 Maybe we can start since you already started. Yesterday we concluded the lecture by having obtain this question. z tem, da je to počke, da je nekaj prvih diskupacij, nekaj boljstvom nekaj prvih diskupacij. Prepoče, da pa bomo občajati, da so nekaj prvih diskupacij na počke, da je to prvih diskupacij. Vse možemo se vzvečati na weba. Zelo smo zbuknili boljstvene distribučje vzvečenje tvojih vsega vzvečenje vzvečenje vzvečenje vzvečenje. Vzvečenje je, da je vzvečenje vzvečenje, ta je zbuknila tvoj stansor, in je zbuknila tvoj stansor, in ki Kijj je, je to zbuknila tvoj stansor, je to zbuknila tvoj stansor, minus The long wavelength, the same object made purely a long wavelength fields, and the omega wijay was sort of aigh gravitational stress enhancement. It was a by g square and then the schematic was omega l k,k delta ej minus vk phi long. kaj, phi, long, delta i, j plus 2 similar terms, where omega, long, i, j, was the smooting of an object quadratic in gravitational fields, which was the i phi of x prime, the j of phi of x prime minus the self energy, the i phi n of x, the i phi n of x. Ok, as we discussed, so this is what we got yesterday, and as we discussed also yesterday, this is not very useful, it's written because it contains short wavelength modes, these are short wavelength that contribute, since it's quadratic, in short wavelength, also in short wavelength field, it can have a long wavelength contribution, which will enter here, so it contributes a lot through here and here, and so it enters into the long wavelength equation. This is how short distance physics affect long distance physics in a linear level. So this is not very useful, because we cannot solve analytically for the short modes, so as we said, it's not useful. As we also mentioned yesterday, we are going to compute a correlation function only along observables, like a velocity of the long mode, or a long wavelength velocity, or a long wavelength density fluctuations. Our description is going to break down in short wavelengths. So we can take the expectation value on the short modes directly at the level of the equation of motion. That is, we can substitute tau j lambda with the expectation value of tau j for tau j lambda, for a given long fixed, for a given long mode configuration, plus add a fluctuating component, which we call stochastic, so this is stochastic component, that accounts for the fact that on a given realization of the universe, imagine that there are many universes, you are an experimentalist, you create many universes, you can see many universes you are creating. Every time there will be, you get some short and long configuration, you compute this quantity, you compute this quantity, you compute this quantity, and so this in every universe will be slightly different, and what you get on every realization is more or less equal to the expectation value that you get for a given long wavelength configuration, plus the stochastic, which accounts for the difference, for the difference that the given realization has respect to the average. That is, not every time the short modes are gonna be exactly the same, so the fact is not totally equal to the one that is on average, but there is a small difference. And the difference is that the small, the difference between the average and the small as you go to longer and longer distances. Why? Because imagine that you are interested in very, very long wavelength modes. Inside a given wavelength, there are many, many, many isolated patches of all the nonlinear scale, one over k nonlinear, about 10 megaparsek. So each region here is a galaxy, we realize another one here. These guys don't talk to each other very much. They're not very correlated. In fact, you can see that they're not at all correlated. And therefore, what happens on a certain wavelength, on a given realization of the universe is what happens due to the, what happens in all these possible patches, many, many. And many, many, the longer we make this wavelength. And therefore, what you get on a given realization, since there are many, many independent patches, is what you get on average, up to a small fluctuation that goes to zero, because in the limit in which the number of independent patches, the number of galaxies inside your wavelength goes to infinity. Okay, so this object, taking this expectation value is exactly, so this is called integrating out the shear modes, this process, the shear modes doing this. And it's what we do, for example, when we do quantum field theory, when we integrate out the shear wave fluctuations in the path integral, keeping a certain configuration of the background field. It's called the background field method. Okay, here there is a certain configuration of the long modes, and we're doing the average of the shear modes. Integrating the shear modes. Here the shear modes are the galaxies, not some fancy small plank and particle, or one TV particle, it's the galaxies, but it's identical. For us the galaxies are infinitesimally small objects. Okay, so it's what, if you open the books, integrating out with background field methods, that's what we do, which is nothing but this. And what is this? This is equal to some function, which is very complicated. Very complicated. Of what? So this expectation value, okay, let me just write. Plus, equal. There will be this stochastic term, I will talk about later. Plus the expectation value will be some very complicated function of everything of which I didn't take the expectation value, I didn't take the expectation value of. That is on anything that is not a shear modes. Because the shear modes I've taken the expectation value, so it doesn't