 I come around Wafa. Only the ones who are attending online need to register. Those who are present do not need to register for it. Registration is only because of using, because those who are connecting via Zoom, they need to be able to connect to the Zoom room. OK, so now I guess we can start. Please, Mark. Good morning. So let me continue from where I stopped yesterday. One small thing I left to say was about the statistics of perturbations. And we know they are very close to Gaussian. It is also a prediction of slow roll inflation in the sense that imagine you have a free field, and which means Ray doesn't interact at all. And then imagine you want to compute the three-point function, which is not being recorded. It was all false, what I said. Giving lectures used to be much easier. Now, even when I teach at my university, I'm always afraid that I forget to click to register. No, no, no, no. Anyway, we're recording with three devices. Shall we start again? Thank you. So we were saying that imagine you want to compute the three-point function. And you know that if you have a free field, this will vanish. Just imagine making quantum operator, right? Annihilation creation operators, and this clearly gives zero. So if these are free fields. And then you can imagine, OK, let me assume that the inflaton is not free because it's interacting with itself. And so how do you encode the interaction of the inflaton? Well, simply you put them in the potential, right? So you can tailor expand the potential. So this would be something like this. Let me tailor expand it around the origin, assuming that I have a minimum in the origin where it vanishes. And then however you see that the interactions are proportional to the derivative of the potentials. And the lower all parameters being small, remember, constrain the derivative of the potential to be small. And so essentially, also, non-gaussianity is going to be small just because these are small self-interactions. Of course, we could consider interactions with other fields. This could enhance the non-gaussianity. And that's something that we will discuss on last lecture. OK, so yesterday I made some claim about perturbations. I wanted to make a quick calculation to support or to justify some of these claims. But what I'm going to do is really a toy model. And it is a test massless field in the sitter. So this is really a toy model for the perturbations. So I'm not going to perturb the metric. And by mean test, I mean that I neglect the energy of this field from the point of view of the expansion of the universe. So you know that when you give the line element for the sitter, we can use the physical time. And in this case, A is equal E to the HT, essentially. So this is how you do it. But you can also use conformal time. You can do many things, but this will be two coordinates that we will discuss. In this case, the scale factor, we write it minus 1 over h tau. And then inflation, the sitter, sorry, is done in this portion here. So tau goes to 0 minus during inflation. And the scale factor grows like this. So why we are using conformal time? Because it's easier to do calculations for fields. So because we will write s, the action of the field, d4x minus g, g mu nu, d mu phi, d nu phi. Massless field, so I didn't add any potential. And you see that's what I'm writing for the action of the field. And now you see that from the square root of the determinant, you get the scale factor to the fourth power. And then from here, you get scale factor to the minus 2. So this will give a half d4x a squared eta mu nu d mu phi d nu phi. If I go to momentum space, this will be a half dT d3k. Then I have a square phi dot square plus k square phi square. Then what I can do, this would be the Fourier modes of the field. And there is an a square also here, of course. Now we like to have canonically normalized field, because you see this is not the canonical normalization. So we are going to define chi a phi. This is canonically normalized. And the equation that we get for chi, that's essentially what I want to discuss, is this a question here? Chi and here, sorry, let me use a consistent notation. Derivative with respect to conformal time. People usually call it prime, so let me just put a prime. So the question of motion for the field chi is going to be the following one. And you can see, right, that this term in reality is 2a squared h squared. And now we can see that there are two regimes. You see from the equation. Remember, yesterday I was telling you that the field is initially inside the horizon, then it goes outside the horizon. And you can see from the equation, because inside the horizon means that the scale factor initially is very small. So this term initially is subdominant. And what we get, we get the same equation as a simple harmonic oscillator in Minkowski. This justifies what I was telling you yesterday, right? And we know what it is the solution here. We take the solution, which is the normalized one in Minkowski, e to the minus i kt divided by square root of 2k. And then when you go out of the horizon, you get this different