 Welcome back and so I we apologize for the weather So the given that there is also wind the coffee break will be Just outside here not on the terrace and the posters are in this hall that goes around this thing So that the poster session will take place there And the weather will be good for the weekend. So it's okay So we have the fourth lecture about dark matter by Francesco there. So today is the last lecture on dark matter and Just to recap what we have seen so far We have is and this is not only something I cover in my lecture, but also in in all the school We know the dark weather is out there. Okay from observations We know that is there is five times more abundant than variants But these observations they do not tell us anything about the mass of the particle physics properties So we don't know what the dark matter is We don't know how big the interactions are with some other particles All we know is that he has gravitational interactions because we observe Gravitational effects on visible stuff that we observe Okay So from a particle physics perspective when you try to propose a model and you try to propose ideas to what the dark matter could be There is not a clear guiding principle, okay, so What we have done in the second and in the third lecture is to focus on a specific class of dark matter candidates that are called Wimps weekly interacting massive particles that are particles with mass between 10 GV and 10 TV more or less and With interaction cross-section with standard model particles of the order of weak scale processes. Okay, so that was the The second and the third lecture in the second lecture we went through the thermal history of these models We saw how these particles gets produced in the early universe and in the third lecture yesterday. We saw Two ideas directing the red detection to detect Wimps today. Okay, so The Wimps was one example of so the Wimps Were just one example of motivated dark matter candidates. So what do I mean by motivated? I mean that these particles have reasons to exist Beyond the dark matter evidence itself. Okay, there are reasons to believe that Wimps must be around today Because of the hierarchy problem. Okay, this is a problem in particle physics and It boils down to understanding why the size of gravitational interactions are So different from the one of weak interactions. Okay, or in other words why the X mass is 125 GV instead of the plant mass and We can discuss that offline if you want about the hierarchy problem But many solutions supersymmetry extra dimensions little higgs They have a Wimp candidate already embedded in the framework. Okay, so it's nice that the single theory solves two problems and so that's why I Called these candidates motivated because they have reasons to exist beyond the dark matter problem itself so another candidate that I think he was Raised during the Q&A session a couple of days ago Is the sterile neutrino? Okay, so the sterile neutrino We also know that Neutrino masses is one reason to go beyond the standard model because within the standard model itself neutrinos are massless and The problem of a neutrino mass requires the addition at some scale of additional degrees of freedom the sterile neutrinos that they are gauge singlets under a standard model and The sterile neutrino Depending on its mass. So if the sterile neutrino is around EV it cannot be the dark matter because remember For a fermion for a fermion that matter mass has to be larger than 100 EV That that's from like observing very small dwarf galaxies using the poly principle and filling the Fermi sphere That's about on the mass. So if you have sterile neutrinos of one electron volt They can be there they can account for neutrino masses, but they cannot be dark matter candidates If the mass is above 100 EV, I don't know between 1 and 10 kV. It's a very interesting region then the sterile neutrino can be a dark matter particle and This is still viable and I will not cover during my lectures just because I had to make a choice. Okay, all you have four lectures and I Decided to speak about another class of candidates so axioms and these are Motivated also because they address another problem of the standard model the strong CP problem that I will review now briefly and They are viable dark matter candidates Okay, so these three are examples of dark matter candidates that you do not postulate Just to explain the observe abundance You attack some other big problem the hierarchy problem the neutrino mass the strong CP problem and Automatically you also have a dark matter particle. So one theory you solve two problems That's why these candidates are nice. Of course, there is no reason why? The observe abundance of dark matter must be explained by some particle solving some other problem. Okay, so this is just To say that I'm not saying that that matter has to be one of these three Candidates, but it's nice. So let's study the axioms today So these are the idea goes back to the 70s late 70s early 80s but when I was a student I remember that People were obsessed with wings and axioms were not really popular and I'm not talking about many years ago but in the last five years, let's say there was a Revival of it mostly driven by new experimental ideas. So people came up with new ideas to discover these particles and So there was an increasing interest on On axioms as dark matter candidates again. Okay, so I decided to speak about these during this last lecture because it's There is a lot of interest in physics, but it's also very timely because It's a topic that if you read the archive every day, it has been discussed a lot Okay, so I will first review the motivation for the axiom. So why we Introduce an axiom to the standard model Then I will describe This particle can be a darker matter candidate So as we will see the range of masses is very different from the wings So these are extremely light dark matter candidates the production mechanism I will do a calculation of the abundance is completely different. So it's something completely different than