of phi as a Heisenberg operator, I can just write down its classical equation of motion, but really thinking of it as a quantum operator. And then it obeys. If I just vary this with respect to phi, I get the following equation of motion. So phi double prime minus 2 over tau phi prime minus grad squared phi plus the mass term equals 0. And we will be interested in the limit where the gradients are small. So namely, this is to be thought either as late times when modes have been stretched well outside the horizon. So namely, tau goes to 0. So you can see this guy is suppressed by spatial derivatives compared to 1 over tau. So we're going to ignore the spatial gradients for simplicity just to see how the mode behaves as tau goes to 0 at late times. So we take, in other words, we can either think of this as taking tau goes to 0 or equivalently, since you see that this, I can just multiply the whole equation by tau squared. So clearly that the parameter of interest is k times tau. Of course, so this is equivalently just k goes to 0, ignoring those spatial gradients. And so now we can figure out the time dependence of this growing mode. And it's easy. Clearly it's going to be a power law. So we can let tau, sorry, let phi goes as tau to a power delta. And again, it's straightforward. You just plug it in. You get a quadratic equation for delta. And the solution, you get two roots that delta is equal to 3 halves, 1 plus minus. This is the famous answer. 1 minus 4m squared over 9h squared. So these roots are real if the mass is less than whatever, 2 thirds h. That's the usual bound. Otherwise you get oscillatory solutions. But for us, let's just focus on this case where m is less than 4m squared is less than 9h squared. That's what we'll focus on. So these roots are both real. So as a result, we get phi at late times growing at some power of tau. We want to pick out the growing mode. And the growing mode, since tau is negative and going towards 0, the growing mode is actually the power of lambda, which is smallest. So it's for the negative sign. So in fact, the growing mode solution, actually maybe I go here so everybody can see. So the growing mode solution will simply be delta minus, sorry, this was to be called delta plus minus. The growing mode is delta minus. So it's tau, sorry, phi going as tau to the delta. We're now delta. I'm simply referring it to delta minus. This is the growing mode solution. And what this tells us is that, in fact, in these symmetry transformations for phi, so these symmetry transformations, I can replace at late times. I can simply replace this tau d d tau. This logarithmic derivative with respect to tau by delta, you see, because I know the mode behaves in this way, quantum mechanically. So in other words, in the symmetry transformations, I can replace tau d d tau with delta at late times. And moreover, if you look here, you see that there's a term here which is minus tau squared plus x squared. I can also neglect tau squared relative to x squared, since tau is going to 0 relative to the special scale of interest. This is the same approximation we made here. And so similarly, I can drop this tau squared x squared. And so at the end of the day, these transformations at late times on the field boil down to the following. So I replace tau d d tau by delta plus, let me write it like this, x dot grad phi and delta k is bi 2 xi delta plus 2 xi x dot grad minus x squared partial xi on phi. So this is how symmetry transformations should act on our field at late times in the inflationary universe. And these you recognize, they're nothing but ordinary spatial dilation and special conformal transformation on the three plane. So these transformations, they're recognized. So this is spatial dilation. And this is just the spatial SCT. So special conformal transformation on the plane. And so this is the well-known result. So asymptotically, the symmetries of the sitter space. So SO4 comma 1, the isometries of the sitter space, all that we've used essentially is that its equivalent has an algebra to the conformal group R3. And this, of course, is an observation that's at the basis of the DS-CFT correspondence. So it acts as conformal transformations at late times. Now, in cosmology, we are precisely interested in computing correlation functions at late times in the inflationary universe. So we're precisely interested in this limit where k tau goes to 0. So this is particularly convenient. So we're precisely interested in this regime where k tau goes to 0. OK. And so in fact, we can capitalize on this to realize that our correlation functions, our late-time correlation functions, should be invariant. So we can use two facts. We can use the fact that, first of all, our correlation functions that involve various powers of phi, they must scale as tau to the particular power delta, where delta, remember, is related to the mass of the field in question. And secondly, we can use the fact that the correlation functions, the x dependence, must be invariant under the full conformal group in three dimensions. And by the way, yeah, so let's just get started. So we can immediately, and by the way, this is an observation