term. So you see that there is really a different behavior. And you see that this is a minus sign, which means that you are going to amplify the mode. That's the amplification due to the expansion of the universe. And the solution, which is the one that gets the right initial condition, is chi is equal 1 minus i k tau divided by square root of 2k e to the minus i k tau. Remember that the time as I was doing before. So this is tau. This is a. This is the scale factor. So early times means tau goes to minus infinite. So if tau goes to minus infinite, the numerator here becomes 1. And you see the usual Minkowski-like normalization. Instead, late time. So if I now do look at this late time, it means that I take tau goes to 0. And so you see now that it's different. You get minus i square root of 2k to the 3 half times tau. So that's what the mode does when tau goes to 0. The phase vanishes. And this term becomes the dominant one. Now, remember that the original field was phi. So phi is chi over a, which is minus h tau chi. Because that's what the scale factor is. And now if I put everything together, I get i square root of 2k to the 3 half h. And particularly what is, we see that indeed the field becomes constant. This is the frozen field that I was telling you yesterday. You see there's no more dependence on time. So there is the oscillation during the inside the horizon. And then the field freezes to a constant value. Now we can ask what is the power of this field. And to get the power, I'm going to do it shorter. But essentially, we can just take the average of the square of the field. Because this is a perturbation. So the average of the field itself will be 0. But here, if you go to Fourier space, you do some algebra, you're going to find what I also wrote yesterday. This is k cubed divided by 2 pi square phi squared. And this is what we call the power spectrum of phi as a function of k. So now let's put these together. So you see that when you take phi squared, you get 1 over k cubed. That exactly counts as with this k cubed. And what we end up with is h over 2 pi squared. So you see that this is a scaling variant and constant. So these are the properties that I was telling you yesterday. Yesterday we were arguing, right? But now we see it from the calculation. So now in reality, we know that inflation is not the sitter. So for example, h decreases during inflation. So this will tell me that the power decreases. So when I do evaluate this quantity, if h decreases, I should evaluate it at horizon crossing. So most that cross the horizon at a different time, they will have different power. But in reality, as I told you, this is not the full calculation. When you do the full calculation, you need to add also the metal perturbation. And you can find very nice reviews, which I will not have time here to cover. But when you add also the metal perturbation, the calculation that I did becomes a bit more complicated. And what one ends up ends up with the things that I was telling you yesterday. So now we would add delta phi, delta g mu nu. We would include the delta phi in the Einstein equation. So this will not be any more a test field. And at the end of the day, you find what I was telling you yesterday. The spectral tilt is ns 1 minus 6 epsilon plus 2 eta. Remember, epsilon is m planck square half v prime over v square eta is m planck square v prime prime over v. And remember that this quantity are much smaller than 1. And r, so the tensor to scalar ratio, are 16 epsilon. Remember, we defined essentially p zeta goes as lambda 1 minus ns and r p gravitational wave divided by p zeta. So just to summarize, this calculation was done for a test field in the center. And the purpose was to show you the transition between the sub-horizon regime to the super-horizon regime. And also to show you concretely that indeed the field freezes. And the power, you see, becomes scaling variant. Now, this is the calculation that we did. And now I'm making some statements without fully proving them that when you add also the metric perturbation and when you also include the relation between them in the Einstein equations, then you'll find some more general properties. You will find that the spectral tilt, the scale dependence of the power, is given by the Zohl-Roll parameters and also the tensor to scalar ratio is given by the Zohl-Roll parameters. Now, during inflation, we use the number of e-folds as a measure of time as well. And we define them n to be the logarithm of the scale factor at the end of inflation, divided by the scale factor with some function of time. So during inflation, a is, of course, smaller than the final result. And so n is greater than 0, because you do the logarithm of something greater than 1. At the end of inflation, a is equal to a n. Therefore, n is equal to 0. So n decreases during inflation. So n is a parameter that is decreasing. And when n is equal to 0, it means the end of inflation. So now, imagine that I give you all these. And I ask you, OK, now tell me the phenomenology for a given potential of inflation. For example, I'm asking you, OK, let's take this