what we have seen in the first three lectures Okay, so let's start from the strong CP problem this is This is a problem of the standard model and It boils down to understanding why CP is respected so well by strong interactions and The best way to introduce this problem is just by looking at the QCD Lagrangian so QCD is quantum chromodynamics is the gauge theory describing the strong force get strong interactions and The QCD Lagrangian, so let me let me write that and then I will comment piece by piece So let's see what convention. Okay Okay, so these Lagrangian describes strong interactions G is the field strength of the glue on so this is the kinetic term of the glue on. This is just the gauge theory Q are quark fields. So Q for me is a vector and Let's consider just QCD with two quarks up and down. So these are the two lightest quarks. Okay So there is a kinetic term for the quark. This is the Dirac Lagrangian So this Lash is the Feynman notation is the contraction between the Dirac matrices and derivative And then there is a mass term for the quarks. Okay, so this mass term is a matrix and I take it to be diagonal and I take the masses to be real. Okay So this Lagrangian describes very well everything we know about strong interactions. Okay at very high Scales at very high temperature the fundamental degrees of freedom are quarks and gluons So in the early universe when the temperature was above 1gV We had the plasma of quark and gluons that were not confined within hadrons And then as the universe cools down we went to a phase transition where quark and gluons Confined into addrons like neutrons protons pions. Yes. So did you mean the covariant derivative? Yeah. Thank you. Thank you Yeah, otherwise, it's a bad gauge theory. Yes. Thanks. Yeah. Yeah, so okay the Sorry, there wasn't yeah Why is all consistent? So This is a vector of 2 Dirac fermion and each u and d they have left and right component so, yeah, it's This this this this vector here is Q indices a flavor index and the Dirac index So you can expand and show that This is a mu q bar u bar u plus md d bar d left and right But I mean the notation are not important. All I want to say is that I'm taking the mass to be real for now Okay, so that's that's important Okay, any other question about the QCD Lagrangian? Okay, I mean don't worry about the notation all I want to emphasize here is that there is a Canary term for gluons a canary term for quarks and then there is a mass. We know the quarks have masses, okay? Okay, so these Lagrangian Describes nature very well described strong interactions very well and He respects CP. So CP is a combination of few transformations Charge conjugation and parity charge conjugation. You just change the charge of the fields Parity you flip the three space directions. So from there to there Okay, so you can convince yourself that each of these operators is invariant under CP We observe the strong interactions to know by the CP. So this is good. However, if we write down the most general Lagrangian If you write down the most general Lagrangian, well, we could write something else Well, okay, let me just Write it this way Okay, so These other two operators I wrote they are renormalizable and so in principle they can be there. Okay, the first one is a Dimensions dimensionless number theta times G G dual here G dual is just the is defined with the Epsilon tensor. Okay, it's just a dual strength field of the glue and The second piece is again the same quark mass term But now I put an overall phase to the quark mass matrix if you remember I said this matrix is real But there is nothing wrong with having a phase. Okay, the theory is still normalizable Now you can convince yourself that both of these operators violates CP. Okay They violates CP Or by the way, T. T is a transformation where you flip the arrow of time. Oh, this. Thank you Thanks. Yeah, so this is a if you want this is a definition of G Okay So CP T is conserved. There is a well-known theorem in field theory the CP D theorem So violating CP is equivalent to violating T You can see that Without going into details You are always free to perform The shawls you can perform transformations on this Lagrangian and you can put the face either all on this operator All on this operator This is just to say that there is only one invariant quantity Which is This difference here. Okay So it's true that I brought here two operators one for The glue on field strength with this conjugate the other one with the quark a phase in the quark mass matrix If you perform carrier rotations on quarks, you can put all the face here or all the face here It's up to you. There is a physical invariant combination, which is the difference between these two phases and this difference Sorry, I think there is a two here because it's a number of but it doesn't matter The only important thing is that there is a physical invariant quantity, which is a combination of these two phases And if you compute any green function any matrix element, it can only depend on this combination Of course, because that's something you observe. Okay, so what are the consequences of? Having this these two new operators or in other words Consequences of introducing CP violation in the theory of strong interactions that it can't just be there These are renormalizable operators. Okay, there is no reason why we should not write them down The consequence is that you induce Neutron electric dipole moment so an electric dipole moment For a particle like the neutron would violate CP because the interaction is D the dipole moment times E electric field and you can do that I will not do the calculation, but There are standard techniques to compute the neutral moment as a function of theta bar and you see that of course It's proportional to theta bar and it's very common to write that in these units So a dipole moment in electric dipole moment is an electric charge times a distance Okay, if you have two point like charges With opposite sign the dipole moment is the charge times the distance So the units of D