 So the last time yesterday we have a look at the cosmological constant, the cosmological constant problem and from today, so today and tomorrow, we'll be devoted to assuming that the cosmological constant problem has been solved, so the cosmological constant in the way that we talked about is basically zero and we will try to explain or make some attempt at explaining the accelerated expansion of the universe and the observations in terms of other things, alternatives. It is fair to say that until now there is not really a compelling alternative, something that you can say, well, it's clearly this because it solves many problems. As I explained at the beginning, all the observations are consistent roughly with the cosmological constant. However, I think it's important to have alternatives, especially because in the future we will have a lot of new data, so it's nice also to compare the model that we have, the Landau CDM with alternatives, and also it's nice to study new effects that could be just be neglected because we are looking at the data in a certain way. Since there is not a clear compelling model, I will talk more about mechanisms and physical effects. I think it's important that you know these mechanisms. If you want to study these sort of things and if you want to have an idea of what people talk when they, I don't know, for instance, talk about trying to constrain gravity at cosmological scales and so on. I will start with the simplest alternative, which is what has been called quintessence. The idea basically is that we are replacing the cosmological constant with the dynamical scalar field, so the action of the scalar field couple to gravity. I'm assuming that there is the standard Einstein inverse action plus a scalar, sorry, we write the full action like that, B of phi. And this scalar field has some potential and clearly this is the kinetic term and clearly the potential, the term is the dynamics. This is very similar to what you have already seen for inflation. So we can, I guess that you did that for inflation, we can extract the stress energy tensor from this action, right? And this is given by the new phi minus g mu nu 1 alpha, the phi squared plus B of phi. And then as you did for inflation, I'm going to assume that the scalar field in first approximation is homogeneous. Okay, so I'm going to focus on the homogeneous solution. And well, it's easy to show that the energy density of the scalar field is minus the zero, zero component of the stress energy tensor. And what I find is that this is equal to 1 alpha phi dot squared plus, so 1 alpha the kinetic energy plus the potential energy. Okay. And then you do the same for the pressure p of phi. And you find that this is equal to the kinetic energy minus the potential energy. And so the equation of state is the pressure, so it's 1 alpha phi dot squared minus p of phi 1 alpha phi dot squared plus p of phi. And this is all the, this clearly can be equal to minus 1 in the cosmological constant limit, so in the limit in which the scalar field is perfectly constant. Okay. So when I can neglect the kinetic term, in the other cases, the equation of state of parameter will be slightly less than my, sorry, more than minus 1. And in general, to have something accelerating close to minus 1, let's say, let's put it like that, we need phi dot squared to be much smaller than the potential energy. So the kinetic energy is much smaller than the potential energy. This is not so dissimilar to what you have seen for inflation. Although in this case, there will not be only the scalar field contributing to the energy density of the universe, but there is also some dramatic component. But eventually in the future, only the scalar field will dominate the universe. So at the level of observations or, let's say, physical effects, what can we say? Let's suppose that w is a constant and that is slightly larger than minus 1. There is no reason to really take it as a constant. In general, there will be some dynamics. If I take the scalar field in a potential, there will be some dynamics and I can study this just like you did for the slower case in inflation. But so what would you expect given that w is slightly larger than the one of the cosmological constant physically? What would have happened in the past? Can you say something? Now, you remember that if I look at the energy density of a species as a function of the scale factor, you remember how the energy density of the species scale with the scale factor, you remember? A1 plus w, right? So the case of the cosmological constant is this one. It's a constant. If w is slightly larger than minus 1, I will have something that is dominating a little bit more than the cosmological constant in the past, right? And the effect will be, well, there will be, of course, an effect on the Q0 parameter that one can study. Another effect that we will have will be on cosmological perturbations, and in particular on the growth of the dark matter perturbations, because as we saw yesterday, we said that the cosmological constant was slowing down the growth of statues. So I expect that there will be slightly a more slowing down of this growth in the presence of a component which has a w larger than minus 1. Yes, larger than minus 1. So something like that, if you study these fluctuations, happens. So yesterday, we draw this plot with all the growth, the growing mode. So in fact, we show this equation, which, by the way, will be