 Thank you. Welcome back. So yesterday we saw basically the structure of geodesics around compact objects. We never discussed they did not need to be black hole spacetimes. We basically just looked at vacuum, spherically symmetric spacetimes. And the only thing we realized was that the structure was non-trivial. There were stable orbits up to some point. And there was a light ring, which defines a photosphere. And if the spacetime continues to be vacuum all the way down, then there is another strange surface, which I'm going to call the event horizon, where basically things seem to be frozen. You throw something in and everything freezes over there according to a synthetic observers. Okay? But we're in the assumption that this is always vacuum. I also discussed a bit of spin. And several of you came to me asking how can it be that black holes spin so slowly when we know the sun or the earth or stars spin above that bound. So if I collapse the black hole, surely I must violate the curve bound. I must create something nasty. I must create a singularity in the spacetime. So before I start my lecture, let me show you an example of what happens if you try to violate this curve bound on the angular momentum. So I'm going to do the following experiment that actually some of you asked. So I'm going to put in the spacetime too much angular momentum, okay? And the way I'm going to do this, there are several ways to do it. I'm just going to throw two particles at one another at very large speeds. It's 95% the speed of light, okay? And the angular momentum in the spacetime is such that if these guys form a black hole and nothing else happens, then the black hole is spinning above the bound. So it won't be a black hole, okay? Very good. So this is my setup. There's two objects here. Actually, because it's easier, each of these objects are a black hole. It doesn't look like it, but they're thrown with some finite impact parameter, okay? And the angular momentum is in excess. Colors are basically amplitude of gravitational radiation, okay? You see, these guys are shot against each other, but they carry just too much angular momentum, and the spacetime and the field equations know this. So they don't collide immediately. Let me replay this, because this is extremely interesting. If these guys merged promptly, they would form something that would be over spinning, okay? So the field equations know about this attempt to violate the bound. So the guys don't merge. They see, oh, wait a minute. There's just too much angular momentum. They are going to zoom out. They give a couple of orbits. They emit the excess angular momentum, where, how? Well, the excess angular momentum is in gravitational radiation, okay? And then it's safe to merge, okay? So here it is. Too much angular momentum. They go around each other a couple of times. They zoom out. They're still going around each other. They release the excess angular momentum. Now they merge. Everything is fine. The final object is a black hole. It's a curved black hole spinning at 96% the curved bound, okay? So everything is safe. This is one possible way that you have to try to generate nasty stuff in the spacetime. But today, I'm not going to discuss how we can do this. I'm going to discuss how we can do a fraction of this. Yes? That's a good question. We suspect this simulation takes two months to do. We suspect that when you fine-tune the initial velocity to be closer and closer to the speed of light then the final spin will approach the curved bound. And by the way, the total energy that these two guys lost is around 30% the center of mass energy. So 30% of the center of mass energy goes away in gravitational waves. It's a massive thing. Okay, but so yesterday we saw that if I take a linearized field, it can be gravitational waves. It can be whatever you want. Then it's described by an equation that looks like this. Okay? I'm going to call these dynamics one. The first thing you should do when you have an object or you have a field is try to scatter the field of that object. Okay? So the first thing I'm going to do is assume that I'm in some scattering situation and I have a field which is a monochromatic wave. So it has this time dependence, some characteristic frequency, omega. Okay? And of course it has some radial dependence, Z of R. If you plug this in the, by the way, again this is generic. I should note that V goes to zero at the horizon and at infinity. Okay? V, if you look at the form of the potential, it's multiplied by GTT. So it's zero at both boundaries. Now the field equation becomes very simple. Actually, I'll write down the partial derivatives as ordinary derivatives. So it's this guy. Okay? It doesn't get simpler than this. So we can solve this. The asymptotic behavior of this guy that I'm going to demand are first, by the way, V again is zero at the horizon. So you can also do a local analysis of this ODE. But it's easier this way. V is zero at the horizon. So solutions are plane waves in this tortuous coordinate, R star, that I defined yesterday. So the solution that I'm going to be demanding is this when R goes to one. Okay? Right? Why this, why the minus sign, there's of course the solutions are plus and minus. But if you look at the time dependence, the minus sign gives you a wave that's traveling to the left. Okay? So the only thing I'm requiring is that my system is isolated. There's nothing coming out of this horizon at some point. Okay? So there's only stuff falling in. At infinity where supposedly I am sending stuff in, then the solutions are, well, a superposition of in going and outgoing waves. Okay? Because V, this potential is zero at the boundaries. So it's also zero at infinity. You can actually, of course, it's a linear equation. So if you do this, transmit the amplitude, you're also fine. Okay? Now, this equation has a potential V which is real and omega independent. I mean, this was just looking at V. And there's no first derivative terms. Okay? So the Ronskin of any two solutions, Ronskin and W, which would be of any two solutions, Z1 and Z2, which is Z1, Z2 minus Z1, Z2 prime, has to be a constant. Okay? So there's a nice conserved quantity here. You can show this is, I mean, it's half a minute. Right? So I'm going to take two solutions. One which is this one. The other, it's a complex conjugate of this one. And I'm going to evaluate the Ronskin at the horizon, 2M, and at infinity, it has to be the same. So I'm going to get some conserved thing. Okay? When you do this exercise, what you find is that A in minus A out is AT squared. It's kind of nice. At least in a non-spinning background, some notion of energy is conserved. Right? The stuff that you throw in, a fraction comes out, another fraction goes down the hole. If you try to repeat, so by the way, this only happens because