 We will have a new lecture about inflation by Manich 살아j. Okay. Thanks. So... What we are going to do in the last two lectures. We are gonna do particle production during reheating, sorry heating and particle production during inflation, so that's the basic idea. We actually started reheating yesterday and we understood that essentially when we have a rate of gamma for any process, v srečenju h to je minus 1, je zelo gama dt, ki je gama over h. Zelo, v srečenju, zelo, kako jazem je vzelo, jazem se, da je proces vsečen, kako gama je vsečen h. Vsečen h je vsečen gama, zelo gama dekrize vsečen vsečen vsečen. in kaj drugi nespojoga je gamma danes na hoditelje, da je svepoje, ki se se vse pristaj začtenja. In, da ga so vse zazavilo, da boiledjamo to, za taj approachedi, se, da bi tukamo, da bi pa vsak padal si, tukaj tukaj s sre stabbedi , da je najbezdi, ki tukaj tukaj, ki je gamma vštot. Tukaj到jemo vse in da bomo delati te vsemčenje, tako kaj ga uživamo ni, of instantaneous decay of the inflaton. For example, this would be the moment where a gamma is equal to h. And if I, for example, do a row of radiation, the instantaneous inflaton decay would give me a plot like this. And the temperature would go as one over the scale factor, because then you have dominant thermal butt. So now let me, yesterday people asked me about an upper bound on the reading temperature, which is actually a very good question. So let me start to address this question. And let me just discuss an example of gravitino production. Now, what is amazing to me, at least, is that we can get very far without knowing very little about supergravity, right? You see, initially you read gravitino production, you say, OK, I have no idea what I'm gonna get, I don't know what supersymmetry, supergravity. But in fact, you don't need, to do this type of estimate, you just need to know the basic rules. And essentially, the basic rule is that each particle is on superpartner, and so the graviton has the partner, which is the gravitino. And then you ask yourself, OK, gravitino will be coupled like the graviton, gravitational, essentially. So in a sense, I can imagine a diagram like this. This will be some gluon. This will be some gluon. This is another gluon. And then here I would have gravitino and gluino. Gluino is the partner of the gluon. And you see, this is like a gravitational interaction of a particle with the graviton, it's just a supersymmetric version. Now, why do I take gluons? Well, here you could take photons, you could take anything from the thermal bath. But here we take gluons, because there is a strong interaction, so there's the strongest vertex we can imagine. Now, here, when you want to estimate a rate, sometimes you can do it very quickly, actually. This vertex is as a coupling constant, which is of order 1. We're not going to do the minutia of the numbers here, we just do estimates. And this vertex has a coupling constant, which is order 1 over n-plank, because this is a gravitational coupling. And then you understand that sigma, the cross-section, will be 1 over n-plank square, that's about it. You know cross-sections, you need to know the dimensions. Cross-sections are areas, so as dimension of 1 over mass square. In natural units, we have the unit of the mass, is equal to the unit of the energy, is equal to the unit of the temperature, is equal to 1 over the unit of time, is equal to 1 over the unit of length. Natural units means h bar is equal to c, is equal to bosman constant is equal to 1. They are lazy, for the lazy people, like theorists, they are beautiful, because you forget everything, you don't need to remember where c, h bar goes. And actually it allows you to do very simple estimates by just knowing this. Because cross-section is one of the area, but you see length is inverse of energy. So you need an energy, essentially to the minus 2. Cross-section is an area, so you need energy to the minus 2. And then you say rate. You see rate, let's look it for here. This is number of processes, dimension 0. This is dimension minus 1, this is dimension plus 1. So how do I get a gamma? Gamma will be sigma times something to make it the right dimension. What do I take? I take the temperature of the thermal butt. You see, I don't know anything about supergravity and I already know how to produce gravitinos. That's pretty remarkable, right? And now we can say, OK, when do I produce gravitinos? I produce gravitinos when gamma is equal h. Again, right? So let me just do it. Gamma is equal h. So t cubed divided by m-plank square is t-square over m-plank, right, the formula for h, forgetting all numerical factors. And then, of course, you see, actually, let me be more clear, because actually that's what I would really do. Let me compute gamma divided by h, and I will be clear in a moment why. So that's the calculation that I will do. Times m-plank divided by t-square, I get t divided by m-plank. So what you learn from this exercise is that this process would be equal to one, only when the temperature is equal to the plant mass, but plant mass is tend to the 18 gb, too much of a bigger temperature. So what we are saying, that this process here is essentially the number of gravitinos divided by the number of photons, because each, this is the probability for a process to occur. And from each process we are producing one gravitino. So essentially, this will tell me the number of gravitinos divided by the number of photons, and we understand that this process is dominated by the earliest possible times. So you see, this will be the moment where you produce most of the gravitinos, because then the temperature is going to decrease. And this gives me the abundance of gravitinos, so essentially n-3-2 is t-reating over m-plank times the number of photons. So this will tell me how many gravitinos that are around. Now, I can ask, of course you could worry about the inverse process, but since, right, this is such a rare thing, we disregard about the inverse process. Imagine this something of the order of 10 to the minus 10, or even smaller, you don't really care about the inverse process, because you have produced so few gravitinos that you really don't care. So this is a process that is always out of equilibrium. It takes place with the probability, the probability for this to take place is t-reating divided by m-plank. Yes? Do we compare to gamma? Because as I say, the question is, why do I compare with gamma? Because the number of processes is essentially gamma dt. And in one, sorry, let