are E, which is the electron charge so the charge of the electron or the proton Times centimeters. So these are the units That they're usually used to express neutron dipole moment and it's proportional to theta So if theta goes to zero, of course the effect goes to zero because we do not have CP evaluation anymore Okay, so let's compare this number with the experimental bound Experimental bound is 10 to the minus 26 E times centimeters. I'm using the same units So I compare theoretical prediction with experimental bound and I conclude. Okay, so let me write that Again, so this is the strong CP problem Okay, we have one dimensionless parameter in the QCD Lagrangian theta bar, which is the invariant combination of Theta and theta of the quarks So this parameter is actually the difference of two phases one appearing in the gluon field strength One appearing in the quark mass matrix and experiments tells us that this number has to be very very small Okay, so whenever we have a small number, we would like to understand why it is so small so There is nothing wrong if you put In the Lagrangian by hand this parameter to zero or even to 10 to the minus 12 Okay, but it's a lack of understanding. We have a small number. We want to understand where it comes from Okay, so this is the strong CP problem and the axion is a solution to the strong CP problem and While it solves the strong CP problem. He also provides a dark matter candidate So that's what we will do today. We will see How to compute the real identity for the axion once I define what the axion is is that clear the problem and what we're going to do Okay So let's see how the axion solves the strong CP problem. Okay, so the basic idea Is the following and by the way This is also known as the Pecci Queen mechanism or PQ and Pecci and Queen were the two people that in 1977 Proposed this mechanism as a solution to a strong CP problem and the idea is The following so the basic idea is that you you promote the theta parameter to a field So in this extension that I'm about to perform of the standard model The theta parameter appearing in that Lagrangian is not a number anymore, but it's a it's a dynamical field That evolves with time. It's a dynamical scalar field. He has to be a scalar and What we will study is The dynamical evolution So this parameter becomes a field We are going to study the equation of motion of this field through the expansion of the universe and we will see that Dynamics will drive theta bar to zero today and that's why we do not observe Violation of CP by strong interactions. Okay Okay, so Just in words how you achieve this I will just outline the few key steps, okay to to get to this dynamical field and then I will We will compute the real density Okay, so of course the first thing you have to do is to extend the standard model. Otherwise We already know the theory. So you extend the standard model by adding new degrees of freedom At high scale and you want your theory to have an a billion symmetry So a billion symmetry like a u1. Okay, just rotation by a phase and This symmetry has to be broken In the vacuum Spontaneously broken the same way the X boson. Sorry the X field breaks that week symmetry Down to E&M. Okay. So here you have something analogous. It's just it's not a gauge symmetry It's a global symmetry But it's a symmetry of the Lagrangian and the vacuum state breaks the symmetry and also you want the symmetry to be Anomalous under QCD and we will see what I mean by this but first let's focus on the first Requirement if we have a global symmetry, which is a symmetry of the Lagrangian, but it's broken by the vacuum state We know from the Goldstone theorem that this theory has to have a Goldstone boson Okay, so this is a famous result From the 60s I think by Jeffrey Goldstone whenever you have a global symmetry broken by a Vacuum state by the vacuum state then for each broken generator We have a mass less degrees of freedom a Goldstone boson is a particle with zero mass and We call this Goldstone boson the axion Okay, and I will call him Phi Okay, so this the Goldstone boson there are also in the literature you may find the axion called a Lowercase a but since a is also the scale factor in order to avoid confusions. I will call it Phi Okay, but you may find the axion called also a so also let's introduce the scale where you break This symmetry I call it f Okay, so I said that the vacuum state of nature breaks this you want symmetry, but at what scale Well, this is a free parameter of the theory that is subject to experimental constraints But for now let's call it f and the idea again is analogous to the electric week symmetry where the X field Breaks electric symmetry at the scale V. She's 200 g V more or less the Fermi scale Here it's an unknown parameter of the theory, but we will see With what we need in order to reproduce the the right density Okay, so the last thing we need to Exploit is the fact that this symmetry is anomalous under QCD. What does it mean? It means that this field Phi I'm using now 8 pi f So I'll give you the result and you have to trust me here But you can show that if the symmetry is anomalous under QCD Then you induce a coupling between this Goldstone boson and the gluon field strength Precisely to the operator gg dual. So this is what I meant when the parameter theta becomes is promoted to a field Okay, now in the Lagrangian, I don't have a number anymore. I have a dynamical field Phi and of course The operator in the Lagrangian has to have dimension four. So these are dimension four I have a scalar field with dimension one I need to divide by a scale and the scale will be the scale where I break the symmetry f, okay But this Operator is very important for two reasons. First of all, it shows us how the theta parameter becomes dynamical. Okay, so if I want to know What theta is today By now, I have to solve the equation of motion for Phi, which is what we will do today Okay in the expanding universe accounting for the Hubble expansion So that's the first Important thing. The other thing is