depend anymore on the shear modes. And what could be? It's everything that is allowed to depend on. For example, and this is a huge amount of stuff, Hubble, Omega-darmetta, the W of the dark energy, M, the mass of the darmetta, maybe the spin of the darmetta. Very, very complicated object. At some point in this long list there will be the energy, how much darmetta is on the point. Also how many barons there are on the point. Well, for the moment we are doing the universal with darmetta, so let's forget about the barons for the moment. And also, for example, how much is the gravitational force in that point and at that time, and many other stuff. So this far show is very complicated, as his name says, and it's apparently daunting. However, we notice, whatever we notice the following, that notice that we are interested instead on computing, on computing long wavelength fluctuations. We're not going to compute tau j directly, but we're going to compute tau j out at long wavelength. By then you see that the only thing that depends on x are these objects. All the rest is constant, doesn't contribute when I ask for long wavelength fluctuation. Contribute only to the zero mode, which we're not going to observe because we're doing short distances, not the zero mode. Therefore, I can show therefore a long wavelength, but a long wavelength at a long wavelength, rho-darmetta of x is the background-rodarmetta times 1 plus delta. We define this delta to be delta rho over rho at a given time. And this is much less than 1 at a long wavelength. And the background energy density does not depend on x. And similarly for this object, it's at a long wavelength, it's very small, it becomes a very big echel wavelength. So this means that since it's small, since fluctuation is small, we can tell or expand. So this object is equal to some f0 of time, which doesn't contribute to any finite wavelength, it's just a zero mode. Plus, for example, some coefficient 1 times delta rho-darmetta of x and t plus another coefficient 2 the plasher of phi of x plus t3, maybe quadratic order delta. I will drop the substitute-darmetta, so delta mis darmetta, delta of x t square plus c4 dI dJ phi dI dJ phi and so on so forth. Basically, I put everything that GR generativity tells me it can depend on. For example, could write here phi of x. Any guess? No, right? Yes, very good. Do you know the reason? Yeah, you can remove it with a local change of coordinate, you go to the local energy frame where phi is zero. So phi is not locally observable. So this cannot depend on phi. Similarly, it cannot... Also it cannot depend on gradient of phi, because gradient of phi, you can also go... you can set to zero with the same change of coordinate also the first derivative of the gradient field, of the gravitational field. What can depend on is the second derivative of the gravitational field because that's the Riemann tensor. That's what Einstein told us. In the local energy frame, the physical effects are associated with the Riemann tensor with the second derivative of the metric. And in the Newtonian limit, the second derivative of the metric is the second derivative of the Newtonian potential. So it cannot depend on phi. It cannot depend on derivative of phi. Can it depend on the velocity at one location? Also not, because the velocity, again, if there is a uniform velocity, I can just go to the local frame and I don't see the velocity. It can depend on the gradient of the velocity. Yes, that can depend, because if there is a gradient of velocity, I see things, for example, are diluting away. So this is a very important step because you see that this function... ...put everything that can depend on GR generativity that tells me it can depend on. And then, since I can tell or expand, I can encode every single term that it can depend on and drop in nothing. And this is why we believe that the fatty theory will be the right description of the system. Because if a generativity is true, we are putting anything that it can depend on. No matter what their matter is, no matter how fancy it is, no matter whether it be a bubbly-coupled nuclear-darmeter, whatever, it's going to be just different coefficients for... ...as you change the darmeter model, you are going to get different coefficients here, but it's going to be the same. It's going to be encoded here. Sorry? Yeah, this we are going to see... ...exactly, this we are going to see in 10 minutes, okay, let's... ...because right now this is maybe nice, but... or maybe ugly, depends on the taste, this but... which will be clear here how many terms should I keep, so I can write all the terms but if I need to keep the infinite term, pardon, OK it is pretty useless, OK so let's see First let me show few comments, let me get your V.Enf Day mi model, so there is an improvement we need to do of these formulas because we needs the most general dependence for Vi če moj da nepoprej za drugo isgeti, da je izgleda se označenja awe. Prokrat, z vmu, na vahi neskrednji vjezak, vzás je z dal počič. Tak jazna ojevaj bila odgleda pravu veči. Maslilo, da nogenu, z noodlesim. Povej, tudi bila. Zestaj, da so počo prohodili, kaj malo klzelo z duža. in izvendim, da bomo podjeli, ki je, je napor, jo vsij vse teške z vsem je začal v to, ko je, tako, nekaj je napor na zelo, začal na zelo, vzeljenje. Selo, nisi, je tukaj za vzelo, če pa sem zelo, da se dodgeš začal, da je, da so dočeli, pa se dočeli, da se od uživnjaj... kaj začal... intensivno... Into... zelo, in ne, ko b wiredo, vsežo delenje kajno linija, učešno kajno linija, vsežo kajno linija je izpradnja, da bi pričenje z pričetnih metar-partikljih in početnih