me imagine, dividing the time into many upper times. So each t is h to the minus 1. Ah, because this is what tells me the abundance, because after the production, the number of gravitino skazes one over volume, the number of photos skazes one over volume, so this ratio becomes constant. So that's the real quantity to look. So the question was, why do I look at the ratio? Because this ratio is the thing that eventually is constant after the production is done. This is called the abundance. OK? Good. One more question. Please. And what about phase transitions? I mean, in phase transitions, for example, or QCD phase transitions, are they also... OK, very... ...on the temperature drop, they have a higher temperature. Very good, very good question. So there is a question about, yesterday I said that the reading temperature needs to be greater than a few MEV. But then the question is, OK, but what about electro-whee phase transition, QCD phase transition that occur earlier? Well, the problem is that they don't need to occur. So the problem, of course, is how to generate barions. And then, OK, we will have to make something maybe difficult if you have only temperature to MEV. But in principle, it's not forbidden. I could be very smart and find a way to create barions, even if I don't go that high in temperature. So the question will be some maybe non-termal mechanism. And I don't know, right? I need to cook it up. Maybe some version of Aflek Dain. I can try imagine using it. But this is not mandatory, right? So if I really want to say, you know, what did the doctor order me? The doctor ordered me greater than two MEV. Then, OK, if I want to live better, maybe it's easier if I also eat wealthy. But if I just say, what is the minimum requiring from the doctor is two MEV. Now, let me ask what is about the upper bound. That's what we are really trying to go there. So this is a cloud. Why do I care about gravitino? I could have made millions of examples. Why exactly gravitino production? Because gravitino are very dangerous beasts. Because once they are produced, they live long. And maybe they live forever. And then there is the risk that gravitino overclose the universe. But let me discuss the situation where the gravitino is unstable. So the gravitino will decay, producing one other super particle. But the decay time of the gravitino is very long because it's gravitational decay. And so this can destroy the elements that you form during nucleosynthesis. Marko, sorry, I read a question from the chat. Does this ratio stay constant or it decreases with temperature? No, this ratio is constant. It says this, you need to compute it every apple time. You add it up. And then you see that this is dominated by the production here. So essentially, what you produce here, for example, is negligible with respect to what you produce here. Look here, I'm doing estimates. I'm not doing the full, exact, integral. I'm just telling you things in a quicker way. So once the production seizes, in a sense, because whenever the temperature drops to a factor of 10, the process becomes 10 times less probable. So this will become eventually completely negligible. And so this ratio essentially, at all effects, becomes constant. The ratio is constant because they both get diluted by the expansion of the universe. And gamma, I mean the amount of light particles in the plasma. No, no, no, I'm keeping details of factors of order of a few because then there would be a g star to the one-third proportionality. You see, I'm just giving orders of magnitude here. But you're totally right. Maybe I should define it divided by the entropy rather than by the number of photons. You're asking me more sophisticated things that they want to do here. I want to give you an order of magnitude estimates on t-eating. And that's all I really need for this. But you're right, I could be more sophisticated, but I'm not. I'm a simple-minded guy. So let me say now, I want to see the decay of the gravitinos. Again, I don't know anything about supergravity, but I know that the graviton couples to a particle like this. So I take the supersymmetric version of this. So here there will be the gravitino. Here there will be a particle. And here it will be the partner. Again, this coupling constant is 1 over m-plank. So now what is the decay rate? The decay rate will be sigma 1 over m-plank square. And then I need to put something that gives me dimension plus 1. So now here I will put m3f to the cube. And that's essentially an estimate on the decay rate of the gravitino. So when does the gravitino decay? It decays when this is t-square over m-plank. And so now I can have a relation between the gravitino mass and the temperature of the universe when the gravitino decays, which will be m3f is equal m-plank t-square to the power 1 third. I want now to see, for example, what happens under which conditions gravitinos decay before big bang-nucleosynthesis. So let me say that I want t, so gravitino decays before bbn. So here I will say m3f needs to be greater than essentially m-plank mev squared to the 1 third. So m-plank is 10 to the 18 gv. Mev is 10 to the minus 3, so 10 to the minus 6 gv squared. And this gives me 10, 1 third. And this gives me essentially 12 divided by 3, 4. 10 to the 4 gv, about 10 tv. So if the gravitino has a mass greater than about 10 tv, it decays before bbn, no problem. I can live my life happy, because the problem is that the gravitino destroys the elements that you form at bbn when they decay. But if they decay before bbn, they won't destroy anything, right? Yes? I'm assuming that the gravitino is not the lightest superparticle, so the gravitino will decay to another particle, which is lighter than the gravitino itself. Otherwise, I could break our parity, but this would, of course, give us a pressure on the rate. OK. Marko, in the chat somebody is asking, not sure I understand why decay occurs when gamma is equal to h of t. Well, so let me try to repeat the argument. Really, when you want to see how many processes take place, that's the definition of rate of a process, you do the calculation gamma dt. This is the number of processes, just how you would end up defining a rate. Now, if you look at an upball time, we just imagine slicing the time in upball, because upball is going down as 1 over t, then this in an upball time will be gamma h to the minus 1. So now I ask myself, when is this number equal to 1, because I'm looking when the decay takes place. This number is equal to 1, when gamma is