 So thank you, it's a great pleasure to be here. I apologize that I missed my flight yesterday. I will try to make it up in the next couple of days. I will continue along the lines of Stas Babak. He told you about, well actually the tough things about gravitational wave detection and generation. I would like to go a bit back to basics, OK? So I'd like to understand exactly what gravitational waves can tell us about gravity. Let me point out that there's reasons to do that. Well, there's actually lots of reasons to do that. I'd like to show this slide. It tells everything we need to know about the future. And that's why I'm very, very happy to be here, because you're the future. So hopefully you'll be telling us something about gravitational waves in 10, 15 years that we didn't know before. We're basically opening a new channel into understanding gravity. And I think Bernard Schultz summarizes very neatly when he said that it's like being death all of your life, and you suddenly have a new channel of perception. There's a lot that goes into gravitational physics. Stas told you, as I said, about the tough part, data processing, doing much field to dig out signals from noise. There's astrophysics. I'm not going into that. I will focus roughly on one item in this list, and that's black hole physics. So my plan for the lectures is the following. I know I missed one. I will try to make up for it. So my idea is to tell you a little bit about regions of spacetime where gravity really becomes important, mostly black holes, although almost everything I say will carry over to, say, neutron stars. By the way, I think I sent some notebooks along. I hope they pass them to you. Let me tell you that some of them are not as efficient as they should be. So the idea was that you understand very clearly how we do calculations or how we can do calculations. You can write it in Python, in C++, whatever, Fortran. Is there a question? Whatever you want. But the idea is being able to follow what we actually do in the field. Very good. Some of the references are these ones. I'll send these slides along as well. There was a recent work a couple of weeks ago that basically summarizes the state of the art in the field of gravitational wave science. So the astrophysics part, the source modeling, and the fundamental picture point of view. By the way, if you have questions, just raise your arm. Now, it's kind of curious that I'll mostly focus on an object that was predicted to exist almost, well, actually more than 100 years ago. This was the first non-trivial solution to the Einstein field equations. It was written down actually earlier by a student of Lawrence. But he was traveling, so he could not publish the results before Schwarzschild. So the morality is always telling our advisors to publish everything very quickly. The reason why I'm going to talk about black holes is not because they were discovered 100 years ago, but it's because mostly of this property. Black holes in GR are unique. This means that if you give me two parameters, mass and spin, I know everything about any black hole in the universe as described by general relativity. The solution, by the way, that I have written down there, it belongs to the Curve family. And it basically describes a black hole with a mass m and the spin a in units where c and g are one. So actually, let me start by telling you I'll always be using these units. And in these units, the mass of the sun, just so that you have an idea of what I'm talking about, is of order 1.48 kilometers, which is 4.92 10 to minus 6 seconds. Whenever we see a quantity that scales like the mass, if that quantity is as a character of time, we know it's of order 10 to minus 6 seconds for something like the sun, OK? Very good. Of course, I think first I'd like to understand if we can actually test the statement that any black hole in the universe is described by two parameters. This will come tomorrow. But I think it's kind of funny, right? That it doesn't look like anything much. But if you think about this room, right? This room alone is described by 10 to the n with then a very, very large number of parameters. So stating that there's billions of objects in the universe and that I know everything about them with just two, it's really an amazing statement. It's really an amazing statement. Actually, I think the content of the statement took decades to understand and took many, many, many, many, many people and I think we're still understanding the content of the statement. Some people rephrase the statement by saying that black holes have no hair and that if you collapse two objects as long as the final mass and spin is the same, then that's it. There's nothing else you can say about it, OK? Of course, the statement, by the way, only refers to stationary geometries. So the dynamics, which I am going to review tomorrow again, may be different, OK? So black holes are very, very simple. They are the elementary particle of gravity and there's actually a nice conjecture about black holes. It's called the hoop conjecture and it tells us that I am not going to actually explore this conjecture any further, but it's important for the motivation for all of this. The hoop conjecture states that if you're able to squeeze matter in a region which is small enough and small enough means that you take a hoop of radius, the partial radius of the object, if you're able to pass the hoop in all directions and the object is contained in the hoop, then the object must be a black hole, OK? So you take me, OK? My mass is of order 75 kilograms. My partial radius is tend to minus something meters. So you can't place me inside the hoop. So I'm not a black hole. That's very good. If you take an electron, an electron is tend to minus roughly, tend to minus 17 centimeters. The partial radius is tend to minus 55, so again, not a black hole. Of course, you can get around this because everything gravitates. So if you boost me or an electron, then suddenly my partial radius becomes larger, right? It's proportional to the Lorentz-Boost factor. So if you take an electron and you boost it to a gamma factor of 10 to the 30, then all of a sudden the hoop conjecture will tell you that it looks like a black hole now. So if you collide matter at very, very