valid, will be still valid in this case, in this particular case, at least in a smaller case. And we saw that in matter dominance, matter domination, delta grows like the scale factor for a cosmological constant. So for SCC, we have a slowing down, and if w is larger than minus 1, we will have even a larger slowing down, something like that. So we will find that the statues are a little bit damper, more damper than in the standard case. In general, this particular quittance fluid does not cluster. I mean, in forced approximations, of course, on large, there will be smaller fluctuations of the order of 10 to the minus 5, but it doesn't cluster at the level of having an energy density contrast that rho of the scalar field over rho field of order 1. This density perturbations, the density perturbations of the scalar field will be always small, and the reason is that if I look at the perturbation equations of this field, this is something like that. So if I expand, so I take the evolution equation of the field, which is the Klein Gordon equation, and I expand it at cleaner order in the field fluctuations. This equation looks like this. So these are the field perturbations. I am assuming now I'm adding perturbations here. So we have a pressure gradient term, which in this case is a corresponds to having a speed of fluctuation of the scalar field equal to 1. Okay, because I'm taking a relativistic scalar field. And so exactly as you show with the Blake, when you have this pressure gradient term, your species, okay, it will be dumped by the Hubble friction, but it will also display oscillations, okay. And so when you go to short scales, by the way, here there will be corrections induced by the metric perturbations. When you go to short scales, or to large scales, the typical behavior will be that these perturbations will oscillate, but will remain small. And eventually they will be dumped by this portion, by the presence of this Hubble friction term, okay. So the scalar field perturbations do not go in general in this case, okay. Good. Okay, there are just to anticipate, there are two cases where these perturbations could go. One is when there are no pressure guardians. So CS is much smaller than one. And the other one is when here there is a source, a gravitational source, which contains many more guardians. And or there is a direct coupling, say a direct coupling with matter, for instance, okay. In this case, the perturbations of this dark energy component will go, but in this case you talk about modified gravity, more than quintessence. Question, yeah, exactly. So very good point. Why does not behave like the matter? I mean, the homogeneous part, yes, is to what you said. The lecture is required to repeat the question. Sorry, yes. So the question was, I said that if W is larger than minus one, perturbations are more suppressed with respect to the lambda CDM case, with respect to the case of the cosmological constant. This is in contrast with what we know, for instance, in the case of the armature where W is equal to zero, and we have a perturbation growth like the scale factor, right? But look at this. So the fact that the fluctuations in the dark matter case growth is due to the dark matter component, right? Is due to the fact that the speed, this CS square here is zero. Okay. So there is no, there is no counterbalance to the gravitational force. There is no pressure. There are no pressure guardians, and therefore you have a growth of fluctuations. Here we are looking at something different. We are looking at the effect of the dark energy perturbations on the dark matter ones. Okay. So if I, now, if I go back in the past and W is less, is larger than minus one, so I'm in this quick test case, I will have less dark matter and more cosmological constant, and so I will slow down more the growth. Okay. Just because I'm reducing the component that can cluster. Okay. Okay. Let us see some shortcomings of a quick test. So as you see, I'm not talking about specific models. I'm not discussing some specific models. I will discuss a particular model in a second. So as we said, five dot squared must be smaller than the potential. So the kinetic energy must be smaller than the potential. If the scalar field is slow rolling, you remember that this is the condition for slow roll. And in this case, I can look at the variation of the scalar field in a Hubble time. Okay. So let's say delta phi in a Hubble time. This will be given by H minus one, roughly speaking, times five dot. And so now if I use this slow roll condition here, I will have that this is roughly V prime over H squared. V prime, while assume that V is a monomial potential, is some phi to some exponent n. Okay. So in this case, V prime can be written like V over phi. And I see that the relative change in the scalar field in a Hubble time will be of the order of V, the potential, but you know, the potential, quintessence is dominating the universe. So the potential will be roughly H squared. If I use the Friedman equation, this will be of the order of 8 pi G3 times the potential because the energy density of the scalar field is dominated by the potential. And so I can write here 8 pi G, okay, H squared phi squared. So H squared simplifies. And I can define one over 8 pi G as a plant mass squared. So let me write this as plant mass over phi. So if I want my scalar field to vary less than the typical value of this scalar field in a Hubble time, and this is typically