that's very old. I think the first people who talked about this were Antoniades, Motola, and somebody else that escapes me in the 90s. And then, of course, Schrodinger with a DFCFT. But one nice paper in particular that discusses this is by Paolo. So if you look at the paper criminally, here's the phone number 1108.0874. He nicely describes the consequences of these symmetries. So let us indeed start with a two-point function. So with a two-point function already, you know that the fact that the spacetime is translationally invariant and rotationally invariant in space, so the fact that you have 3D spatial rotations and translations. So just that fact, of course, tells you that the two-point function, so we're only going to consider equal time correlation functions, so phi of x1 at tau, phi of x2 at tau, already you know that this quantity is simply, just by these symmetries, must be in general terms, it must be a function of the distance x1 minus x2 and time, tau. But of course, we have more than this. We have the full conformal invariance plus the time dependence. So in fact, the spatial dilation is enough to fix everything. So just the fact that it must be spatial. So what you know is the following. So their correlation function, x1 tau, x2 tau, by conformal invariance, by dilation invariance, I'm sorry, it's enough. It must be a function of the distance to the 2 delta. And the rest you fix by knowledge of the time dependence of the fields. So this must go as tau to the 2 delta. And that's it. So just based on symmetries, the two-point function is fixed up to an overall constant, which of course is just a normalization of the terms. Ah, it doesn't matter. This is completely fixed. As long as it's preserving the symmetries, as long as it's preserving the symmetries of the background, then it's fine, yes. Otherwise, it's completely independent. Yeah. Oh, good question. No, so you just have to be careful in taking this limit. So let's just put the minus here. So in the limit where m goes to 0, you see delta goes to 0. And what you get at the end of the day, if you start expanding this, so you see, first of all, the answer becomes scaling variance. And there's a correction in delta, which goes as a log. As a log of x1 minus x2, which is what you expect. Yeah. Yeah, but this is the assumption we make all the time when we do inflationary computations. Now, of course, you can also have people have considered departures from what we call the Bunch-Davies state, the adiabatic vacuum. As long as these are not too violent departures from Bunch-Davies, these departures will be erased as tau goes to 0. In fact, most of the time when people start fiddling with these states, they find huge departures, which in fact lead to a breakdown of inflation. So you have to be quite careful about the kind of departures you're allowed. So if you want, that's just for this discussion, assume Bunch-Davies, just to fix ideas. The point I want to make is just that even with this assumption, irrespective of the Lagrange and I wrote down, I could deduce the two-point function without calculating Hankel functions or anything. And it becomes powerful now when we look at three-point functions and higher-point functions. So for two-point function, it's kind of trivial. So let's indeed move to, ah, so I want to stress, yeah. So at the level of two-point function, there are no further constraints. So if you want the special conformal transformation acts trivially, okay? So everything is fixed by dilation. At the three-point level, on the other hand, I have the following. So let's consider again three fields at equal time. Now again, by virtue of the fact that I have three powers of the field, the answer will go as tau to the three delta. And now I want to first write down something that's dilation invariant. Well, there are many things I can write down now because I have three points, so I can construct two distances. Okay, so in particular, so I can write three distances. So delta, I'll write first the answer. So I can write this, x2 minus x3. But if it were just about dilation invariance, there are other things. I could multiply this by an arbitrary function of ratios of these distances, right? So if it were just about dilation invariance, you could consider ratios like these, okay? Et cetera, et cetera. All the various ratios. So by dilation invariance, it's not enough, but in fact, once you impose special conformal transformations, you show that this function must be a constant. So that in fact, this is the unique answer. So even the three point function is fixed completely by symmetries up to a normalization. And so on and so forth. Of course, as you go on to higher and higher point functions, there are more freedom, but they're fixed by conformal invariance. At the four point level, you're forced to have these cross ratios function of the cross ratios, and so on and so forth. Okay, so the last thing I wanna do is to actually, because this will be helpful for the next lecture, I wanna describe these transformations in momentum space. So let's write down the momentum space version of these symmetry transformations since we're gonna make use of