model here. Let me copy the relations upstairs. So ns 1 minus 6 epsilon plus 2 eta r 16 epsilon. And then I tell you, OK, here is the model of inflation. Let me take the simplest. This model is now ruled out. But let me just discuss this, because it's the simpler one to do calculations. I just tell you, I have a massive inflaton field with quadratic potential. Tell me what is the value of ns for this model and what is the value of r for this model. So what you do right first of all, you draw it so that you get a feeling what's going on. And now you see, naively, it seems that you say, oh, but the potential becomes steeper when I go higher up. It doesn't look flat. But remember that these are the flatness parameters that you need to evaluate. So for example, if you compute epsilon, this is equal m-plank squared divided by 2. Then here you have the derivative of the potential, so m squared phi divided by the potential itself. So you see this gets 2 m-plank squared divided by phi squared. So you see that for this potential, the greater phi is, the more flat indeed the potential is. It doesn't because flatness, remember, you need also to take this into account. If you take eta, now you do the second derivative. This is m squared phi squared. No, m squared, that's it. A half m squared phi squared, you get the same result. So you see here that the inflation ends. If here is phi, here is v, inflation essentially ends when phi is the square root of 2 m-plank. So phi comes from here, goes down. And when phi reaches square root of 2 m-plank, that's the moment that inflation ends, because the slow-roll parameters are no longer smaller than 1. Now, but what you need to do, essentially, you need to evaluate these relations at some given moment during inflation. Because this is the end of inflation. So this is the value of phi at the end of inflation. But the value of phi when the C and B modes were produced was higher. So this is the value phi C and B. And the question is, what should we put here for this value phi C and B? The first thing that we can do, we can solve the equations of motion. And this will be left to you as homework. So the first thing that you do, you need to find phi as a function of n. And this will be a homework that I will post later. I will guide you. You just look at the equation of motion for the scalar field, and you will be able to find phi as a function of n. For this potential, what you find, phi nearly equal 4n plus 2n-plank. And in the homework, I will guide you to solve this. It's not very hard. When you see n is equal to 0, indeed, you find that phi is square root of 2n-plank. But when n is greater than 0, phi is bigger. So this tells me, the farther away phi is, the bigger is the number of e-foil. It means the earlier I am during inflation. And now, if you just now take this number and you put it here in the formula for epsilon and for eta, and then you put this formula here, you will get a phenomenological prediction for this model. You have phi as a function of n. So you can compute the slow-roll parameters as a function of number of e-foils. And so you can compute these two quantities as a function of e-foils. And what we get, we get the prediction ns is 1 minus 2 over n. And r is 8 over n. So now, you see that essentially the question still remains in the sense, which value of n should I use? These are the predictions, but the question is, what? So is it clear? We use n as a measure of time. So we give the value of the inflaton as a function of number of e-foils. And so we can do the phenomenological prediction as a function of the number of e-foils. If you change the potential, you can just use the same machinery. You will just get some different functions here. But once you solve the homework, you will see that you have a tool to do it for any potential that you want. But now the question remains, and maybe you have heard that n is equal to 60, but why, right? So that's what we have to discuss now. Do you have any questions so far? Yes, please. I see that you just have the same scale as one scale. Should we be concerned about correction of the theory? Yeah, very much so. And in fact, there's a lot of problems in these models. Because, for example, if you try to embed them in string theory, it is really not simple. And even if you just try to have a consistent theory of quantum gravity, you don't know whether you can trust your potential for these field values. There's a huge literature about it. So indeed, it's a concern that people address. What do you mean by acceptable? No, this model has been ruled out, but the latest CMB data. But it's not by much, but this model by itself is ruled out. Very important. When I told you, I was making delta phi. These for tau went to minus infinity. This was e to the minus ik tau divided by square root of 2k. The literature, this is called the Bunch Davis vacuum. It essentially is the same vacuum that you would write in Minkowski. This is the least energy vacuum. And so that's what people typically assume. But for example, you could have a model where you would say, no, my vacuum initially is populated. So