metar-partikljih vsežo je izpradnja kajno linija. In zelo si, da bilem svoje vrloženje in vratuečenje v roju, se vzelo, da bi se početnje in so početnih metar-pranje o veči, da bilem svome Cientadat tensor. Mi ba ga idij, phi. Gdje je to... Vekne. In, to, ki spodobem poj They sko tez, sem da prаются. Če je idij je phi, tudi menex? Z pravdeni poj, menex1. Če je idij phi menex? Esteli sem bilo nisem možno, je i v držah i jaj, ba je to z njad. OK, da sem tudi izdelno bas, pomembna da je z delta. Sreče, da je delto of x. Ovolj, izgledaj, delto of x otjez v regioni, od vseh regioni, v odrkaj kano linijar. Zato kano linijar je nekaj nekaj, kaj nekaj, nekaj nekaj kaj, kaj nekaj. Zato delto of x je nekaj nekaj, kaj nekaj nekaj, kaj nekaj nekaj nekaj, plus the derivative of delta x1 in the range of distances that I care about, which is 1 over kano linja. And then maybe if there is a correction, which is d2 delta over kano linja square and so on so forth. So we see that we don't need just to put the field. In general the configuration of delta in all this region will matter, but the configuration delta in all this region is well given by a telorespacho in the derivatives around the origin, which is the origin where the point of the composition is. So here we should add also, therefore, derivative, suppressed by what? Suppressed by kano linja, the range of interest where I care, derivative over kano linja of delta of x, or derivative over kano linja, for example, of the i dj phi, maybe times derivative over kano linja, sorry, no, sorry, no, derivative over kano linja, sorry, derivative dL, maybe dL d i dj phi in things like that. So in general here I should include in the function, which was very complicated, so we go in the second line, all derivative divided by suppressed by kano, special derivative by suppressed by kano linja. OK, so we have a derivative expansion. This begins to show you, your companion, what was kind of the suppression that these terms have, because the more, you see, each derivative is going to be a small correction as long as this is going to be evaluated, we are going to be interested in low wavelength, right? So derivative along wavelength field goes like k over kano linja, it comes like kano linja, so this expansion is an expansion in powers of kano linja, and the next term is smaller than the first one if k is less than kano linja. So clearly this expansion that we are doing is going to break down k over the kano linja, but it's going to be very rapidly convergent if k is much less than kano linja. OK, now if you carefully noticed what you learned about last week about the mass function, you realized that actually the time it takes in a lot of form, OK, so this is done. Next, we have to do one more improvement. You realize that the time it takes in a lot of form is not very, very fast, it's not. The timescary form is about an upload time, and during this upload time the field has gone a bit, the certain point that became the halo has gone a bit around, has traveled some distance, and in particular the long wavelength field that affect the formation of the halo has changed because the timescale of the long fields is of order Hubble. How do we know that? You remember last week, Ravi explained that the linear solution, for example, in a universe filled with a matter but to good approximation doesn't change much, is aot times the initial linear perturbation goes like this, so it's proportional to this, a function of k times aot, and so delta dot over delta on the linear solution goes like k, goes like Hubble, and these are long fields, these are linear solutions, so this will be the long field. The long field changes by order 1 in upload time, but also the shuffle fluctuation that collapses into this black hole takes an upload time to fall. So when I write this formula the halo's mass, will depend on the time of observation or also the time, maybe an upload time earlier, half an upload time earlier, and therefore should I put here rho at the current time or rho at x at t minus an upload time. Clearly, in general, will depend on all these times, all the times during which the formation of the black hole occurred. So we improve the formula by saying that actually this very complicated function not only contains all this field, but evaluated on the pass like cone, all this field must be taken on the pass like cone of the point. But since the object, but the meta did not move much, did not move much, just took a long time to form, but since didn't move. So the pass like coin is actually, is maybe like the past tube of radius k no linear to the minus one, but Hubble, order Hubble length, Hubble inverse length, time length. So it's a tube that stands inside of order Hubble time and only stretches in the special direction of k no linear. So therefore, when I take this complicated formula and I begin to try to respond, I should actually evaluate all this field on the pass like cone. So this formula becomes, oops, okay, sorry, f. So now, f very complicated, f very complicated. When I try to respond, therefore I need to include all earlier times, so becomes integral over some t prime up to time t and each coefficient is going to be a function of t prime. So it will be some kernel, kernel one, for example, between t and t prime, times, for example, a placian of phi at time t prime and at the location x prime of x where the point that ended up here at x was at the time t prime. At time t prime, x was at time x prime of x at time t prime, was here at the point. So we have to write the dependence on the field at the location. So we call this x fluid of x. And this kernel tell us that if you put is how