large energies, then you may succeed in getting something that looks like a black hole, OK? Actually, this was tested a few years back. And I like the experiment a lot. So the experiment was the following. I take two stars, OK? Of course, they are stars. They're not black holes. The compactness is 10 times smaller than that of a black hole. So you need to either put 10 times more mass or squeeze the objects 10 times to get a black hole out of them. So people started colliding these guys. And if you collide them, so you see the calculation would be at Lorentz-Boost of the order of 10, gamma equals 10, there should be a black hole somewhere in the spacetime. So at Lorentz-Boost of 2, so by the way, these stars are special stars. They are made only of scalar fields. And at Lorentz-Boost 2, you get this. So just a scalar field, these stars pass through one another. There's some changes in the scalar field and that's basically it. But something funny happens when you increase the center of mass energy. This is what happens at Lorentz-Boost. Sorry, let me see if this plays. Lorentz-Boost 4, we were expecting something to happen at Lorentz-Boost 10. And actually, we get a black hole already at Lorentz-Boost 4. So even smaller than the threshold predicted by the hoop conjecture. Actually, the hoop conjecture also tells you that it should generically form black holes if you fluctuate matter at large enough amplitudes, right? As long as the amplitude is large enough that you satisfy the hoop conjecture, then you're going to collapse to a black hole. So the conclusion of this very, very short story is that black holes are super. They are super, guys, and they are super objects to be discussing in this fantastic venue. They are super compact. They are super dark. Or I think we think so. They are super simple. And therefore, they are ideal laboratories to test fundamental properties, right? Actually, the very notion that we have black holes in the universe should be challenged and can be challenged using, I hope, these three properties. And we're going to be discussing that. OK, so my framework is going to be GR. So Einstein tensor equals to some stress tensor. Everything or almost everything I say will apply to any theory you like, or at least the generalization should be obvious. Well, at least a fraction of what I'm going to say is obvious. The other fraction is terribly difficult, OK? Of course, we're all unique given a sufficient amount of constraints. This theory is unique in the sense that if you try to build a theory which has at both second-order derivatives no extra fields and satisfies a coordinate reparameterization invariance, then it's the only theory you can build out, OK? So it's unique in some sense. There's a nice property of the Einstein tensor, and it's called the Bianchi Identities, divergence is 0. And this means that this derivative is 0. And this basically encodes the equations of motion. So the equations of motion, which in Newton's theory you had to prescribe, right? It's not enough to tell me what the gravity law is. I also need to specify how objects move given a force law. It's basically built in the field equations. Of course, we still need to work this out. But if you work this out for, let's say, point particles, then you get the geodesic equation. And the geodesic equation tells you that the second derivative with respect to some affine parameter satisfies this. Yes, I will be using this. By the way, one of the notebooks implements this. And I show there something very simple, which is called the Birkoff theorem. If you can't open, just let me know at the end. Which basically amounts to saying that there's only one solution of the field equations, which is spherically symmetric. And this is known while you can change coordinates and put it in the form of the Schwarzschild solution with some R, which is an aerial radius coordinate. Schwarzschild is the most general, so it's a very symmetric solution. Very good. So the reason why I'm even discussing geodesic motion is, well, if I'm going to study black holes, I need to understand how light and matter moves around black holes. So what I'm going to show you now is exactly how we can do this, right? Now, of course, in principle, the only thing you need to do is to, well, take your geometry, find the Christoffel connections, solve the geodesic equation of motion, and they're done. There's an easier way, and the easier way is to recognize that there's a Lagrangian four-point particles in the spacetime, and it's basically this. So for instance, if I take the Schwarzschild geometry, the Schwarzschild geometry in these coordinates, I'm going to write it in this way, m is the mass, this is the two-sphere. Then I can write the Lagrangian, and the Lagrangian, by the way, this is 0 for light, no particles, and minus 1 for time-like motion. Then I get the following. I get that 2l is minus f t dot squared plus 1 over f r dot squared plus r squared theta dot squared plus r squared sine squared theta phi dot squared. And now it's kind of trivial to see, well, it's a textbook example, right, to get equations of motion and to get spatially conserved quantities. You look at the spacetime, actually, it doesn't depend on time nor on the azimuthal coordinate phi. So if you write down the Euler Lagrangian equations, then you'll find that there's two conserved quantities, the energy, which I'm going to write in this way. So by the way, f is this guy. There's a conserved energy, there's a conserved angular momentum, and now the only thing I need to do is to write down the normalization condition, this one. And the normalization condition gives me that 2l, let me call this delta 1. It's either 0 or minus 1. And so this will be r dot squared equals f e squared over f minus l squared over r squared plus delta 1, which I'm going to call the effective potential for geodesic motion. Very good. Again, you can do this for any spacetime, of course, especially if the spacetime has enough symmetries. You can use the Lagrangian approach. This is well known. Now you can start thinking about, well, OK, what's the features of this motion? Well, you start thinking about circular motion. And by the way, one of the reasons why we like circular motion so much is that if you take a