what I would like to see if I don't want big changes in the equation of state, sizeable changes that are already rolled out in the equation of state, then you see from this relation that the phi should be larger than the plant mass. So I have what are called trans-plank and changes in the field or super-plank and let's say field variation. So sorry, field variation. The value of phi must be larger than the plant mass. Have you seen this in inflation? Did you discuss about that? No? No, okay. Well, okay. So in principle, there is no theorem for beating that, although it seems difficult in concrete models to realize, to have this in model. For instance, in string theory, it seems difficult to have a phi mass larger than the plant mass. A manifestation of this that has become fashionable or discussed a lot since a few years is the swarm-planned conjecture that you have probably heard about. The first of this conjecture that has been applied to inflation and also to quintessence tells you that the potential, the first derivative of the potential of the scalar field should be larger than the potential itself divided by plant mass. And this is exactly in string theory. And this is exactly the same as saying that the phi should be smaller than the plant mass because it's exactly the opposite of of what I showed you here. If you want a smaller. Yes. Yes. So two things. So first of all, maybe you don't know, but Vafa gave a lecture about the swarm-planned on Tuesday, although he didn't enter too much into detail. So I'm not sure if he mentioned it, but anyway, just to say. Another thing from the audience, from the zoom, I think phi dot square is very it's very bigger than zero, not V. Sorry. Yeah. Phi dot square is much bigger than zero. No, sorry. No, I think no, but I think the answer is no, it's much smaller than V. It's also bigger than zero. Let's see if maybe. Okay. You write again. Okay. Okay. Well, not the camera, but I think you can say decline. Ah, sorry. Okay. So, okay, this is the first thing that I wanted to say. The second thing is the following. Can you see now? If you, if you look at, so by using, by using what? By using this equation, if you imagine that this potential is simply m square phi square potential, so the square field as a mass as you have probably seen in inflation. So in this case, this is H zero square today. So I'm using this equation here. I put m square phi square. I rewrite eight by G as m plus square. And I find that since phi must be larger than the Planck mass, I find that the mass of the scalar field of the quintessence field must be much smaller than the Hubble rate today, which is 10 to the minus 33 electron volt. So this is a super light field. And well, let's say yesterday we look at the quantum correction to the cosmological constant. Also quintessence, if it is coupled to other fields, not only stand among the fields, but maybe other fields that we haven't seen yet would receive huge corrections. And so it would be difficult to, to understand the smallest of this mass. Okay. The same if I put a lambda to the four coupling, I can do the same calculation. And I find that lambda must be much smaller than 10 to the minus 120. So it's a, it's a very strange uncoupled, very weakly coupled field, let's say. Okay. So it seems to be to have a very unnatural potential. Except as someone mentioned yesterday, except if you impose some symmetry on the scalar field. For instance, the typical example is if the scalar field is a pseudo Nambu-Goston boson. In this case, you know that the, well, the potential, the potential is something like that. And, and because in the limit where, so, so a Nambu-Goston boson is a field which is massless. So where, which realises shift symmetry. Okay. Epsilon Nambu-Goston boson is, is the same except that you are, you are softly breaking this, the shift symmetry. And therefore the correction to the, to the mass of this scalar field can be controlled and can be, can be made as small as, as you want, if you want, in a, in a naturally, natural way. In this case, however, just as what, what we discussed before, if you want the mass of this scalar field to be, to be small, you also need the f, sorry. Yeah. You, you need f to be much larger than the Planck mass because you want phi to be larger than the Planck mass. And again, this seems to be difficult to, to achieve in a, with the concrete, with concrete models. Okay. I think now I have to make a choice. Do you prefer that I talk about tracking fields or dark energy perturbations? Yes. Well, because in this case, there would be no mass. It would be perfectly shift symmetric. EGoston boson, right? Yes. Like, yes. There is a symmetry which is softly broken, like, yeah, like, like the axon in, in QCD, for instance. But, but, yeah. If it was a ghost on boson, it would have only derivative couplings. It would not have a potential because it should satisfy the symmetry, the shift symmetry. So you would expect the Lagrangian to, to contain only terms like d phi square or d phi to the fourth, et cetera. So what do you prefer? No, it's difficult. Yeah, I know. I know. You prefer the, the cluster perturbation? Okay. Anyone, anyone? No. Okay. Let's talk about that. So we mentioned, in fact, it's connected to this. So let's say clustering princesses. And in the north, you will also have the tracking field. So, so we said earlier that this is the equation satisfied by perturbation of the scalar field. C squared, k