that I said before that when you break a global symmetry You get the massless Goldstone boson. Well, not quite here because the symmetry is anomalous It's not an exact symmetry. It's broken by quantum effects. And so you induce a mass for the axiom so This field is not really a Goldstone boson because the Goldstone boson is massless It's called a pseudo Nambu Goldstone boson pseudo because it's almost massless up to these corrections here. Okay so you can also do a calculation for the mass and I will give you just the result here and Actually more or less so the mass of the axiom is given by This combination in the numerator you have the mass of the pion and the decay constant of the pion So this is 135 MeV. This is 92 MeV. So this is Qcd scale physics, which is hundreds of MeV and in the denominator You have the breaking scale of the pitch a queen symmetry Not surprising that the effect goes like one over F because the mass comes from this operator So if you send F To very large values you reduce the coupling between the axiom and the and the gluons and this mass from counter-accusative effects So the effect has to go down as you increase F and You can put numbers and I think it's one electron volt For a decay constant of 10 to the 7 gv Okay, so the last thing we need to do is to study. Yes Yes, oh Yeah, yeah, yeah, sorry, sorry, sorry There's only one F. Of course. There's only one F. There's only one F and Yeah, you check them the dimensions and what I wrote was wrong. So yes, it's phi over s alpha s alpha s is the Fine structure context of strong interaction. So it's g square over 4 pi for strong interactions Thank you. Okay. Okay. So you see that now we will do the rally density calculation and I anticipate the result Axiom dark matter is for F There are again some caveats here, but it's quite safe to say that What we will compute today is the following result the axioms can account for dark matter is F is bigger than 10 to the 10 gv So we are talking about the symmetry breaking at the scale much higher than the one we probed by experiments And you see from this equation and the axiom mass is at least merely electron volt 10 to the 7 divided 10 to the 10 10 to the minus 3 million electron volt So we are in a completely different mass range than what we saw the first three lectures where masses were above 10 gv, okay, and Let me also say that These small masses are possible only because the axiom is a boson. You remember the second lecture I think I say that for a fermion the masses to be larger than 100 ev For a scalar the limits is much weaker and it's like 10 to the minus 22 ev. So for scalars and For bosons in general going down to such a low masses is not a problem Okay, but this is something completely new with respect to what we have seen the first three lectures Okay F This F here Oh Fp. Oh These are just so these are parameters of QCD So MPi is the mass of the pion And Fp, Fp is the pion decay constant So it's something it parameterized the matrix element for the pion decay into leptons and neutrinos like mu plus and then a neutrino and It's we measure that by measuring the pion width so it's Well, I mean in some sense if you know character is pions are also got some bosons and so F is the Scale associated to the pions, but Yeah constraints on F. Yes, so One constraint is that if we want the axiom to be dark matter F has to be large enough But after doing the calculation and arriving that result I will comment what other constraints we have on F Because we have like limits from still looking at stars the way a star cools Experimental limits as well. So there are yeah, so we'll see we can chat offline, of course But my goal for today's lecture is to go through the calculation very dense and then we can discuss more Other questions very good. Okay We need one last ingredient and then we can go and compute axiom dark matter So I was a bit sloppy here when I say that the axiom gets them out. It's true. The axiom gets a mass This is a mass at zero temperature. Okay, so that's the axiom mass in the universe today as We consider the axiom in the thermal bath of quark and gluons This mass is temperature dependent Okay, so if we want to track the evolution of the axiom field as the universe expands We need the expression for the axiom mass at any temperature So I will give you this is It's actually there are still some there is still some debate about the axiom the actual the actual value for this mass You can either estimate through QCD non perturbative calculations The best way nowadays is to do lattice simulations and there is still some disagreement, but To sketch this calculation is not crucial the precise value. So I will just give you the result and just to make sure Our notation is Consistent I call this M zero This means mass at zero temperature. Okay, so now I will give you an equation for M Phi so M Phi as a function of T and roughly speaking This is equal to M zero When we are below the QCD phase transition scale, so lambda QCD Say 200 MeV So once I go through the QCD phase transition the axiom mass is frozen to its value at zero temperature once I go above the QCD phase transition then I have to account for these effects and So I and then of course this is T bigger the lambda QCD bigger and You see that the mass decreases as we consider universe at higher temperatures it decreases as a power law and I define this Behavior with this the following exponential index and over two. Okay, this is just a convention and Let me give you So as I say there is really not a consensus about the value of n but the number worth keeping in mind the range worth keeping in mind is between 6.8 and 7 Okay, so once you take Half of these the axiom mass decreases as the temperature to the minus 3.5 more or less Okay, so this input will be very important because we want to solve the evolution for the of the axiom in the early universe So we need to know the mass at any temperature and this is the result It's I think it's a challenging calculation to do you have to do computer simulations on the