small mass, of course, there's no point particles, right? It doesn't exist. So there's higher order corrections in all of these. They scale like the mass ratio of the object. I'm not going to that. This is called the self-force problem. But this is enough for me. If you throw this guy into a black hole spacetime, it's going to move according to this effective potential. But if you now take into account gravitational wave emission, which it does, it's moving. There's some very quadruple moment that Stas Wabak told you about. What you find is that very, very quickly, the orbit tends to circularize. So gravitational wave emission circularizes motion. Very good. Now, circular motion means that, well, our lot. So we have two conditions. If it's circular, then the radius is constant. And the derivative of the radius is also constant. So it's a one-minute exercise to show that if delta 1 is 0, actually, I can write down. Well, I have derivatives here, never mind. If delta 1 is 0, these two conditions are only compatible if r is equal to 3n. This means that there's a very special location in compact objects with some people called the light ring, some people called the photon sphere. Photon sphere, because you can generate the spacetime by rotation. I mean, this was only an equatorial cut. If you rotate this, you get a sphere. So there's a very special location where I can try a laser beam. And the laser beam, basically, it's a high frequency beam. So it's kind of a null particle. It will orbit the central mass in a circular motion. It does need to be a local, right? Anything whose surface is below 3m will do this. Of course, again, I mean units where gc is 1. So there's a g over c squared here. OK, question is, is the motion stable? It's not. So the motion is unstable. Actually, you can compute easily the time scale if you basically perturb this equation, right? So you do r equals the circular radius 3m plus some delta. And your r dot squared will be, well, will be the potential at the point 3m, which is 0, plus r minus rc times v prime at the circular point, plus 1 half of r minus rc squared and the second derivative. And what you find is all the usual stuff, right? That the solutions are exponential. So this and these are 0 by definition of circular motion. This is basically delta. This is also basically delta. So what you find is that the solution is an exponential delta equals e to lambda t. And lambda is this number. It's a little prime over 2 dot squared. I mean, actually, this is kind of funny. I never see this in textbooks. But it's a textbook calculation, right? If you do the calculation in this corner, then you're going to get a tau here. So you need to use a t dot to change. This is t dot. You need a t dot to change from tau to t, but that's it. This is equal, by the way, to 1 over 3 square root of 3m in the Schwarzschild geometry. This is also called the Lyapunov exponents of null motion. But you see, it's really just solving the genetic equation. So if you want to put numbers in, then 1 over lambda is 1.5 milliseconds m over 30 solar masses. So for the event that LIGO saw, the first event, then the time scale, the instability time scale is of order a millisecond. So if you try to put light orbiting a lago on a circular motion, and the lago is 30 solar masses, it quickly, quickly meaning 1.5 milliseconds, will disappear. Either go off to infinity or fall down the hole. Of course, you can also repeat this. So this was basically just circular motion. You can also ask the question, well, OK, let me just take a lantern and shine photons on a buckle by buckle. I mean this geometry. Where's the geometry? I mean this geometry. I shine buckles here. And let me see which I shine photons here. And I see which photons actually go back to infinity to some target I might have. And which ones fall down the center. And if you do this, well, it's kind of easy to the calculation. It's in the notes. The idea is really to rewrite this as a following. So I'm going to rewrite the effective potential in this way. I divide by the energy. And I define an impact parameter b. And this I can rewrite in the following way. 1 minus 1 over r squared minus 2m over r cubed. So it's clear that there must be a turning point, meaning there must be a point at which this is 0. So if I shine something, eventually it has to come back, right? Just looking at the large v limit. Now if I define y in the following way, and I rescale my impact parameter as this, then I get the following equation. r dot squared equals to 1 minus 1 over y squared, 1 minus 1 over y, v hat squared over 4. And this is basically a cubic. Excuse me. Excuse me. y cubed is 1 minus v hat over 2. 1 over y square root of 1 minus 1 over y. 1 plus b over 2. 1 over y square root of 1. You can reduce this to a cubic equation. It's very simple. And you're going to find the following. If b, if impact parameter b, larger than 3 square root of 3m. So at a large distance from the central object, I throw in a photon. If the photon has an impact parameter larger than this, then it's not absorbed. So the absorption cross-section of a black hole, well, absorption cross-section for particles, for null particles, is this i v squared, which is 27m squared. This is for light. Of course, you can repeat all of this for a time-like motion. You can throw this microphone into this geometry. And you're going to find that radial motion, you're going to find the following. Again, just geodesic equation, you're going to find that dt is e over f square root of e squared plus f delta 1 dt. Actually, this is more general than just radial. It's general motion. And of course, what this tells you is that the coordinate time to reach r equals to m diverges. Something well known, but it means there is something kind of funny going on. And the reason why I'm giving you this kind of straightforward example is that it shows immediately that you cannot get an observational proof. You can never get an observational proof of buckles. So whatever signal, gravitational wave, electromagnetic wave, neutrino, whatever you get, it can't be proving anything about the existence of the r equals to m surface. Just because it takes an infinite amount of coordinate means this coordinate time is actually the time I measure in