much the placian of phi at time t prime affects the tau ij at time t at time t. So the all t prime matter plus, again, plus another kernel, for example, t, t prime, maybe, for example, d2 phi square of x fluid t prime, again, and so on, so forth. OK. So this is the most general formula compatible with the symmetry of the problem. Yes. We're trying to write an equation. Thank you. Yeah, please ask questions. Sorry, I forgot. So we're trying to find, remember that, we're trying to find an equation for the long wavelength mode. But since a short wavelength mode affects long wavelength mode in a linear level, maybe I should explain this. If you take a field of x and you take it quadratic, take the long wavelength energy, you go to for a space, it's a certain number k. OK. This is equal to the integral d3k, d3x, e to the i kx, phi of x square. This is, but the problem of Fourier transform is the convolution of the two Fourier transform, so phi of k minus q, phi of q. So all q's, all q's that goes to infinity contribute. This is non-zero for k goes to infinity. So product of short wavelength field contributes a long wavelength. So this means that in a linear level, short wavelength fields affect the long wavelength fields. OK. Therefore, if you want to write an equation for the long fields, you need to know something about the short fields. OK. So what they do is this tau, a j. I'll finish this. I didn't finish it. How many? How many you're asking? OK. How many do you bear? Are these class renormalization and this class or maybe next? We're going to see all of these in this class, next class. The cutoff will be of order k no linear, but the result will be cutoff independent. Can you ask me the same question when we talk about this? Yes. Because it's like in quantum determinants. First you write the equation and then you study the renormalization. And then you study the renormalization. Right. We do two steps. Yes. We're going to do this later. So I'll discuss later. I won't tell you the answer, but I cannot have time now to do it completely. But let me just say that in quantum determinants first you were Maxwell Lagrangian and then we studied renormalization later. Right. So we're going to do the same. First we write the equation and now we're writing the equation. And then later. Yeah. To finish your answer. OK. So we saw that tau, a j, how short modes affect long distances is through tau, a j. OK. But this tau, a j we need to write know what it is. And it's very complicated to know what it is. That's what, in a sense, what when you study the mass function you are trying to understand what it is, but that's, it's very complicated and then it's very hard to be extremely precise. But this is an extremely precise answer. That is, everything that generativity can tell us, it can do is this. And since I'm interested in the long wavelength, the fluctuations are very small. So I begin to tell a respond. But the most general to tell a response is this one. OK. Now you might still be slightly unhappy. But the fact that, in principle, there are infinite terms in this stereo response show. OK. But then we will see later that for precision at a given, for a given precision of the calculation I will need only a few terms. So the number of times we need is small or is finite given the precision that we want to reach. And this we are going to see later. So that will be the philosophy. OK. Let me see, check. So this is the crucial passage why the fetifil theory is right. Right. Because it's right in the most general dependence on the show mode that we are. At the cost of such coefficients we are general. Now, notice that these are not just coefficients. These are time dependent kernel. That is, this dependence on two times. OK. For example, this is not so unfamiliar. For example, as I told you we are going to do something similar to electric materials. And remember in the electrics you had the electric cost which was a fashion of omega which, for example, I don't remember precisely, but the polarizability as a fashion of omega in full space was epsilon of omega times the electric field of omega which in the space means that p of t is the integral in t prime of the polarizability between t and minus t prime e of t prime. These terms are the analogs of in the electric material of the electric constants. Now, the fact that they depend on omega means that in real space they depend on their kernel, the fashion of time. And the answer you see also in the electric material the answer is not local in time. That is, the polarization time t depends on the electric field and time t prime. Here the stress tensor time t is the time t depends on the total tensor time t prime. Now, the only difference is that since in in electromagnetism time translation is symmetry, deception can depend only on t minus t prime, only the difference in time. Here, because time translation is broken, it's actually is t minus t prime comma t, so it depends independent of t and t prime. So this is not so unfamiliar. But the statement that this object depends tau j depends on the field at earlier times is, we call it, nonlocality in time. That is, the short wavelength object do not depend locally in time on the long ones. But we have locality in space. That is, the short wavelength object depends on the field at the same location or in a very near bi of the same location. Where the location is the location of the field that the added time t prime. I mean, this fluid can be given a I mean, I can give an expression for x fluid. x fluid depends on x and t and t prime is nothing but x minus the velocity the path he did from t prime to t. Ok, so this is the