 with questions as I lecture. If there's anything that doesn't make sense, just go ahead and ask. It's quite possible if it doesn't make sense, it's because of jet lag. So ask me, and I'll try to explain it better. So I'm going to give four lectures in a series. And I decided to do something a little bit unconventional. So instead of giving all four lectures on the details of inflation and how to calculate quantum perturbations and so forth, I'm going to condense that material down into the first two. And then in the last two lectures, I'm going to tell you what are some of the problems with inflation, what are some of the controversial aspects of it, and how does it fit into the bigger picture of theoretical physics, string theory, cosmology, and so forth. So if you want, the first two lectures will be about how to do research on inflation. The last two lectures will be about what to do research on. What are the interesting questions? So a little bit more precisely, in lecture one, I'll review cosmology. And I'll describe the problems which led Alan Guth and others to propose inflation in the first place. So that will be today and possibly later today, depending on, well, this would be the first lecture of the lecture now. And this lecture will possibly be later today, depending on Vitor's success in arriving. So in the second lecture, I'll describe how fluctuations emerge during inflation and eventually lead to structure. So when people say structure and cosmology, we mean galaxies, stars, galaxies, clusters of galaxies, the interesting stuff. And then in lecture three, which will be on Friday, I think three and four will both be on Friday, if I'm not wrong. So in lecture three, I'll talk about some controversies and challenges for inflation. And in lecture four, about the big picture, which for me means the strength theory, the landscape, and how inflation fits into that. Any questions so far? So let me ask you this. How many of you have followed a course in general relativity? And how many of you have not? Well, I won't assume much knowledge of that, but I will use a little. All right, so let's start off with a definition of what is cosmology. So cosmology, it's the study of the universe on the largest scales in space and in time. So this means what was the origin of the universe? What's the fate of the universe? And what's its large-scale structure? And if we want to understand the answers to these questions, what was the origin and fate of the universe? What's its large-scale structure? Inflation affects all of these very strongly. So it plays a key role in determining the answers to all of these questions. Now, we believe that the universe on these large scales is described by general relativity, hence my question. So in general relativity, as almost all of you know, one tries to solve Einstein's equations, which take this form. Now, throughout these lectures, I'm going to use units where h bar is equal to c is equal to 1. So if you want to restore the units, you just need to analyze the dimensions of the quantities you're looking at and multiply by the right number of powers of h bar and c to get the dimensions right. And there's always a unique way to do that. So you don't lose anything except when you save some chalk in doing this. And g here is Newton's constant. It sets the strength of gravity. If g equals 0, there's no gravity. T mu nu is the stress energy tensor. And g mu nu is the Einstein tensor, which can be written like this. So r mu nu is the Ricci curvature. And it's a function of g and its derivatives. r is the Ricci scalar, the trace of r, where trace means you can track the indices with g mu nu and then sum. And again, throughout these lectures, I'll use the standard Einstein summation convention. So if there's an index repeated, this is supposed to say mu nu. Let me write it a little bit better. If you see a repeated index, then it's summed over. And I won't write the sum. And g itself is a function of the coordinates x mu. Now, so g mu nu is the metric on some manifold. And I'll only be interested in four dimensional space time in these lectures. So g mu nu is a metric on some four dimensional manifold m. And because we're talking about cosmology, so we need to include both time and space. And so g has signature 1 comma 3. OK, so this means one direction, and that'll be time, has signature plus. And three directions, which will be space, have signature minus. This is a convention. You could choose it to be 3 comma 1 instead. And it makes no difference, except for some minus signs and various equations. But this way, if you want. Distances in time are real, and distances in space are imaginary. Anyway, that's the convention. And so this is described as, people say it's a Lorentzian signature, because it's the same signature as Lorentzian spacetime and Kasky spacetime. And it's pseudo-Riemannian is the term that mathematicians use. A Riemannian manifold would have 0 comma 4 or 4 comma 0. Anyway, so it's a spacetime. And more importantly, we have to be able to write it as a line times some three dimensional Riemannian manifold. OK, so this is time, and this is space. If we can't do that, it's not a cosmology. That's my definition of cosmology. So this M3 is some Riemannian 3 manifold. So it's Euclidean signature. And it has some fixed topology, which cannot change. It's slightly bigger. OK, all right, I'll write it larger. There's a theorem. This is a theorem in classical general relativity. So it's probably not true in quantum version. But in classical general relativity, there's a theorem it's due to Gerrach in the 70s. It's called the splitting theorem, which says that, first of all, whenever you have a Cauchy problem, so whenever you have a well-defined initial value surface that you can evolve forward and backward in time, the spacetime must split like this. And secondly, the topology of this spatial part can never change. It's basically because you can take the data on some initial time slice, and then you can evolve it forward. If the topology changed, there would be some kind of singularity. It would be