 Yeah Okay, please take your seat. All right, I would like to to thank Matt that Sorry, it was not working Thanks, so this morning We discussed cosmology and some of the fine tuning problems Which seem to exist in our in our universe specifically we talked about the horizon problem and the flatness problem with the curvature problem So the most widely accepted explanation for those tunings for those puzzles Is inflation in this lecture? We'll talk about inflation as a solution to the horizon and flatness problems and Also as a mechanism for generating the perturbations that we observe on large scales Actually on all scales, but in particular on large scales where they're easy to test So first of all, what is inflation? Well, we talked about this morning how when W the equation of state is less than minus a third Then the expansion of the universe accelerates. So one of Einstein's equations Says that if roll plus 3p is less than zero Then a double dot is positive and the expansion of the universe will accelerate There's a constant there so What is inflation? It's a period in which the expansion of the universe accelerates. So it's a period in which W is Actually close to minus one. Let's write it as minus one plus delta where delta is greater than zero and it's of order a Few percent something like point oh five ish. It's exact value won't be so important So if you recall energy densities Redshift like a To the minus 3 1 plus w So for example with w equals 0 like a to the minus 3 and so with w minus 1 plus delta Then we have a to the minus 3 delta with again delta close to 0 so The energy density of inflation. So this is row inflation does decrease as the universe expands, but it does so slowly Much more slowly than any form of ordinary matter or radiation Okay, so that's more or less what inflation is. It's a period of exponential expansion Characterized by some small parameter now people don't normally Write a power law dependence like this because they don't assume that w is a constant and we won't assume that So rather than writing The scale factor is a power of t Let me write it as follows. I'll write a of t as some initial a at the beginning of inflation times e to the h t minus t initial and If h were exactly constant This would be deciders based on this would be the solution to Einstein's equations with purely vacuum energy with a vacuum energy Which doesn't change at all. So it's like delta equals zero All right, so h dot equals zero That's like delta equals zero. That's w equals minus one But I'm not gonna assume h dot is zero in fact, I'm gonna assume it's not zero But small in fact specifically I'll assume that h dot over h squared is small So what does this mean? It means that the change in h With respect to time h dot over Roughly one humble time Isn't very large. Okay, so if you if you have H at some time and you wait a hubble time the fractional change in h will be small And that's what this that's what this equation means All right, so so so that's what inflation is inflation is a quasi exponential expansion Where the scale factor is e to the h t but h is gradually decreasing. So it's not quite exponential It's of course accelerating As long as this derivative is small So this will be a period of accelerated expansion Now let's see what happens to the particle horizon. So if you recall the particle horizon Satisfied this equation. So this is the equation for a null geodesic So it's the trajectory that a massless particle would follow and Let's integrate this from some initial time When inflation has begun and this is the scale factor up to some time t So if we do this integral and let me put a an i subscript on h here i for inflation get this Which I can rewrite where n of t is just h t minus t initial and is called the number of e-folds Because As n increases A in a gets multiplied by e okay, so it's like rather than doubling a you multiply it by e So it's an e-fold So n is the number of e-folds and since h is almost constant n increases roughly linearly with time So as soon as n gets bigger than one or two we can neglect this term And we find simply one over A i h i In other words this co-moving particle horizon Barely moves during inflation it increases a little bit and then it just freezes in place This is very different from the situation with radiation domination or matter domination or any kind of decelerated phase of the universe Where the co-moving particle horizon grows without bound here instead it approaches this maximum value So in co-moving coordinates light barely moves during decider expansion You might be interested though in what this is mean physically what happens to the physical distance so the physical particle horizon is a times delta r and so that goes as e to the n actually just over h i so The physical particle horizon is initially one over h as usual But as time passes and increases it grows exponentially So put in very simple terms what's happening is that inflation is blowing up the universe It's making the universe expand so rapidly that light just like everything else gets carried along with the Hubble flow So even something moving at the speed of light Effectively doesn't move in these co-moving coordinates. It's effectively frozen. Remember the co-moving coordinates are the ones These are co-moving coordinates right so that the You're measuring it in these in these units without taking into account the scaling Of this space. So this is let's say zero. This is pi. This is still zero. This is still pi So light barely moves in these coordinates because the space is expanding so rapidly That there's more and more distance to cover if you want to move a distance pi or something and come over coordinates Okay, so light is just frozen in these in these coordinates but This horizon is going exponentially in the physical coordinates because the spaces So what inflation does is it takes an initial Hubble patch of size one over hi And it blows it up into an exponentially