expectation value or the fact is the sense. Rather complicated but not infinitely complicated, ok? I mean, as you will see, the same structure I mean, this is described that met a halo with arbitrary precision, ok? So that couldn't be too simple, ok? Please ask questions, I mean, it's good. Ok, now the only thing I need in the finding is delta tau. Ok, what is delta tau? As you said, was the difference between on a certain region of size of order k inverse between the expectation value and what happens really in reality in a given realization because of the short modes. So delta tau is a stochastic field that has a extremely localized correlation function. Delta tau is something which has value zero. In fact, it's the difference between the field and the expectation value and correlation function of variance or higher profession which are however localized in space. I mean, we can put arbitrary ones. So, two objects are correlated only if you define it on top of each other, otherwise they are uncorrelated and the size is can only at cube. So, it defines the theory. Later we're gonna solve for it. It is interesting, however, to get some familiarity. Is the pace good or I'm going to slow? Can you shake your hand if I'm going to slow? No, shake your head, sorry. I'm going to slow? No, like this. Ok, fine. Ok. I'm going to fast? No. Ok, so I'll not go faster. Ok, so just to gain intuition let me skip this part and let me just talk about perturbation theory how to do with this part. Ok, so now we have equation now we have equations let us solve let us solve for them. So, this is the second section of the notes. I mean, if I will I will have time and I'll come back to the part I skip in session one which is how to do perturbation theory with this theory and renormalization. Ok, yes. I think within the theory this is the only answer because anything smaller than is local. So, for us canolina is a pixel and this is the if there is a correlation between two different pixels is because of the long mode but then the long mode is in the theory then it's not been taken the expectation value over. Yes, this answer is equivalent to have a small correlation of the size of the canolina the correlation length of the canolina it doesn't make any difference for us in general with the counterthreads. Ok, thank you. Ok. Ok, how we do perturbation theory? So, now we got some equations which has which are a repair. So, we have equations. What are them? Laplacian of phi. This one is Poisson equation equal our equation for the long modes x squared along delta long. This is the how perturbations in gravitational fields are affected by perturbations in the density then I rewrite the continuity equation and then the momentum equation equal and then there is the object we draw before where for this we have the form you look there. Ok, for that we look there. So, this is a function of delta tau and all the Laplacian of phi long bla bla bla. Ok. Now, you should notice that this equation contains all the long modes the short modes have gone. Ok. I've gone because the short modes enter in tau j and have substituted them with these kernels. So, at the cost of the coefficient this equation don't contain any long mode. So, can be solved in the long mode. I mean they are solved. I want to solve for the long modes they are closed, they contain all the long modes we can solve them. Ok, let's start to solve them. Now, you can see that they are nonlinear for example this is ok, this is a nonlinear term this is a nonlinear term because so we are interested in expansion so we are interested in small fluctuations it is not very interested sorry at nonlinear which is where we can do fluctuations are small where we are focusing fluctuations are small So, it is very intuitive that probably a good way to solve this equation is to solve telorespanding in the smallest of the fluctuations Ok. So, remember that by the way that rho long is 1 plus delta long times rho background so rho long contains the fluctuation delta rho over rho So, it is very intuitive that we are going to solve them iteratively in particular in fact just to see ourselves that it is nonlinear so let me drop let me drop for a moment the terms that come from tau j for the moment we neglect tau j ok, we do if it wasn't there we can remember to put it back later and the first the equation takes the following we go the equation takes the following a h delta l prime plus ah sorry, let me first say one thing sorry from this equation one can take the equation of motion for the vorticity vorticity it is defined as omega long i equal epsilon ijk dj of vk now, you see, if you take the question for momento you take a dj and then contract with epsilon ijk you are going to get the question for omega which is the following d in dt plus h minus forgetting the stress tensor w i long equal epsilon ijk dj 1 over a epsilon kmn vlm m omega n ok, now once is that now, we know that modes, as you started with Ravi fluctuations are constant side Hubble scale and then as time goes on, they become shorter in Hubble and they start to grow, but they start very very small and they start to grow so, at early times, when I solve this equation the nonlinear terms v is a small perturbation because in the homogeneous universe velocity is zero this is small at early times and the solution to this term is that w the vortizija goes to zero vortizija goes to zero in the linear regime is quickly driven to zero and then there is a nonlinear correction but the linear correction is proportional to omega as well which is small so, the solution is driven to zero so omega i l is quickly driven to zero so, in this equation