impossible to evolve forward or back. So the proof is really that simple, just that you can apply a time evolution operator and evolve back to that initial surface. And that doesn't change the topology. So that's a good question. So there's a famous class of examples, which are what you learn first when you study this topic, which are the so-called Friedman, Robertson, Walker, or sometimes Lumetra gets added, cosmologies. And these are cosmologies where one makes a very special and restrictive assumption that, let's just say, space has two properties. It's homogeneous and it's isotropic. So the motivation for making these assumptions is that they are nearly true in our universe at the largest scales we can observe. And I'll come back to that quite a bit later. So when we look on the largest scales we can see today, we see a universe which is nearly homogeneous and isotropic. What's the meaning of these terms? Homogeneous means the universe is the same at every point. And isotropic means it's the same in all directions. It's spherically symmetric. At first sight, you might think these are redundant with each other, but they're not. For instance, imagine you had a universe which is full of an electric field that's always pointing in some particular direction and has constant amplitude. So that would be homogeneous. At every point, it's the same, but not isotropic, because this electric field chooses a direction. Conversely, you could have a universe which is spherically symmetric around some point, like where we are. But where the density or some other quantity varies with radius, that would be isotropic, at least at that point, but it would not be homogeneous. So these are two distinct features. In fact, it's rather difficult to test this one because after all, we're sitting at one point and we don't really have access to a full volume. We have access to our sky, which is telling us more about isotropy than it is about homogeneity. However, it would be rather unreasonable to believe that the universe is spherically symmetric and centered on us. And so the fact that it looks isotropic from our vantage point is a pretty good reason to believe that it's also homogeneous. And in fact, there are some tests of homogeneity that go beyond isotropy. Which, if we have time, we can discuss. That the curvature can change? Yes. That's right. Yeah, good. Yeah, so fixing the topology of a three-manifold, it has a pretty weak effect on the curvature. So for example, if M3 is a sphere, a three-dimensional sphere, then for any function on a sphere, there exists a metric for which that's the curvature. So absolutely anything can be the curvature of a metric on S3. It can be positive everywhere. It can be negative everywhere. It can be positive someplace is negative. So the topology doesn't have a very strong effect on the curvature. But it does have some effects. I'll actually come back to this in one of the last lectures. But that's right. So fixing the topology certainly doesn't fix the metric. It also doesn't fix the Ricci scalar or the Ricci tensor. But it does have some implications. And one of those implications is that there are actually only four possibilities for which it's possible for space to be homogeneous and isotropic. People often say there are three, because they forget one. There are four. So if we want to make this assumption, M3 should be an element of the set S, where S is four manifolds. So there's S3, the three sphere. Where should I put the three? I'll put it here. There's a manifold, which is the one people forget. S3 mod Z2, there's R3, and there's H3. So what are these? S3 is a hypersphere. It's a three-dimensional analog of a sphere. So it has finite volume. And if you move in any direction, you'll come back to where you started after a while. It has no non-contractable, sorry, all loops are contractable on S3, just like they are on S2. S3 mod Z2, this is also known as RP3. It looks just like S3 locally. It's also compact, finite volume. But it has some non-contractable loops in it. And you can think of this as S3, where you identify antipodal points. So if you go halfway around the world, you're back to where you started, instead of having to go the whole way around. R3, that's just Euclidean three-dimensional space. The usual standard, flat, three-dimensional space that you're used to. And H3 is hyperbolic space. So this one, of course, has infinite volume. So does this one. Hyperbolic space is a space in which areas and volumes grow the same way, at least for spheres that are larger than a certain size. If you're familiar with the Poincare disk or the Lobachevsky plane, or I think it has another name, those are two-dimensional hyperbolic spaces. This is three-dimensional hyperbolic space. Anyway, these are the four topologies that M3 could have if it's going to be homogeneous and isotropic. Now, you might ask, I mean, there are lots of other possible topologies, infinitely many. For instance, you could take R3 and make it into a torus by just identifying some directions. So you could have a three-dimensional torus, T3. But this can never be isotropic. And to convince yourself of that, imagine a two-dimensional torus. So you can draw a two-dimensional torus by taking a plane and then identifying this line with this line and this line with this line. So the identification is straight across. So these two points are the same. These two points are the same. And also these two points are the same. And these two are the same. I suspect all of you are too young for this, but if you've ever played the game Asteroids, you're a little ship. You shoot bullets. There's some rocks you're trying to blow up. And when your bullets exit here, they reappear here. They exit there. They reappear there. So they wrap around. That's a game played on the surface of a torus. OK, that's a torus. And if you stare at this picture, you can see that it's not isotropic. Because if I start from here and I shoot a bullet in that direction, it comes back and hits me after however long it takes to travel that distance. And that might be the same as this direction