my vacuum is different from that. And then the results will be different. And people have also studied this as well. But let me just say that this is the simplest choice, and it fits the data. So that's why this is probably the most popular one. Arbitrary in the extent that is the simplest choice, and it fits the data. So always in science, if you have something, you can make it more complicated. The question is whether you should or you should not. But indeed, there is a huge literature on this. Yes, please. Very good. Think about where we're at now. Two points. There is a phase that I didn't discuss here about quantum diffusion. And if phi is very, very large, just the addition of quantum perturbations will become more important than the classical motion. But this happens far, far away. So it doesn't concern us. So our question is phenomenology. And that's what I'm going to address now. What is the value of n? Because you see, when it starts, it's a badly posed question in the phenomenological context. Because essentially, most probably, you cannot see experimentally when it starts. So maybe the question we need to ask is the scale that we see in the CMB. When were they produced? Were it produced when the inflator was here, here, here, here? And in a sense, that's exactly what this question is about. What value of n should I put when I compare with CMB experiment? That's what we're going to discuss now. Marco? Yes? Some questions in Zoom about m, little m. Can you explain again what is little m? Little m? Yeah. Little m is just a parameter. It's the mass of the inflatum field, right? So imagine I have a generic potential. And if I have a generic potential, I can ask, what is the mass of the field from the potential? By definition, m squared is a half the second derivative of the potential. Evaluate it in 0 if I have a minimum in 0. So if this is the definition of a mass, if I have a potential which is only the quadratic term, that's why I'm going to write it. And then there may be other terms, but this is the mass term. Now, if you look for this model, if you just look at the normalization of the perturbations, m is of the order 10 to the 13 gv. But for our discussion, m is just a constant parameter. So OK, we understood that it's very important to know the value of n to do any phenomenology. And now the question that comes about is which value of n should I choose when I compare with the CMB data? In order to answer this question, I must do a little regression. And I need to speak about reheating after inflation because we are going to see these two are very much connected. So during inflation, volume goes as essentially a cube. This is how much the volume grows. So this will be e to the number of e-folds to the cube. So e to the 3n. And we are going to see that typically we expect n, we will see, n could be order 50 to 60. So during inflation, you see that you devolve the space grows by a lot. So if there are particles at the onset of inflation, these particles just get diluted away and you don't see them anymore. So everything that is left after inflation, essentially when this simplest version is the inflaton field, so everything you see and also everything that you don't see was created by the decay of the inflaton. Maybe not directly by the decay of the inflaton, but maybe the inflaton decays in matter, let's say, where matter has just been standard model fields, or maybe decays in something as any decays to matter. We don't know, but essentially everything that we see in this scenario needs to start from the decay of the inflaton field. So this is an extremely important process, right? And we have essentially reheating. I was mentioning yesterday, but let me say it again. For me, it's a very broad word, and it is all processes from decay of the inflaton to establishment of the hot thermal bath of Big Bang theory. So right naively, if you think about the Big Bang, that if you ask everyone on the street, either they say what? Or they will say that the temperature was infinite at early time. But this is in the context of inflation. This is an extrapolation. We will say, no, at some point the temperature was created because the inflaton decays, right? That's the basic idea. But from here on, this would be the standard Big Bang theory. And here instead is reheating. So that's what we need to discuss. There are some requirements, not very stringent, but nonetheless, one is the temperature. So it should be greater than about 2 MeV because you need a Big Bang nucleosynthesis. And also then you need to have a cold dark matter. Let's say dark matter. And then you need a barionosymmetry. So these are three things that need to be taken into account during your reheating process. Now, why do we speak about the reheating? Well, reheating is the most unknown cosmological period. As I told you, essentially if I do an hour of time, right? This is one second. This is when BBN, Big Bang nucleosynthesis, starts. And then we know very well the history, right? We have radiation domination. Then we have matter domination. We know what the equality time is. Then today we have