this next time. So we have in real space, as we've seen, let's just do dilation. And I'm gonna let you do the SET as an exercise. So in dilation, we saw that delta D of phi in real space is just lambda delta plus x dot grad on phi. So let's see how that acts in momentum space in Fourier space. So let's write phi. So I'm gonna write phi in Fourier space. So let's do it directly here. Well, in fact, let's do it here. So I have that phi. So if I write it in Fourier space, so phi of x is D3k, e to the i k dot x, phi of k, i sub k of eta. And so now when I apply this operator, this operator x dot grad, so if you want x dot grad on this object, the x dot grad will only act on the phase. So that's the only one that carries x dependence. So I'm gonna have D3k. So phi sub k of eta comes out. And then I have here x dot grad on the phase. So it's a simple exercise to convince yourself that the grad, of course, brings down an i k. Meanwhile, x is equivalent to idk. So at the end of the day, or one over idk, this guy is the same as k ddk on this. So this is the same as k dot partial partial k. And now I can integrate by parts the ddk, and you see that the ddk is gonna act both on the phi, but also on the k. And when it does, it gives me a factor of three because it's the divergence of k. Okay, so I get, at the end of the day, this is equivalent to minus, by integration by parts, minus d3k of three plus k dot ddk on phi sub k. Eta the i, little nap. Okay, and so at the end of the day, in momentum space, the transformation amounts to the following. So therefore, in momentum space, delta D of phi of k, oh sorry, I switched to eta, I meant it's tau. Okay, sorry about that. Hopefully that's clear. I kept switching in my notes as a style for conformal time. And so the variation in momentum space is simply the following. It's lambda times delta minus three minus k dot ddk on the four year mode. Now y minus three, it's obvious, minus three, why? Because you just lower the dimension, right? So if you want to really, if phi has a dimension delta, then its Fourier transform must have mass dimension delta minus three because of the dk. So this is just telling us that in Fourier space, this field has a mass dimension delta minus three, as you'd expect, the weight of delta minus three, and then the k to dk scaling as usual. Okay, so now we wanna use this fact, what we're getting at, again, which will be useful for next time, is to now consider how a correlation function transforms under this dilation. So let's consider a general correlation function, which is O of k one k sub n, so a general endpoint function. And the various k's here, they correspond to different fields in the correlation function, and each of these guys can have different weight, so different deltas, okay? And we wanna see how does this transform under dilation. So first of all, since what we know is for sure this correlation function is gonna be invariant under spatial translations, so by spatial translations, we know that this will be proportional to a momentum conserving delta function, so by spatial translation, we know that this object O must be proportional to a delta function conserving the sum of the momenta, okay? And we don't really care about this delta function aside from the fact that it comes along for the ride, so we're gonna define, to make our life easier, we're gonna focus on the amplitude, the pre-factor that multiplies this delta function. So I'm gonna write it as follows, we're gonna define, so O, so this correlation function, k one through k n, oops, is gonna be by definition a convenient two pi cubed, a momentum conserving delta function delta, I'm gonna call it capital P, capital P will be the sum of the ks, and now the correlation function by definition O prime, okay, this is kind of a trivial thing, okay? But it just means the correlation with a prime is a correlation function without the delta function quite simply, okay? And we wanna ask how does the prime correlation function transforms? And so P again, just to write it, so where P, capital P is the sum of the ks, chock, chock, chock, chock, chock, chock, chock. So let's consider how this object transforms under dilation, so delta D of this guy. So delta D of O is delta D of this delta function P, times O prime, okay? And now we just have to be careful about how these different pieces transform. So by definition, this object, which is an object that depends on nk's, will just be summing over the variations of each field participating in the correlation function. And so it's simply gonna be a sum from a equals one to n of this operator, of this particular field variations, but sum for each delta in each k. So this will be sum over A of delta A minus three minus k sub A dot DDK sub A, okay? Acting on the whole thing. So acting on the delta, delta three P times O prime. And the reason I do this is just to focus on how this operator will act on the delta, because at the end of the day, what my goal is is to write this as delta times the variation of O prime. So we just wanna fish out how delta gets transformed, okay? So let's do it. So this is a linear operator, so I can just do the chain rule. So I get two terms, so I get minus, and let's see how I wanna do it. Ah yeah, so the delta minus three, of course, doesn't care, right? It's just a number, so let's just focus mainly on this guy. So I have minus sum over A one through n of k A dot DDK A, sorry, like this, acting on the delta times O prime, okay? And then I have the same thing now, but acting on the O prime.  