equal to h. So essentially, when we do these estimates, we always compare gamma of a process with h. And when the two things are equal, then it means this is the moment in which the process occurs. Of course, I could do all the integral, in principle, I should do these, maybe sometimes gamma depends on temperature, I need to be careful about that, but in doing this estimate, you will see that what I'm doing here for this particular example, I can just do gamma divided by h. As I say, it's always very hard, because we all prefer to have exact calculation in front of us, but here I don't want to do exact, I just want to do estimates, and I'm doing things which are small, tiny mistakes, but at the end the results are the right orders of magnitude. Yeah, okay, that's a very good question, which I will not enter too much into details, but essentially, this is something, so this will be a super particle, and this will be a particle, for example, of the standard model. So this could be a photon, or this could be even a gluon, a quark, and this can generate either electromagnetic shower or hadronic shower. This thing can have the energy, because this could be, for example, 100 gV. This thing could have the energy to destroy the nuclei that you have formed at bbn. And so now we will see that all the calculation of bbn will be disrupted by this, and since we don't have an observational evidence for this, that's okay. We need to say that this process needs to be suppressed. Is it clear? So now, what is the point? Now you need to do very... So if you have this, no problem. Done deal, let's move on. But if instead you don't have this condition, then you need to do detailed calculations on how many gravitinos you have and how many nuclei you destroy. And these I cannot do quick estimate. These you need to do detailed calculation. But you see that you will always end up with a limit on how many gravitinos you have, right? Because essentially you will have a problem if you have a thousand gravitinos, or a million gravitinos, or a billion gravitinos. I don't know, but it's clear that the limit you get is proportional to the number of gravitinos. So the limit you have will become an upper limit on this quantity here that controls the number of gravitinos. So now let's see a figure. And here I will not be able to discuss the details of the figure, I just want to give you the flavor. So this is from a paper by these Japanese authors. And you can see that essentially here is a gravitino mass, here is a limit on reading temperature. These are four plus, because these are different versions of choice of supersymmetric particle masses. So let me not enter in detail, I'll just move one of them, but it's just to show you that it's very model dependent. But the concepts are not model dependent, so these lines here are the limit exclusion from BBN. So you see if the mass of the gravitino is of the order of 10 TV, then the limits disappear, for the reason I was just telling you before. If instead the mass of the gravitino is lighter, you need to include the limits. For example, here you don't want to destroy too much deuterium, for example. And notice that the limit is an upper limit on the reading temperature. And for example, notice that the limit could be very strong if you have in this limit here, for example. Here it could be as small as a T reading smaller than 10 to the 5GV, which of course is a very, very strong limit. So is it clear? So these limit, of course, are again not mandatory from the doctor, because maybe supergravity is not realized in nature, or maybe the gravitinos are very heavy and they decay early, or maybe the gravitino is the lightest particle, and then, of course, you need to consider different type of limits. But in a sense, this gives you a flavor of what traditionally people have in mind when they put limits on reading temperature. Okay? I'll do a stop sharing. Okay. Okay, so. Now I want to make, I want to drop one of the assumptions I've been doing. And I ask myself the question, is reading really instantaneous? Remember the plot that we were doing, right? This was time. This was, for example, a row of radiation. Here we say the inflaton decays. And so you have something like this. And now I want to question this picture here. Is this really what takes place? Now, why do I want to even question this? Because let me imagine that I want to remember that H is proportional to rot to the half divided by M-plank. And remember that H goes as 1 over T. So if I want to compute T-reeating, so let's say that this is the time that here I call T-reeating. And I want to divide this by T-end. Remember, this is the end of inflation. Now I can do estimate by looking at this scaling. This will be raw end of inflation divided by raw re-eating to the half. This will say, let's say that we disregard the change of energy during inflation in this estimate. Remember, for a thermal butt, this will be proportional to the re-eating temperature squared. And then the bottom, the top, with something that we discussed in one of the previous lectures, it will actually depend on the amount of gravitational weight that we observe, but that was essentially the result. Now let's take T-reating as an example, 10 to the 5GV. This was one of the limits that we saw from the previous figure. Then you put in this ratio here. And you'll find that T-reating is about 10 to the 23 R to the half T-end. I don't know what R is. R, remember, is the tensor to scalar ratio, that we don't know how much it is. But notice that the time of re-eating could be even a factor of 10 to the 20 greater than the time of end of inflation. Maybe you could say that inflation completes in 10 to the minus 30 seconds. Re-eating will complete in 10 to the minus 10 seconds, 20 orders of magnitude more. So you see that there is a huge separation of time between the end of inflation and re-eating. Re-eating could be happening on a timescale, which is 10 to the 20 times greater. Given this observation, we need to ask ourselves, can I really reasonably say that re-eating is instantaneous, because look how much time it takes to decay. And then, in particular, I can also think just based on very simple knowledge of what a decay does. If here I have the number of particles, I know that they exponentially decay. If you have a decay, n is equal to n0 e to the minus gamma t. Let's forget about the expansion of the unit. I just make this a simple discussion here. And if I ask myself, when do I produce the maximum amount of