which means that neglecting the stress sensor the vortizija is going to zero now, we are going to see that the stress sensor actually can create some vortizija but the vortizija is a small effect so, the stress sensor will be always an important but small effect so, which means that we can base our perturbation theory set in the initially the vortizija is to zero ok, so we can set to zero which means, if there is no vortizija that all the information of the velocity is equivalent to the information of theta which is d i of v i ok, because d i is equal to d i over laplacena of theta this is wrong, if there is vortizija but it is correct if there is no vortizija so ok, so our equation all v here whenever there is v here a very high order in perturbation theory this is correct v is d i over laplacena of theta ok, so the question that we get are the following so, we can write them as a k r e h delta prime long plus theta long equal minus intake in d 3 q ok, ill drop to per cube you can find them in the notes alpha q k minus q times delta long of q theta long of k minus q this is the continuity question and the other one is a h theta prime long plus h theta long this is the momentum equation this is the force of gravity and then there is no linear term coming from the v grad v v grad v there which is minus d 3 q theta q k minus q theta long ok, minus q theta long of q ok, you see that and these are all fields evaluated at k prime means d in d a and h is equal to 1 over a d a in d tau where tau is conformal time so d tau a is dt over a ok, so this is the question where alpha alpha I mean, one can directly derive the alpha alpha k q ok q is equal to k plus q dot k over k square and beta of k q is equal to k plus q square k dot q over 2 q square k square ok, so this is a question and nothing but the writing in full space this is a question that they are nonlinear in particular there are quadratic terms implies that there are terms which involve convolutional fields because the Fourier transform of a product is the convolutional of Fourier transforms ok, we are going to solve them in perturbation theory that is a linear level we are interested in long wavelength k much less than k nonlinear so delta long v long vilo vilo everything is very very small in the relevant units so sorry, which are all dimension lists so now we are going to solve them there is a linear equation delta prime long plus theta long equal 0 and then ok, there is a h and then there is a h theta long prime plus h theta long plus 3 half h0 square omega meter over a delta long equal 0 these are all function of k and a ok, all the variables are function of k and a and the solution this you solved last week with Ravi or if you don't remember you plug it in Mathematica that's an easy trick and the solution reads the following the delta linear and we call one the linear long of k and a is equal to a a growth factor d of a over the growth factor time 0 a 0 means present time times is to constant variable long of k where delta s long delta s long of k prime this is a Gaussian variable so all the two point function is present and the two point function is equal to 2 pi cube delta 3 of k plus k prime times the initial power spectrum of the meta by the way has been normalized this is the power spectrum at time a 0 so this is the linear evolved initial power spectrum linear evolved taking a present time 2 present time 2 a 0 so the shape of p11 of k0 is this one this is k this is p11 of k0 let me just plot k cube p11 of k0 p11 of k0 it's a function that goes a bit like this and it has a break at about 10 to the minus 2 h inverse mega parsec that's a meta tradition equality and then it becomes over the one around some number between somewhere between point 1 and 1 between point 1 and 1 flatuation becomes over the one so if I focus on long wavelength k cube p okay that's our basis of the idea why we started this ok this is the linear solution how do we go find the linear solution please ask questions I mean ok let's do the second order solution second order how are we going to solve the second order the equation has a linear structure this term here and then on the right there are quadratic terms so if you solve perturbatively means that what we are going to do is like if we have we are going to plug back the linear solution here and solve for the second order solution that is we write delta of k and a as the linear solution of k and a plus a quadratic solution per k and a and so on so forth and similarly for theta and the question satisfied for delta 1 is this one this is the question for delta 1 these are all ones you can put all ones because it's the question for the one for the linear field and the question for the quadratic field is this equation where here I put the quadratic field the second on the field and here if I substitute that expression and I want to go up to second order the only ones terms that are surviving are these ones but then you see because if I put 2, 1 here the overall fluctuation is of order is cubic maybe I should say that delta 2 we assume that delta 2 is of order delta 1 square that is it's an iterative solution delta is this clear? so do you see that the second order solution the question for the second order variable is the following structure there is some complicated differential equation differential operator acting on a variable delta 2 delta 2, delta 2 doesn't matter and equal now you see I know the solution for delta 1 these are already solved so on the right hand side there is something that is known some function of k of x and t of k and t or a so the solution to this kind of equation means that the second