if it's a square torus. But if I shoot it along a diagonal, it may or may not hit me. But if it does, it'll take longer. Or it may not hit me at all. It may miss me. So different directions are not equivalent. And it's impossible to put a metric on the space that fixes that. It's just intrinsically not isotropic. So that's why there are only these four possibilities. And again, everybody tends to forget RP3 and say there are just these three. This is referred to as a closed universe, open, flat. Good. Now, it's important to say, and this was what the question was asking, that even if the universe has one of these topologies, it does not mean that it's homogeneous and isotropic. So the implication goes the other way. If it's going to be homogeneous and isotropic, it has to be a member of the set. But being a member of the set does not guarantee that it's homogeneous and isotropic. Because you can always put some matter in. Let's say you have your spherical universe. You can always have matter be concentrated in one place more than it is in another. And if you do that, the metric will respond because of Einstein's equations. And so the metric itself will not be homogeneous. There will be more matter in one place than another. OK. And another thing to say is that if the topology is not one of these, you can still be very close to homogeneous and isotropic. Because you can just make the universe very, very, very large so that, for instance, you'll never see this wrapping around phenomenon because it'll take too long. And in that case, you won't be able to distinguish this torus from just flat space. So it's not really motivated observationally to assume that the topology is one of these. Because you can always assume the universe is big enough that you won't notice the anisotropy induced by this other topology. Anyway, so that's FRW cosmology. And in fact, what does inflation do? So I'm going to come back to this in just a few minutes. But what inflation does is it blows the universe up and makes it really large, so large that all these effects of the topology can easily be far beyond our horizon and not observable. So inflation is a way of being sure that you don't care about the topology. Once it happens, if it lasts long enough, at least, it makes the universe so large that you'll never be able to measure the topology. And it simply won't matter what it is. Everything will look flat. But for now, let's assume the topology is one of these. So the universe is homogeneous and isotropic. And let's write the metric that goes along with that. OK, so three possible spatial metrics here, all of which are homogeneous and isotropic. One of them corresponds to the closed cases, one to flat and one to open. So k is an integer that's plus 1, minus 1, or 0. And d omega 2 squared, that's the metric on a two-sphere. So it takes this form. It's the standard round metric on a two-sphere. Now, to see that this metric actually describes what I said it did, let's take the case k equals plus 1. Then if I define a new coordinate psi as the inverse sine of r, then you can see this is going to become sine squared. And this simply becomes deep psi squared. So we just get this metric. And that is the metric on a three-sphere. Sorry, ask it again. Is it a square, did you say? Yeah, it's a square. D-sigma, the three? This one. Sorry, sorry, that's a square. Sorry, yes. Thank you. And this is also. OK, yeah. So this is the metric on a three-sphere. It looks just like the metric on a two-sphere, except while this d phi squared is replaced by d omega 2 squared. And notice that the range of r is from 0, the range of psi rather is from 0 to pi. Now, if k is equal to minus 1, then we can define sinh of psi equal to r. And this becomes a plus. And the metric becomes hyperbolic sine. So now you can see this fact that I mentioned that areas and volumes grow the same at large psi, psi bigger than about 1. Maybe you can see. So this grows exponentially with psi for large psi. So the size of the two spheres here grows very rapidly as psi increases. So that's hyperbolic space. And of course, for k equals 0, we don't have to do anything. This is just flat space. All right, so those are the three spatial topologies for homogeneous and isotropic cosmology. Now, I should have said this metric is a three-sphere with radius 1. Can people see the bottom of the board? Can anyone not see the bottom? So this is a three-sphere with radius 1. So when we multiply it by a, let's say with k equals plus 1, this is a three-sphere with radius a, a of t. You just make it bigger by a factor of a. It's a not a squared because this is ds squared. It's a distance squared. Now, that means that we can think of this metric as describing a space, which in that case is a three-sphere, which expands as time passes. So it's an expanding universe. Well, I shouldn't say it necessarily expands. It's size changes. We'll see in a moment that, well, typically it expands. And that in fact, it will be almost impossible to have a of t constant. But in any case, it's size changes with time. Let's just put it like that. Now, you might wonder if k is equal to 0, you already have an infinite volume flat space. What does it mean for that to expand? It's a question we're thinking about. It might become a little more clear in a few minutes. It does mean something. But in a way, it's convenient to think about this closed compact universe. It's a little bit easier to picture. And you can always make it so large that you can't distinguish it from flat space. So that's, I think, if you find that confusing, you can always imagine that the universe is closed. And so it's really a finite volume thing that's growing with time. So now if I take this metric, and by the way, you should convince yourself that this metric suffices to describe a cosmology where the space is homogenous and isotropic. It's not very difficult. If you take this metric and you plug it back into Einstein's equations, which are way over there, the result is pretty simple. It has to be because there's only one function in the metric. There's only this a of