huge region and you can imagine that this might solve the flatness and horizon problems because the flatness and horizon problems were why there are so many independent horizon volumes in our past and why those horizon volumes looks so similar and looks so flat Well, if in fact there weren't so many Maybe there was only one Then you can understand why they are look the same. It's because really they're all just coming from one initial patch So let's see if that can work the way we express the flatness problem We said we know that omega k today is Less than about 10 to the minus 2 and if we extrapolate this back very far in time We discovered that this means that omega k in the past So at some early time t initial has to be extremely small end of the minus 60 if we go back to the plank time How can we explain why this number is so small? There was no easy explanation without inflation. You just had to put it in by hand Let's see what happens if there's a period of inflation well omega k is K over a h squared and during inflation h is roughly constant and a grows exponentially, so So this becomes a initial To the end and so omega k It was like e to the minus 2n It decreases exponentially. So it's not hard to explain Why omega k is small at least not if n can be reasonably large and we'll see in a minute exactly how large it needs to be Okay, so we can make a mega k small Even if it was order 1 to begin with this is its value to begin with If it's order 1 to begin with we can make it small by just having large enough n and Physically, what's going on? well remember That you can think of a mega k as the square of the ratio of the radius of the universe To the hubble to the hubble length, right? So this is the radius of the universe divided by the radius of One hubble patch so the reason this sorry The minus 2 I wrote it upside down So the reason this goes to zero Is that the radius of the universe is blowing up exponentially because inflation is making the space get larger and larger It goes as again is e to the n Well, nothing happens to the radius of a hubble patch h is more or less constant Okay, so so that's what's happening physically Now how big does does n need to be? Well, so let's say that omega k initially is about order 1 then if we want Omega k now If we want to make a k now To be less than 1 then we need This quantity to be greater than 1 It's 1 over omega k now Times omega k then So we need this one to be larger than 1 So how big is this? Well, how big is this? So I can write a naught as Ai so that was the scale factor at the beginning of inflation at t equals ti Times e to the n So that's the scale factor at the end of inflation times a naught over a E just equal to this So I've just written a naught multiplied by and divided by a e Now What is a not over a e? Well, I forgot to mention it this morning That when you have a gas of radiation in the universe in thermal equilibrium at some temperature and the universe expands The temperature scales Simply as 1 over a so it's not actually entirely obvious Why this should be true if you have a black body spectrum of radiation and you grow the universe You still have a black body spectrum. That's the part that's not entirely obvious, but it's true and The temperature scales as 1 over a so t Is proportional to 1 over a So this ratio here I can write as T at the end of inflation divided by t now and so using that and Just plugging into here. I find that e to the n Should be greater than h inflation over h now Times t now Which is 2.7 degrees Kelvin. That's the temperature of the CMB Divided by t at the end of inflation. How big is this right-hand side? Well, we know observationally for reasons that I'll hopefully get to by the end of this lecture that h inflation must be less than about 10 to the minus 5 times m-plank That's something we know from from observation if inflation happened it must have had an h less than this h naught is About 10 to the minus 60 Times m-plank. What about t at the end of inflation? well, it had to have been at least h inflation and How do I know that well again as we're going to discuss during inflation there's a Quantum effect Which creates a temperature? So even if you sort of start in the vacuum with no particles it doesn't last the space Produces radiation with the temperature. That's a water h And so the temperature at the end of inflation certainly can't be lower than that it can actually be higher than that quite easily Depending on what happens at the end of inflation, but it certainly can't be lower I'm not keeping track of changes in each for this. So this is just h during inflation which I'm treating as constant Yeah, I'm gonna come back and and talk about slow roll and and so forth But for this analysis, we don't need it. It's just it's h during inflation which which doesn't change very much during inflation Yeah, it might change by a factor of two or three or so, right? But but but for this at this level of squiggles it really it's changes. Don't matter. Yeah, thanks for the question Okay, so Right, okay So so so we'll put in the largest possible value for hi ten the minus five times on plank the smallest possible value for te and And what we get then is about ten to the twenty seven and So if we write that in terms of e or just take the log it says that n Should be greater than 62 And so again, this is sort of the worst possible case in the sense that if I made h hi smaller This number would be smaller if I made te larger it would also be smaller. So If n is bigger than 62 that's enough no matter what The Hubble constant in a temperature, but in many models. It's not necessary for it to be quite that large so Again, so what did we do here? We said we want to make omega k in the early universe very very small It's so small that Although it increases from the early universe until now it's still less than one today As we know is the case and as we said this morning that requires that omega k in the early universe