Okay, so this last term is of the form that we want.  And so we just have to massage this first term, and it's very easy because of the fact that delta only depends on the sum of the ks, okay? So first of all, okay, so since it depends only on the sum of the ks, each of these derivatives is equivalent to differentiating with respect to P, okay? So this is just DDP. And now the sum of these derivatives acting on this guy, right? It's just gonna be the sum over k because DDP on delta P is just independent of N, of A, sorry. Okay, so this will just give me P. Does that make sense? Yes, just sums up. So I have minus, as a result, minus P. Let me see where am I? Yes, so minus P dot DDP, acting on the delta function and this whole lot times O prime and then plus the rest. So minus this stuff. The delta alpha was just multiplying the whole lot. So I just shoved it into this term. No, no, no, delta is not the Laplacian. No, no, no, it's just the weight. It's just the mass dimension. Yes, so it's just each, if you want, I think of these as a collection of massive fields, each with different mass. The delta for each of them is what we wrote down. So it's three halves. So delta sub A will be three halves, one minus square root one and four, A squared over nine, H squared. It's just the weight, the mass, if you want, of each of the fields participating in the correlator. Yes, I know in Russia, because my post doc writes Laplacian as triangle and we have endless fights about this, but okay. Now here it just means weight, okay? Now, okay, so we're basically done. So of course this correlation function, at the end of the day, we think of it as going back to real space. We integrate over momentum, so I can integrate by parts, this DDP, and then I'll get two terms. The DDP can act on the P, or it can act on the O prime. But notice that if it acts on the O prime, it will be what I'm gonna be left with will be P times delta, which is zero, you see? So the only place where I can get something on zero, if it's this DDP after integration by parts, hits on the P, which will give me, of course, a factor of three. So this gives me plus three, okay? All this worked for just a three, but okay. So is life. O prime plus the rest, and the rest is this. So it's minus delta. Yeah, let's do it. So it's just sum over equals one to N. Let's see, how do I wanna do it? That's right. Yeah, so let's write it as follows. I can write it as follows. So I can write as this guy times delta A minus KA dot DDKA. Cause that's the term that depends on A gets summed over. And then the factors of three, they just summed over so I get minus three N from this. And all this multiplies O prime. Okay, so after this long song and dance, the end result is the following. It's not very profound at the end of the day, but it's worth going through it. At the end of the day, now I can extract how O prime varies. So at the end of the day, the variation of O prime, which I ascertain from this relation. So now I just drop the delta functions everywhere and I ascertain that O prime transforms as follows. It simply transforms as sum from A equals one to N of delta A minus KA dot DDKA minus three N minus one. And this is multiplying O prime. Okay, now just to philosophize on the result, it's very easy to see. So remember, since we're working in momentum space, every guy in here has a minus three because of the D3K. So since I have N fields, I get minus three N. All right, that's the minus three N. But then I stripped off a delta function, which would have given me another minus three, so plus three. Okay, so at the end of the day, this guy must have this effective weight dimension. And I'm gonna leave it to you as an exercise to show the following. As an exercise, you can repeat this for the SET. So repeat for delta KI. So namely, you can work out how the SET acts on momentum space first on the field, like we did here. And then you work out how it applies to a general correlation function. And then on the primed version thereof, and the answer is the following. So it's delta K sub I on O prime. By the way, can you see? Is it two? Is it okay? Feel in the back? Yeah? No? Okay, I write the answer there. So the answer is the following. It's very nice. Is it on your balcony, watching the sea, compute. And you find the following. Delta KI is O prime is I, sum from A equals one through N of two delta A minus three, D, DK sub I, D, DK, A, I, plus K, A, I, two derivative D, K, A squared. So this is actually Laplacian in momentum space.  So please do this tonight. The copies are due on my desk tomorrow. Just kidding. I'm very tempted to talk about my alternative to inflation from Atheist, but I'm gonna restrain myself for tonight.  I'm allowed to talk about whatever I want after that. Okay, thank you. That's it. Question?