particles? If I do dn in dt, this is equal to n0 gamma e to the minus gamma t. This is the moment in which you produce a lot of particles, essentially. So it's not true that you produce all the particles at the end. You produce a lot of particles at the very beginning. So it's not true that here the temperature is zero. Certainly this needs to be modified. Am I clear? Let's discuss how this needs to be modified, and let's discuss whether this is a big deal or a small deal. At the end of the day, we will find that this is a small deal, but it's really non-trivial. No, it's not a good motivator to tell you this. But I want to be honest, but we will see that there is really a lot of non-trivial interplay in answering to this question. There is a question from Zoom. Since gravitinos themselves are massive, we will have a short period of matter domination epoch after eating and before bbn, right? How could that affect the usual hot big bang era? OK, this could be, again, it's model dependent. It depends on how many gravitinos you produce, but the amount of gravitinos I was considering was always very small, so they don't come to dominate. But you may imagine a different situation where you could produce many more of them, but the limits on reading temperature I was telling you here make sure that the gravitinos do not come to dominate. Now, I'm still making one assumption, in which you can find a discussion in the literature. I will post you the papers in the post-election notes about thermalization of the decay products, because this is a different issue, right? I'm telling you decay is not instantaneous, so this picture is wrong, but I will still assume that the decay products thermalize instantaneously. The inflaton keeps decaying in a continuous way, and the daughters keep thermalizing quickly. This is not trivial. This has been discussed in a few papers, and I will give you the literature about this. But for simplicity in this lecture, I'm going to the conventional idea that the decay is low, and so the thermalization is much faster on the time scale of the decay, which is a reasonable thing, but needs to be studied. Now, let me look at the equations that we need to solve. So, let me first write the equations and then I will discuss them. Rho phi plus 3h plus gamma rho phi is equal to 0. Rho dot gamma plus 4h rho gamma is equal to gamma rho phi, and then rho phi plus rho gamma is equal to 3m-plank-square h-square. So, the first equation tells me that the inflaton deplets because of the expansion of the universe and because produces radiation. The second equation tells me that rho gamma, the energy of radiation, gets produced by the decay of the inflaton, and then it gets diluted by the expansion of the universe. This equation instead is the usual equation for a borate in a two-fluid system, inflaton and radiation. Gamma, I don't mean just photons, so, they mean all the particles in the thermal bath. Notice, for example, that if I have gamma is equal to 0, then the first equation will tell me rho phi proportional to a to the minus 3. And from the second equation, I will get h is equal to over 3t. This is just the usual assumption that we have done so far. If the inflaton doesn't decay, the inflaton is oscillating and is behaving like matter. This is exactly what we saw yesterday. But now there is the fact that the inflaton is decaying. Now, we can analytically solve these equations by doing some approximations. The approximation we are going to make, for example, is gamma is much smaller than h, because h is the one at the very start and the decay is very slow, as we saw. And rho gamma is much smaller than rho phi. So, the inflaton, we are looking at this situation here, where a lot of inflaton quanta are decaying, but the inflaton is still dominant. There is still a lot of inflaton left. So, the inflaton eventually will decay. So, re-eating will take place here. Here, re-eating completes, because here rho gamma will become greater than rho phi when almost all of the inflaton is decayed. Here we are in a situation where the inflaton starts to decay and the inflaton produces a lot of gamma, but there is still a lot of inflaton around. So, that's what we are trying to get out of this equation. And we can do some approximation, as I say. For example, we can neglect this term here. We can neglect this term here. So, what we say, we say that we have essentially inflaton dominating, behaving like matter. And now, since we know this, this is our consistent with the approximation that we... So, if you make this approximation, you get this equation here. And then you can put this equation here and you can just solve it, because this will become a differential equation in terms of something that you know very well. So, this is a couple system. The approximations are, we are very early. So, gamma is much smaller than h. OK, then I see the inflaton be as matter. Then, there is much less energy in gamma than in the inflaton. So, h is 2 over 3t. So, you see, I know this now. And now, I can put them in this equation and I can solve this equation. Is it clear? So, let's see the result. I'm not gonna solve this. This is our homework, actually. I will ask you to solve this differential equation. But let me just solve it. Let me write down the solution. You see, this becomes a differential equation where we know everything, essentially. We know h is 2 over 3t, and we know rho phi. So, let me just write the equation for simplicity. One second. d rho gamma in dt plus 4 2 over 3t rho gamma plus gamma, then rho phi is just a constant divided by volume cube. So, 1 over a cube is equal 1 over t squared. So, this will be t squared. Is equal gamma over t squared. You see, this is a simple differential equation that we can just solve, OK? No, no magic here. The result is rho gamma is equal 6 fifth m plank square h and gamma a to the 5 halfs minus 1 divided by a to the fourth, where a is equal 1 at the end of inflation. And this is also the upper rate at the end of inflation. You see that this is very different from what we had assumed before. Notice that when scale factor is equal 1 at the end of inflation, rho gamma is equal 0. That's correct. And now, this equation, it takes a second to do it. Just imagine doing the derivative, as you find the maxima of this function here. You find that there is a maximum at a nearly equal 1.5. So, let me just do the plot. You just can do this very quickly. So, essentially, you would have something like that. Here you have the scale