there is no first derivative and V is zero at the boundary. So you can write these sort of expressions. If you add spin, there's only a slight change. You can do this. I mean, it's very simple, by the way. You take, instead of a partial, you take a curve, like a scalar fill in. This equation basically remains the same, except that there's a small change in this potential here. And the change that it makes is that instead of this solution here, okay, you have something like this. So the horizon behavior changes very, very slightly, where this quantity, I don't know the name of it, VAR5, whatever, is omega minus M capital omega. And M, remember, is the index of the spherical harmonics. Okay? So now you can repeat the Ronskin analysis. You can repeat this stuff, right? And what you get is this. A in squared minus A out squared is omega minus M capital omega over omega AT squared. So it's kind of impressive that even though, of course, so if there's no spin, if we have a Schwarz-Lwackel, we recover this, okay? But now, if you throw a wave which has a frequency omega smaller than this stuff, okay, then what this tells you is that what you get out is larger than what you sent in, right? This number is negative, okay? So if omega smaller than M omega, we get super radians. Of course, I haven't computed the quantities A in and A out and AT. I just know they satisfy this relation. But I sent you a notebook which is called Scattering Scalar, where this is done, okay? And if you plot these guys here, you're going to find, so the plot of one minus A out squared over A in squared looks like this, okay, as a function of frequency for a Schwarz-Lwackel. The notebook, I think does curve as well, but it's fine. So there's two interesting limits here. The first is the high frequency limit. This is basically the cross section we saw yesterday. So you're testing those to three square root of 3M, okay, roughly. The low frequency limit goes like, so this curve here at small frequencies goes like 16 omega squared, which tells you that the low frequency absorption cross section of Wackels, sigma, is basically the area of the Wackel, okay? This was something that Das Gibbons and Mathur found out in 1999, I think. Very nice result. It holds for a wide class of strictly symmetric Wackel geometries. Very good. So I was only scattering a monochromatic wave. I would now like to understand, so you see, I am imposing something which is stationarity. So I'm forcing the system to be stationary. This constant frequency wave, I'm driving the system. I'd like to know now what's the response of this phase time to an arbitrary fluctuation. So I'm going to take this equation now, and I'm going, instead of assuming this, I'm going to perform a Laplace transform on it and try to solve it and see what happens, okay? So I'll introduce a Laplace transform. By the way, there's always, I think some of you may think, oh, why am I doing this? I could just take a computer and solve this by whatever method I can. But it's never a good idea, well, now it's a good idea because we know what's going to happen. But it's never a good idea to just use brute force. Methods such as Fourier transform or something, anything that simplifies PDs, gives you very, very general properties of the system, something that requires sometimes years to do using computers, okay? So this is the idea behind what I'm going to be doing. Okay, so I'll introduce a Laplace transform, meaning this guy here, okay? So I go from time to some s domain. And now it's very simple, right? I multiply this equation by e to the minus s t. I take care of second time derivatives by doing a double integration of my parts. And what you get is the following. Wow, you basically get this equation with some source term that gives you initial conditions, right? So you find, wait, I'm going to promote this to ordinary derivatives. And I may, so this is on purpose, I, instead of s, I'm using omega just to make it look like the previous one. So I'm defining s to be this, okay? So you find a nice, very simple, ordinary differential equation, but now it's driven. So there's a source term. The source term depends on the time derivative of initial conditions, okay? And on the initial field itself, right? So the only thing we need to know now is how we're going to solve that kind of equation. So again, there's several ways to do this. I'm going to use the standard Green's function approach. And the idea is very simple. You solve this by using the homogeneous solutions of the problem and then basically integrating with the source term, okay? So the equation describes sourced fluctuations. So I'm going to define a wave function psi L and psi R. And these wave functions are the ones, so I'm going to impose on the final solution something that I want my system to do. For example, if I'm here and I'm throwing this chalk at the object, the black or whatever, okay? What I'm going to be demanding is something that looks very reasonable, that at large distances, only field that is outgoing must be present. There was nobody at infinity throwing stuff in, only me here, okay? And at the horizon, there's only stuff going in because I'm the person producing fluctuations. Again, this can be a scalar field, but it can be gravitational waves generated if this, so this important. The source term, okay, which is only connected to initial conditions, can perfectly well describe a small buckle falling into a supermassive buckle. This is what Stas called an emery, right? An extreme mass ratio in spiral. This is how we do the calculations, right? So the small guy is a fluctuation in the big guy. All of this applies. The source term is a bit different. You can get it by, well, by looking at the source term of point-like particles in the field equations. This is done in a notebook I sent you called infall light, whatever, variation of constants. There's a weird name. Anyway, everything is worked out there, including the solution to the equations. Okay, very good. So I'm going to then define this psi L in this way. Psi L is the one that behaves nicely on the left. My left has the horizon, okay? And of course it has some behavior which I don't control, so I'm fixing the behavior at the horizon. At large distances, it goes like in that form, right? And I'm going to define another wave function which is a nice one at infinity. Psi R, which is this, E to the i omega r star, when r star goes to plus infinity, and something at the horizon which I don't care, okay? The general solution to the problem, given these two homogeneous solutions, is psi, is psi R integral up to some point, right? So I'm going to find of the source term i psi L over the Ronskin plus psi L integrate from r star to some point of the source term psi R over the Ronskin, the r star. And then of course plus homogeneous