be extremely tiny But inflation just does that it makes omega k Extremely tiny and tiny enough if the number of refolders at least roughly 60. Okay. Any questions about that? All right, so this solves the curvature problem It also solves the horizon problem Which should not be surprising? Because they're very closely related so to explain that we need to make We need to add one ingredient or it's not really one ingredient, but we need to Discuss the behavior of this particle horizon in a little more detail So, let's take the simplest possible month excuse me simplest possible model so We'll say that inflation begins at t equals ti and Then it ends at t equals t and Which would be? T i plus n over h And then immediately after inflation ends the universe becomes radiation dominated And then it's radiation dominated all the way from t and until t not So the first part of this is reasonably realistic In most inflation models at least the universe does become radiation dominated right after inflation ends But of course in the real universe Radiation gives way to matter and then matter gives way to vacuum energy or dark energy So I'm just ignoring those two. I'm ignoring matter and and vacuum energy Just to make it easier And then let me make one other simplifying assumption Let's assume that we can see we can observe the universe at t and In the real universe When we look at the cosmic microwave background, we're observing the universe at decoupling when Electrons are recaptured by protons and the universe becomes transparent But in this toy universe, we're gonna see all the way back to the end of inflation Okay, so the end of inflation is called reheating And so what I'm doing is identifying reheating with last scattering and again. I'm just doing this to try to To make the picture a little simpler Okay, so now let me show you what's gonna happen and then I'll justify it So this morning I drew a picture which showed how the particle horizon Behaves with time. I'm gonna draw a slightly different picture. Well, it's almost exactly the same So the way to think about this picture We're here. We live at time tau naught, which is t naught. I'll explain tau in a second We look back in time along light cones Right, so we look back with photons which propagate along these null geodesics and then we observe the universe in here Actually really we observe it only on this the rim of this cone, but in principle we could also see inside So anyway, we observe the universe back at this time time tau and now The horizon problem is that without inflation so with just radiation domination This slice here contains many Disconnected regions right so if we were to go back further And assume that the universe was radiation dominated the whole way Then this part is disconnected from this part is this this part this part. Okay, so there's many Disconnected Hubble volumes in our past like huh and yet they're all at the same temperature Of course, this picture is wildly out of scale because there should be something like What was it 10 to the 30 or something of these guys along this line? Okay, so so in fact these two lines should be extremely close together So now what happens if we add inflation, so let me keep this at the same level and Here I'm gonna put The other two lines This one is gonna be not quite as far up and Then this one it turns out is gonna be right here Very very close. This is town not and so now when we draw the past like on it's tiny and It fits easily inside one Like on on this tau initial surface So so our entire past light cone is Contained within the future of just one point on the initial surface As opposed to the future of many points as over here All right, so now let's see how that happens all I need to do is calculate these towels to justify this picture that I drew So let me tell you what towel is first of all towel is just the integral D a over over sorry dt over a and we might as well define it from t initial Up to some time t This is called the conformal time and it's equal to delta r So if I draw pictures where the vertical axis is tau and the horizontal axis is r Then light rays are at 45 degree lines Because tau is plus or minus r is the equation for one of these light rays Okay, so it's more convenient to draw things in terms of tau than in terms of t Now we just need to calculate how much what are these values of tau here in these two scenarios? Well, so we've already done it for the case of radiation domination Maybe I need some warm room. So let me go back over here the meaning of the number of e-folds. You're asking about yeah Good. Yeah, so I Just erased the scale factor, but let me write it again. So a is some a initial e To the h t minus t initial So this is the radius of the universe you can think of this and And n is just this exponent So it's just how large this exponent is at time t And so it tells you how big is the how big is the radius of the universe And how big is it compared to where it started at the beginning of inflation? It's bigger by a factor of e to the end so each each time 1 over h during inflation the universe increases in size by a Multiplicative factor of e after one e-fold. It's 2.7 times bigger after 2. It's 2.7 squared, etc So it's growing exponentially with time Let's see so you're asking about that number 62 over there Yeah, so that that's the that's the number so this is if you want n of t here here and This and over here. I wasn't very clear. This is an n of t and Okay, so it says that at the end of inflation there should have been at least this many e-folds of inflation Yeah, that's the total number Thank you Okay, so so let's calculate what happens to this tau with and without inflation It's a no inflation. We know that tau is well first a is So in this case with no inflation the universe is radiation dominated forever so it Expands like t to the one-half and a at t initial should be a initial so that's the right pre-factor And so if we compute tau We get And