squared, delta phi equal to metric perturbations. And that the value of this speed of fluctuations of the scalar field is crucial to determine whether it's clustering or not. Okay. So let's take a Lagrangian which is so now I'm just considering the scalar field Lagrangian. And I will assume that it's a function. I don't specify it yet, but we can specify it of phi, the scalar field, and x, where x, okay. Sometimes this is called the K essence. And if it's using inflation, it's called the K inflation. And if it's using mechanics, it's called K. So you can, you can write down a stress energy tensor also in this case. And you, you can write it like that. And you can interpret this as the stress energy tensor of the scalar field of the, sorry, of the perfect fluid, which is this one. Okay. If you identify capital P, which is the Lagrangian with the pressure, and then you have to identify, so the pressure is capital P. And the energy density is 2 pi comma x. When I say comma x, it means derivative. So derivative with respect to x over P minus P. And finally, u mu, so the full velocity of the, of the fluid is this one. So from here you can, you can reconstruct the, the, the equation of state as we did before by taking this divided by this. But you have an extra freedom. Now, if I, if I expand my scalar field in, in perturbations around the background like I did before, and now I also will expand the x around minus five dot squared. I, I'm going to find, well, actually, let me give you directly the, the, well, okay. So I'm going to find this special. So this is the background value of the first derivative of P with respect to x. And then we have a kinetic term for perturbation like that. And then I have a term which is absent in the case of a standard quintessence field, which is, which is this one. And then other terms without, without derivatives. And, and you see that now, so the, usually in the, in the standard quintessence case, so standard quintessence case is P is equal to x one alpha x minus one alpha x minus v of i. Right. So the second derivative of P with respect to x is zero. And I find that, that, well, I find a harmonic oscillator with the, with oscillations propagating at the speed of, at the speed of light. Okay. Here there's one. However, in, in this case, I find that the speed of fluctuations will be given by what is in front of here, which is P of x divided by what is in front of the time, time kinetic term, which is P of x plus x 2x P of xx, which is this term here. Okay. So in the limit where x P xx is much larger than P of x, C squared is much smaller than one. And now my, my, my dark energy field that can really cluster. And, and what can happen in particular when CS is, is of the order of 10 to the minus 6, 10 to the minus 7. So when the, when the speed, when the sound horizon of these fluctuations are, are smaller than, than, I don't know, a few megaparsec or something like that is that you really see that the dark energy field clusters along with, with our matter and contributes to, to the structural formation together with our matter. Okay. There are people who are building simulations to, to try to, to simulate that and so on. So you can, you can treat also this, this possibility. Questions about this, this one? Well, but it's the same. No, it's not. It's the same. Maybe the only thing that I would say is that I should put a minus to be future directed, but otherwise it's the same. I think it's fine. Yeah, sure. So, so, no, I think not. So I don't remember exactly the formula, but so I know that the dark energy fluctuations are suppressed with respect to the dark matter ones. In fact, how do you remember the formula for dark energy fluctuations when the speed of the sound is zero? Well, it's something, well, okay. Anyway, it will have the size of the dark matter fluctuations, but with an extra suppression that I think that I don't remember, but I should check. One plus W. Yeah, I don't know if it was this. So at the end, okay, depending on W, you have some sizeable perturbation, but not something dramatic that could explain. Yes. In any case, I don't see why it should go in that direction of explaining there's your attention and also this perturbation go very, at very high time as, and as we heard this morning, we would like more to look for a solution at early time. So, yeah, okay. Now let me enter in a, in a new, in a new area. So now, now we start. So until, until now I consider a scalar field, which is the coupled from, from a standard particle, from the standard model. But let us start looking at, at coupling to matter or if you want, I will start introducing scalar tensor theories. So the possibility that this scalar field modifies gravity mixes with gravity modifying the standard gravitational dynamics. So, so we would like to, to study if the scalar field can be coupled to, to matter. If it is coupled to matter, it would be safer that it couples to matter in the same way because we have very tight constraints on the equivalence principle. And we know that, so we have seen that the scalar fields are typically very light. And you know that the light, the scalar field is changing, it changes a fifth force. If you don't know it, you will see it very soon. And so in order for objects to fall in the same way, we would like all the objects to couple in the same way to this, to this scalar field, okay. So there is a, a traditional way of introducing this scalar tensor theories, which