lambda domination. And then before, right, we have inflation. And we can at least test some part of the inflation through the CMB. But then, reheating, we really have very little, if known, direct observational reach to it. And so this is certainly the most unknown part. And what there is uncertainty associated to that. And as we're going to see, this uncertainty reflect on the value of N that is the number of efforts. So let me precisely discuss this. So what I want to emphasize now is that there are uncertainties in reheating. I want to show that this reflect into uncertainty in inflationary predictions. Even if I know the model of inflation, suppose really I knew that this is the right model of inflation. Still, I would have some problems in making predictions due to the fact that there is this uncertainty in reheating. That's what we're going to discuss now. So you know that Planck gives the most accurate CMB measurement on its own. So Planck pivot scale is chosen to be 0.05 megaparsec to the minus 1. So Planck gives a result. You open the abstract of Planck, and Planck tells you millions of numbers. For example, what is the spectral tilt? What are the limits on tensor to scalar ratio? But then these numbers depend on the scale, as we discussed. And Planck typically gives these numbers at these pivot scales. And so the question that we ask is at which value of n, at which if n was this scale produced? That's the question. So let me do a cartoon to visualize what we are saying. So here we have the scales. Here we have time. And now I need to put a few lines. This is what I call the time tk. Here is what I call the time tn. Here is what I will call the time reheating. And here is what I call today. So this is today. This is an end of inflation. This is end of reheating. And here is when the model left the horizon. So let me just do inflation. During inflation, this would be h to the minus 1. Remember, h grows very slightly, but it's essentially very close to constant. And then here there will be some power law. And here instead is what physical scales do. And this is the Planck pivot scale. So this is the wavelength. That's what it did. So the scale has re-entered in our sky. I'm neglecting, here I should take a small amount of energy density, our dark energy domination here. But I am neglecting this part. And that's essentially the calculation that I need to do. And the number of e falls this year. So what I need to say, I need to find, essentially, given the value of lambda, I need to find. And given that today I observe the scale here, I need to trace it back. And I need to find what is the value of n. So here is the heating stage. And here is radiation domination, matter domination. And there will be also lambda domination for a little bit. So what are the unknown parameters that I would like to know in order to really do this plot? So there are some unknown inflationary parameters. One is rho k, which is equal nearly the inflaton potential. This is the energy density at horizon crossing. Would tell me the energy of the universe at this moment here. Since the inflaton is going very slowly, I can neglect the kinetic energy of the inflaton. Then another parameter that I don't know is the energy density at the end of inflation. This is no longer equal to the potential of the inflaton, because at the end of the inflation the inflaton is moving faster. And these are instead the unknown reading parameters. We is essentially a rho reading. So is the energy density at the end of reading. And also I don't know the equation of state. So let me say this is an effective equation of state during reading. Remember, we discussed yesterday the equation of state. What we say, for example, if we have radiation, this is equal 1 third. If you have matter, the equation of state is equal to 0. If you have a cosmological constant, the equation of state is minus 1. But what was the equation of state during reading? And why should it be constant? It probably was not even constant. But we could imagine an average equation of state. So if I want to consider an average equation of state, what was the nature of the dominant source during reading? And I will discuss different possibilities. So now I need to have horizon crossing in this plot here. Horizon crossing is defined to be when lambda divided by 2 pi, which is a over k. Remember, this would be the usual relation that we have for Minkowski between the wavelength and the wave number. But now lambda grows as the scale factor. And when this is equal h to the minus 1, this is the moment. You see there is one horizon crossing here. And then, of course, there will be the horizon crossing today when it re-enters. And then remember the number of e-folds that we just defined was this one here, a and e to the minus n. So the scale factor was smaller, and this was, therefore, positive. Now, you can show. And this will be another homework. n is about 55.6 plus 2 logarithm vk to the half divided by 10 to the 16 gv plus logarithm 10 to the 16 gv rho n to the 1 fourth plus 1 minus 3w 12 1 plus w logarithm rho reading rho n. So you see, if you know