I can rewrite this in a slightly nicer way as 1 over h initial a initial times the square root of t over t initial Where h initial 1 over 2 times t initial So that's without inflation. What about with inflation so then we have a different Dependence for a so a is a I e to the h t minus ti As long as t is less than t and and greater than t initial and after that It's a Ie to the end so that's its value at the end of inflation times t over t and to the power one and now let's compute tau so tau is Computation we just did times a small term and then Dropping this term here, which is small It It picks up an extra term So there's this guy this first term here and then plus Something very similar to what we have there. So if we go to times Close to today, so this is much bigger than this we can neglect this as well and then I can rewrite this as follows and The condition I derived for solving the curvature problem Which is somewhere over here This one here if you do a little algebra you find that it guarantees that this term here is less than one Okay, so basically what it says is that there's very little conformal time Compared to the conformal time between T initial and T end So compared to the amount of conformal time that passes Here there's very little up here Okay, so it says this amount of conformal time is much less than this or less than this doesn't have to be much less And since the snarl razor at 45 degrees that guarantees that this past like on is smaller than this future like In other words, it guarantees that the part of the reheating surface We can see is smaller than the part that's contained within just the future like on of one one point Okay, so so the same condition that solves the the flatness problem Solves the horizon problem. It means that every everything we can see in the universe came from one Initial point on this surface at the beginning of inflation questions about this. Yeah, what did I win? Oh, what if you wait, sorry, what if I wait? Yes, good question. Yes, thanks Yeah, if you wait it depends on your future. So if your future is radiation dominated you will see more and more of the universe because Well This gets bigger and bigger and eventually or I should run over here, sorry eventually You'll see more than one horizon volume and you'll see sort of past the beginning of inflation Right, it's it's it's pretty intuitive that the further away you look the further back in time You're looking and that's what's happening here if you wait longer. You can look further because this surface is sort of further back in time, so you're seeing further back in time and And And yeah, eventually you see past the beginning of inflation and then you should see whatever crazy stuff was there before inflation started That's if your future is radiation dominated or any Decelerating phase, but in fact the universe today is accelerating not decelerating because of dark energy Which means we're entering a new phase of inflation and it turns out that that Screens us from ever seeing more than we already see Even if we could wait a trillion years We would barely see anything more than we already see In fact the CMB sky will get darker and darker exponentially And not only and so not only will it get darker it won't probe a larger region So we live at the best possible time to be doing cosmology and any more questions. Yeah, right? So in the real world Well, there's there's there's still a fact though that as you as you wait You'll see a bigger and bigger piece of the last gathering surface So you're seeing things that are farther away and therefore eventually you will see Predebations that were there before inflation started, right? So so yeah, you're seeing the universe at a fixed time You're right, but you're seeing it on larger scales and that effectively means you're looking further back in time in this sense Right, so this is just How it's the integral of this So it's it's it's tau as long as t is less than te. All right, so so I'm just performing this integral here and In this line, I've integrated up to a time t which is less than te. So I'm only integrating that part there And in this line, I get the contribution from here, which is mostly this piece Plus whatever I get from integrating that. Okay, which is which is here. So so Yeah, so this so this is the scale factor at any time and This is the tau corresponding to it at any time. Okay, so here it is I mean, you can just take this if you want, but this is this is a for te less than te. This is for te bigger than te The question was whether there's a constraint on and the number of e-folds from initial perturbations If the initial perturbations are very large, then they may prevent inflation from beginning at that time So in other words when inflation begins the initial perturbations can't really be larger than order one in some sense and then You might need a few extra e-folds to get rid of those, but it would just be a few remember I assumed that omega k was order one. That's already an order one perturbation. So not really I've pretty much put that in Yeah, I think I'm answering your question today. Yeah There was one up there Case one. Yes Case one is no k equals zero. I've ignored curvature everywhere Except okay, I'm doing the same thing I did this morning So as long as omega k is small you can ignore it when it gets to be larger than one you shouldn't ignore it anymore But it's never larger than than than order one in any of these calculations Yeah If you want to include it you can it just makes the math uglier and it would take longer to write on the blackboard But it won't change anything It's just roughly constant during inflation. We're gonna come back to that right now. Yeah, it changes slowly okay, so So inflation solves the horizon and flatness problems in the sense that if you take a universe without it You would have to tune the initial conditions very precisely to get what we see today But if you take a universe that does undergo