my detector. So my detector has to measure an infinite amount of time before anything has had time to go to the 2m surface and come back. Very good. So this is actually the summary of what I was saying. There's a 2m surface here, this black circle. The 2m would be, this is a non-spinning black hole, geometry, would be my event horizon. There's the photon orbit here, the light ring, this 3m surface where this laser beam can orbit in a circular motion. And then I didn't, actually it's written in the notes, but it's quite easy to do the same, do circular motion for time-like particles. You're going to find, of course, that at large distances you recover everything that Newton had to say about it. So circular motion is possible, motion is stable, so that's a good thing to know. So here, everything we knew from Newtonian gravity holds, if you try to place an object on a circular motion too close to the object, you're going to find two things. First, beyond a certain place, circular motion is possible, but it's unstable. So this is called the innermost stable circular orbit. You are not supposed to find anything within that radius because if you have an aggression disk, if you have a star, a small star orbiting a black hole or anything that's here, it doesn't need to be a black hole. Quickly, beyond this point, that motion becomes unstable. Anything will make the star plunge into the central object. So this region should be empty of matter. We are not expecting anything here. This is good. This is good because some of the things we want to test actually is the presence of a photosphere of this light ring. This is a very special relativistic effect. Newton's gravity does not have this. And we also want to test another region called the Ergo region. I'm going there in a minute. The Ergo region is very, very unique. It's a region around spinning objects where you can have negative energies. It's also called the static limit because anything that enters the Ergo region, this dashed line over here, cannot remain at rest with respect to outside observers. So if you take a huge powerful rocket, you can always stand still here, stand still with respect to me. However, it doesn't matter how powerful your rocket is. If you enter the Ergo region, you are always, always forced to co-rotate with the geometry. OK? Very good. So geodesics is all there is. Actually, it's all people do with telescopes. You just follow ray tracing. You just follow light rays from stars or accretion disks. And they give you these beautiful images. This particular case is a star. It's supposed to be something that the event horizon telescope is going to measure. But my point is that you look at this and say, wow, these guys are going to see the horizon of vehicles. Actually, it's called the event horizon telescope. But this is false. You see, if a light ray has an impact parameter smaller than this, smaller than this, it has to plunge in the vehicle. And actually, the way it plunges is asymptoting the light ring. So you throw a light ray with this impact parameter. It's going to spiral infinitely many times around this region here. OK? So any photon that you throw in that enters the photosphere, the light ring, is not going to escape. What this means is that this picture, it's not a picture of horizons. It's a picture of light rings. OK? This circle here is not, I mean, this black stuff inside is not because there's an horizon. It's because there's a light ring. OK? So photons are not probing horizons. It's silly. Nothing can probe an horizon. OK? Of course, now you can. There's several effects. There's special relativistic Doppler effects. There's gravitational reddening of light when you are in a collision disk. So all sorts of stuff play a role. Back beaming, forward beaming. If you have an equation disk around a spinning buckle, you'll get all sorts of nice effects. For instance, this is what happens when you look at a buckle surrounded by an equation disk, and you vary the angle with which you're seeing this. So at zero, you're looking at the black hole equation disk head on. At 90, this system is adjoint. And then you see you get these very nice interstellar light shapes for the same. This is really just lensing Doppler and gravitational redshift. But again, these whiteish regions are not horizons. They're just light rings. Very good. So I didn't tell you about Kerr yet. Kerr is very peculiar. So this is only mostly non-spinning. If you want spin, then typically, in the equation I wrote down in the second slide or so, the angular momentum of your spacetime is that rotation parameter a times m. And if you want there to be an horizon in the spacetime, then, well, you look at all the metric components. You require them to be regular. And you will find that this quantity a is bounded. a is smaller or equal to m if there is to be an horizon in the spacetime. If the rotation is faster than that number, a equals m is called the Kerr bound, then there's a naked singularity in the spacetime. Naked singularity only means that you fall in the black hole. You can probe regions where gravity has measured by, for example, the Kretschmann scalar. So you take this guy, an invariant number, r a b c d, r a b c d. Everybody will agree on this. It's an invariant. And you can measure that guy to be as large as you want. You can come back and tell me everything about it. This is a problem because, well, we don't know how gravity behaves in regions where curvature is very large. So we try to avoid this. Well, nobody actually knows how to avoid this, by the way. But we try to avoid this by conjecturing that this bound always exists, meaning this is sometimes called cosmic censorship conjecture. What this means actually is, well, several things. It means that if you do stuff to your universe, and, well, that stuff is always going to form a black hole, meaning amount of rotation is always smallish, OK? It's always small enough. There are some counter examples. I don't think I'm going through them. But these counter examples require very fine tune, actually, infinite fine tuning in the initial conditions. If you're willing to tolerate infinite fine tuning, then you can produce negative singularities. Otherwise, in most circumstances, we can't. So it seems that this