order it means that delta 2 is given by the integral of x and t you get the integral overall space time point of the gris function between the retard gris function between an x prime of the source evaluated at space time x prime so these are four variables they are four dimensional variables so that's the way that's the way the we solve a differential operator on a fashion equal some known term and in particular the gris fashion satisfies the following equation the same differential operator on the retard gris fashion differential operator respect to variable x the retard gris fashion is between x and x prime is equal to delta 4 of x mu minus x prime gris fashion is the solution to the problem of the differential operator on the gris fashion equal delta 4 we are with some boundary condition with boundary conditions in particular let's apply this logic problem attend ok in particular in our problem the gris fashion the differential equation the gris fashion satisfies the differential equation these are two first order differential equation prime prime which I can combine in one single second order differential equation and the second order differential equation for the gris fashion for the gris functions gris function is the following basically takes a one can take the a derivative on the second equation a substitute theta with simply substitute theta with delta prime so we get the following equation minus a square h square second derivative respect to the scale factor a prime so there is a different gris fashion for each Fourier mode so I can write this operator the one we wrote there you can write it in full space and there is a different gris fashion for each Fourier mode but the k since the equation is linear k just goes around for the ride it doesn't matter and then there is minus a multiplying plus a h h prime these are all functions of a times the a of g of a prime plus 3 omega meta a zero square over 2a g a prime depends on k by k in principle depends on k in principle depends on k but we see that there are no k in the equation equal delta a prime ok, so this is the differential equation for the gris function and the boundary condition are g a prime g retarded is always the retarded gris function so all gris functions satisfy the same differential equation what we mean by retarded are the following boundary condition associated to this differential equation g retarded equal zero for a less than a prime so this is the gris fashion that tells us the effect at time a of a source at time c prime j is the source and clearly it's retarded because there is no effect in the past from a source put in the future ok, now for lambda cdm this solution has some you can solve it exactly is some hypergeometry function that you can get it by put it into mathematical for example and solve it or some other or opening some book of differential equation it doesn't matter, it's some some term but you can see that given that there is no k in the equation then the gris function will not depend on k and this is important so the effect this means the effect for each k the effect at the source at time c prime at time a is independent of k is the same for every k ok, so now the gris function we plug back here and we get the second order solution delta 2 delta 2 long k a is some slightly complicated expression which I write it once for completeness also for demystification it is there is the integral of the gris function up to a from zero of the gris function between a and a tilde a tilde square h of a tilde d prime of a tilde square at times 2 times the integral d3q of beta of q k minus q and then there is delta s of q delta s of k minus q delta s is this exactly Gaussian variable whose stupid function is this k plus integral up to a of d a tilde of g a tilde times 2 a tilde square h square of a tilde d prime of a tilde square plus 3 a0 omega m d a tilde square over a tilde times integral in d3q of the other interaction alpha of k q k minus q of delta s of k minus q times delta s of q so this is the solution a second order you see there are two terms alpha and beta so there are in a sense two sources and the gris function multiplies both but with different independent coefficients so this is alpha and beta and this is one independent coefficient multiplying the gris function this is a different independent coefficient multiplying the same gris function in front of everybody because indeed the result is the gris fashion acting on the total source but this structure is useful because you can see that the solution has a k dependent term but not dependent on a and this term is doesn't depend on a of k and q but depends only on a and the prime so as we saw here the integral integral in q is equivalent to integral in space so the sponsor to the source should have been an integral in space and in time and this is the integral in time and this is the integral in space but the two integrals factorize you can do the integral in time and then you can do separately the integral in space and this is due to the fact that the gris function is k independent ok now this is a very useful simplification one can notice there is a useful simplification that now we are going to use so similarly one can find the third resolution again what are the equation for delta 3 delta 3 plus theta delta 3 equal the source which now will contain delta l 2 plus also delta l 1 delta l 2 and one iterates I give you a reference where you can find the third resolution in the notes but I will not write it yes ok this is exactly ok the quick answer is that you see by special that each delta tau was delta plus delta square plus delta cube delta is smaller and