is, which is the following. So to couple, to, to put a function of the scalar field in front of the Ricci, the Ricci scalar. And then to write the, the, the action for a scalar field like the one that I showed you before. At the same time, I can add the action of some matter component that I assume to be minimally coupled to gravity. So to couple, I assume that matter, whatever it is, it couples to the gravitational metric that I'm using here. So it seems that the scalar field does not couple to matter in this case, right? I just put it, I have just put it in this gravitational action. I can derive the, while the generalized answers equation in, in this case, by varying this action with respect to the, to the metric. So if I vary s with respect to g mu nu, then I am going to obtain some modified answers equation that you are invited to check in one of the exercises. So the left hand side of the answers equation is clearly modified instead of having the instant answer. You have the instant answer times the function of phi. And then some, some terms that depend on the second derivative of the scalar field. Sorry, there is a question from Zuma. Yeah. Why can the scalar field couple to matter field? In this case, could you just say one more? So for the moment, it seems that it doesn't couple, but you, you will see that in fact, I can choose a different metric. I don't know what to look. I can choose a different metric to describe my system. And, and in this metric, you will see that there is a direct coupling of the scalar field to, yes. Yeah. So in this case, there is no direct coupling between the scalar field and, and matter. On the other hand, I'm using a gravitational field or gravitational metric, which does not satisfy the usual Einstein equation. And therefore, and therefore, at the end you have, you have a coupling between matter and the scalar field. In fact, it depends on, on the frame. It's called frame that you use to describe your system. Sorry, there is another question from the same person. Does a direct coupling violate the equivalent principle somehow? Yeah. So let me maybe answer that later because it's something that I partially said. Okay. When, when, when matter couples minimally to the metric, okay, we say that we are in the Jordan frame, but I can change the frame and I put a frame in, what's the name of in, in quotation marks because it's not really a frame, but it's, it's, it's a choice of a field to describe the, the gravitational field. So I can go in this so-called Einstein frame by making a field definition which, which you can find out yourself by using, so if you, if you go to Walt's, so the Walt's book on general activity, your book, and you look to append X, D, World tell you how the Ricci curator scalar changes under such a field definition or such a conformal transformation. It's called, okay. So the Ricci scalar written as a function of a metric, till the metric G mu nu will be given by some Ricci scalar written in terms of the metric G mu nu without a tilde. And then there will be some factor of omega, which is, by the way, omega will be, will be a function of the scalar field here with the derivatives of omega. And if you do a convenient choice, which is to choose omega square equal to f minus one, where f is exactly this function, you see that I can go to a form where the action is written like this. So now I'm using the Einstein frame metric. And let's say, roughly speaking, the scalar field, we have the scalar field action, we have the standard form, p of i Einstein. But matter, of course, will be coupled to a metric G mu nu Einstein and the scalar field. Okay. So just to be clear, this is the transformation that I'm using here. So when we speak about frames, it's a bit of a misnomer because, in fact, what we are doing is just a field definition. Okay, we are just describing the gravitational degree of freedom either with the G mu nu or with G mu nu Einstein in Einstein frame. And depending on the frame where we are, we can say that in Jordan frame, matter couples only to the gravitational metric, but the laws of gravity are modified. And in Einstein frame, the laws of gravity are the same, but matter couples directly to the scalar field. Okay, you see. So at the end, observables must be the same, independently of the frame where you look at them. But you can have a different description. Any question? Yes, there is a question. Will field definition not cause problems with quantum loops and renormalization of one action compared to another? No. So as long as the field definition is invertible, the physical description is exactly the same. And so then, of course, it may be more convenient to use one field, one metric with respect to the other just for particle computations. For instance, in general in cosmology, most of the observations that we see, that we make, we can say that they are made in the Jordan frame. Okay. The reason being that in the Jordan frame, test particles follow geodesics. And this is what we are used to when we make observations. And while, for instance, in Einstein frame, it's true that gravity is the same, but masses seem to be varying with the scalar field. So the interpretation, it's a bit more difficult. But in principle, the two descriptions are absolutely the same as long as the field definition is invertible. Yes. Why? Why so? Because, well, as long as I consider test particles, they are coupled in the same way to the scalar field. So