bound is OK. The problem with this bound, by the way, is that if you look at the sun, then the sun, it's not a black hole, by the way. But this number for the sun is larger than this. So for the sun, a over m, which is the pure number, is of order 1.1. For the Earth, it's 10 to the cubed or whatever. And if you tell a child to take a spinning top and put it spinning on the floor, then a spinning top will have a over m of order 10 to the 18. So it's in this kind of sense that black holes spin very slowly. Of course, the black hole spinning this slowly, this extremal black hole, it's still a very relative, well, it's actually an extremely relativistic object. And it's spinning at the speed of light on the surface. There's a nice exercise you can do with curved black holes. Very simple exercise. So let me take two minutes to do it. Remember that Kerr has a GTT component that's different from GRR. And you can compute what an observer at rest sees. By the way, can you see my handwriting back there? Or less, not really. No. I'll try to enlarge it. OK, so this observer at rest will have a four velocity so it's at rest, so it has to be ut, 0, 0, 0. You compute the covariant guy, it will be GTT, u of t. And the square of this guy will give you this very simple example. We'll give you this, ut squared, GTT. Now, you have to remember something very basic about special and general relativity, which is observers are always time-like. So we get a square here. I don't even care what this guy is. This is a square, this is positive. So whenever GTT changes sign and becomes positive, if that's possible, then we have a problem. The problem being no observer can be at rest. This is called, so when GTT is larger than 0, we call this the Ergo region. No observer can, in fact, remain at rest with respect to distant stars. So you're dragged along always. Actually, in the past there were, well, in the past and in the present, but there were some ideas to extract energy away from black holes using the Ergo region. And the idea is very, well, more or less simple. The idea is that if everything is forced to correlate with the geometry, once they pass this kind of critical threshold, then I can use this to my advantage. So I can build a machine, a very simple machine, like this one. So this is basically my generator. I lower these are some metal shafts. I lower this device in a way that this metal circumference is within the Ergo region. It's forced to correlate with the black hole. And of course, it's going to power whatever I put in here. It's not very different from extracting energy away from a carousel using just friction. So Ergo region gives you some sort of friction. It does that for particles. I hope to have time to show you that it also does that for fundamental fields, where we call it super radians. So if you throw waves into spinning black holes, they are going to be amplified by spinning black holes. Another idea, this was the original Penrose construction, was that well, an alien civilization, if they were sufficiently advanced, could basically use exactly the same thing. So they build their entire world around the spinning black hole at the center. It's a very weird construction, let's see. And the idea would be, to those metal shafts, would basically be garbage that they'd be throwing in the black hole and using it to extract rotational energy away from the black hole. So you throw your garbage in, it's forced to co-rotate, and your extra, well, of course, you're slowing the black hole down, then when it's sufficiently slowly spinning, you move one to another black hole in the universe. That's the idea. OK, so the question is, this is all very nice, but we need a spinning black hole. So generically, it would be surprising not to have black hole spinning for the same reason that the earth is rotating around the sun. So if you take any form of matter, you place it around the black hole, it destroys, usually titled disruption, the material. Those guys have some angular momentum, they start falling in, and the black hole, the only thing that it does is eat the angular momentum that it was given. So you can do that via accretion. You can do that by throwing two black holes with angular momentum at one another. So these are called mergers. So on the average, it would be extremely unlikely that you don't have spinning black holes. I don't know if you'd answer it. Oh, you're allowed to, so in all these processes, I don't know if that's what you're allowed to do that, as long as you don't violate the area law. So the entropy always increases in the process, meaning that the area of the black hole is not decreasing. The mass is, you can get away with decreasing the mass if you make sure that the area either remains constant or increases, and you can show that it does here, right? So actually, you can extract up to 29% of the black hole mass in this way if it's extremal, if it's spinning at exactly A equals N. Very good. There's a different sort of process. It's always the same thing, okay? This was Penrose's idea of extracting energy, again, using the same kind of thing. You throw a grenade into a black hole, the black hole is spinning, you force the grenade to explode inside the region, right? If you do that, then you're able to find a way, not all grenades will do this, okay? But certain grenades when they explode into, they will emit one of the small pieces will carry an energy larger than the initial grenade had, okay, and this way you're extracting energies in a somewhat uncontrollable way. It's not a huge amount of energy, it's up to 20% or so of the incoming energy, but still, it's an energy extraction process. Very good. And this kind of closes my geodesic motion around compact objects. I would now like to discuss a little bit the counterpart to particles, and those are waves, okay? My waves will be fundamental fields, scalars, vectors, gravitons. Give me two minutes to erase this. Of course, in the end, what I'm looking for is understanding what kind of extra information can, for instance, a gravitational wave tell me that a point particle can't, right? I already know that I'm limited by, say, the photons here for particles, and the question is, am I limited by that when I use a gravitational wave or a scalar wave or