What's good job to Boris back again with another video. So we're gonna check out wrestling counters to buy BDW J Forever man. Did I say that right BDW J for up? Yeah, I did We're gonna check this out. We haven't done a wrestling counter in quite some time When it continued the series or whatnot for you guys appreciate all love and support Let's get right into this video Right on the temple. It's time.  No Watch out, that's right. So I know let's get back to this match The scoff right there, you know, he might he might have been Control of the wrist shows underneath Looking for that pop over DDT, but oh That was cool. I like that I see a lot of Darby Allen's left in this video. Oh, no Oh From the disciples of slash Whoa Oh my is impressive. Wait, what did what did it come to? All right, Shinsuke Silver and green as look at this from Shazam from Australia hands Oh no a style clash. Oh my gosh Oh into a whoa, that was impressive. Whoa Wait, what the oh H was man She's He's matches just from the clips itself is fantastic because you know, we're here watching, you know, when you're in that ring You'll know as well as I do. Oh my gosh. Wow counter countering the leg Whoa, this was the the best pacing for tessa blanchard. Okay, you know, the argument could be made Sancho that you go for a quick one early Oh Let me go and take exception to that. Oh my goodness.  Yeah, I did Oh, look at this sunset flip Look at that.  Okay rolled right through. Okay London connects London high risk Oh Here's a springboard going up Oh Oh rest in peace man rest in peace got off Let shane get to his feet rick of body charge again. What matches this Nice on to his feet train Miguel filling in the space The different set of rules that they have to deal with and moose may have given a big advantage. Oh, man He's so Way out of it Great technique there and then reversed into a whole of our own These ladies are having a lot of fun counters and counter out there Yeah, today and look for that detonation kick was well scouted. Oh, this is amazing Tony storming just using all of that strength to crush the life out of storm, but Storm of the council Oh when you're beaten the cramp is that rvd second shoulder block? Oh Oh Oh, look for the cut.  That's on some superhero. Shit that that was That was cool Oh got him right into it. He was just waiting for it. Boom. That's it That was cool, man That was that was that Enjoyed this video, man. I definitely wanted to check out Wrestling come to definitely going to check out any other parts That he's dropped. I'm just working my way through the video So if y'all want me to check out some more of bdwj forever wrestling clips, let me know down below You know, I will Definitely got to come back to the his channel and check out some of the wrestling clips He's able to create compilations, but appreciate all of his support row to adk appreciate y'all can give me See y'all next week


Hey there friends, thanks for checking in. I feature a lot of budget guns. And every time I show a budget gun, someone says you are missing this specific Ruger. How can you feature all these budget guns and not include that one? And I thought about it, so it was available locally. I was able to pick it up, never been shot, got it for a killer price, and a lot of people do because it certainly is a budget gun. Now if you recall back, Ruger's introduction to the single stack nine millimeter handgun craze that was going on about 15 years ago was the LC9. And then it had long trigger pull, hammer fired, super long trigger pull. People complained about it. You'd pull that trigger and drop that muzzle downward. People were shooting low, left and right. So then they came out with the LC9S. The LC9S, which is a striker fire, had a trigger that was very light. So it went from overly heavy to overly light. This is what I said about the LC9S 10 years ago. The trigger is extremely light. Matter of fact, it kind of reminds me of, you know, a four pound 1911.  