they will behave exactly in the same way. If I take one apple and one pier, I don't know. Now, two different objects, as long as I neglect the self-gravity of the object. So I'm in the test particle limit. In fact, they will fall exactly in the same way. In general, not. And this is due to the fact that it's only in GR that once you include the gravitational self-coupling, all just fall exactly in the same way. In this particular theory, but this goes a bit beyond. In this theory, the non-linearities of the self-couplings of the scalar field, if you want, the gravitational self-coupling of the scalar field, will induce, you say, strong equivalence principle violations. And that could be observed. But they would be more observed for compact objects like black holes, or stars, or things like that, than on small objects on Earth. So if you consider only test particles, these test particles satisfy the weak equivalence principle. So we are assuming that they do infect. In fact, we could break this assumption and we could say, well, there is not a single Jordan frame. Suppose, for instance, that there are two quotation marks of Jordan frame, one for their matter and one for standard model. And the standard model and their matter particles are uncoupled. Then, well, as long as we look at the standard model, we will not see equivalence principle violation. But we could serve them by looking at, I don't know, alos of their matter or things like that. Which is exactly what Maria showed yesterday. She was considering a model where their matter is coupled to the energy, but not the rest of the standard model, I think. So this would be an example of a different coupling to matter. But still, since we observe only in lab, we look at the full of objects of standard matter, we will not see equivalence violation. So as long as you consider test particles, it's fine. But then when you start looking at, this is why it's so interesting to look at the strong field regime of objects, like for instance, black holes or things like that. Okay, so one comment is the one that I just mentioned. Then let us look a bit at these two different descriptions and the physical, the various physical effects that we could have. So as we said, so there is the Einstein frame. And so in Einstein frame, gravity is standard. However, particles do not follow geodesics. So for instance, well, here I write down the conservation equation for the stress energy tensor of matter in Einstein frame. And this would be sourced by, would have a, well, in Einstein frame as a source term that depends on the scalar field and the coupling to the trace of the stress energy tensor. Okay, and also the scalar field, instead of having the standard ligand equation, will have a coupling like that. So in Jordan frame, these couplings are not there, but gravity is modified. In Einstein frame, there are these couplings and gravity is the same as a GR. Okay, so as you see, the scalar field couples to the trace of the stress energy tensor. Okay, well, maybe the only thing, so this factor of A prime over A appears so many times that you want to define a parameter that I will call beta at the moment. Okay, and in Jordan frame matter for those geodesics, let's write it here. So in Jordan frame, we have the mu t mu nu equal to zero. The Einstein's equation had been written before and the Klein Gordon equation is modified because the scalar field mixes with gravity. So we have three main effects. One is the presence of a fifth force. So if I write the geodesic equation in the Jordan frame, I find x i dot dot plus gamma 00i, where gamma is the Christoffel symbol with the 00 lower indices. And let me be in the non-relativistic approximation. So I use, well, I use a metric ds squared equal to minus 1 plus 2 phi dt squared plus a squared t 1 minus 2 psi dx squared. So this equation would give me something like that, which looks like Newton's law, Newton's law. However, phi does not satisfy the usual gravitational equation. So for instance, does not satisfy the personal equation, the usual personal equation. What satisfies the personal equation is phi in Einstein frame. Okay, because it's only in the Einstein frame that gravity is a standard. So in fact, I can write this geodesic equation as phi in the Einstein frame minus beta, sorry, which is the parameter that I introduced before in the Jordan frame. And I can find this equation by using, if you want, by using the geodesic equation in Einstein frame, which tells me that phi in Einstein frame is equal to phi in Jordan frame. Well, let me write the opposite, the other one, plus beta phi. And now I write down my personal equation but using the Einstein frame metric. And I assume that there is a static source of mass m here. Static source of mass m, the solution of the personal equation will be that phi in the Einstein frame is equal to minus gm over r, so standard Newtonian potential. On the other hand, so now I would like to, so I found the profile of phi in the presence of a static source of phi in the Einstein frame. Now I want to find what the solution of small phi is, so the scalar field is in the presence of a static source and then reconstruct phi in Jordan frame, because it's the one that I observe, for instance, in galaxy class there in the last research, typically. So my scalar field satisfies the Klein-Gordon equation, as we said before. And as we said, it couples directly to matter, as you remember, in the scalar field