t. And what you find are two equations, one that depends on the first derivative of a. And another one, which depends on the second derivative of a. So t. OK, so t mu nu is a diagonal matrix. It has to be diagonal because any off diagonal component would indicate some kind of anisotropy. If these components here were non-zero, so maybe I should remind you that the 0,0 component of t is the energy density. The space-based components are pressures. These components here, along the top row, t is a symmetric matrix, so it's the same down here. These represent flows of energy. If you have a flow of energy, the universe is not isotropic because the flow is pointing in a particular direction. If you want, this is like a spatial vector. And you can't have any spatial vector non-zero if the universe is isotropic. Similarly, these represent anisotropies as well. And the diagonal elements have to be equal because otherwise there would be some, again, anisotropic component, something that selects out the different directions. So the spatial part of t has to be proportional to the identity matrix. That's the only rotationally invariant three tensor. OK, so t mu nu must take this form. And so we just have a row and a p. And this is how they appear in these equations. Dot means derivative with respect to time. And what else? This is the 0, 0 component of Einstein's equations. And this comes from the space-based component. Remember, there's many of Einstein's equations because there's a mu nu index here. But in this case, they reduce to just these two. Now, with some algebra, we can combine these two equations and write the following, that rho dot is minus 3h rho plus p. And that equation follows directly from the definition of rho and p, and then conservation of energy. It actually just says, well, it says that p is minus dE by dV. So dE is minus p dV. That means dE by dT is minus p dV by dT. And E is rho times a cubed, or at least it's proportional to that. So take our spherical universe. The volume is proportional to a cubed multiplied by rho. And that's the energy. So that's E. V is proportional to a cubed with the same proportionality factor. And now, if you just plug this in, you'll get this equation here. So the two Einstein's equations together enforce conservation of energy and momentum. Another thing to notice is this h, a dot over a. Note that it has units of 1 over time, which is 1 over distance since we said c equal to 1. OK, good. Now, what do these equations tell us? Well, among other things, they tell us that it's very hard for the universe to be static. If you want a dot equal to 0, you need h to be 0. So this has to vanish. And so that's a very particular condition on rho. And then if you want it to remain static, you need a double dot equal to 0. So you need rho plus 3p to vanish. So you can arrange this to happen. Rho generally is positive. So we need k to be plus 1 in order to satisfy that. But then we also need p less than 0, which is strange. You can do this if, for instance, the universe is closed and rho consists of two components, ordinary matter and something called dark energy, which has the property, or let's say vacuum energy, which has the property that p, p vacuum, is minus rho. So if I choose rho vacuum to be positive, p is negative. And then it's possible to make this happen. So there does exist a static solution to these equations. But it's extremely unstable. If I add just a little bit of rho, then this a dot will become positive and the universe will start to expand. If I decrease rho by just a little bit, a dot will be negative and it'll start to collapse. And that's only in the homogeneous case. So the homogeneous case is akin to this. It's like balancing a ball on the very top of an upside down parabola. Just the slightest perturbation will make it fall. If you allow inhomogeneities, it's even more unstable because inhomogeneities tend to grow and invalidate this whole analysis. So it's pretty much impossible to have a static solution to these equations. Cosmology must be time dependent. By the way, you all know the famous story that Einstein invented the cosmological constant so that he could find a static solution because he needed something like this to find it. He didn't notice that it was unstable. OK. Good. Now another thing to say about this. Ordinary matter and radiation have rho and p positive, or at least non-negative. So this means if that's the only thing in the universe that a double dot is less than 0. So if you have a universe that's expanding a dot greater than 0, its expansion will be slowing down. And this is completely what you would expect. Gravity is an attractive force. You have a bunch of stuff in the universe which is flying apart. That's what it means for the universe to be expanding. That stuff is attracting itself. And so its expansion is slowing down. Just like if you throw a rock in the air, its velocity, well, its speed decreases because it's attracted back towards the ground. So a double dot is negative. And what this implies is that the universe has a finite age, which was a shocking realization back in the day. So if we have a dot positive, so here's, I'm plotting a versus t, a dot is positive, and a double dot is negative. It's curving down. That's what it would be if a double dot were 0. So it hits this axis at some finite time, which we can call t equals 0. And at that moment, a is 0. The universe has 0 volume. That's a singularity because if we were to compute the curvatures, we would find that they depend on time. And they depend on inverse powers of a. So they blow up when a goes to 0. When a curvature blows up in general relativity, you can no longer trust Einstein's equations. So you expect that Einstein's equations will be modified by something. So we don't really believe that we can trust this analysis all the way down to t equals 0 at this singularity. But if we do trust it, well, it tells us that there's this point where the volume vanishes. OK, and this is, of course, called the Big Bang. That term was coined by Fred Hoyle, who was an eminent cosmologist. He made