the evolution of these, here you know the evolution. It depends on the inflationary model. Here you know the evolution if you know the equation of state. And here you certainly know the evolution. So if you know the equation of state here, you will know the slopes. Then, if you know these times, in order to know these times, you will need to know what was the energy at that given amount of time. Then you can redo this plot. And so you can take a given scale on your sky. This will be the length that you measure in your sky. And you can trace it back to this moment, and you can compute the number of e-folds. That's what this relation is telling us, that if you know the scale, this is true for the pivot scale. And this seems a bit mysterious, but if you do the homework, I'll try to guide you. It is not particularly hard to see this. Essentially, it really means that we just need to take this number here, and we need to trace it back through this history until we find this value here. And you can see here, there is this magical number that is almost 60. That is essentially what justifies our choice. You need to just do the algebra, and it's not particularly difficult in a sense. But what I want to discuss with you, you see, imagine that you see there are the first line. There is these terms that are uncertainty in the inflaton model, because we know the scale of the inflaton at horizon crossing and the scale of the inflaton at the end. Instead, the last line is a term that encodes the physics of reheating. This is the equation of state-during reheating. And this is the energy density at the end of reheating. And so this is uncertainty due to the fact that we don't know exactly what went on through reheating. And so I want to now evaluate this term, because imagine that this term is always 0.5. I don't know what it is, but I don't care. If I see that this term would be 0.5, who cares? I don't know what it is, but essentially, this doesn't really play any role. But imagine that instead this term is 10 or 15. Then it plays a big role, because out of 60, out of 55, you have an uncertainty, which is a big number. So now I want to understand what this term can be and whether it is important, whether it's something that I can forget about it or whether it's something that I need to keep on the back of my mind. And so effective, I mean that imagine that the reheating, there was maybe a species that dominates and is massive. So at that moment, W is equal to 0. Then the species decay, then W becomes 1 third. Then another massive species take over. W is equal to 0. And then you would compute an average. So essentially, it's an effective number. Just tell me what is the average behavior during the reading. This is, you see, since I don't really know anything, this is the simplest thing I can try to do. At least I will learn what was the average thing going on at that moment. And I want to discuss with you two extreme cases and what is this value here for the quickest possible reading and for this, OK, maybe I should keep this. I should evaluate this term for the quickest possible reading and for the slowest possible reading. Let me call this term delta n is the number of e-folds that is associated to the physics of reading. Now, what is the quickest possible thing that you can imagine? The inflaton decays immediately. And therefore, essentially, rho n is equal to rho reading. And so this term is equal to 0. So if the immediate decay, this gives me delta n is equal to 0. Essentially, it's obvious, right? The immediate reading means that this time coincides with this time. Inflaton decays right away. Then these times will be the same time. And therefore, there is delta n is equal to 0. So delta n is how much you accumulate during this period. OK? Now let's instead discuss what is the slowest possible decay. And let me assume, right, that there was a cartoon that I already drew about the potential of the inflaton. So this is phi v. And the inflaton at reading is oscillating about the minimum of the potential. So heuristically, we can imagine that the inflaton does some oscillations if I add these t and phi. And maybe the amplitude of the oscillation is decreasing due to the expansion of the universe. This is the end of inflation, right? The inflaton arrives here. If the inflaton decays right away, OK, that's done. Delta n is equal to 0. But if the inflaton stays around for a long time, the inflaton will keep oscillating. It could do billions, or I don't know. A lot of it, we will see how much. It can do a lot of oscillations. And it does not decay. It's just there that it keeps oscillating. And its amplitude will, the oscillations will decrease because of the expansion of the universe. But this essentially will just go on for a long amount of time. And the equations that govern these are these equations here. And it's essentially, let's take just a massive inflaton. So let's just take the potential for a massive inflaton field. This would be m squared phi is equal to 0. This will be 1 half m squared phi squared. And so just if you look at the first equation, you recognize almost the