solutions, okay? This is a general solution to any second order of the, you can do, you can show this just by taking second derivatives of this guy and showing that it satisfies this, right? It's called variation of constants. Good. I haven't imposed the boundary conditions on that solution, right? I want that solution to only have outgoing waves at both boundaries. And the way I'm going to do this is by adjusting the coefficients in this homogeneous guy, okay? And the end result is this one, it's psi R, okay? So this final solution guarantees it's still of the same form, right? But you use the homogeneous terms here to tweak a little bit the boundaries in your integrals, okay? That's the only thing I've done. And now of course, at least if the source terms are fine, you can check that when R goes to infinity, this guy goes to infinity, this integral goes to zero and the only thing you're left with is a solution that really is outgoing on the right. And when R goes to minus infinity, this term drops down and you only get indeed a term that goes down the hole. So this looks fine. LIGO and most detectors we can use is placed very far away, right? So for LIGO, so for detectors, the only thing we care about is this term because this term usually is very, very, very small. So psi R, which I know, this guy I know it's this one at large distances, okay? So this you can write immediately like this, integral of this guy, minus to plus infinity of I psi L over omega dr star, okay? So it's just an integral at the end. The only thing we need to do is to actually just compute an integral. Of course, being by being in frequency. You give some frequency, right? You have to span the frequency range and that's it. By the way, the runs can between these two solutions, you can compute the runs can at infinity and you're going to get this w is 2i omega a n. And you can do that analytically, right? No, no, no. These are not causing normal modes yet. Okay? This is just a solution to a very generic problem. Okay? So the numerical calculation does the following. That I sent you in the notebook scattering scalars. You give an omega, say 10 to minus three, you solve this equation, the homogeneous version of this equation, to find a in and a out, okay? Right? You have now a in, a out, and the solution psi L. So you start from the horizon, you integrate, you find psi L, a in, a out and you do the integral. That's it. That's what the notebook does and then it sweeps across all frequencies and eventually it transforms back using Bromwich integrals to the time domain. Okay? That's it. Then you found the gravitational wave. So the result looks something like this. This was done, of course, many years ago, in the 70s when Vishweshwara was doing this sort of experiment. This is what he found. So I still haven't specified this I, this initial conditions. Vishweshwara and many other people, like the notebook I sent you, give, well, some arbitrary function, usually a Gaussian profile. Okay? And the Gaussian depends on the width, the amplitude, and that's it, right? And the location of the Gaussian. And what Vishweshwara was finding was always the same thing. So you should read this in reverse. Time is flowing to the left. Okay? So what he found was the following. Well, I'm sitting somewhere. I see, so it's a Gaussian. I see the first peak of the Gaussian passing through me. And if you're in Minkowski's space time, that's it, right? Things travel on the light cone. I turned this laser on. I turned it off. That's it. But in curved space time, because of curvature, you have backscattering. Okay? Actually, you have backscattering. You have another interesting thing, which is when the pulse of your initial data reaches what looked like the horizon, okay? Then things were vibrating somewhere. And he found this dumb sinusoid behavior. This is nothing special. What's special is that if you vary, well, if you vary the amplitude and the location of the initial data, you always find the same thing. It's kind of surprising that the ringing frequency and the dumping timescale is always the same. Okay? This is the weird thing somehow. Actually, you can repeat this experiment. You can repeat this experiment. It's also in the notebook I sent you with a small black hole, so not a Gaussian, something that looks astrophysically interesting. A neutron star or a black hole fall into a supermassive guy, right, into the center of the galaxy. Right? So this is shown here. This is shown, where is it? Here. Okay? This is for a star that starts from rest and then slowly falls in. And this is something that is thrown out of LACL at close to the speed of light. And you can compare this against full numerical relativity calculations doing equal mass collisions of LACL, such as one I showed you. And the pattern is always the same. There's always some peak, so the stuff reaches the maximum. The detector would jump out when it sees this guy. And then the fluctuation is exponentially dumped in the ringing way. So people call this the ring down or the quasi-normal mode decay, but it's always here. Okay? So the late-time stages of what LIGO sees every time, well, they look like a ring down. Okay? They look like this. The parameters are always the same. So if you have a black hole of mass M, then the sign of the wave is this sign. Okay? So it's sign of 0.374, old numerical calculations seem to say this, T over the mass of the black hole in these units. Okay? So a black hole like with the mass of the sun would have a frequency of over, I don't know, two kilohertz or something. Okay? The dumped exponential always looks like this. So it's always the same values, the same constants. Okay? Yeah, yeah, yeah, I'm going there. Yeah, yeah. I hope to be there. Okay, so of course the first thing we should try to understand is where is this coming from and, well, how can we use this to actually do science? Right? Okay, so if you have a neutron star, things become messier very quickly because this is pure space time, right? Remember what we did yesterday and today is pure vacuum. Okay? If you have a neutron star colliding with another neutron star, then, well, the initial stages are the same. They haven't touched each other, so this precursor is going to be the same or this one, but then they touch on another and before you see the ring down, you see a lot of dirt effects because there's mass ejection, there's huge nonlinear effects in neutron stars. The final state always looks like this. If it's a neutron star, then the frequencies are different, right? So if the final state is a neutron star, the oscillation frequencies are different. If the final state is a black hole, then the oscillations are the ones here. They are the same as these ones. It doesn't really matter how you created the final state. If it's a black hole, it