major contributions, very important contributions. But he hated this whole idea of an expanding universe. He fervently believed that the universe should be static. And he believed that to the point that he went on a kind of publicity tour where he would give public lectures ridiculing the idea that the universe could be expanding. And he made up this term just to make fun of this idea. He was more and more sort of ostracized by his contemporary physicists. But he became quite famous. He wrote books. So if you see people going around ridiculing ideas in modern cosmology, eminent cosmologists doing that, you might remember this story. It's an interesting cautionary tale. Anyway. OK, so that's the Big Bang. So now, to help us solve these equations, let's take this continuity equation that's written over there. And let me make an assumption that the equation of state p over rho is constant in time. If I make this assumption, then I can rewrite this equation like this just by dividing by rho. This is delog rho by dt, delog a by dt. So I can immediately integrate this and get that rho is some constant times a to the power minus 3 1 plus w. So now let me do some cases. w equals 0. So this means p equals 0. And this is usually called dust or pressureless dust. So it's like a bunch of particles that aren't moving. They don't exert any pressure on the walls of the container because they're not moving. So with this value of w, we find that rho scales as a to the minus 3. w equals 1 3rd gives us rho scales as a to the minus 4. That's the equation of state of a gas of photons, of radiation. So this you can understand, this a to the minus 3. It's very simple. You have some number of massive particles. And a cubed is the volume. So if the number is conserved, then rho will decrease like 1 over volume. Rho is n over v, where particles divided by volume. Similarly, this a to the minus 4 you can understand because if you think of a gas of photons, the number is conserved as the space expands. So the number density goes like 1 over a cubed. But each photon red shifts as the space stretches. So the wavelength of each photon increases like a. And that means its energy goes down like 1 over a. So each particle loses energy as the space expands. So there's an extra factor of a. And finally, if we choose w equals minus 1, so this is p equals minus rho, then we get rho is proportional to a to the 0. It's constant. So w equals minus 1 is what's called vacuum energy because it's an energy density associated to each cubic centimeter of vacuum. It doesn't change as a expands. So the total energy is increasing if rho is positive as the universe grows because its volume is growing. The energy density is remaining constant. So that's a very weird form of energy. Good. What about h? Oh, yeah. Yeah, w bigger than 1 is perhaps possible, but it's pretty weird. So let's see. If we plug it in here, it just means it doesn't look odd here if we just tell you that a red shifts very, very rapidly. But it's a very stiff equation of state. There's sort of more pressure than energy when you include the C's. So there's no ordinary form of stuff that has that equation of state. But whether it's really inconsistent, it's not entirely clear, I would say. The speed of sound is often related to w. So naively, at least, w bigger than 1 would give you a speed of sound that's greater than 1. And that means faster than light, hm? Is that perfect? Yeah, exactly. So it depends. It's really dp by d rho that matters. That's right. So if it's a perfect fluid, it would have speed of sound that's greater than 1, and that's certainly problematic. But there might be some other stuff that has speed of sound less than or equal to 1 and still effectively has p greater than rho. OK. Now, in our real universe, we don't have just one type of stuff. We have matter, which is non-relativistic and effectively pressureless. We have radiation. And it seems we have vacuum energy. So the actual universe contains all these components, which means, oh, I should do one other thing. Sorry. I should solve for a as a function of time before I say that. So we have that h squared. Let's take a to be t to the d. So then h, which is a dot of r, is d over t. So h squared. And now, if I plug in these scalings for rho, I have some constant a to the minus 3 1 plus w. And so if I invert this and solve for a, I find that a is proportional to t to the 2 over 3 1 plus w. So in the case w equals 0, we have a is proportional to t to the t thirds. Here, we have a is proportional to t to the 1 half. And here, it looks like we have t to the infinity. Well, actually, what we get is e to the t times some constant h. t to the infinity is sort of like an exponential, just like t to the 0 is like a log. OK, so those are the scalings. That's how the universe grows, if it's filled with a fluid or filled with a substance that has these different values of w. Notice that for both matter and radiation, it's t to a power less than 1. That means a double dot is negative. So the universe is expanding but slowing down. But for vacuum energy, it's e to the th a double dot is positive h squared. And so the universe is accelerating. It's expanding more and more rapidly. Yeah, yeah, I forgot to say. Yeah, so everything I've written on the board is valid. Well, everything I've written on the board is valid with k equals 0. Some things are valid also with k not equal to 0, but yeah. Here, I've ignored the k over a squared term, just for simplicity. We know in the universe we live in that this k over a squared term is much smaller than the row term, at least today. And it was therefore much smaller, and I'll come back to this in a moment, going all the way back into the past, except possibly in the very early universe. So it's pretty good approximation to ignore it in our universe. And I'll do that in some things I'm saying just for simplicity. Anyway, so our universe contains a bunch of things, not just one type of stuff. And so the actual dependence of A on T is much more complicated, not much more, but it's more complicated than this. But it's easy to work