But then what Ruger did was they took the heavy trigger and then the super light trigger, meshed it together and came out with a budget handgun that models the LC9S. But it's priced much less. And here we have the Ruger EC9S. EC, I think, stands for economical. That's just what I think. Many of the same features, except for a little bit of a heavier trigger pull, which is good, the LC9S trigger was too light. And it has integrated sights. So they don't have three dots like the LC9S. Okay, integrated. It comes with this piece here. No real ejection, seven rounds, single stack mag. And it does have a magazine disconnect, which irritates me, but it is what it is. Ruger is a bit on the overly safe side. Okay, they stamp all this stuff, read the owner's manual, you know, check to be unloaded or what some of the things they say. But the EC9S is a single stack nine millimeter handgun. They got rid of the LC9S and the LC9. This is their model MSRP's for about 350. But if you search hard enough, you can pick this up for in the mid-250s, brand new. And this is brand new. And I paid less for that, less than that locally, okay? But it offers quite a bit. I have some experience with all these models. I don't have video footage, but that's okay, I will assume. Seven round magazine. It has a thumb safety right there. A 3.1 inch barrel with an integrated barrel bushing, inter-gull, and there are the sights. Serrated black, serrated black, both which way. This is Sebel, I'm gonna show that glass-filled nylon, okay? It's a form of polymer glass-filled. But the trigger is not super light. I would guess right around five and a half pounds, right? I'll get a true measurement on that with the reset right there. So it's heavier than the LC9S, which is basically a copy of this, but a very light trigger with three dot sights. And it's much lighter than the original LC9. So they balanced it out, came out with this, and now this is the only one of the three that they produced. And they made a budget priced, very much so. And another thing they did was the finish on the slide is much better than what was on the LC9. That one rusted out. I had one, I showed it, it was my buddies. He would bucket carry it and that thing was so rusted out. So this is it. Here we have it unloaded, EC9S weighing in at 16 and five, eighth ounces. But now we have seven rounds right here. We're looking at 20 and three quarters ounces. Here's the carry round. 21 and an eighth ounce loaded with eight rounds. Let's disassemble the Ruger EC9S. If I have that orange piece in there, that will allow the trigger to pull, but yet not lock the slide back. And we are clear. The first thing we're going to do is take this gate right there, push it down, that will expose the pin. Okay, a little plastic piece, you push it down. Then we need to move the slide back just a hair. I've got this little wrench. Some people use a punch. You go in the opposite direction, push the pin out. I got it partially out, but that's the pin. And then we'll pull the trigger, and off comes the slide from the frame. Here we have a dual captive recoil spring, guide round assembly, and a 3.1 inch barrel with the integral muzzle bushing. That's what that is. It locks up really tight. Then we'll put it back into place. It's the same process with the LC9S, the EC9S. It's very identical for less money. That pin's not all in. We'll just push it in a little bit further, and we'll raise that gate. I don't know if I have it in all the way. That doesn't look like it is. There it is, and let's raise that gate. And there we go. ECS9 is reassembled. Here we'll take a look at the grip texturing. Not bad. It does have the pinky extension for three finger contact on the grip. And I think a lot of people, well I know a lot of people pocket carry, and you can see it fits in there really nice. And it's life. So the EC9S certainly has its place. LC9 was super popular. Then people moved away from it because of that long trigger pull. The LC9S, because it was striker fired, became popular. Then people moved away from it because the trigger was too light. I did an accurate measurement, five pounds even. Five pounds even. With a reset that is almost full position. Hit that tension right there. No over travel. So that is the Ruger EC9S. It is budget. Definitely budget, right around 250 bucks. And maybe 270 or so if you shop around a little bit. But it's a lightweight single stack, nine millimeter handgun that is Ruger's balance between their previous models, and it is budget price. So I'm gonna get this to the range. See how well it does. A lot of the single stack fans still out there. I know people, my P365 does everything I needed to do. We have to acknowledge some people like to pocket carry a single stack, especially now in the warmer weather. So here is the Ruger EC9S. Definitely a budget price gun. And from what I can tell, pretty nice. I have quite a big experience with the other two models. Not so much the EC9S, but it's not that much different. I think it'll be just fine. If you like videos like this, please subscribe. And share. I always appreciate thumbs up button. Thanks for watching and you guys be safe.