lattice and the lattice have some resolution power and computing power so Different collaborations were finding slightly different results, but the overall pictures will not be affected. So the calculation I'm going to do now Will not be affected by this this range here. So Yeah, you can you can trust this at high temperatures But around the scale where T is of the order of lambda QCD you cannot rely on perturbation theory and so When T is much larger than on the QCD you can definitely use effective with your techniques, but Around the QCD first transition the only way to go is it's a computer As I will show so these are now we will go through the derivation and the way it works is that this is not Particle dark matter in the sense that we have seen in the first lecture. So these are not like single particles It's a condensate is a it's a bosonic condensate and as I will show it behaves like cold dark matter So from the point of view of visualization Formation of galaxies it behaves precisely as as the cold dark matter have been talking about until now So that that's okay There was another question Well in the in the early juniors now Now the this is also something that is debated at the present time But for the purpose of right density calculation, there was the condensate and we can trust the calculation today structure formation is It's still not known, but but it's cold dark matter. I will derive this result. Sorry. There was another question Yeah, so I would say that the physical intuition is that QCD is asymptotic free So if you consider strong interaction at very high energy scale or short length scale It gets weaker and weaker and weaker. So that's properties called asymptotic freedom So if you consider the universe at a very high temperature, then strong interactions are weak And so the action gets much through strong interactions. So that's why you can see why The the axiom mass decreases with temperature because the interaction gets smaller and the interactions are what give the axiom the mass Very good. Okay So yes Exactly so there is a connection between these interaction here because the mass comes from this interaction and And you see that here there is an energy scale the temperature. So it's precisely that Yeah, but this is a weakly couple. So this is true for like for example electrons in the in the plasma Yeah, yeah, yeah, but this is this is a different story It's not like the electron coupled to the photons in our universe to get the D by mass proportional to T This is different. It's it doesn't come from that type of perturbative. This is a non-perturbative effects okay so let's let's go now the last thing I want to do is to compute The evolution of this field and compute how much energy density in axioms I expect today, okay So if I want to know how the axiom field evolves Well, we know we do that we write down the equation of motions and we solve them So I can delete everything I will leave this last backward because it's good to have the axiom mass Okay evolution of the axiom, okay, so We consider the evolution of the Isotropic and homogeneous background you can also perturb this equation to deal with perturbation But I'm interested in raw not on Delta raw. Okay, so let's discuss the evolution of the fundamental mode of the axiom. So it's a boson and The equation of motion for this I'm sure many of you are familiar with this result This is a question of motion for a scalar field in an expanding universe And I'm sure we will see this a lot next week when we do inflation and It's the same same same story. Okay so What do we see from this equation? This equation is the equation of an harmonic oscillator is Phi double dot the second derivative of the field. It's equal to minus some frequency times the field itself There is a damping term. Okay, so this is It's a damped harmonic oscillator so the harmonic oscillator with a friction term and Of course as I mentioned already for whims if you write a paper on an axion you want to produce some Careful result you give this equation to a computer and you solve the equation So that that's the way to go. Okay, but since this is a lecture on a blackboard We can do some approximation and get a feeling of how the final result depends on the Fundamental parameter of the theory, but the right way to do this is is of course by By solving the equation numerically Okay, so since we are about to see an approximate solution Let's identify Two opposite regimes for this field So the first one is when t is very large How large we will quantify in a second, but if t is very large now look at the last blackboard How much is the how big is the axiomus? Zero right if t is large The axiomus drops as a power law and there is a t large enough when I can't forget about the mass Okay, so if t is very large then We can ignore the mass and then We have only Hubble friction. So we look at this equation We forget about the mass. We forget about the oscillating part and we get the Hubble wins So it's what keeps the axiom stuck a Solution of this equation without the mass term is Phi equal to a constant. Okay, if you Put if you forget about the mass and you look at this equation and you plug Phi equal to a constant That's a solution for the equation. So the axiom field doesn't move until The mass kicks in so at very large temperature the axiom field does not evolve. It's stuck. It doesn't move Now as we go to lower t now we quantify in a second We will quantify what large and small means, but I'm just giving the qualitative picture Okay, then we will actually compute when this happens for lower t Then the mass the mass kicks in Okay Because now look again at the equation Temperature gets lower mass gets bigger at the same time Hubble is going down as the universe cools down So there will be a time when the effect of this term in the equation becomes important. Okay, and so what we See and we will see in a second are What we see and we will see that in a second is that the axiom field will start moving Because now this term will make it move and it will move