out. If the right-hand side of this equation contains a sum of terms like this, it's not difficult to solve. And what you find is that as the universe expands, it goes through various phases. So this here has the most negative power of A. These are the three components. This has the most negative power of A. So the smaller the universe is, the more important this term is. As the universe gets smaller, radiation gets denser, like matter, but also each photon gains energy. So this dominates the early universe to expect them to be radiation-dominated and to grow like T to the 1 half. At a certain point, radiation red shifts past where matter is, so if there's some radiation and some matter, at a certain point, matter will come to dominate. And from then on, the universe will grow like T to the 2 thirds until if there's any dark energy or vacuum energy, that comes to dominate. And then it will grow exponentially. And in the history of our universe that we know of, well, those three periods, we know very well. There was a period of radiation domination, then a period of matter domination, and now we live in a period of what seems to be vacuum energy domination. Hello, what time do I finish? OK. OK, good. So any questions on this? It's pretty strong. It's not true for just about anything in detail. So for example, if you have some massive particles which are moving relativistically because they're hot, then their W is not 0. It's more like a third. And then as they cool, they start to behave as pressureless dust. So if you really want to do this properly, you should take all that into account. But it's not very difficult to do for most forms of matter and radiation. But it does make the solutions more complicated than this. OK, so very good. So let's talk about causality. In general relativity, light is affected by gravity and it moves at finite speed. So let's work out how a ray of light or any massless particle would propagate in this metric. So again, I'll take k equals 0. Actually, let's write it in spherical coordinates. And let's consider a radial light ray. So a null, a light particle follows a null geodesic, assuming it doesn't scatter off of anything. A null geodesic has ds squared equals 0, 0 proper length. So this means that dt is plus or minus a dr for radial geodesic. If I integrate both sides of this, divided by a, and integrate both sides of this equation, I find that delta r is the integral dt over a. So this is the distance in these coordinates that light moves. If I integrate it from 0 up to some time t, it's the distance that light moved from the big bang up until some time. So for instance, suppose the universe was radiation dominated all the time. So if a is t to the 1 half, then when you integrate this, you'll get something like that. And so what you see is that this distance grows with time. And it's finite crucially. There was a question. Can you raise your hand? Sorry, I'm having trouble seeing. Ah, there you are, yeah. t times h over here. Yeah, you're asking about this equation, maybe? Or this one? Here. Yeah. Good. So h in this equation is 8 pi g over 3 times rho lambda. OK, so this is the energy density of the vacuum times this constant. That's what h is. And if I plug in a to the left-hand side, take two derivatives, and then divide by a, then it satisfies this equation. Is that clear? Yeah, this is not true for W not equal to minus 1. Yeah, no indeed. And in fact, it's very important that it's different when W is minus 1. That's the whole point. We'll get there in a minute. Thanks for pointing that out. OK, so assuming radiation domination for the moment, we find that light travels this distance after a time t. Now, notice that this coordinate r, it's the thing that's multiplied by a. So if I did k equals plus 1, this would be a sphere of radius 1. So you can think of this coordinate r as kind of like some tick marks on the universe. And as the universe expands or shrinks, tick marks remain in the same positions. OK, so delta r is some distance like this, about the same distance here. So this coordinate r here, or this distance delta r, is called a co-moving distance. Co-moving, co-moving. It's not a physical distance. It doesn't change. It's the same delta r in these two pictures. It doesn't change as a grows. a times delta r is a physical distance. It's the distance you would measure with a meter stick or something. OK, so this is the co-moving distance that light can travel after time t. Now, what does it mean? Well, it means that, again, imagine the universe was a sphere. You might complain that I neglected the k over a squared term when I calculated this. That's true. But never mind that for the moment. Imagine the universe is a sphere. If we get to large enough t, this delta r will reach pi, which would mean that the light has traveled half way around, or it'll reach 2 pi. Eventually, it'll travel all the way around. So in a universe which is decelerating, if you wait long enough, light will have propagated everywhere around the whole universe. Or if it's an infinite volume universe, it will propagate an arbitrarily large distance. So what does this mean? Well, it means that more and more of the universe gets connected to itself. So imagine that the universe, when it was born, was not in thermal equilibrium. Suppose it was hotter in one part and colder in another part. For two regions to come into equilibrium, so let's say this is hot over here, cold over here. For two regions to come into equilibrium, there has to be an exchange of energy. There has to be time for particles to propagate from one region to the other. So that can only happen at a sufficiently late time. Now, hold that thought. Another comment about H. If I take the physical, this is, by the way, it's called the particle horizon. So this is called the delta R. It's called the particle horizon. So if I look at this A times delta R and plug it in over here, A times delta R is t to the 1 half times 2 times t to the 1 half. So it's 2t. And that is 1 over H. So 1 over H is two things. First of all, it's the age of the universe, or well, twice