as a source term due to matter. So this is a delta function and the solution of this equation is that the scalar field will, yeah, sorry, it's a 4 pi r and then there will be some yukawa suppression if the scalar field is as a mass. So as you see, the fifth force is attractive, there is the same minus sign in front of the gravitational force. It can be suppressed if the scalar field is a mass. Let's assume that the mass of phi is over the h0, so the Hubble rate today. In this case, I'm going to combine this here with this and I find that phi is equal to minus 1 plus 2. I'm going to write it like that, g r, where I define alpha. Alpha is a dimensionless quantity, sorry for defining so many. So my Newtonian potential is enhanced by the effect of the fifth force, by something which is always positive, because there is a beta here and a beta here. So this gives me a beta squared and then I just rewrite this beta in terms of alpha, which is dimensionless. So I want to define some dimensionless constant. So of course, if I had kept my mass, then I would have had some some modification of gravity, which are scale dependent, if you want, because I would have a dependence of r in here. Then the second effect that I have is that in general, in these modified gravity theories, the two potentials, so the potential that enters, that enters in the 0,0 component of the metric and the one that enters in the g ij, the special component of the metric, are not the same. And in g r, they are in the absence of stress energy, yes, stresses in the energy, momentum tensor of matter. And this can be seen quite easily by, yeah, so if I take, okay, yes, if I consider the conformal transformation between the Jordan frame metric and the Einstein frame metric, and I perturb it by using these expressions, I will find this equation that we already found, so I will find that phi is equal to phi Einstein plus beta phi from the g 0,0 component. And if I take the g ij component and I look at the perturbations, I find the relation between the psi in the Jordan frame and psi in the Einstein frame. And in the case of psi, there is a minus psi for... Filippo, let me ask a question from Zoom. Could you please repeat the argument that allows you to neglect the Yukawa suppression and to put the mass of the scalar field over the Hubble? Yeah, so in general, it is not the case that I can suppress this mass. However, I would like to find a model where I have sizable effects on large scales. So in this case, I need to set the mass of the scalar field, which is smaller. And, yeah, and so I will consider this case from now on just to be more specific, but of course, I could keep and I could study the scale dependence of this model in principle. Okay, so if you look at the relation between the Jordan frame and metric potentials and the Einstein frame and metric potentials by knowing that in the Einstein frame, these two are the same because I'm assuming G.R., I see that there is, well, I see that phi is equal to psi minus 2 beta phi, sorry, plus 2 beta phi. Okay, and I can also find the psi from here and I see that psi, I write it here, is equal to minus 1 minus 2 alpha squared Gm over R. Another question from Zoom. Yeah. In Einstein's frame, why is psi equal to phi? Could you please say it once more? Yeah, this is what, so if you take Einstein's equation, G mu nu equal to 8 pi G mu nu, you take the Gij components and you take in particular the traceless part of this Gij components and you assume that there are no anisotopic stresses, you will find this equation, box, sorry, Laplacian of phi minus psi in Einstein's frame equal to zero. Okay, or in general, is this equation equal to anisotopic stresses? Okay, and if there are no anisotopic stresses, phi and psi are the same. And in fact, you have an exercise at the end of these lectures where you are invited to work with this equation in the in the non-relativistic approximation. And you see that, well, I wrote down the components of the Einstein tensor in the non-relativistic approximation and you see that these are given by this, this expression. Okay, in fact, what you really find is this equation here, V i, V j minus one third delta ij Laplacian. Okay. Okay, the third effect that you have is that in general, G Newton can run, can depend on time, but this you may have expected it because you remember that in my Einstein equation, I had in Jordan frame, this coupling, F i r, no, in my, sorry, in the action in Jordan frame, I replaced, so in GR, in GR, the Einstein neighbor term is the Ricci tensor divided by 16 pi G. And in this, in this theory, this one over 16 pi G is replaced by a function of phi. And if phi varies in time, of course, also your G can vary in time. Okay. Okay. So in general, this is given by 2 a prime over a, so 2 beta phi dot. So if phi depends on time, also the Newton constant depends on time, which of course is something that is constrained. Ten minutes, no? No, it's finished. Okay, okay. Okay. Then I can stop here. It's true, no? No, no, it's true. Yes. So next time, I will show you, I will quickly show you what are the constraints that you have on these theories. And then we will talk about a way to avoid these constraints on short scales, on solar systems scales, while maintaining possibilities of modifying gravity on various scales. So given that there is the discussion session soon after the coffee break, maybe I would post on the questions to that. Okay? Okay. We're back in