white at low frequencies? So let me take the following simple action. Just gravity and a scalar field. So my action will be this. I'll try to enlarge this, okay? I'm placing a master mu for my scalar field just because I'll use it eventually. I don't think today, but I will kind of try to understand what massive fields can give us. But the idea is just to start with this very, very simple framework, simple scalar. Now, if you vary the action, you're going to get the following equations of motion, the Klein-Gordon equation for the scalar field, and Einstein's equations with some stress tensor described by the scalar, mu, sine mu minus one four. Now, solving this, it's extremely difficult. I don't know of any general solution to this problem. Well, any solution that's non-numerical. Of course, you can evolve this numerically. We know how to do this more or less well. But I'd like to understand the basic, more important features, yeah? Sorry, here, yeah, these are the standard kinetic term for a scalar. So I'd like to understand if there's something I can understand that doesn't require very special and particular numerical solutions. So I'd like to understand if there's generic statements I can do about such theories or others, not relying on the full solution to this problem. Now, if you think about our universe, my goals of masses between four solar masses and 10 to the nine solar masses, these are huge, extremely massive guys, right? Any process that you see, accretion disks, gamma ray bursts, astronauts around vehicles, they're really, really a small amount of matter close to these massive guys, right? They're a very, very small fluctuation around these guys. So the most natural kind of approach to the problem is by assuming that your field psi scales like some small quantity epsilon, okay? And you try to solve the field equations order by order in this parameter epsilon. Now, what you're going to find, of course, is that your equations amount to do in the following. At zero further, you're solving this. At first order, you're solving this, so this is order zero or the one, and now you're going to get order square corrections to the Einstein field equations. So your background metric should get corrections proportional to epsilon squared, but I'm going to stop before I get there, right? So we know how to solve this. I mean, we have actually the most generic solution to those equations. It's the Kerr family in vacuum, okay? So it's vacuum. So the only thing I need to do is try to understand this equation in a fixed background. So now, in this approach at least, I'm taking all the variant derivatives in a fixed background solution of the first equation, okay? So this guy, you can rewrite it as one over square root of minus g, and I'm going to call this my equation A. I don't know if I need it. I mean, in the blackboard, I may not need the numbers, but still. Oh, sorry, this is my equation B. Now this is way easier. The problem is linear. It's true that I may have a complicated background, but still it's completely linear, okay? So I'm going to do the following. I'm going to expand my solutions. So I focus on spherical symmetry. If it's spherical symmetry, I know a basis on the sphere for the angular dependence, and I'm going to expand my field as psi equals some sum over multiples of some field, phi, which depends on D and R. The one over R is just convenience. Really not important. And the spherical harmonics, a function of theta and small phi. So small phi R is a multiple coordinate. Small theta is the angular coordinate, okay? By the way, this exercise is done in the notebook scattering scalars, okay? And if you plug this in, so the only thing you see, the only thing we need is to solve this guy, okay? And we are sure that this is the most general solution to the problem, okay? I'm just expanding on a set. The spherical harmonics are a basis for scalar fields, so I'm fine. I'm losing nothing now. Now, if you substitute all the derivatives in a non-spinning background, so the partial geometry, then you're getting this equation. D2 R star two of your field phi minus D2 DT two of your phi minus V phi equals to zero, where V is some effective potential. It's F L L plus one over R squared plus two M over R cubed. F is the GTT of Schwarz, okay? One minus one over R. And this R star coordinate is defined in the following way with the help of my original R coordinate. It's one over F, okay? So this is fantastic in the sense that now I have only, well, it's still a PDE, but now I only have time and radial coordinate and I can even get rid of time doing a Fourier transform, okay? So it's fine, it's still linear. It's all very nice. And the only thing I got was an effective potential. Now, this is a scalar, so you could ask, well, this is all nice, but I mean, scalars are very, very special. Can I do this for vectors or tensors? And of course the answer is yes. So of course you can take the theory, so these scalars, you can take Maxwell theory. So you can't expand the vector potential components in scalar harmonics. They don't transform under rotations in the way you'd like, but you can take scalar spherical harmonics and build vectors out of it, right? So for example, you can build a vector which is just the gradient of a spherical harmonic. You can take the angular momentum operator and build another vector. By the way, this is zero, d theta of y ln by y ln and this is minus i r sine of theta. This is a covariant derivative on the sphere. Zero i r over sine theta, okay? So basically you can take operations on scalar harmonics and do exactly the same thing and basically find a basis for vectors. So you can expand now any vector you'd like on a sphere using vector spherical harmonics. So explicitly you can do the following. You can take any vector you'd like. If your background is fairly symmetric, it doesn't need to be a buckle, okay? Can be a star, anything you'd like. And you can expand your vector in you as this. A sum over multiples of two guys. One is a of t of n r over sine theta, d phi of your i ln minus a dr sine theta derivative. With theta of y ln. This is one component, you'll find there's vectors always have two different sets of harmonics. And the second, which is a polar, is this guy. Function, generic