equation of the simple harmonic oscillator, right? If there was not this term, this would be the equation of the simple harmonic oscillator. But then you see that you have a friction term. Remember, in the simple harmonic oscillator, a friction term would be something that is acceleration is equal minus gamma velocity, right? This would be a friction term. I teach mechanics at a steer, so I remember about this. But what is the point here that this gamma depends on the inflaton itself? So it's a bit more complicated system because, you see, the friction is proportional to the above rate. And the above rate depends on the value of the inflaton. So it's a system which is a bit more complicated. It's not just constant friction. However, you know, since we all have done physics one, that we expect damped oscillations, right? Oscillations that are decreasing with time. And now we can ask ourselves, OK, how does it really decrease? And in order to do how it really decreases, we can just do the virial theorem. We can apply the virial theorem. Neglecting the decrease of the amplitude during a few oscillations. Because the amplitude is decreasing with the time scale h to the minus 1. And h at this point becomes much smaller than m because phi is smaller than Planck mass. So the above rate becomes much smaller than the oscillation rate. So the inflaton oscillates with a time scale m to the minus 1. And the amplitude decreases with a time scale h to the minus 1. So the amplitude decreases adiabatically. It decreases very, very slowly. So during a few oscillations, we can neglect the decrease of the amplitude. And we can use the virial theorem. And remember that the virial theorem tells you that the average of the kinetic energy is equal to the average of the potential energy. This will be for the simple harmonic oscillator, for example. So if you have saw, remember that the pressure, what is the average of the pressure, the pressure is exactly 1 half phi dot square minus v, which is a kinetic minus potential. And this is equal to 0. So in this model, during oscillations, w effective is equal to 0. Now the question of state keeps varying during the oscillations. And whenever the inflaton is at a minimum, the equation of state is like the one of a vacuum. And then it's the one which is dominated by kinetic energy. So the equation of state here is w is equal to plus 1. And here it is like if it was a vacuum. So w is equal to minus 1. The equation of state of the inflaton keeps oscillating between minus 1 and plus 1. On an average, it's equal to 0 due to the virial theorem. So essentially, the inflaton behaves as matter. So now you see the two different scenarios. First of all, the first immediate inflaton decay. So you immediately produce the thermal bath. Delta n is equal to 0. And then there is this other decay that is extremely slow. And for a lot of time, the equation of state of the universe is w is equal to 0. So now let's evaluate this term, assuming that now we understand that w is equal to 0. So we have delta n is essentially 1 over 12 logarithm rho reading divided by rho n. And now let's take the slowest possible decay. This thing could be of the order of MeV. I told you that you just need to have a thermal temperature. It's sufficiently to have the BBN thermal bath. And this could be as high as 10 to the 16 GV. These are energy density, so you need to put to the full power. You see, this is really the slowest possible ever decay of the inflaton field. And this ratio, if you compute it, it will give the number minus 15. So essentially, it's delta n. We don't know what it is. But it's something between minus 15 and 0. So now you see that having a term that is delta n, which is something between 0 and minus 15, it's huge compared to 55. And if we now go to even the plank paper, so let me see if I can show it. I'm sure we are all familiar with this. But nonetheless, this was the plank paper, the latest, I guess, plank paper. And you can see that these are the same predictions that I was telling you for the primordial tilt and the tensor to scalar ratio. And the ellipses or the half ellipses will be from the CMB data. And now you see that they studied a lot of different models. But you see that one inflaton model is not a point. And here they evaluate the phenomenological predictions for n that goes between 50 and 60. So they take something. They say, OK, you see 55 is the central value, in a sense. And then we think reasonable to have an uncertainty of order 10. And so you see that's what we get. So imagine this was the model that we discussed a moment ago. You see now it's ruled out, m squared phi squared. Yes, yes, yes, yes. OK, so no, this is the point. OK. And you can see that every given model is not given as a point, but it is given as a line because of these uncertainties during reading. And you see that it's sizeable. So I think I justified why we take delta n is equal to 60. And then, however, I think I try to make the point that there is uncertainty associated to this. How much time do I have, 10 minutes, still? I think I started a little later. So OK, I'm going