inflation You don't have to do any tuning you can assume that there are order one variations in density and temperature on the initial slice and order one Curvature and so on and inflation gets rid of them for you So it solves those problems But you might wonder whether it's too much of a good thing after all The universe we live in is not homogeneous or isotropic. I mean just look around right this room is neither homogeneous or isotropic fortunately and Even if you look on on larger scales So even if you look at past the scale of the Milky Way There are galaxies in certain places They're not smoothly distributed everywhere and even if you look on very large scales on the larger scales we can see what you find is that fluctuations in density divided by the average density are something like 10 to the minus 5 or a few times 10 to the minus 5 So a few parts in a hundred thousand now If inflation lasted for too long then whatever was there initially gets just erased with exponential efficiency And so you might worry that there would be no fluctuations in density at all Or that they would become that you would have to very carefully tune and the number of e-folds To get this to turn out right But it turns out that's not the case and the reason it's not the case is related to Hawking radiation So you've all heard that if you have a black hole It has a horizon and it radiates Particles, it's not actually black It glows very faintly One way to understand this is that if you're standing or rather hovering near a black hole You have to accelerate Outwards to keep from falling in Just like standing on the floor you're actually accelerating upwards at 9.8 meters per second squared That's what an accelerometer would read So similarly to avoid falling into a black hole You have to have a rocket which is which is thrusting in in this direction So there's some Rocket which is pushing you that way and that keeps you from falling in and when you accelerate you measure particles This is a fact about quantum field theory that accelerated observers or accelerated particle detectors register the existence of particles even in a vacuum Okay, so what does that have to do with anything? Well, if we were in exact to sitter space, so if H did not depend on time at all during inflation Then there's a whole slew of coordinate systems you can write For to sitter space time. Let me actually write the black hole coordinate system first so this is a Schwartzel black hole So that's the metric for a Schwartzel black hole We won't ever need it again, but notice that this coefficient is zero and One I mean this coefficient is zero as well and there's one over it when r equals rs rs is the radius of the black hole The sitter space time so that's like inflation with h exactly constant has a metric Which can be written like this These are not the frw coordinates But they look a lot like a black hole in the sense that when r is equal to 1 over h This coefficient vanishes and so does this one when r is less than 1 over h Everything is normal So it's like an inside-out black hole. It describes the interior of an event horizon And if you want to avoid falling into that event horizon Then you have to accelerate towards the center. So you'll measure particles So this is a hand-wavy way of saying that in the sitter space. There's a temperature a quantum temperature Just like there is for a black hole And how big is it what can what can its value possibly be? Well, there's only one answer because Decider space time only has one parameter in the metric actually there's only one answer here, too. It's one over our Schwarzschulm There's only one answer here. It's h. So t to sitter H actually divided by 2 pi if you want the 2 pi, but basically it's h Okay All right now when black holes Hawking radiate You say measure a particle outside There must be an effect on the black hole by conservation of energy right if a particle has escaped the black hole lost that mass So it got smaller and it also deformed in its shape if you measure it in some particular position Then well, it has some momentum the black hole gets a kick, etc. Okay, so Hawking radiation has an effect on the black hole and similarly The center radiation has an effect on the universe It changes it so In order to calculate effects like that We need a model For what is making the universe undergo this exponential expansion? We need to know why The scale factor behaves like this, so what's the density that's driving it and so let me present the sort of simplest model for that It's a scalar field fine, it's called the inflaton because it's the field that causes inflation and It's Canonically coupled to gravity so here. I'm going to write an action and I'm going to do some variations if you're not familiar with This kind of manipulation I'll try to talk you through it, but I just need it to define some to find some equations of motion Okay, so that's the action. This part is the Einstein-Hilbert action. This is what gives you gr This part is the action for the scalar field and It's coupled to gravity because there's a g me anew here and there's a square root of minus g here Okay, so there are terms that involve g the metric times phi the field and that means the fields and gravity talk to each other Yeah, here. I'm not going to discuss the standard model or the Higgs or anything You can try to make the Higgs itself the inflaton. It doesn't work very well, but it's maybe possible And you can couple this to this this can be a new field that's coupled to the standard model That's fine as well as long as the coupling isn't too strong if it's too strong It'll mess up inflation if it's not too strong. It's kind of good actually it helps it reheat, but yeah But I'm not going to discuss the standard model Right now. I just want to explain how this can cause inflation. That's a good question It doesn't have to be a scalar field Scalar fields are the simplest sort of fields in some sense. They don't have