has to behave in the same way. Okay, so my idea now is to very quickly see if we can understand what's going on and if this is a property of black holes or if there's other space times that behave in the same way. So this is where quasi normals and tails come in. By the way, curved space is very special, right? Because if you look at the Green's function of Minkowski, really there's delta functions in 3 plus 1 and the delta function forces everything to travel at the speed of light and on the light cone. Everything out, the light cone doesn't really get to the receiver. In curved space time, things travel inside the light cone. The reason is curvature. So you have an instantaneous propagation, but then a wave that was going this way can back scatter off the space time curvature and reach receivers at later times, right? So you always have a signal. It doesn't die off immediately as it does in Minkowski. And now the question is, is this because of the curvature of the space time curvature at large distances or is this because really we have horizons in the space time? Well, at large distances we saw that psi is this, right? Some convolution with a source term and an integral. That's it. But we need to do, so we found this by doing the inverse transform, right? So we need to go to the time domain. And the way we go to the time domain, that is why it's going this. Psi of t is right over 2 pi integral of, in omega of this guy, okay? Of course numerically, you just do it on the real line and that's it. So you integrate over omega and that's it. But you can, I think there's a lot more to gain if you look at the, what you're doing, right? So you have your omega space and you're integrating this guy across this real axis, right? But you find if you look at the structure of these functions in the complex plane, you're going to find that there's weird things there. Well, not weird actually, there's things there, there's structure. And the structure in particular means that there's poles here in the complex, in the lower half complex plane, okay? I'm going to be calling these poles quasi normal modes. At these poles, the run skin, that guy, w over there is zero, okay? That's why it's a pole. Of course if it's zero, if you look at this solution, the solution is this. The runs can be between this guy and this is 2i omega a in. So a zero of the runs can means really that there's frequencies for which a in is zero. There's frequencies in the complex plane for which a in is zero. And this is kind of nice, right? Because it tells you that if you search over the complex plane, you will find frequencies where a wave that behaves nicely at the horizon, it's only in going at the horizon, also behaves nicely at the other boundary, at special infinity, right? So this starts looking like a vibrating string. This starts looking like an eigenvalue problem where I search for the characteristic frequencies of a system and the way I do this is by imposing correct boundary conditions, right? Or if you want, it starts looking like the eigenstate of the hydrogen atom, right? You do the same thing. Except that it's in the complex plane. The reason why it's in the complex plane is because we can't have, you see, we can't have stationary states here. We can't have real frequencies doing the job. This system is dissipative. We're talking about a system that can lose energy to infinity and that can lose energy down the hole. So we could never find, we could never find a pole in the real line, okay? But anyway, so you could try to take care of this integral by closing the integral, right? And doing contour integration. What you find from this would be that your original integral here will be a contribution from this large semi-circle and, of course, the residues from whatever poles it finds in, right? So your integral will be time inversion. It will be a sum of our poles. The large semi-circle. Actually, in black hole space, in our particular case, there's also a branch cut at omega equal zero here that you need to take care of. And because it's at omega equal zero, this can only mess with the very late-time behavior of fields, okay? Actually, I mean, now it's easy to see that the residues at the poles always have to produce something like an exponential of some characteristic frequencies. So your damp sinusoid comes from these ones. These guys contribute a power law tail. So this is exponential damping and this is a power law. Oh, I have it here, okay? So this is what I'm doing. I'm closing the contour. Actually, I have an example of the power law. So this is an example that you also have in the notebook. You can do this. When you compute the field, a scalar field, gravitational wave, whatever, okay? You will find an initial burst. This is the prompt contribution. Usually it comes from the large semi-circle, okay? Then you find the ring down. This is a log lean scale, okay? So you clearly see that this is an exponentially decaying sinusoid, right? So it's decaying exponentially. At very late times, you're going to find that the branch cut makes a contribution, okay? And the field, instead of decaying exponentially now, decays polynomially, t to the minus something, okay? We have never seen this power law in things that LIGO cares for. Actually, I'm curious to know if there's processes where you see this. This only comes from the back scattering of space time curvature. So you'd find this in neutron stars, right? It's only because of the curvature of space time on large scales, okay? So far, all the simulations that were done for LIGO are black holes, small black holes falling on large black holes, equal mass black holes and so on. I don't think the numerical precision is still enough that we can go down six orders of magnitude and find power law tails, but they have to be there, buried somewhere in the simulations. Okay. So if these guys come, so if the exponential damping comes from the poles, now I'd like to know how can I compute these poles, right? And there's, well, this is an industry, by the way. It's still an industry of techniques to compute buzzing all modes. From the circle, there is, yes, yes. Oh, no, no, no, no, no. So this exponential, oh, it's here, sorry, this is here. No, no, this would be the delta, I have an example that, very simple example. It's a delta function for the potential that can solve everything. I hope I have time. But no, this, the sum of our poles is the one that gives you an exponential of a complex number. So a fraction gives you the exponential damping. The other gives you the sine function. This is it. Okay. How to compute UNMs? Well, it seems straightforward, right? So if we're looking only for poles of this, of the Green's function, and the pole happens when A in is zero, the only thing we need to do is start at the horizon, integrate outwards, and