to finish one more thing that I want to discuss. Sorry, maybe I should stop share. I want to now start a little more to discuss about reheating, because I think I proved the relevance. It's a question about VK, maybe not. Before you erase, they can repeat what was VK. VK is the potential of the inflaton when the model left the horizon. So we are going to discuss reheating, making initially some very simple assumptions. And then we are going to simply make it more realistic by dropping the assumptions. So the first assumption I want to make is instantaneous inflaton decay and instantaneous thermalization. By instantaneous, I don't mean exactly at the end of inflation. I mean there will be some decay rate. So there will be a moment when the inflaton decays. So let me just compute what in this context will be the reheating temperature. Now, when you have a process in the time t, which is of the order inverse h, the number of processes taking place integral gamma dt, which is of the order of gamma over h. Because delta t is of the order of 1 over h. So gamma is the rate for a process. Notice that gamma has mass dimension plus 1. Because you see this is number of processes. So this is dimensionless quantity. And this is mass dimension minus 1 in natural units. So gamma is the dimension of an energy. This is the decay rate of the inflaton field. And the number of processes that take place is the integral of gamma in one upper time. This is the amount of processes that take place in that time, which is parametrically gamma divided by h. So you see, by looking at this relation here, we can really say the decay happens when gamma is equal h. Typically, h is a function of time. h initially is greater than gamma. At some point, h decreases. At the moment when gamma is equal h, we say this is the decay of the inflaton. And here, we are going to say that the inflaton decay is instantaneously, immediately at that time. So now, we can also find the temperature. Because if the inflaton decays and you have a thermal bat, because we are saying that there is instantaneous thermalization of the decay process of the inflaton, we are just going to write gamma. Let's do gamma squared. h squared is equal to rho divided by 3m Planck squared. And this will be the rho of the thermal bat. So the idea is that, at some point, when gamma is equal h, all the inflaton decays. This is just a very sudden approximation. And the decay products thermalize immediately. If you do so, then this will be 1 over 3m Planck squared, a pi squared 30, if I remember right. G star t re-eating to the 4. This is the number of relativistic degrees of freedom, which is of order 100, at least in the standard model, is of order of 100. And essentially, from this relation, we'll find that, parametrically, the re-eating temperature is of the order of gamma squared, gamma and Planck squared root. So you know the decay of the inflaton. From the decay rate of the inflaton, you can immediately find what is the re-eating temperature. And the figure, the plot I was telling you, is the one that I was telling you before. So here is time. Here is when there is the value gamma is equal h. So here is the temperature. Initially, there is no temperature. And then, all of a sudden, you form the temperature. And then here, the temperature goes as 1 over a. And this value here is the re-eating temperature, which is parametrically gamma and Planck. So this will be that you see the inflaton. Here is oscillating. So there are the inflaton oscillations. And then, all of a sudden, the inflaton decays. And it's going to produce the thermal. Probably, I will stop here. Do you have any questions? Yes, please. Now, this will be a different scenario. And for example, this will be what you want. Sorry, Martin, can you repeat? Yes. So the question was, I was doing a special case in which the inflaton was oscillating about the minimum of the potential. But then the question is, what about if I have a potential, for example, like this, so where the inflaton rolls, and then it keeps rolling? And then in this case, you would have a different scenario where here you have kination. And here, the question of state is steady plus 1 because you just have kinetic energy. And this is a plausible scenario, but I was just more simply minded, imagine that there is a minimum in the potential. Maybe here is OK, but in principle, one could imagine something like that. Or could imagine that the inflaton decays into a field that is massless, and so it gives all kinetic energy. Yeah, it's possible. You see, it's part of the uncertainties related to reading. Yes? You said that we want reheating to happen before we bend local synthesis, but there is also, is there also an upper bound energy? We'll discuss this tomorrow. OK. There is depending within models. So if you assume some models, for example, there's an upper bound for not overproducing gravitinos, for example. Depends on what the model you have in mind. But the lower bound is really robust. The upper bound depends on the model. Thank you. OK? OK, so I'll end up here, and I'll see you tomorrow. Thank you. Thank you.