any spins You don't have to worry about spinning disease. That's easier to write on the board They also can have Expectation values which don't pick out any direction in space so they don't break isotropy or homogeneity But it doesn't have to be a scalar there are models of inflation where the thing that drives inflation is not a scalar But I just wanted to focus on the simplest possible thing to so how to show how it can happen So let's figure out what team you new is So that we can decide whether this can actually Give rise to inflation So we can find team you knew the action with respect to the metric and just the part of the action which Depends on the scalar the other part would give us Einstein's equations if we do this variation So if we do this we get get this And there's an equation of motion for the scalar which is Okay, so it's a little complicated But we're going to assume homogeneity and isopropyl. Yeah, not with this sign you mentioned So I think this is right with with this sign. This is mostly minus So I think it's right. All right, so so So these look a little complicated, but let's assume as we did for the metric homogeneity Meaning that phi is only a function of time and not a function of space Then A lot of things simplify so row which is t zero zero is just one half Phi dot squared Plus v of Phi And P Which is minus t ii with with no sum is One half phi dot squared minus v of Phi and all the other components of the stress tensor are zero And so w p over row is One half Phi dot squared minus v divided by one half Phi dot squared plus v Okay And so if we want w To be close to minus one It means we want Phi dot squared to be small compared to v If Phi dot squared is small compared to v then this is almost minus one. It's a little bit more positive than minus one So that's what we need to arrange. We need to arrange that Phi dot squared is not important Compared to v and the equation of motion becomes just Phi double dot plus 3h Phi dot Plus or minus plus v prime You prime means dv by d Phi equals zero if you look at this equation of motion It's like the equation for a particle if you think of Phi is like x It's like the equation of motion for a particle subject to a force minus dv by dx Here's its acceleration with a friction term Which is proportional to h right? This is like a positive friction term That's the right sign for a friction term. It would slow the motion of the particle down And that's also encouraging because again, we don't want Phi dot to be very large We don't want the kinetic energy to be very large compared to the potential and so it's good to have damping That's going to slow down the velocity of the field if this is damped enough then Phi dot can be very small and And so it's possible for v to be more important than Phi dot and for it to stay that way for a while Okay, so we want something like a very an over damped Harmonic oscillator or something and in fact we can make it literally a harmonic oscillator or almost literally if we choose v To be one half m squared Phi squared Okay, so this is the famous quadratic inflation and So then v prime is Phi. So this is almost exactly a damped harmonic oscillator The only thing that makes it not quite is that h depends on v Because remember h squared is equal to row and sorry h depends on Phi because h squared is equal to row and Row depends on Phi over here. It'll be nearly constant. So it'll be nearly like a damped harmonic oscillator Good so now Let's see if it's possible To make w close to minus one so is it possible for Phi dot squared to be small What do we want we want that we want to be able to ignore the acceleration term because in an over damped oscillator You don't accelerate And we want to be able to ignore the kinetic energy and we also want h dot over h squared To be much less than what I already mentioned this. This is really the condition that we need This is the condition that says that h is changing very slowly during inflation. So we want all those things This is a comma. Yeah, these are up here. This is also a comma. Yeah, it should be much less than both of those things We should just be able to neglect this term in that equation of motion So we can check whether there exists a solution like this in a tricky way. We just assume it exists And then plug it back in and see whether it solves the equations if it does we found it so in other words we can check if it's self-consistent to assume these things and Checking that is going to take a little bit of algebra So maybe I'm going to Skip over a little bit You can do this very easily yourselves What you end up finding if all the for all these things to be true what you need is that you need v prime over v squared So this thing is usually defined to be epsilon and you want this to be much less than one dimensionless You need this and you need v double prime over v one over 8 pi g So this is defined to be eta and you also need this to be much less than one So if if both of these quantities are small Then all these things can be true There exists a solution where all these things are true at what rolls what rolls? Yeah, it's not a very good word It means that phi is changing with time slowly So you're supposed to think of a ball rolling down a hill, but there's no rolling There's no in our moment of inertia, but there's no angular momentum. It's just sliding really should be called slow slide But for some reason it's called slow roll. Don't ask me why Okay, so Yeah, you can check with just some algebra that if these quantities are small then If you plug this back in here You'll find a solution which satisfies all these conditions. Okay, so what this tells us is that if we have a potential v For which epsilon and eta are small then the universe can undergo inflation So first of all Let's see whether this can be true for m squared phi squared so If I take v to be one half m squared phi squared Then v prime is m squared phi So v prime over v squared Times one half