search in the complex line for a frequency omega such that A in is zero. This is really trivial. It's the way we search. I mean, it's the way we look for the eigenstates of a string or the hydrogen atom, okay? However, this is, well, it's slightly more involved in this, in this context because the poles are in the negative, the negative half of the complex plane, right? So the imaginary part of this frequency is negative. It's a good thing that it's negative because the time dependence of the mode is this, right? So because it's negative, we are guaranteed that the fluctuation is exponentially damped in time. If it were positive, then your space time would be unstable, okay? We have not found instabilities for massless fields in blackout space times. So it's negative. The problem is it's good on the time domain. It's bad on coordinate space. So the wave functions you're looking for diverge exponentially at both boundaries, okay? Right? Which is a bad thing. So there's ways to deal with that. One of them is basically to use asymptotic expansions of the analytic asymptotic expansions of the wave function at the horizon and infinity. So one, expand analytically, the wave function goes to boundaries. What I mean is the following. So you impose, for instance, you go to the horizon and you impose that psi L is e to the minus i omega r star, one plus a in plus a one, a two, right? You impose this. Now you see you factored out already the dangerous term. You don't need to worry about growing exponentials because you put it by hand. So it's done. You insert this in the ODE, you get a one, a two, and so on. And now you can start integrating numerically from a finite distance, okay? You do the same at infinity. This would be, well, similar stuff. So you start from, not from a very large domain, you start from a finite domain where you got rid already of the divergences and you start integrating towards a common point. This is called shooting. This is what Chandrasekhar and Dottweiler did in the 70s. And this works, it works for, so you're going to find solutions. The solutions, so let me, you're going to find a bunch of solutions. Well, not that many with this method, but you're going to find some. For instance, in units where the mass is one, the first solution looks something like this. 0.374 minus 0.089i. Okay, it's one number. Actually, it's the easiest number to get with this technique. But there's others and you get it with this technique. So this is for, remember that we did a spherical harmonic analysis. So there's always this L, L plus one. So this is L equals to two. But even when you fix L2, okay, you find the other solutions, right? And it's common to order these numbers by increasing magnitude of imaginary part. Why? Well, because this controls how fast the mode dumps out, right? So modes with a very large imaginary part will dump away very quickly. We don't care about them, okay? So you basically do omega n and some people call it n equals zero n equal or whatever. This guy is a fundamental mode, okay? The technique is not very good if you want to extract n larger than two. You can do it, but it's a lot of work. Another technique which is still, I think, the best one, and both of them are in the notebook I sent you. It's called QNM's ICTP. Goes around the problem of having to integrate DOD. So it reduces the problem to a three-term recurrence relation, which we put in the continued fraction forum. And this was something that Lever told us how to do. And the idea is very simple. So you write down your wave function in the following form. So again, you factor out all the divergences that you might have in the problem. R to the 4i m omega e to the 2i m omega r minus 2m. And then you do a series expansion in this way. Of course, Lever had a lot of experience with molecular hydrogen in quantum mechanics, so he knew somehow this was going to work for Lakers. I don't know how. But if you plug this into the ODE, so in, well, I erased it, into the homogeneous version of our equation, you're going to get a three-term recurrence relation for these coefficients, AN, okay? So you're going to find that alpha 0, A1 plus beta 0, A0 is 0, and alpha n, AN plus 1 plus beta n, AN plus gamma n, AN minus 1 is 0, where alpha, beta, and gamma are coefficients. I'm not going to write them down there in the notebook, okay? And now you still need to worry about does the series converge, okay? And in fact, there is one representation of this series that does not converge, and you need to worry about it. So you need to make sure that your ANs decrease as n increases, okay? So solution converges if AN decreases with n. This is called the minimal solution, and I show in the notes, which are not online, but I'll send them, that when you do this, then you reduce the problem to this form. AN plus 1 over AN is equal to this, minus gamma n plus 1 over beta n plus 1 plus alpha n plus 1 AN plus 2 over AN plus 1, which is equal, oops. This gives you an infinite continued fraction for the coefficients, okay? So if we plug N equals 0, we find the following. A1 over A0 has to be equal to minus beta n over alpha n, we knew from this. This is the first term in the recurrence relation, and it has to be equal to beta n minus alpha n gamma 1 over beta 1 minus alpha 1 gamma 2 over beta 2 minus, and this continues, okay? So in the end, so the moral, in the end what you need to solve is this infinite continued fraction, okay? Remember all these guys, alphas, betas, and gammas, we know them. They're only a function of the local mass and the angular index, L, okay? So it's quite easy to solve Mathematica. The code I sent you is Mathematica, right? You get basically arbitrary precision for basically an arbitrary number of modes omega, okay? You can get them with 20 digits, 30 if you want, right? So the structure of these modes is the following. What is this, no? I didn't do anything. It's updating my Skype without my permission. So this is the structure in the complex plane of our solutions, omega, so this is the real part, this is the imaginary part in units of 2M, okay? And what you find, so focus on L equals 2, by the way, this is for gravitational fluctuations, so gravitational waves. And the modes have the following structure. This is the low, so the relevant mode is down here. This guy is the relevant mode. The lowest imaginary part is the mode that takes longer to disappear, right? So what the line goes is, if it sees, it should be this mode, okay? But then you find others. Imaginary part keeps on increasing. There's actually a mode with zero real part. We still, well, there's a mode with zero real part, large imaginary part. At large, at large index N, so when the imaginary part is really, really, really large, we find that, of course, these modes we think don't make any difference