like By following a monic oscillator with a dumping term, so there will be harmonic oscillations But the amplitude of this oscillation will be reduced with time as a consequence of this friction term Okay, so these are the two regimes a very large Temperature the axiom doesn't move lower temperature the axiom starts oscillating with an amplitude. It's reduced with time so the real question now is What is the temperature where we can interpolate these two opposite regimes in other words at what temperature? the mass the effect of the mass starts becoming important and the way to estimate that is a Hand waving wave of course if you solve the equation numerical Then you see this behavior as an output of your numerical calculation But if you want to get an estimate of the order of magnitude of the temperature where this happens Well You compare 3h with m Okay, in other words you are comparing two timescales Okay, Hubble is a timescale actually the inverse of Hubble is the timescale characteristic with the expansion M is also a timescale M is a frequency and it's the Frequency or equivalent to the period the period of these harmonic oscillations. Okay, so At very high temperature The timescale of Hubble definitely wins. There is no hope for the axiom to move when these two timescales Becomes equal Then you start to see oscillations. Okay, so the way you get The the answer to this question is by comparing two timescales or two rays if you want so let's do that for the reason for the tree because of the tree in this equation, so it's just and let For the estimate I'm about to do we can forget that because three is equal to one to pi two. Yeah Okay, so Hubble So I call T. Osh Osh Osh oscillation the temperature where Oscillations begin so Hubble at this temperature is given by the usual value T square over m-plunk up to order one factors Now I look at the previous equation. I compare it with the mass This is m0. You remember the value of the axiom at zero temperature is m pi f pi over f Okay, so that's the zero temperature mass. Of course. I don't have to forget the temperature dependence Okay, so now in my theory F is a free parameter It's the only free parameter I have because once I specify f the axiom mass is something that I can compute the zero temperature axiom mass So if I look at this equation, the only unknown thing is T. Osh that appears here and appears here so I Can solve Let me do that in this blackboard so in the interest of time. Let me just give you Yeah, let me just give the result in this form and the power is zero point one eighty five Okay, so this is the result Okay So you see that I put a number for F that it's in the interesting range as we will see in a second What I want to emphasize is the dependence on F is actually pretty small because there is this small power Index and this zero point one eighty five is a combination of these n and so I think I put n equal to six point eight to get this number and The real important number is this one. Okay, it's one point eighteen g e v okay, so this equation tells us The roughly speaking the temperature scale where oscillations begin. Okay, so the axiom field is stuck all the way to few g e v and Few g e v is the temperature of the universe when the universe start when the axiom starts oscillating Okay, so if we want to make a plot of the value of the axiom as a function of time So this is Phi and This is T So T here is the time of the Friedman-Robertson-Walker metric so at the very beginning Phi Doesn't move Abel friction. So this is let's call this Abel friction. Hey, so let me I Forgot what I have to drive afterwards. So it's good if I this is the time and I identify Oscillation time, let me use a different color So at the very beginning the axiom field is stuck and it doesn't move because of Abel friction When the time gets to Tosh so time Tosh is the time corresponding to this temperature I can solve the Friedman equation and find a relation between time and temperature Once I get here, I start oscillations and These oscillations are like this I'm exaggerating of course because on a blackboard I cannot do a better job, but So this is the evolution of the axiom field in the early universe It doesn't move all the way to a given temperature because of the effect of Abel friction Below the temperature the oscillations can begin and you see this oscillation and by the way the frequency of this oscillation is the axiom mass and The reason why the amplitude gets Dumped with time is because there is a friction Abel friction Okay, so what we are left to do for today, and we have 20 minutes so I'm confident we will do that is to Understand how to describe the energy density stored in these oscillations all the way to today So we can make an estimate of raw axiom today Okay, because this is an harmonic oscillator with friction So he loses energy because of friction, but there is energy density stored in that And so we want to describe the evolution of this energy density, okay So I will leave this plot here because it's useful. I leave also the equation of motion So first of all somebody asked me before Is this called dark matter or not? So let me start by showing you that this energy density stored in the oscillation behaves exactly like cold dark matter Okay, why energy? Density so the energy density stored in the field Is the sum of the kinetic energy plus? the potential energy Okay, it's just an harmonic oscillator. Okay You can convince yourself that The period of this oscillation is much much shorter than the damping time Okay, so you can apply the virial theorem also to this this oscillation and the virial theorem Will describes will describe the average so if you write the total row as Kinetic plus potential The virial theorem says that the average kinetic energy Over one oscillation is equal to the average potential energy So this is just So we we mentioned the virial theorem in the first lecture. I think and the virial theorem was for Newton force one over R square harmonic force is one is not one over a square is Proportional