factor. Here you start from a is equal 1. This is the end of inflation. This thing goes up and then it goes down like this. And it goes down as a to the minus 3 half. Yes. Why do I neglect? No, I don't neglect. This thing here is what here becomes this. No, no, no, look. OK, look, let me go slowly, because this goes maybe too quick. You agree with the three equations? Where are you? OK, do you agree with the three equations? So, here I neglect gamma with respect to h, because gamma is very small compared to above rate, because otherwise all the input would decay immediately. So, this is our regime. From here you immediately get rho phi is equal 1 over a cubed. Period. Then here you neglect this rho gamma. From here you get h is equal 2 over 3t. So, here I get essentially rho phi is equal c over a cubed, which is equal c divided by t square, another constant. And this is exactly what I'm writing here. So, once I do this approximation, this equation I solve it exactly under that approximation. You see that I don't neglect the right hand side, because this is what gives the production. OK? So, let's appreciate this result. There is a sudden production, and this is very, very quick, because this is a is equal 1.5. So, this is immediately right after inflation. I reach a peak immediately after inflation. And then the rho gamma decreases as a to the minus 3a. After the production, after inflation has decayed, we know that rho gamma goes as a to the minus 4, right? Once the inflaton has completely decayed, then I don't have the production anymore, rho gamma goes as 1 over a to the 4. Notice that here instead, this is decreasing much more slowly, because there is the combination of two effects. The universe is expanding, so rho gamma is decreasing, but the inflaton keeps decaying, so it keeps giving quanta of gamma. So, if the inflaton stops, at some point the heating stops, and this will go down, so this is inflaton totally decayed, and then there will be a change of slope a to the minus 4, which is much, much faster decay. Is it clear? Now, there is this phase here, right? If you instead did instantaneous inflaton decay, that's what you would assume for rho gamma. So, this is instantaneous phi decay approximation. Is this figure clear? So, what we did before, remember what we did before, before we said the inflaton decays all of a sudden when gamma is equal h, and then the radiation will dilute. So, that's what we were doing before, right? Remember the figure that we were doing before? The inflaton decays all of a sudden and then it dilutes radiation. But now we see that the things are very different. The inflaton starts decaying immediately, so there is an immediate growth of rho gamma, then there will be this intermediate period, and then rho gamma will go down as a to the minus 4. Notice that this is super conventional with imperturbative inflaton decay. I'm not inventing anything, I'm just dropping the assumption of instantaneous inflaton decay. So, this is the correct calculation to be done in the case in which the inflaton decays perturbatively. Any questions? It's interesting to compare, right? How big is this height here versus this height here? Let's try to compare. This was the mistake that we did up to 30 minutes ago, right? Up to 30 minutes ago, we were imagining this is how it goes, so we would say this is the maximum possible energy that radiation had. Now, instead, we see that this is the maximum possible energy that radiation had. So, let's discuss what is the ratio between this number and this number to see whether we were doing a big mistake or a small mistake. Is it clear? Let's do this. You see, we have already this. We know that A1.5 maximizes this. So, out of there, we get... Please. ...decay only in radiation. Is all the possible that the... ...decay in some non-relativist matter? Oh, this will change the picture, but what I'm assuming is that the inflaton decays in things which are relativistic and they thermalize. Now, imagine the inflaton has a huge energy. So, even if you produce something which would be 100 gV, this would be like massless from this point of view. You need to produce something that is super heavy to see the effect of its mass. For conventional particles, even if they are heavy for our standards, maybe they are bad. Oh, okay, now it's on again. Okay, good. Is it clear what I've done so far? Is it clear that we are trying to compare this height against this height to see what the mistake was? Okay, so let's just say A of order 1.5 in this calculation here and we find that this rho gamma max is 0.4 m-plank square h and times gamma. So, this is this value here. Let me now instead go back to the assumption of instantaneous inflaton decay. So, back instantaneous inflaton decay. And so, we go back to saying that the decay takes place when gamma is equal h. This means that rho gamma instantaneous is equal 3 m-plank square h square is equal 3 m-plank square gamma square. So, this quantity here instead is 3 gamma square m-plank squared. So, this is the mistake that we did 30 minutes ago and this is the result, the correct result that we now we know is true. Let's compute the ratio between these two numbers. So, I'm really computing the ratio between this height and this height here. Rho gamma max divided by rho gamma instantaneous is 0.4 m-plank square h and gamma divided by 3 gamma square m-plank squared. 0.4 divided by 3 is 0.1, as we all know, 0.1 h and divided by gamma. And we see, so the ratio, again, 0.1 h and divided by gamma and this could be much, much greater than 1 by many orders of magnitude. Because remember, h is the rate at the beginning of inflation but remember that we were considering a situation where gamma was, the decay was taking place on a timescale which was 20 orders of magnitude greater. So, you could imagine that this could be some 10, 20 orders of magnitude. So, here the board doesn't give justice to this. We would need to have a board, which is enormous to get a feeling, but you understand that this could be many orders of magnitude greater than this. Is the answer clear? Remember, I'm not doing anything fancy. Standard perturbative inflatum decay. I prove to you that the assumption that most books do makes a mistake of some 10, 20 orders of magnitude on the maximum amount of energy in the radiation. This is the assumption that you make and this is instead what you should have done. Now, given this, the question is, why do we always do instantaneous inflatum decay? Is it all wrong? Because now