the age of the universe in this case. And it's also the distance, the physical distance, that light can travel from the Big Bang until now. So it defines two things, the size of the universe. The size of the universe that's affected by one point. And also it's age, roughly. It's a factor of two. All right. So now let me explain the horizon problem, cosmology. So let's suppose that the universe was radiation-dominated all along. So let's just ignore matter and dark energy. Putting them in would change a few numbers a little bit, but otherwise it would have very little effect on this discussion. So here is t equals 0. This is a singularity, so let's make it spiky. And here is today. Today is usually denoted t sub 0. So if you consider yourself sitting at some point today and you ask, what parts of the universe can you see? We can draw some lines back towards the Big Bang. These wouldn't really be straight in these coordinates. It's t to the 1 half, so they should have some curvature. But we draw them back to this t equals 0 surface. And what we can see today, in principle, at least, is all the stuff in this region of the universe. So there's some delta r here. That's the co-moving distance that light traveled. So it traveled from here to here, traveled delta r. On the other hand, let's consider some very early time. Let's call it t initial. So this region here has access only to this part of the initial conditions. And this region here has access only to this part of the universe and so forth. Delta r is much smaller back then. So what you see is that the state of the universe at this early time is affected by many disconnected pieces. It splits apart into many disconnected volumes, which have not had time to communicate to each other. So imagine we can measure the temperature of the universe at this time. Would we expect it to be the same everywhere along here? It depends on what you believe about the initial conditions. But if there was any inhomogeneity in the initial conditions, if there was any variation in temperature or density in the early universe, it wouldn't have had time to thermalize. Because this region and this region and this region, they never talked to each other. They're completely disconnected causally at this point. And in fact, in cosmology, we can do precisely this. We can look back and we can measure the temperature using the cosmic microwave background of the universe at an early time. And what we see is that the temperature is almost exactly the same. It only varies by one part in 100,000. It probably varies by less than the temperature in this room. Why would we or would not? Why would we? I'm not sure what to expect. So no one really knows what the initial conditions should have been. And indeed, when we come back to the sort of controversies about inflation, this is one of them that we'll discuss. But well, it would be odd if the initial conditions were almost perfect but not quite perfect homogeneity. So what we're going to learn is that inflation provides a mechanism that essentially, no matter what the initial conditions were, produces a universe that looks just like the one we see. So without inflation, you'd have to assume that the initial conditions are exactly right to produce what we see. You wouldn't have any mechanism to explain it. But we'll come back to this quite a bit. So that's what it sounds like. But when you work at the math, that's not what you find. So let's not actually go to the Big Bang. Let's go to this early time because the Big Bang is a singularity, so it's a little hard to talk about. So then the universe is not really a point. It has finite size. And what we can see is that these regions can't talk to each other until a much later time. So for example, this guy and this guy can only communicate way up there. So if we go all the way back to the Big Bang, it is confusing. It's a singularity. It has zero volume. It's hard to know what to say. But we probably shouldn't do that anyway. We should go back to some very early time, but not quite all the way back. Yeah? No, I mean, I wouldn't expect any of the equations I wrote to be valid literally at t equals 0. In particular, general relativity is almost certainly not valid. So we know from quantum mechanics that there should be large corrections to it before a certain time. So yeah, I wouldn't trust it at t equals 0. But I would trust it at some early time here. So at least there's no reason not to. OK, good. So how many in this picture, it looks like about n and r10, but how many of these volumes are actually there? Well, of course, it depends on what we choose for t initial. So let's choose the Planck time, which is 10 to the minus 43 seconds. The motivation for choosing that is that we certainly can't trust these equations before that. So that's as far back as we can go. The Planck time is the time when, well, the energy density of the universe will be of order of the Planck density. It's the time when large corrections to general relativity certainly kick in. So we certainly can't trust GR before this time. If we go back to that time, then we said that the particle horizon grows like t to the 1 half. So delta r of t0 divided by delta r of t Planck is the age of the universe today. Anybody know what it is in seconds? You can remember it like this. It's 10 to the 10 years, roughly 10 billion years. 10 to the 10. And a year is pi times 10 to the 7 seconds. So it's about 10 to the 17 seconds. OK, so 10 to the 17 seconds divided by 10 to the minus 43 to the 1 half. So this gives us about 10 to the 30. That's linear scale. If we want to know how many volumes are in there, then I should cube this. So there's about 10 to the 90 causal volumes, or Hubble volumes, let's just say, inside this code here. Remember, this is really three dimensional also. OK, so there's about 10 to the 90 different regions in there. Actually, if you include matter and dark energy, you get more like 10 to the 80. But it's a very big number. So why do they look all the same? There's another problem, which we're going to see is very closely related, called the flatness problem. 