times one over 8 pi g is Let's see. So there's a one half and then we have m squared phi Squared and then we have one half squared phi squared squared and so we should get to over phi squared And one over 8 pi g Is otherwise known as m plank squared. This is the so-called reduced plank mass Okay, so this is epsilon So you see that if you want epsilon much less than one Then phi has to be greater than m plank And so this is a so-called large field inflation model. It's an inflation model in which the infal time field phi Must be larger than m plank That's not necessarily a problem. It may be it's controversial whether it's an issue or not But in any case that's what the math tells us So now let's try to compute n at tn to the total number of e-folds Did we get out of this? So the number of e-folds the integral of h dt So here I'm going to keep the time dependence of h so I'm writing it as an integral Okay, because that's really what what defines n and Now what do we mean by t end? Well inflation will end when epsilon is no longer small So let's say it ends when epsilon is equal to one Okay, so that's a bit arbitrary of that definition, but it won't matter much if we choose it to be one or one half It won't make much difference. Okay, so then and Is the integral up to the time when epsilon is equal to one starting from Some t initial of h dt now Again I can go through the algebra First of all we can rewrite h dt as h d phi Over phi dot by a change of variable. So now we're integrating in phi from some initial phi up to the phi where epsilon is equal to one And h is proportional to the square root of v Which is proportional to phi So if you go through this it's ends up being 1 over 2m plank squared That's the integral of phi d phi. It's phi squared over 4m plank squared Evaluated from my initial to phi end, which is the phi where epsilon equals one So where's epsilon? Oh down here. Yeah. Yeah, sorry. Yeah, it's positive for this one, but you're right That's really the condition. Thanks Okay, so what is this phi end? Well, it's the phi where epsilon is equal to one So it's phi and is the square root of two and so what I get here then is Phi initial squared over 4m plank squared Minus one half So you see this definition of phi and contributed just this one half if we change it to epsilon equals one half instead of one It would have been something slightly different, but it barely makes a difference And so if I want n greater than 62 This implies that phi initial should be greater than about 16m plank Roughly, and so what happens in the solution? here's our potential and Here I'm gonna draw via five versus five there's some value here, which is root 2m plank and Over here also, which is minus root 2m plank In here, there's no inflation so oops in this region in here epsilon is greater than one and There's no inflation but here epsilon is less than one and also over here and So here there's inflation and actually here. There's also inflation you can inflate on either side of this thing It's symmetric so it doesn't matter and If you start far enough up like down here at 15m plank or so Then by the time you roll down to here the universe will have expanded by a factor of e to the 62 If you start further up, then it'll expand by an even larger factor, which is fine in five minutes We can talk about perturbations No problem. Yes Okay, so let's talk about perturbations so so far everything has been completely homogeneous and Well, it'd be bad if everything was completely homogeneous because then There'd be no stars or galaxies or people and we need to generate some perturbations So really the key is going to be this formula for n over here so Let me give you a preview of how this will work you see here that n depends on The initial value of phi the further away from the origin you start the more inflation you get So imagine that in some part of the universe during inflation. There's a random jump But this was the homogeneous value of Phi but somewhere in the universe and Phi gets a little bit of a kick so not everywhere but in one Hubble volume Phi gets a little bit of a kick either up or down this potential Just randomly If it's moved down its potential then inflation will end sooner If it's moved up its potential then inflation will end later in that region So what you'll end up with is a universe which ended inflation either earlier or later in some part and If you end inflation earlier Then that part of the universe is effectively older than what's around it and Since density decreases with time during radiation domination or matter domination being older means being less dense So you produced an under density and Exactly the same thing works in reverse If inflation ends later Because you fluctuated up the potential Then that part of the universe is effectively younger and therefore it will have higher density So if these fluctuations are happening randomly all the time during inflation You'll end up with a patchwork of regions where The density isn't exactly the same as you move from region to region. Okay, so that's really all it is It's just that this fields Cannot be perfectly homogeneous If you remember a little bit of quantum mechanics It's not possible to have a definite value of x because if you have a definite value of x then p would have infinite uncertainty In quantum field theory the scalar phi is like x. It can't have a definite value It must have some uncertainty in its value. Otherwise it would have infinite momentum Which would mess this up in a different way So it's impossible for phi to have a perfectly well-defined value if you measure phi you won't always get the same answer You won't always get the classical average value Yeah, sorry, why is quantum mechanics not not Yeah, so so the second part of your comment, which was that quantum fluctuations are small It's true and we're gonna find out in a minute how big they are. They are small The first part of your comment, which was that you can't