we need to understand what is the effect of this on some real thing. For example, on the gravity in abundance. Because before, we computed the abundance under this assumption. And under this assumption, here, we found N3f divided by Ngam is equal to t, rejting divided by Nplank. But now, this was done following the wrong thermal history. If I take the right thermal history, what do I get as a ratio? Maybe I get something which is enormously greater, because this could be many orders of magnitude greater than this. And the answer, in fact, is no. You essentially get the same result. Nearly the same result. Mako, I interrupt you with a question from Zuma. The question is, which is referring, I think, to these three equations, which is the underlying theory action in parentheses, from which these differential equations are derived? Yes, very good question. In a sense, what we are really doing here, we are just doing essentially a conservation equation on the energy momentum tensor of these objects here. So what we are really applying, we are applying d mu t mu nu, in the case in which gamma is equal to zero, this will be equal to zero. And then you will say that separately, radiation goes down as 1 over a to the 4, and matter goes down as 1 over a to the cube. If you instead consider the interaction between the two fluids, that's what is account by this factor of gamma. So you see that essentially, the decay of the inflaton grows the energy of gamma, and the decay of the inflaton decreases the energy of gamma. So this will be interaction of these two fluids, and this instead is the Freeman equation, the zero-zero Einstein equation. Other questions? OK. Maybe I can ask you the question, because the answer is not really trivial. I'm telling you, even if this is many orders of magnitude greater than these, when you compute the amount of gravitinos from the wrong history, from the approximated history in the right history, you essentially get the same result. Why do you think this can be the case? It's not obvious, right? This will be one of the homeworks that I will guide you to solve. But anyone wants to take a guess? Yes, we should look at the ratio between what and what. Very good, very good, very good, very good. You see, this is the quantity that we really are after. So it's true that here you produce a lot of gravitinos, because remember that the production of gravitinos is proportional to the temperature. If this is a lot of many orders of magnitude greater than these, you will produce a huge amount of gravitinos, really a huge amount of gravitinos. But the inflaton has yet to decay. And so the inflaton is still in its bag, a lot of will be photons. And so the inflaton, during this period, this is 8 to the minus 3, instead of 8 to the minus 4, because the inflaton keeps producing a lot of photons. And so eventually, as she said, this thing needs to be computed at the end of re-eating, not here, because what you produce here will be diluted by the following production of photons. And you know, when we studied the BBN limit, in this time scale, BBN is here. So we don't really care how many gravitinos we have at this moment. We care how many gravitinos we have at this moment. And the amount of gravitinos that we have at this moment is related to the amount of gravitinos that we have here, because we need to study N3 half divided by N gamma at the end of re-eating, because this will be N3 half divided by N gamma at the time of BBN. Now, I will show you a plot now to simplify this. It doesn't mean that this is completely never useful. In fact, among the first papers that studied this, they discussed the production of super heavy dark matter. I don't know if you remember the name Wimpzillas. This is a rocky cove that made this fantastic name, super heavy dark matter. Imagine that you have a particle, this is raw, but imagine that you have a particle, let's say that this would be raw to the 1,4, the temperature, that has a mass here. If you have a super heavy particle that has a mass like here, you will never be able to produce, if you do the instantaneous decay approximation, but if you do the correct thermal history, you will produce this super heavy particle. In a sense, if you have a particle that has a mass between T max and T re-eating, then if you do instantaneous decay approximation, you will not see the production of this particle, but in reality you have production of this particle. So let me make it clear. The energy density of radiation here is much, much greater than what it will be here. But here there is still the inflaton around. So the thermal energy here is much greater than the final re-eating energy, but there is still the inflaton around and this needs to be taken into account. So imagine that you have, before I show you the plot, let me tell you another possible way in which this could be relevant. So what we are saying, right? We are saying that here there is the inflaton field and the inflaton decays into radiation and then there is a particle x that gets produced out of the thermal bath. This is the decay of the inflaton and here you could imagine that you have a particle that is produced in this way, for example. Gamma, gamma goes in x, x and you could imagine that the cross-section for producing this particle is temperature to the power of n. In the case of gravitinos, the cross-section was 1 over n-plank square. So for gravitino n is equal to zero. But imagine, for example, that you have a different particle for which n is greater than zero. The greater n is, the more you are sensitive to the temperature. And if you have a theory which has n greater or equal to 6, which you need to have special models, but you could imagine that, then in this case, most of the production takes place here. So if n is smaller than 6, the dilution wins and most of the production takes place here. If n instead is greater than 6, you will have production here. Let me... This will be the homework. I will give you some hint to solve the homework. But let me just show you the plot of this result. And if I forgot how to do it, just a second, ajuta, zoom, OK? So this is one model, one calculation in one given model. I will not enter into the details of the model. This is essentially time and this is the abundance of gravitinos. Why is the number of gravitinos divided by number of photos normalized to the value that you have under the instantaneous decay approximation? And you can see that this is the right thing that you will need to do. The black