function, f, t and r, y ln, h, t and r, y ln, k, t and r, y ln, k, t and r. Y ln. So these guys are here only because of this construction. That's the only thing I've done, okay? Taking gradients and angular momentum operators. Those two sets transform differently when you do a reflection symmetry. This is sometimes called the axial sector. And this is the polar. Some people prefer to use magnetic and electric components. It's all the same. If you plug this decomposition in the Maxwell equations, again, the reasoning is always the same, okay? You get an equation that looks like this. Well, now this is Maxwell. You forget about this, you're solving this in vacuum. You get Schwarzschild and Maxwell in Schwarzschild background, okay? Now if you plug it in, this is done in the notebook, then you're going to find the following equation. D2, the r star two of some wave function psi. For the axial sector, actually, psi is A, okay? For the polar sector, let's see the notebook, is a linear combination of H and F, okay? But the equation is this. Very nice because it's exactly, well, it looks exactly like the equation for scalars where your potential now is F. This is really just the centrifugal term. It's Darren Ninkowski, and then that's it, okay? Psi, if you want to know, psi is A for axial, and this combination, r squared over L, L plus one, DT of H minus DR of F for polar, very good. So Maxwell equations or the Klein-Gordan equation really, at the end, look the same. The effective potential is very, very slightly different. Instead of the two M over R cubed, there's nothing here. You can repeat this for tensors. You do exactly the same thing. Now, of course, you only have the Einstein field equations. You expand now your background metric in your metric in the background plus some small deviation, what I'm calling, so your background here would be, for instance, the Schwarzschild metric, okay? And now you have to understand what you're going to do with this guy. Now, in exactly the same way that we built vector harmonics, you can build tensor. You can just repeat the procedure, right? Apply angular momentum operators or gradient operators, and build tensors for the harmonics, okay? And what you're going to do that is the following. You're going to get, again, a set of harmonics, one called the polar harmonics, the other called the axial. Of course, you're using exactly the same thing. You're using symmetries. You're using a decomposition on the sphere. Again, details are in the notebook and the way you actually massage the Einstein equations are in the notebook, they look like this. Okay, this is the explicit form of the decomposition in a certain gauge called the regi-wheeler gauge, okay? These are the axial components and these are the polar components. By the way, these, of course, are also the two polarizations of gravitational waves in the same way that these are the two degrees of freedom of light, right? Two polarizations for Maxwell equations. Very good. You can rearrange this. You can plug them, plug all of these in Einstein's equations. It's a lot of work. The axial, you can do it in five minutes. The polar, I don't know how anybody managed to do it, okay? I think the person must have gone insane when they did it. They actually left academia, but. But so, the result is the following. You get a very, very simple, so for axial, you redefine your fluctuation H1, you redefine it as psi, and you get a wave equation that looks exactly as before, but the effective potential is here, okay? Right? So the only thing that changes, from one case to the other, is really just the effective potential. So you see scalars, vectors, and tensors, okay? There's actually, so this is for the axial sector. The polar is a bit more, well actually, it's way more complex. The effective potential is a lot more complex than this, but we can show, and I don't think I'm going to have time, you can show that they have exactly the same content as far as scattering, oscillation frequencies, whatever is concerned. I'm not going to show this, I think, but you can do it, okay? How do you define? The salinity. Oh, you don't. For now, so the question is, how do you define the stress tensor of these guys? You don't need to do it, so all you need to know from this approach, it's that it's quadratic, at least in my fluctuation H. And then you don't need to worry about it. You need to worry if you're going to define energy flux in gravitational waves, okay? For here, to solve for the fluctuation, you don't. This is a higher order in epsilon. The only thing you need to do is solve this guy at linear order in epsilon, okay? And I think Stas Babak, more or less, told you already how to handle the effective stress tensor for gravity waves, okay? Very good, so the result of all of this is the following. Do I still have time? Perfect, perfect, how much? 10 minutes, okay. Now, the result of all of this is the following. So, you get a equation that looks like this. For all massless fields we know of, well, scalars, vectors, and tensors, they look like this. Let me enlarge this stuff, okay? And your spin, your field spin is encoded in the effective potential V, which you can write as V equals F, F is one minus one over R times, Jiminkowski centrifugal term, L was one over R squared, and then plus one minus S squared to M over R cubed, okay? And S now is zero, one, or two. It's a nice, a neat, compact form to describe whatever you want. Actually, this is kind of interesting because if you look at this effective potential in the limit that your angular quantum, so this is really just the eigenvalue of spherical harmonics, in the limit that L goes to infinity, the effective potential that you get from here is exactly the same that you got from geodesic motion of point particles. So there is a kind of correspondence between the two limits. It's called, of course, the icon or geometric optics approximation we received from here. So at high frequency, high angular number, waves and particles should behave in the same way. Very good. So this tells me very generically the dynamics of fluctuations. Let me see what happens if something happens for static solutions. So I'm going to take the time dependence to go away. Well, static only means I get here and this