So you can tell everybody what you're doing. Tax in the bottom, your tabletop. You don't have to put it in. So it doesn't stick to whatever it is that it's sitting on. Yeah. It'll drip off, and it'll really stick to whatever it is under it. And it's easier to pick up and tilt if you have a little look under it. I was gonna, should I put it down? The safety pins, push pins. Push pins. For sure. Or if you don't, it looks like this, and it's hard to get your hand under it without messing up your edges, and it will definitely bond with it. I really do, so I'm kinda nervous about it, and we're gonna see if it works out. We're gonna learn together, shall we? We shall. Pro tip, use double layer of gloves so you can take one off when you wanna pick up some of your plug-inables. That's not a word. That's not a word. Like our, this, our torch has been resined to its, it doesn't come on anymore. It's waste. It won't, I wonder if I just hit it to the ground. No, because it's just done, because it's resined together.  Blow dryer. Blow dryer. So yeah.  All the things. Virotex Lite, they were out of the large boxes at our neighborhood, Hobby Lobby, so they're little mini boxes, just like this. Make sure there's no debris in your resin apparatus, because then it'll get on your piece, and then you get to pick it out. And it will probably most likely mess up whatever it is that you just did. And it'll always happen on the part that you like the most. We're using one of these that tells us how many parts we're pouring, it makes life easier. We use the push pins. Make sure that they go in between where the wood meets. Well, a lot easier to push in, take out, and it's away from the edge, so you can scrape the edges. Excess. Yep. Go ahead and push them all the way in. That way, you're sure that you have a level surface, because it's super important that it's level. It is not level, all your resin will rush to the lowest point, because it is liquid, and that's what liquid likes to do. So everything that you did will run off and adjust. Also, today we're doing a dip tick. If you don't know what that is, super simple. It's two canvases that are one painting that we have, this larger one, and nine by 12. I'm not really sure this is nine by. We're just gonna make this whole thing. What is that, 16 ounces? Gamershine. Mix it, two to three minutes. Everybody says three minutes, but it's just mix it. Scoop the sides, the bottom. It gives tiring, but some people use power tool, but they just have more bubbles to work out, which is fine, because a torch or heat gun will take care of all the little bubbles. The most important thing is make sure it's mixed fairly together, because if you don't, sticky spots, that will never, and it's super frustrating, because then you have to either throw the piece away, or you have to scrape that off. That leaves a weird spot, and re-pour it. Our Instagram account, because she wanted to know how much flow trawl to add into a resin piece, and I'm gonna let you know this horrible idea. We use flow trawl for our acrylic pours, but don't use it in your resin pours, because you're only gonna make a mess for yourself. It won't work. And also, they had another problem where I think they just added too much paint, because she ended up with a frothy, chunky pour. It's hard to explain, but I think the problem is, that she added too much acrylic paint. I don't know that you can add too much ink, I'm sure you can, but it's harder to do. But with acrylics, you have to be very careful, because they're thicker, and it'll change the scientific makeup of your resin. It's science on all levels, I'm sure. I don't know the science, but I know that if you add too much, it's gonna change how it sets, and you don't wanna go there. We are gonna use some fun colors today. Everything that we're using today is acrylic. As far as the colors. I think they're all medium body acrylics. But this one is by Art District. It's just a regular titanium. And we're also using, the rest of these are Amsterdam. This color is a mystery. Sky blue. There's so many names on here for this one. It's number 551. The one you just used? Number 557. The green is awesome. The green is very opaque. There he used to get it. I know. The purple is number 568. You can use this teal yet. It's number 522, which is a very vibrant color. And then the green is not by Amsterdam. It is by Mural Paint Marker Company, and the color is volatile organic compound. Number 2178. No, it's called Slime, which is way more appropriate than what I first said. It's this color, and the bottle looks like this. We got this at a local art shop called Aisle. Aisle, this is Aisle, right? I haven't seen anything like this at a Hobby Lobby or Michael's or Joanne's. You can probably order it online, but. It's good paint. If you want more information on these paints, just message us below. Message us below. Comment below or send us a message. I always put the resin in the cups first. That way I know how much pigment I need for the amount of resin I've chosen to use