trust quantum mechanics in a super planking regime Right, so I referred to this controversial issue that phi has to be bigger than a plank The counterpoint is that although the value of the field is bigger than a plank the energy density is way below on plank So there's it's not clear that there's a problem there Nothing obviously breaks down in effective field theory or even in perturbative quantum gravity So it's not at all obvious that there's an issue and it's because the value of the field is not Really something that that feeds directly into the action, right? What's in the action is the integral of the derivative of the field squared It's the kinetic energy or v of phi and all of those terms are much less than m planks to the fourth so this means that quantum gravity corrections at least at the perturbative level are all very small and Quantum field theory corrections are perfectly fine. It's a free field theory so Anyway, I don't really want to delve too far into that because we don't have to do this I just picked m squared phi squared because it's the simplest example but we could choose a more complicated potential in which phi doesn't have a super planking value and it wouldn't affect Much of anything else that I'm saying But your second comment about quantum fluctuations being small is It's very prescient So let's find out how how big there Okay So, yeah, let me do this since we have five minutes. Let me do it very quickly in a quick way Right so so I said that if there's a random variation in phi in some Hubble volume Then that will either delay or advance when inflation ends in that region of the universe So how big would you expect? Delta Phi to be this is the fluctuation in Phi with respect to this classical Homogeneous background that we've been describing. How big would you expect this to be? per Hubble volume per Hubble time Well, you probably know the answer. There's only one scale It has to be of order H Possible answer because there's no other scale in this problem Well, there's M, but M has to be much less than H. We didn't we didn't show that but it's a consequence of these slow roll conditions So So Delta Phi has to be of order H All right, so so what does this mean? Well, we have that formula for n over there And if I just ignore that one half It's one over four and plank squared times Phi squared So Delta n is one over two in Plank's grid Times Phi Delta Phi and Plugging in these values. So during inflation Phi is of order 10 times M plank or so Delta Phi is H. So I get a few times H over M Plank and I Told you before that first of all observationally this quantity must be less than 10 to the minus 5 Less than our border 10 to the minus 5. So that's an observational fact But it's also true for the reason that you pointed out that this had better be small because otherwise indeed quantum mechanics does break down if H is of order in Plank then the curvature of this Of this universe during inflation is of order in Plank. You can't trust classical gravity all bets are off Nothing here would make sense. So we must be in a regime where H Which is the inverse radius of curvature inverse radius of this of this universe Should be H should be smaller than M Plank. So H over M Plank is small And then it's true that that indeed this is a small fluctuation. So Delta n is much less than one Okay, so the typical change in how many e-folds inflation less is much less than one Now, how do we translate this? How do we translate Delta n into something observable? well We have our background metric It looks like this and now Let me replace a with a plus Delta a like this And then I'm just going to keep the first term not the Delta a squared term. So this Delta a Can be a function of position as well as a function of time And it describes the perturbation away from The Delta equals zero is the classical background that we started with Now How big is this Delta a over a? Delta a over a I Can rewrite as a dot over a Delta t So this is just H Delta t and H Delta t is Delta n Because remember that n is the integral of h d t so this Delta n is exactly without the two is exactly This perturbation in the metric here. I can write this another way I can also write this as h Delta phi over phi dot. I'm just using Differentials phi dot is d phi by dt and So this is another standard formula For the size of the perturbations And now what we've seen is that for this m squared phi squared model So this is true for m squared phi squared That this is a few Times h over m plank. Okay, so the bottom line is that These fluctuations in phi Which are of order h give rise to fluctuations in the metric Which are of order H Delta phi over phi dot or h over m plank in the case of m squared phi squared So it's not oh, I should warn you this is only for m squared phi squared So in a model with small field for instance, it would not be h over m and like h would be much smaller And this could still be the same size. It's because phi dot would be smaller But but for m squared phi squared, it's it's it's that simple and yeah, so so So so long as there are any variations in the value of phi and As long as they're of order h and not say much smaller or much larger Then you expect to find perturbations in the metric of about this order and in fact these perturbations in the metric can be directly translated into perturbations to the temperature of the cmb or perturbations in the density of of dark matter or For that matter clusters of galaxies and so forth in large-scale structure. So you've been learning about large-scale structure Where there's a power spectrum a primordial power spectrum of perturbations which then evolves in lots of interesting ways That primordial power spectrum is what this is producing So if I had a little more time I would translate this into momentum space and show you how it produces a scale invariant or nearly scale invariant spectrum of perturbations, but but we're out of time so I'll stop there. Thank you