for LIGO, okay? But 10 years ago, there was a conjecture by people using the Bohr's correspondence principle to argue that modes with large N were connected to the quantization of the black hole area. Well, there's a very, very interesting story behind that, but a fraction of the story was because people suspected, because of numerics, that the modes have the following structure. Logarithm of three over eight pi minus two N plus one i over eight, okay? Back then, we had no idea that this was a log of three, but somebody took the calculator at 2 a.m. in the morning and was interested. The story is very nice. Anyway, the person suspected, because Baconstein had suggested this, that this might, there might be, the high dumping limit might have a logarithm of an integer, okay? So the person started to see, we knew the number, I think with seven decimal places back then, and the person said, oh, my god, this log of three. I usually try the log of two, it didn't work. Log of three, yeah, I got it. And of course, a couple of years later, we showed analytically that it is a log of three. So this is a large N behavior. At large L, so fixed L, large N, okay? At large L, you can use WKB and you find that M omega is L over three square root of three minus i N over three square root of three. And if you were paying attention to yesterday, then you're going to realize this is really just the null geodesic properties. So this is what I was calling L omega, so the frequency of a photon in the circular orbit, and this was my parameter lambda, the time scale that the photon takes orbit in the light ring, okay? So I think it's a nice, it's a nice way of thinking about it. You don't need to think about it in this way, but it's a nice way to think about these modes as modes that are trying to live on the light ring on the photosphere, but it's an unstable motion, so eventually they decay. And the imaginary part of the mode only gives you so this time scale really is connected to the instability time scale of the null geodesic, okay? Of course, you don't need to think about it this way. Neutron stars don't have light rings, but they have quasi-normal modes. My t-shirt has a quasi-normal mode, you see? It's dumb. I mean, it's not a normal mode. It's dumb. There's no light ring here, believe me. Okay? Still, there's a quasi-normal mode. Still, there's no connection, obvious connection, in non-compact spacetimes, but it's very useful to think about the connection in ultra-compact spacetimes. There were nice conjectures by York and others in the 80s trying to explain the evaporation as a large end behavior as well. I don't know where it stands now. There are some attempts. I don't think we... Well, I certainly don't, but I don't think we still understand fully the problem. Okay, so I think... How much time do I have? 20 minutes? Okay, so I'll skip this part. And if I have time, I'll come back. So there's a nice example where you can solve everything analytically. The example is in my notes, so if you want go over them. So it's a potential with a delta function. So the potential is a delta function. The initial data is a delta function. You can solve everything very nicely, very concisely, and you see all the structure here. You see the prompt response, the light cone response initially followed by the ring down that's generated when the potential is ringing. Okay? But I'll go over it if I have time. Okay, so let me show you then why we care about quasi-normal modes and why we care about ring down and so on. So I'm going to start doing an experiment that I do in the notebook, except that now I'm going to put a movie. Okay, I'm throwing a point particle into a black hole, and I'm going to plot the response. Okay, this side, the gravitational wave that Leica would see. And it looks like this. So this is my star or small black hole, whatever. It's falling into a large black hole. This point here marks the light ring of the big guy. Okay? This plot is the gravitational wave form as a function of time. Okay? And this is just a plot of the effective potential. Okay? V. It's not a coincidence, of course, that your effective potential has structure at the light ring. So this peak is very close to the light ring. It's exactly at the light ring when L is very large. Okay? Very good. And what you see is that, let me light the star fall again. Okay, the stars fall in, just a small precursor here. As soon as it crosses the light ring, it excites the modes. So you see the burst of radiation and the ring down seem to be excited really when the star crosses the photos here of the black hole. As we saw yesterday, it's impossible that anything here is related to the horizon. It would take an infinite amount of time to reach us. Okay? So can't do. So it seems like the only thing we're seeing in life is, is way more related to light rings and photospheres than horizons. Okay? Very good. No, no, in this limit, no, no, the calculation is in the notebook. So the calculation is this, the calculation, which is in the notebook, but it's this. So we solve this equals some source. The source term now describes a small point particle falling to a black hole. So you have to project the components in the tensor, the spherical harmonics and eventually get a nice source term. Actually, very simple. So we're solving this and this is what we see here. Oh, actually, actually, I'm glad you asked. So if this is not trivial, okay? If you, you, if you go here, this is a time domain waveform, something that numerical relativity would also give you. Okay? If you go here and you try to fit for an exponentially damped sinusoid, the parameters you get are these ones. You can do it. You have the data in the notebook. Okay? These are the parameters. So this is nice, right? The calculation we did using quasi normal, so the quasi normal calculation, which doesn't know about sources, agrees very nicely with time domain waveforms. We don't know about quasi normal modes or shouldn't. I mean, they, they go on the real axis. They don't care about complex planes, right? So this is nice. Actually, tomorrow I'm going to explain that there's something we don't understand. There's something is, no, I'm going to explain tomorrow. But anyway, so in black hole space times, these, the quasi normal modes agree perfectly with this. Actually, in numerical relativity, we use tables of quasi normal modes to compute the mass and spin of black holes of the final black hole because it's harder to get numerical space times and extract. Okay. So for the next 10 minutes, I would like to explain why there's such a fuss about ring down in quasi normal modes and how we can use it. Actually, tomorrow I would also like to discuss a little bit how we can test if the object is a black hole or not using quasi normal modes. The first thing that I think