to R. So the virial theorem holds, but it's a bit different It says that the average kinetic energy over one oscillation is equal to the average potential energy over on oscillation So with this in mind What really want to do is to the compute zero over dt So how does row evolve with time? Well, row of all with time. Well, we just do the calculation We just write down one of five square plus Here I haven't done anything. I just applied the definition of raw now I start to derivatives So this is five dot Phi double dot plus m square phi dot phi I Collect phi dot So in the the first step is just applying the definition easy Second step also easy derivative time derivative chain rule First step I factor out a factor of phi dot and now I look back at the equation of motion and I know that the term in the square bracket is minus 3 times Hubble times phi dot Just because of the equation of motion during the evolution the action of course respects the equation of motion 3h Phi dot square Let me write that q times Phi dot square over 2 Again, I haven't done anything deep here, but this is the kinetic energy. Okay, so this is Ro K and Since the kinetic energy is equal to the potential energy, of course average over one oscillation Twice the kinetic energy is also the total energy Because is kinetic is equal to potential so k plus v is like qk because k is equal to v okay, so I end up with minus 3 h Ro Phi, okay So this equation shows that the energy density stored in this oscillation is not constant over time why because The oscillations are damped. There is a friction Hubble Hubble is annoying If it gives friction to to the action field and so this equation in this board here shows you how the energy density dreads shifts with time and you can show that if If you have an equation Phi dot minus 3 h Ro Phi the solution is Ro Phi Redshift as a to the minus 3 okay So something that redshift as a to the minus 3 is not a relativistic matter or I didn't Equivalently is something with zero pressure. Okay, so this is a cold condensate that behaves like cold dark matter and so It's a viable dark matter candidate now The last thing I want to do is Well, we know that Ro Phi behaves well in the sense that he has the redshift behavior Which is the way we want But how big is Ro Phi today? Okay, and we want to make sure that Ro Phi today is Precisely giving an omega-dark matter of 27 percent Okay, so that's the last thing we do. I still have 15 minutes Okay, so there are two ways to get to yes Decaying so that's not a decay. That's just reducing the amplitude of the oscillation But it's not like having a decay term in the in the in the Yes Yeah, the amplitude decreases with time. That's correct. Yes. Yes It doesn't disappear. So the amplitude of Phi today, I will do that now Yeah, yeah, so what I want to do is to compute Ro of Phi today Okay, this is Ro Phi today and Ro Phi today is K plus Ro V Today and by applying the VR theorem again, this is the result So it's twice the potential energy Because of the VR theorem kinetic and potential are the same So it's the mass today square times the amplitude today square the mass today. We know it's Today the unit the temperature of the universe is way below 200 MV. Okay, so we know this What we don't know is 5 0 today. Okay, so that's what we will do today. We will do now and And So there are there are two ways to get to this answer one is to solve the Differential equation numerical as I mentioned before and that gives you an optimal for 5 0 but what I want to do is to use an approximation which gives the right order of magnitude estimate for Ro Phi and This approximation is the following so This goes back to Classical mechanics Okay, now for mechanics Forget everything. I told you about the axiom. So I give you a Lagrangian like this Lagrangian in classical mechanics not in field theory. So this is all your Lagrange equation in classical mechanics So it's kinetic energy minus potential energy Lagrangian is L minus t minus k minus v I just multiplied this Lagrangian by the cube of the scale factor. Okay now you Can compute the oiler Lagrange equation for this Lagrangian and again I emphasize This is no field theory. This is just classical mechanics techniques. Okay, I give you a Lagrangian Where this is the variable the coordinate the court. This is the coordinate Phi. This is Phi dot I Can derive the oiler Lagrangian equation and you can show that You get this Okay, so the equation of motion of the axiom is the same as the one derived by Using classical mechanics and mix to this Lagrangian now the key point to realize here is that The axiom So the these are harmonic oscillator with a given frequency m so the mass of the axiom is the frequency of the oscillation because omega square appears in the force term and The parameters are changing with time how ball is changing with time m is also changing with time But they're changing slowly compared to the time scale of the oscillation That's something we can check and I'm giving just you the answer because of time, but For a system like this where the parameters are changing slowly. There is something called the adiabatic invariant adiabatic This is something you you you find in the for example in the textbook by Landau least its volume one okay, so classical mechanics and if As it is the case for this equation The parameters appearing in the question are changes slowly with respect to the Timescale of the system, which is the inverse of the mass, which is that the oscillation the frequency of the oscillation the period of the oscillation Then this adiabatic invariant is Is the finest The integral over one orbit of P Phi D Phi where P Phi is the conjugate momentum With respect to the variable Phi and now I look at the Lagrangian P Phi it has to be a cube Phi dot I Hope I got it right Yes, okay okay, so The way you do this calculation is you compute the adiabatic invariant and then You impose since this is a constant through the evolution of the field You impose that the value of this adiabatic invariant I That you compute