is zero. Okay? So the only thing I'll be solving is this guy. One minus two m over r, this is f. Psi prime, prime minus v psi equals to zero. This is this equation when time derivatives go away. Very good. So the question now is, of course, are there solutions to the equation or not? Can I find non-trivial solutions to the equation? Now there are several ways you can do this well at least three interesting ways. I'm going to use one of them. I'm going to assume there is a solution. I'm going to multiply by the complex conjugate of that solution, psi star, and I get the following. I integrate from the horizon from the two m surface to infinity and I get the following. Integral between two m and infinity of one minus two m over r. Psi prime, prime minus v psi squared. Oh, there's a psi star here, okay? This has to be zero. It's the, well, it's the field equation. So I do an integration of my parts in this term and I get one minus two m over r, psi prime psi star. This is a boundary term that I get from this integration from parts. Minus integral between two m and infinity of one minus two m over r, psi prime squared minus integral two m, v psi squared. This has to be zero. Now, of course, what we have to worry about is this guy, the boundary term. But if you look at the behavior of this equation, when r goes to infinity or to two m, you're going to find the following. At infinity, psi is either r to the minus l or r to the l plus one, twice trivial. I mean, it becomes flat space. At two m, then psi is either constant or a log. Okay? So any regular solution that you may have is going to make the boundary term go to zero. Okay? So now no boundary term for us. So, well, the rest is more or less simple. We look at the effective potential. This has to be zero. Okay? This is clearly negative outside the horizon. One minus two m over r by definition is positive. This guy is positive. So the only thing we need to worry about is the potential, v. Okay? Now, for scalars, if you look at your potential, v, this is it, v is positive definite for scalars. So conclusion number one, there are no static solutions to the field equations outside of local space time. None. You cannot get any regular solution outside of local. For scalars, for vectors, it's interesting that something happens. For vectors, s, so this guy is one. This term is zero, but we're still after this guy. Now, l is larger or equal to zero. So, v is larger than zero for vectors. If l is larger than zero. Okay? Right? For tensors, exactly the same thing. For tensors, v is larger than zero for tensors. If l larger than one. Okay? You got two pieces in the tensor case that might cause you trouble. And actually, there's a reason for them to cause trouble because there are solutions we know, actually, we know exact solutions to all the ancient equations. In the vector case, if l is equal to zero. Okay? The exception is charge black hole. These are called Rice and Nordstrom black holes. The exceptions for the tensor case, we also know them because you see now, this fluctuation is doing something to your space-time coordinates. The l zero and one are really just redefining the mass in the case of l equals zero and redefining momentum, angular or linear momentum for l one. And actually, when you do this, you're going to find that your solution, which was non-spinning initially, is describing a slowly spinning curve black hole. Okay? And so this exhausts all the possibilities that we have. Of course, this proof is only at a linear level. I never showed that the right-hand side at full nonlinear order also satisfies these properties. It does. Actually, the proof, you could do it. It carries on using the same technique, but I haven't done it, okay? But it does. And this is part of what's called the no-hair theorems or the uniqueness theorems, right? There's no solutions. There's no non-trivial solutions which are associated to global charges of the space-time that describe new black holes, okay? In fact, these no-hair theorems were generalized for the case of scalars. The theorem, there's two theorems and I'm going to try to show you, I hope, that you can go around one of them, but the theorem as follows. Any stationary, any isolated, of course, stationary, regular local, in the Einstein Klein-Gordon theory, okay? With a time-independent boson, which is actually what we've been doing, has to belong to the Kerr family, okay? Time-independent boson. Theorem number two, the same thing, an isolated stationary, regular local in the Einstein Klein, with one real scalar, so I drop the assumption that it's time-independent, but I introduce an extra assumption that it's a real scalar field. It could be complex, okay? Now, in either of these cases, the solution has to belong to the Kerr, has to be a Kerr-Bakkel. Trivial scalar field, geometry, Kerr. Now, why is this so? Well, because if you have, I mean, I think at least heuristically you can think about this in the following way. You have a time-independent scalar and you're trying to somehow support this guy in a space time where it's impossible to support matter. There's no surfaces. It's vacuum all the way down, right? So it's impossible to support anything. Now, in the second example, you try to get rid of that. You say, well, okay, then let me give it some time dependence. Time dependence brings pressure, right? Pressure terms. So maybe I can get away with it. In fact, you get away with building self-gravitating solutions. There are self-gravitating solutions of scalars if they are time dependent. These are called boson stars, right? Or oscillotons, whatever you like. But can you place them around the Bakkel and build a time stationary solution? And the answer is no. You might even get away with the support at the surface, but if the field is real, then the stress tensor, which I, I'm so smart that I erased it, the stress tensor will also be time dependent. And if it's time dependent, the space time is basically radiating gravitational waves. It cannot be stationary, okay? There are ways to get around these theorems. So there are solutions we know of which actually have scalar hair, but they have to violate these guys. It's a time dependent boson, and it's a complex scalar, okay? And I think for now, I'll stop. Thank you.