10 to the minus 43 seconds. Sorry, I'll try to write bigger. OK, let's talk about the flatness problem. So let me put the curvature term back in. So we have h squared. And now it's convenient to take this equation and divide both sides by h squared. And this is called omega total. This is called omega k. So we have 1 equals omega total plus omega k. Not that omega k is negative when k is positive. That's some annoying convention. But anyway, it looks like this. Or omega k, let's put absolute value signs, is 1 minus omega total. And by the way, this omega total contains all the rows. So it's got row matter, row radiation, row lambda, if there is one. We know from observation that this can be no bigger than about 10 to the minus 2. That's why I said we can ignore it. In the universe today, this term is at most about 1%, about 10 to the minus 2. So now let's go back. That's today. So let's go back to this time t plank and see what happens. What happens to omega k? Well, it scales like 1 over h a squared. And for radiation domination, a is t to the 1 half, h is 1 over t. So this goes as t. So if I go back, excuse me, if I go back, and I evaluate omega k at t equals t plank, it's about 10 to the minus 60, which is certainly a small number. Now, what does that mean physically? Well, there's a nice way to think about omega k. So if I write a h as a divided by 1 over h, can't stop me from doing that. So this here, this is the radius of the Hubble volume. It's effectively the size of the universe, the physical size of the universe that you can see. And this is the radius of the whole universe. It's the radius of curvature. Remember, in the closed universe, a was literally the radius of that sphere. So this is the ratio, a h, is the ratio of the radius of the universe divided by the Hubble length, divided by the part you can see. So if you want, it's the size of the universe divided by the size of the observable universe. OK, now a h is 1 over the square root of omega k. So the fact that observation tells us that omega k is less than 10 to the minus 2 tells us that a h is greater than about 10. OK, so what we know today is that if the universe has a curvature, if it's a sphere, the radius of that sphere is about 10 times bigger than the observable universe, than our Hubble volume today. 10, because it's the square root, it's 1 over the square root of omega k. And now, if we go back to this time, t-plank, it tells us that the radius of the universe back then, so this is a h today, if we go back to t-plank, then it would be about 10 to the 30, square root of 1 over 10 to the minus 60. So we would say that the size of the universe back then was 10 to the 30 times in linear scale, the Hubble length back then. Or if you cube it to get a volume, you would get 10 to the 90 Hubble volumes per universe. OK, so it says that the universe back then was much, much, much bigger than this causal size. So there's another way of looking at this flatness problem. So how did it get so big? How did it get so flat? And again, you can say this either way. You can say, what set this number to be so tiny, this omega k back at this early time, to be so tiny? Or if you're looking at this way, it says, what says that the total size of the universe, which is at least this big, is so much bigger than the characteristic scale at that time, than the Hubble length at that time? OK, so these are two ways of saying the same thing. And these problems are very closely related. I mean, you'll notice it's exactly the same number, or pretty close anyway. What it really says is that without inflation, the universe had to be very, very flat and very, very homogeneous on scales much longer than c times t, where t is its age, much longer than the light crossing time of the universe, the distance light could travel back then. OK, so I should finish. And unless there's any last questions? Yeah, thank you. Yeah, that's true. I cheated a little bit. Yeah, so when I derived the scaling of a, I assumed k was 0, and now I'm taking it not to be 0. Right, so the only time that will matter is when this term becomes large compared to this term. In other words, if we keep this term, but it's small compared to rho, it won't change very much the solution we found for a. So it'll be pretty close to t to the 1 half. At sufficiently late times, that will no longer be true, because this term will grow larger than this one. So in the late universe, this t to the 1 half is not correct anymore, and there's a slightly different solution. But in the early universe, it doesn't matter. And since the universe today has this small compared to this, it really doesn't matter much. But you're right, we could do a more careful job where we take both terms into account. It wouldn't change anything. This number wouldn't change at all. Not this one, but this one, just because it's completely irrelevant in the early universe. But yeah, thanks for the question. Yeah, oh, yeah, that's a good question. I would say, yes, there is a difference. They enter Einstein's, yeah, the question was if there's a difference between vacuum energy and cosmological constant. So Einstein added a term, I erased it, but he thought of it as a term on the left-hand side of his equation, so part of the geometric, you know, part of the curvatures, which was lambda times T mu nu. And so that was the cosmological constant. If you shift that term over to the right, then it looks like it's part of T mu nu. Of course, it's the same. So in that sense, they're identical. However, vacuum energy is not necessarily fixed forever. It's something that can change. For instance, you might have multiple vacua in a theory, multiple stable phases, which have different vacuum energy. So you can have theories with multiple possible values of the vacuum energy. But strictly speaking, there's only one cosmological constant. So they're not exactly the same, but as long as vacuum energy doesn't change, they play exactly the same role in the equation. OK, so I think we have a coffee break. If anybody has more questions, you can talk to me then. So thank you.