at the time when the oscillation starts what they call to us Is equal to the value of I today because it's constant Okay, and that's a way to relate the axiom Amplitude at the very beginning of time when you started oscillating to the action of today Okay, so since I have five minutes left. I'll just give you the answer, but if this is straightforward calculation Okay, so the adiabatic invariant is 8u m Just m Phi square, okay So the way to get the amplitude of axiom oscillation today is to compare this value at The time of the oscillations and today so at the time of oscillation the scale factor. I call it aosh cube The axiom mass Well, I know it's temperature dependence and I know an expression for this temperature So I can plug this expression and find the mass at that time and this is The initial value of the axiom field, okay And today This is the scale factor today a zero cube As already say the axiom mass today is m zero the one the very top of the other blackboard and Phi zero Square is the amplitude of this the axiom today so this equation Gives me a way to evaluate Files actually we I can write this I Can multiply by m zero here and by m zero here and I divide by a zero here And what I have on the right hand side of this equation is the axiom density today It's m square Phi square is one after potential entity times two virial theorem again And here I know everything because the ratio of scale factors is the ratio of temperatures We know how the temperature evolves the C and B temperature. So it's something I can compute and since I'm almost out of time, let me give you the answer and I Will comment on the answer so you can compute After few trivial steps the contribution of omega. Let me call it omega axiom So this is the energy density stored in this oscillation of the field And I normalize this as zero point 27 which is the number I like because it's a number a measure times Phi I over F square times F to the power of 1.18 So that's something if you do just a few trivial steps you can drive. So to conclude We have a way to compute energy density stored in this oscillation You see that this is something we never mentioned but the final answer will depend on the initial condition of the axiom So I have an equation of motion and in order to solve the equation of motion and find the value of the variable today I need to know the initial condition. I know it was at rest. So this is a second-order differential equation The time derivative was zero at the beginning. That's something I know the initial amplitude is something I don't know and it depends. Well now there are a lot of There is a lot of other things to say because it depends if this Symmetry was broken before or after inflation if it was broken before inflation then inflation Extended I mean the universe expanded exponentially and so now within the up all horizon today We only have one value of Phi I and so this value It's reasonable to take it to be order what Phi I over F over there one, but There are other reasons to also prefer the smaller value, but anyway I Think it is reasonable to say whether you have paycheck wean broken before after inflation This ratio is naturally over the one unless you have some additional reason to take it to be small and there are reasons so this equation tells us that the interesting range for F is 10 to the 11 gv, okay and What else to say so this me this contribution is called misalignment Because The key of these mechanisms that the axiom field was displaced by its minimum And so it could oscillate as the universe evolved so misaligned because it was misaligned with respect to the to the it's it's Value corresponding to the minimum energy So there could be other contributions coming from the network of domain world's strings that format the face transitions these are only there if paycheck wean is broken after inflation and They can have an impact as big as this and there is still debate about the size of these contributions, but I would say that the last two things I want to say is that 10 to the 11 10 to the 10 or bigger are interesting value for axiom dark matter Phi can be at most over the F up to 2 pi factor so F cannot be too small but you can make F larger if you make Phi over F smaller and That's something that requires some justification, but you can and Somebody before I mentioned other bounds on F So there are bounds on F for example because if the axioms are there They could be a way to cool down the Sun so axiom can be producing the Sun and they just escape the Sun because they're so Weakly couples to us so by studying cellar structure or even supernovae the way supernova explosion works in the present of an axiom We found bounds on F, but they are smaller than this so the interesting region for axiom dark matter F larger than 10 to the 10 is viable There is no experiment telling us that No bound telling us that we cannot live in this region F equal to 10 to the 7 would be excluded By this consideration of stellar bomb 10 to the 10 10 to the 11 and bigger value of course the bigger you go The safer you are because you decouple the axiom so smaller values of F are a problem and As I mentioned at the beginning the reason why I wanted to give this lecture is because the idea of axiom dark matter is old But new experimental ideas Have been put forward in the last five at most ten years To test a much wider range of F so the previous ideas were focused on the axiom wind of Around this 10 11 10 to 11 gv value because of this equation, but as I said There is another unknown quantity here So the range of F where you can get that matter is actually bigger And if you want to know more about these ideas come and talk to me So I'm out of time. I Stop here. I thank you very much for your question both during the lecture and after lecture It was a lot of fun for me. And as I say that would be around also next week not just this week Way more relaxed because I don't have to prepare the lectures at night So if you see me around and you want to talk more feel free to talk to me. Thank you