one is really what takes place. And however, the orange one is what you get in the instantaneous decay approximation. And you see that eventually the result becomes the same. This is another maybe more complicated way to show that what you produce here essentially gets diluted by eventual production of radiation from the inflaton. Here instead is a different example in which there is a model where I have a cross section for the production of the particle x. There goes a temperature to the 6. It's a model where I exchange a mediator of spin 2, a heavy mediator. Then you will see that the black curve is the production in the real case and the orange curve is the approximation. And you see that there is indeed dilution. You produce a lot of this particle here, then this gets diluted. But eventually the final result is about an order of magnitude greater than the one in the instantaneous decay approximation. So in a sense, the message I want to give you from all this discussion is that this is what really goes on, at least provided that this is the right picture of the world. But eventually, depending on what processes you are really interested in, this may or may not be relevant. Is it clear? Questions? Is something being created that is indicated to gravitinos? Or do we have some other gravitinos that help? No, in this case, you produce a lot of gravitinos. That's what I've discussed so far. You are complicating it more than what I've done so far. Here you produce gravitinos from the thermal butt. So these are the gravitinos that are produced here. And you see that in the instantaneous decay approximation, instead you are producing zero gravitinos. So at these early times, there is a lot more gravitinos than in the instantaneous decay approximation. But eventually, this gets diluted, because eventually the instantaneous decay calculation catches up. So the second case is a different model, in which there is a particle, x, that gets produced in a cross-section that depends to the sixth power of the temperature. If you're interested after this lecture, I can tell you what the model is, but the model is just one example. I just want to focus here on this temperature to the sixth dependence. For this model, I produce a lot, and eventually the final result is greater than what I would have in instantaneous decay approximation. The y-axis is the same quantity. It is the ratio between the abundance divided by the naive abundance. Yes? Well, I see the dilution, because you see the problem is that here we decided to plot this ratio, which now I realize maybe is not the most convenient thing to plot. But the crucial point is to understand that eventually this thing become together. Oh, the second one, again, here, again, you see the dilution. Here you see, if there was no dilution, you wouldn't say, okay, you see I produce ten order of magnitude more than the instantaneous decay. But there is dilution, but the dilution is not erasing everything, and so there is some leftover, because there is a bigger dependence on the temperature. So if there is bigger dependence on the temperature, you are more sensitive to what takes place here. It's true that we still, there will be some dilution, but it's more strong this effect here. It's just mathematical calculation, and once you solve the differential equation, you will see this. Because the inflaton keeps producing photons. You see, you have the thermal bath here, which is huge, but there is the inflaton, which is even huger, and the inflaton keeps producing photons. So what you produce here gets diluted. Does it get diluted a lot or a little? Depends on what is the coefficient here. And I'm telling you that if you have n equals 6 or greater, then this effect becomes important. Marco, there is a, just somebody is asking, if you can put the reference to reproduce this calculation. I will put the reference, these calculations were done in a paper of mine, and I will add the reference, and then I will post it. Is the cut-off scale of this model? As I say, I am willing to discuss the model in private. After all, right now, this will be time wasted for this lecture, because it's just an example, not really the main focus of the lecture. If you care about the model, I'll be super happy to talk later. OK, how much time do I have? 15? OK, fantastic. One more question. Ah, OK, very good. The question is that the production of X particle reduces n gamma. Yes, but we are discussing cases in which the production is always very limited. For example, remember that the gravitinos here was 10 to the minus 10. So essentially I extract a little energy from the thermal butt, but I can neglect it in these considerations. I want to show now a different type. Now I will move to these final 15 minutes, and tomorrow lectures will be from the computer, so I can show you a bit more things. I'll try to go slowly, but I think I'm done with more introductory stuff, and I just want to discuss some example where it's easier for me to use the computer. So what I want to discuss, I want to discuss a completely different way of having a reading, which is called non-perturbative reading. So far, when we computed this decay rate, this gamma that was entering in this calculation, this was the perturbative decay rate. This is the rate that I could imagine if I have a particle in Minkowski and just diluted that just decays. However, the inflaton is not really a collection of independent quanta. The inflaton is more a condensate that keeps oscillating. And the fact that the inflaton is a condensate that is oscillating should have some effects that go beyond the perturbative decay that I was telling you. When you do Feynman diagram to do the decay of a particle, you treat the particle as an isolated quantum that just decays. But now the inflaton has equivalent oscillations, and in this paper you see now very old papers, people realize that this can actually be important. And I'm gonna show you an example taken from Leti's simulation, which is probably the simplest example. You see there is a quadratic inflaton potential. Maybe this is true at the minimum. I don't know what the inflaton potential is during inflation, but here I care about the potential during reading, so there will be massive inflaton potential. And then there is another scalar particle, chi, that represents matter. Of course, this is a toy example, I just do calculation. But this represents a particle, chi, that is produced by the inflaton. And what I will show you, I will show you some