I was worried when I started my PhD was, this is all nice, but black holes, actually somebody yesterday told me, why do I care about this? You're doing these black holes in vacuum. I mean, the universe is real. There's matter all over. I don't know. I don't care about no hair theorems. There's no isolated black holes in the universe, period, right? And this is true. I mean, it's an important question. Why should I even care that the Kerr family is the unique solution to vacuum ice and field integration? We don't have vacuum, right? So there's only some notion of, you know, currentness in any vlog in the universe. So the thing that you should worry is, well, I have galaxies. I have dark matter. I have cosmological constant. This is going surely to mess up all the frequency calculations that I do. So anybody in numerical relativity, they don't put dark matter. They don't put cosmological constant. They don't put magnetic fields yet. So how can I trust anything? But you see, the way we used to compute these numbers, there's two things that kind of reassure us. The first one is that at least in the large L limit, the numbers that we're getting were only related to the light ring. Okay? Something is happening at the light ring. So, and because anything beyond the ESCO, beyond the innermost stable circular orbit should be empty, things fall very quickly when they reach the ESCO, this kind of reassures me that, well, I am not going to find matter close to the light ring. So it should be fine. Okay? It doesn't mean we're fine. And honestly, I think there's a lot to go here. But at least, you know, it should be a relatively empty region in space-time. But the other thing you notice is how easy it is to compute these frequencies. I mean, it really takes a few minutes once you're in the scheme. Right? So you can start messing around with your space-time. You can start adding charts. We know how charge black holes, the geometry of charged black holes. You can start adding a cosmological constant. We know some black hole solutions immersed in the cedar universe. You can add magnetic fields. We know a little bit, not much, but we know something about that. You can add accretion. You can, well, all of this, by the way, is done by hand. But anyway, you can also add a bit of dark matter outside the black hole. If you do this calculation for reasonable parameters, you're going to find that the frequencies change. These numbers are going to change. Okay? For reasonable values, they change according to this column. Okay? So if you put matter, a lot of matter, if you put 10% of the black hole mass in the innermost circular orbit, so it's a lot of mass in a very small region, okay? Even then, the frequencies change by roughly 1%. Okay? Dark matter and cosmological effects really seem to have a very small effect on ring down. Nobody yet actually studied the full waveform as seen by a detector, you know, in our universe, with cosmological constant, with dark matter, and so on. We don't even know if power law tails are present again. So there's a lot to do, I think, here, but it takes a lot of work, of course. You do find something, so this table basically summarizes why people think we're fine, okay? But if you actually do the calculation, there's something that the table doesn't show, and I'm going to discuss tomorrow, which is if you start putting it. So one of the things we did was put a spherical shell, so let me show you what we did. So we know how to do this. We place a black hole here, mass m, okay? And it's very complicated to get solutions of the field equations for arbitrary matter distribution. So we took a spherical shell of matter, mass delta m at a radius r, okay? And now you can compute, you can fluctuate this geometry, you can compute modes. And you find, so let me give you numbers, okay? If delta m is 10 to minus 2 of m, so 10 to minus 2m, and the radius, let's say, is 100m, okay? The change in the frequency is rather small. It's over the 10 to minus 2% or so. But if you fix the mass, this is unbelievable, okay? If you fix the mass and you put the radius of the shell to larger and larger distances, then you're going to be discovering that the change in the frequency is increasing. So the change, the way that the frequency changes, so the frequency changes more and more as the shell radius is taken to larger distances. It can't be, it can't be because the surface density is decreasing. I fix the mass, I increase the radius, I have a very dilute material, okay? The reason why this happens, I strongly believe, but I don't know, is that the modes, so if you think about preservation theory in quantum mechanics, you have to do this inner product with the ground, with the background wave functions, right? And if you think about the wave function of the quasi-normal mode, of the unperturbed quasi-normal mode, it goes like this at large distances. It's an outgoing wave. And remember, there's a plus sign, okay? But quasi-normal modes have a negative imaginary part, so these guys grow exponentially, right? So you're projecting a fluctuation into something at some radius that's growing exponentially with that radius. So it's exponentially sensitive to where you're doing your, your preservation analysis, right? And I think this, so you can get changes of 100%. If you place this radius sufficiently far away, you get changes of more than 100% if you want. And I will leave this puzzle here for today, because tomorrow I'll have a bit more to say about that. Okay, but so we're fine about the environment. So the next thing is doing science. And the science I want to be doing is using the uniqueness theorem. If molecules are characterized by two parameters, mass and charge, I must be able to use ring down to find those parameters. In fact, I must be able to use ring down to test the conjecture that they only depend on the parameters. And the way we do this is the way you do this with the string. The modes of the string are here. Okay, this, I mean, this is a high school calculation, right? The frequency of a string which is made to be fixed at the ends is an integer times the velocity of waves on the string over two times the length of the string. Okay, and the modes are these ones. So I can, let me start with a string. I look at the string, I detect the vibration modes of the string. And if I see the first mode, N1, okay, meaning I measure the frequency of the first mode, I can use this relation to compute the length of the string. Okay, and this is nice. Now I hear the sound of the string and I know the length. If I see the second mode, so if I see N2, what can I do? Well, I can repeat the calculation, I can recompute L. If L agrees with the previous one, I'm fine. I go home and I say, okay, we have this string and it has this length. If it doesn't agree, I have problems. The