the term is smaller than the first so it should not matter as much as the first because it is smaller so if I am not going to be very precise probably I can neglect the second or the term and keep only the first term ok that is the logic and in fact you will see in a second that for each order in which I go in a calculation there is a certain number of terms I keep in tau which is required by normalization in fact we are going to do the calculation in the calculation I mean I think it is simpler to see it in the calculation then there will be a general rule ok so so summarizing the solution at our solution as the following form so there is a very useful simplification that is to a very good approximation this integral in time can be done and they take the very very very very simple form delta 2 of k and a is equal to d of a square remember the solution was going like d of a over a0 to the 1 here goes simply to the square and then the integral d3q1 d3q2 of delta 3 of k minus q1 minus q2 times f2 of q1 q2 delta of q1 a0 delta 2 delta 1 of q2 a0 and similarly delta 3 of k and a is d of a q, so by the way notice that we collapsed the sum of those two terms into one term whose prefector is this one simply the graph of the square and this f2 is some simple function f2 schematika is 1 plus q dot k over q2 things like this some function of k and q with the products you can find it in the reference but it is some simple polynomial function I mean fractional function and the cubic similarly the cubic it very nicely goes like the cube and now I have linear fluctuations f3 of q1 q2 q3 delta 1 let's say that when I don't put the time variable means that this is the present time delta 1 of q2 delta 1 of q3 so the solution takes the following rather simple form in other words we can do explicitly the integral in time this was one time integral this was two time integrals simply the integral in space ok now I can begin to compute the nonlinear correction to the two point function so the power spectrum pok is the correlation function of let me check that I didn't skip anything of delta k delta k and delta is sum over n delta n times sum over m of delta m and we are working assuming that delta is small indeed we are working we know that we are working all the long wavelengths so in effect we computed only up to cubic order so this term is this contribution from delta 1 delta 1 remember that this is ok space k prime k so there was delta 1 delta 1 which is the linear power spectrum this is delta function ok plus k prime p1 1 ok but then if I go to the at 40 quarters in delta I get one term which is delta 2 delta 2 ok k prime ok and apart these I call it delta 3 ok plus k prime the delta 3 is there by translation invariant the universe is translation invariant it is not time translation invariant but it is special translation invariant and this we call it p22 of k p22 because I am taking 22 and then there is delta 3 I can take the third order on one side the linear order in the other side and this I call it delta function ok plus k prime p31 and there is another term which is delta 1 delta 3 which is p13 ok so I get the sum of three different terms two of them are really related by permutation and each one what it is basically for example p22 I write two terms of this guy and I do remember that these are Gaussian variables so I take the expectation value so let me derive the diagram ok I will just write this and then I will continue next time so let's write down what is their form ok how do you compute p22 so p22 is the expectation value as we said delta 2 delta 2 so we plug the stuff in so we get dA is A over dA is 0 to the fourth then we get integral dQ1 then dQ1 prime dQ2 prime f2 or Q1 Q2 f2 or Q1 prime Q2 prime then I get delta of Q1 delta Q2 these are all delta linear delta Q1 prime delta Q2 prime linear and now I take the expectation value of this object but the expectation value these are all numbers so the expectation value is here now these are Gaussian variables these are Gaussian so since they are Gaussian this term here I use big theorem and do all the possible contrasts so I can contract this with this or this with this and also I can do the other contrast when I take the expectation value this with this and this with this so this is equal to this one gives me delta 3 or Q1 plus Q1 prime you are one or Q1 and the other gives me delta 3 or Q2 plus Q2 prime you are one sorry these are Q1 these are Q2 now when I plug this expectation value here Q1 prime and Q2 prime eliminate I can do the Q1 prime Q2 prime integral setting Q1 prime equal to Q2 prime so this is equal to DA over DA0 squared integral in D3 sorry here I forgot I forgot delta function OK minus Q1 minus Q2 times delta OK minus Q1 prime minus Q2 prime so this becomes integral so this two delta fashion kill two of these integral one of these one can be killed for example the Q2 integral and the other remains gives me delta 3 of K plus K prime delta fashion momentum conservation and then I get integral in DQ of F2 of K minus Q Q2 P11 of Q P11 of K minus Q so this is the result of P22 and similarly there is an analog result for P31 OK tomorrow we are going to see how next class is first day I suggest you go through the algebra quickly algebra yourself and in the next class we are going to evaluate these integrals and see what is the result and then you will see that the stress sensor is essential to get a result that makes sense and actually is correct OK one question if you think I am going to slow or too fast please come to me here privately you tell me so that I can adjust a bit the pace or if it's fine also just tell me it's fine OK thank you