Leti simulation. There are some cause available on the market. This was the first code, which is the simplest one, it's the only one that I can use. So that's why I put this one there. And in the movie, we will see a slicing. So what we will see? We will see x, y plane. So this is z suppressed. And on the vertical axis, we will see either the value of the inflaton or of the field chi. So we will see that the inflaton is oscillating, initially. And then during the very quickly, you will see that the particle chi gets excited. And then you will see that also the inflaton will get excited. So now this is the movie. So on the left, you see the inflaton. This is the first half oscillation, one complete oscillation. And you see, just after already the first oscillation, you already start producing the field chi. And you see that the field chi, this is the fragmentation of chi, this is just quanta. So this is particle production of the field chi. And then you see that the field chi produces quanta of the inflaton field. So what we see here is an extremely quick production of particles that takes place not after 10 to the 30 times. It takes place right after at the end of inflation. Due to this coherent effect, that was discussed, that I tried to argue a second ago. Maybe let me finish 10 minutes and then I can collect the final questions. So the lattice simulation that I showed you have essentially, let me show it again, the first stage, this production of quanta of chi can be done from linearized calculation. And this can be done analytically. It goes under the name of parametric resonance. And I can discuss in private how to set up the equation to do it. But then you see you end up in a regime which is completely nonlinear. What you see here is highly nonlinear, it's not small perturbations. And these can be understood only from the lattice. And we can try to make a sense from what we see from the lattice simulation. The first period can be done analytically. And essentially the basic idea is that you have a mass for the field chi that depends on the inflaton field. You see the mass of the inflaton is m, is a constant parameter, so the inflaton is oscillating with frequency m. And instead the mass for chi is the value of the inflaton field. And the inflaton field, remember, is a huge vev. In this model here is essentially comparable to Planck mass at the end of inflation. So the inflaton is having a spanning a huge amount of value. So the mass of the field chi changes by a huge amount. And its effective frequency changes, is evolving in that way. There is, yesterday there was a nice talk and it was mentioning also this question here, one of the students talk. Essentially you can show analytically that when this quantity here essentially is greater than one, then you have a lot of particle production. And what is beautiful, you can see it here, here is the analytical calculation of the growth, the occupation number of the field chi. You can see that the production happens at step. Why the production happens at step? Because remember that phi is the mass of the inflaton. So when there is the mass of chi, when phi is very away from the origin, then chi is very massive and is not produced. But whenever phi swings to the origin, remember phi is essentially oscillating about this potential, so phi keeps moving, whenever phi swings through the origin, here m chi is equal to zero at this moment here. And this is the moment in which chi is effectively massless and you produce a lot of it. And so you can see that the production takes place in steps, which is exactly the moment in which phi crosses the origin. There are two effects here. The first effect is that whenever phi is equal to zero, you can indeed have a great omega. The second effect is that this is not just once, this is a periodic production, and so you can get some resonance effect. And what we see here is indeed the effect of the resonance. This is the first stage that can be done analytically. After that, we can only follow the lattice simulations. And this would be the spectra for the inflaton field and for the field chi as a function of time. This is like rainbow colors. So you see that the first field that gets produced is the field chi. But then you saw that the field chi backreacts and rescutters and destroys the coherency of the inflaton field. You saw from the movie chi goes produced first, but then you see chi is completely homogeneous, so it backreacts on the inflaton that becomes homogeneous, and then the process stops, but you end up with these distributions. These distributions are very far from thermal, and nobody has been able to compute the thermalization in this type of models. People argue what it could be, but there is no full calculation of thermalization, because the lattice after some point breaks and it leaves you a state which is very far from thermal. These, I'm sorry, the colors are not very good, but this shows the equation of state of this plasma, which is neither matter nor radiation. Remember, so this is evolution of time, this is the equation of state of the totality of this plasma. Remember yesterday we talked about effective equation of state during reading? Remember, matter is zero, radiation is one-third, and you see that you get this plasma that at this moment has an equation of state that it is in between. So you see the complication of all this business. If you want, I have not prepared the questions here, but at least the first stage can be understood analytically, and if you want, after the lecture, I'd be happy to show you if anyone is interested about the analytical calculation. OK, I will stop here and I take questions if there is time, or we go to lunch and we do the questions after lunch. Yeah, there is one question from the room. So since the radiation perfect fluid can be modeled, I think coming back to these equations on the board, since the radiation perfect fluid can be modeled by the Lagrangian of the standard electromagnetic field, can we model re-eating with some specific coupling of the inflaton field with the electromagnetic one? We can, certainly if we know what the inflaton is, we can put it in a Lagrangian and then we can work out the equations out of that Lagrangian. Here, I try to be, you see, as model independent as possible, so here I'm ignoring the microscopic description of the model, but if we have a concrete model, even better, for the point of view of doing calculations. Yes. OK, let's thank Marco for the lecture. OK, thank you.