problem being it can't be a string, right? It certainly cannot be this string, okay? And so the idea is to do precisely the same with Wacknells. Okay, the idea is that we measure, so we measure one mode. One mode, by the way, means this. One mode has the decay timescale tau and the ringing frequency F, okay? These numbers are a function of mass and spin. So measurement of one tau and one frequency allows me to take mass and spin, okay? And measure, well, I hope measure of two modes will allow me to test the prediction that these are curved Wacknells. And of course, this kind of game was played decades ago. There was even a conjecture, not in Wacknells. There was a conjecture by Herman Weill, so he was looking at drums, so not Wacknells, but systems that resonate, okay? And he started realizing that the spectra, the quasi, they're normal modes, but the spectra of the normal modes was related to the area of the membrane that's vibrating, okay? In particular, there was a nice asymptotic relation between the area of the drum. It doesn't matter the shape, okay? The area and the asymptotic values of the quasi, of the normal modes, okay? So this was left as a conjecture for decades. The conjecture being that I can know, I can have knowledge about the entire surface that's vibrating, not only the area, but also the shape, just by listening to the modes of a drum, okay? And in fact, there's a very beautiful paper by Mark Koch asking the question if one can hear the shape of a drum in the sense of, if I know all the spectra, can I know the drum that's playing? And the answer was, was the following, no. The answer was no on the basis of a counter example. These are two different drums, by the way it's a very funny article, but these are two different drums, so these are two-dimensional memories of course, okay? And they have exactly the same spectra, so kind of puts a rest to the conjecture that we can listen to the shape of a drum. And then you might worry, well, then we have a problem. If this works in two-dimensional membranes, how the heck can I be sure that I can listen to the shape of a black hole, okay? But of course, I show you this to also tell you that in GR we're safer. We're safer because we don't have the freedom to tweak the boundary of black holes because in GR at least we still have something close to no-hair theorems, right? So we don't have freedom enough to change the black hole boundary and come up with black holes that have the same spectra with completely different shape, okay? The area is the same, but the shape is, so this is still fine. This for the area, this is kind of a lower moment of the conjecture. The full conjecture would be that not only the area, but the entire shape of the membrane. Yes. The area is safe. The area is still safe because it's the same area. I think this is proven for two-dimensional membranes, yeah. For black holes we're fine, again because of no-hair theorems, and in fact you can do this. Now, if you try to do this in the real universe, you will find that these guys, tau and f, of course, come with some error, okay? I mean, I'm going to show you a waveform for LIGO and it comes with a huge error. So you need to worry that you're going to try, you see, you're going to have, you're going to put down two frequencies, let's say, you measure two modes, you're putting down two frequencies and you want to test GR, but if the error is too large, let's say this frequency comes with this associated error and this other comes with this one. You need to make sure that the error is small, the measurement error is small enough that you can distinguish the two frequencies and the two dumping times, right? So of course, your measurement needs to have sufficiently high quality, okay? So we did this kind, we did the following. We populated the universe across cosmic ages and we let black holes collide, okay? You need to know how to grow black holes, it depends a lot on how you actually form them. Some of them are going to merge collide in the way I showed. This is going to generate gravitational waves and ring down. Some of this ring down we are going to see, some of it is just too weak, okay? This plot shows you what a future detector called Liza is going to be seeing, okay? We have not yet decided on the exact design of Liza. So the different points here are for different ways of building the device, okay? And the different curves here are for the different ways of actually forming and growing black holes, okay? From a very pessimistic assumption to a very optimistic. But the final message is even in the most pessimistic way of growing black holes, even in the most pessimistic way of building the experiment, Liza is going to be seeing at least one event per year where these guys are clearly distinguishable, okay? So at least one event per year, we are going to be able to use two ring down frequencies to test general relativity. So this is nice. So I'm going to finish with something that I would like to continue tomorrow. And this is the following. This is not, I mean, we're discussing ring down and gravitational waves, but it's not a dream, right? It happened. And the waveform is here. This was seen in the Hanford detector in the north, in the north of the U.S., this in the south. And if you make some effort, it doesn't take a lot of effort. You see some pattern here. The pattern, I hope Stas told you everything about it. It's two guys. We don't know yet what. I'm sure everybody else would say they're two black holes. But we have two compact guys. They are orbiting around each other. They lose gravitational waves. They lose energy because they emit gravitational waves. So the frequency has to increase just the atomic analysis, okay? When the frequency increases, the velocity increases, and it's a simple calculation to see that the amplitude of the wave also has to increase, okay? So what you're seeing here is the objects getting close by. At some point, they collide here, they merge, and then the signal disappears, okay? Now, we use this detection to come up with the numerical relativity waveform, basically, with the signal that best approaches, that comes closer to this noisy signal. And that theoretical perfect signal is this one, okay? The final stage here is a clear ringdown, not with these parameters, because this wackel is spinning at roughly 0.67, A over M, roughly 0.67, okay? So it's not exactly these numbers, but we know the numbers, okay? And it all looks very nice here. It doesn't look as nice here, but still. It looks sufficient evidence that what LIGO sees are wackels, okay? But I hope by now I convince you at least that this is a strong argument that LIGO sees photosiers, okay? And tomorrow, I'd like to discuss what happens if there's no horizon here, okay? And I think I'm going to stop now. Thank you.