per color. Make sure you mix, especially with acrylics, your, look how fun that looks. By fun, I mean creepy. Mix your pigments in super well, because if you don't, you'll also have problems with the resin setting properly. You'll end up with chunks of paint in and amongst your resin. And that's not fun, because it will also set differently. And it may set matte instead of the super gloss that the resin is. How opaque do you want your white? You can still see through it. This is the first time you guys are going through this, but that's for the first time as well. So this is gonna be our first negative space. Negative pour and dirty cup negative. I guess it doesn't have to be super opaque because this is white. Should we use just a little cup? Yeah, because you can always mix in more color. You cannot take it out. So it just makes a little out of time until you get. Yeah, but what's gonna be interesting with this is, it doesn't mix like acrylic paint when you mix them together. The color is not the color in the resin. Let's get there closer, should we get them closer? What we're doing is just gonna put you all closer. From light to dark. I want this to be completely white on the outside. It's very crazy looking. So now we wanna do white? Yeah, what I'm trying to accomplish is to just all negative space here. We don't want any color out here. That's what the goal is. We'll see if that ends up happening that way. What normally happens is what resin wants to happen. That's how resin rolls, does what it wants. When you're working with this much white resin, you need to make sure that your gloves don't have any of the pigments from when you mix your colors in because obviously when you touch your white resin areas it's going to transfer some of the color. So that's why we're doing as far away from where we pour the color parts first. Did you put orange in there? Nope. Huh. We're just together without them. We're just gonna put them in there. I guess we can do the air gun or the hairdryer. I think we need to just pour some clear in that little gap in. Because the clear's gonna show as white anyways. Or we can just line it with some color. Yeah, and you have plenty of colors. So the heat gun's gonna do more than one thing. It's gonna pop the little bubbles that are in the resin that we mixed in from making sure the two parts are thoroughly mixed together. And it's also going to help us to make sure that the two parts of resin are overlaying on each other. And this is what gives us cells and lacing and things like that.  I think we should blow it. Be cautious when you're using any kind of heat that you don't leave it in one spot for too long. Because then you'll scorch your resin in odd portion of your heat. Not really, that's not good. And the heat also, if it's starting to set after a couple of minutes, it also reactivates it. This time you can go through and see if there's any raw patches or air bubbles or if the blow dryer put anything into your piece that you don't want there. Smooth these. Where's it gonna go? It's set, so you can put the heat reactivated. It gets hot, so it'll sit in there, it'll set up. It's really easier to work with, time frame stuff. Try to do the gold spray paint thing. That's what I wanna do.  I think we should just do it. Time ever trying this. It looks like gold leaf. It's wet, like this top part is the gold. Pumping out.


For the CircuitPython Parsec today, I wanted to talk about reading an analog input on a CircuitPython device. So here I have an itsy-bitsy M0, and plugged into it is a soft potentiometer, but the things I'm doing would be the same if you were using a regular potentiometer, a force-sense resistor, anything that is an analog input. I have a power and a ground pin connected to this soft pot, and in the center is the signal pin, which is actually acting as a voltage divider. So here you can see, as I adjust the position on this soft pot where I'm pressing down on its little resistive material, I am changing the analog output voltage that it's sending, which is being read by the analog input on the itsy-bitsy, and then we get a nice smooth graph there, which you can see is correlated to the LED brightness. What's going on when you want to use an analog input? We're importing, the most important things are boards, so we get pin definitions for some of these analog pins, and the analog IO. Then I'm setting up this soft pot object, it's a variable that is the analog IO on board A5, which is where I have this plugged in. Then I'm setting up some LED stuff so that we can drive its brightness. And then in my true loop here, you can see I am reading that soft pot value, and just giving that a nice neat name. So analog value is whatever soft pot dot value is. And then I print that in my serial monitor, and I'm also then driving the LED duty cycle from that. And so that is how you can read the analog value of a potentiometer, a soft pot, or other analog device in CircuitPython on an analog to digital conversion pin. And that is your CircuitPython Parsec.