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?  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?  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.  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.  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.  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.  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?  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.  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.  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.  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.  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.  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.  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.  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?  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.  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.  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?  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?  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.


hey you guys so if you haven't noticed yes I'm wearing the same clothes as my previous last few product reviews because we do a lot yeah we're doing a lot so going forward with the oldie but goodie product reviews we have the Jamaican mango and lime locking gel now you guys I was able to locate this product at my local fries was just which is Kroger's in the Midwest and South so let's just jump right into this this is six ounces it says on the container starts and maintains lock and twist naturally long lasting hold no buildup key no buildup because this is a jelly like consistency now I've dipped in this a few times I used it to lay down JB's edges I'm so sorry I used this to lay down my child edges and they also used this to set my twist I've been wearing my hair you guys since we've been quarantining I haven't changed much I've been wearing my hair and twist for like two to three weeks and I've been using this product so it smells really good one of the few product horizontally I would prefer a product of no sense unscented products is right at my alley but this product does have a nice sound smell to it so it's the color of a mango this is the consistency like a mango color and it is a jelly like consistency extremely affordable and when I mean extremely affordable I got this on sale for under five bucks when it you cannot look when it so what I'm gonna do with this product I'm looking for a cute spring like look now y'all I cannot flat twist to save my life you know those flat twists they do where they port their hair in the front the back you know it's all the way down symmetry no we can't do that I can't do that I'm just gonna be honest with you now I'm gonna attempt attempt to do my version of flat twist and then we're gonna add perm rods to get this curly curly look okay all right you guys so I'm taking my hair down I'm gonna go ahead and moisturize my hair I'm going to use the spray bottle I'm applying some water in larger sections and then we're gonna be using the curl the funding cream that I purchased from Sally's a couple of months ago I still have about five Georgia that stuff you guys along with the Jamaican mango and lime locking gel I'm going to go ahead look I'm not gonna flat twist I'm not even gonna try it I'm gonna go ahead and sit my hair and medium to medium small twists and I will be applying perm rods in about two more days I decided to leave these twists in for three more days that is one of the ways that I ensure that I get a very defined twist out I leave the twist in for a couple of days okay and I also ensure that my ends are properly detangled y'all I'm just about done here twisting my hair down I will apologize in advance my child is gonna be in some of this footage I'm showing you how I'm going to be adding the perm rods it's gonna a little tricky to record with my son home all day so here I am three days later I added the perm rods but only up to halfway of my hair now as I start to take down these twists they it is a little frizzy and I do think that's a combination of the product along with me not using any oil to help to unravel I felt like at the time I didn't need oil because I was touching my hair and it felt moisturized it actually felt a little oily so I just didn't think that you know I just didn't feel like there was a need to add more oil to my hair because I didn't want it to be heavy so I'm just gonna continue to separate my hair and unravel okay see how it's a little frizzy or you can see the frizziness on this side I'm going to I still think it's great definition now I'm not gonna deny that great different definition excuse me going to fluff up my hair you know me I like big hair so this is what the final results look like yeah a little frizzy on the ends but that's okay I'm still gonna rock it I'm gonna turn around to the back yeah for a city honestly you guys I could correct this I think I know what I could do next time use a heavier leaving conditioner with the lock and gel and oil to unravel but that is it you guys all hope you enjoyed this video thank you so much for watching take care bye


But after this stopway, this journey will take 45 minutes. This stopway will also connect to Ghangriya. Ghangriya is the place which is the entry point of the national interest. This stopway will be developed with a value of approximately Rs. 2,430 crores. This change will be a family member who will provide security and stability to the people. The development of this important foundation will increase the religious poverty. This will increase the economic development in the region. Along with that, there will be many opportunities for employment. After this, they will visit Adi Guru Shankarachari's Samadhi Stal. Before the mix-up, we have also communicated with the Shreemiks. So, we are showing you a few moments ago. The Shreem Jeevi who has been there for 12 months, has been working there for the first time in the season. The Prime Minister met them and talked to them. In 2013, when the disaster came here, everything was destroyed. After that, in 2014, Prime Minister Narendra Modi, who was the leader of this company, came to work here. And the Shreem Jeevi also keeps working here. And also tell you that through the big helicopters of the Air Force, the machinery has also been delivered here. As you can see, everything has been done there.


What is this what's going on? Oh, we're gonna record it on the overhead. Of course. Sorry.  I spit on your So all the tests are green. Can you confirm all tests are green? Yeah, that's only green boxes No, red or yellow boxes. No, that's just the lesions Boxes submit to see Q whatever that means commit Q. I don't so it's now queued up for commit. Oh Boom How we can aggressive use that as the end card or something Okay, well now that the end's done with let's do the start Right Yeah, this is our third go at recording this episode first time went well didn't it was really awesome good episode We were so high energy So the energy is going to be all good things come in threes as we try and remember all of the off the cuff Funny things we said during the other records and we'll crowbar them in Whether you like it or not here we go off you go because we have funny go on go on So don't tell me through your disgusting kerning first, right? Oh the Kerning. This is not getting any easier Third time I've seen the screen and look what is this and you could drive a plane down that and My personal this is just how chrome layouts this font. Oh, so we're saying chromes bad. Yeah, excellent Okay, I did that's what I did. I fixed chrome you fix chrome one of all of chrome I have I now have broken it is still broken. Okay, mostly but the part that now works I didn't work before I fixed up you now have code in chrome. I have code in Chrome Well, all right most like so tell me for it again So I Work on a side project which doesn't really matter, but basically if you know you have an image It's not square, but you kind of want it to be square You can either crop in or you can make it bigger and fill in the the the empty parts with a nice color So you see you have an image that is the correct shape You are making it square, but then you are making it square within like yeah, it's called letter boxing I did the thing. Yeah, it's very Instagram Of course it is it's for Instagram because they do their images square apparently you can also do five by four ratio But who does that who who does that? Okay, so yeah So the part that I want to talk about is that in this app. I used web assembly because sometimes that's fun Let's go. It actually wasn't necessary. I think but I thought was quite fun And just to remind everyone this is how you use web assembly you fetch Is it required that you go that high when you say Assembly the fun fact I only do this in English. I apparently never do this when I speak German When you speak German and it's one tone only or not in us, but not funny Well, you said but not funny and Right so assembly assembly here we go So you fetch your web assembly binary and just throw it into instantiate streaming That gives you back an instance and that instance has the functions that you can call This is what assembly and so I wrote this code and I developed and it was fun And so, you know in Chrome it was fine Yes, like let's test on other browsers and then Firefox went Nah, this doesn't like and I was like Firefox you stupid wait, so you okay, so you've chosen to to use a green cross there In you know in Unicode there is no Cross in a square that is red only the green one. There's a red by itself But no, there's no there's no equivalent to this in red That's horrible. Yeah, I was really sad and so I was like I could just use the free floating acts, but it looks Should you CSS like he rotates filter? Oh, I should have used that. There you go. I'm so I'm so disappointed myself now You could have done it between one of the free takes we've done Anyway anyway, so um, yeah, Firefox wasn't working. I was like What did it do wrong now? Why is Firefox of working like can they just fix it like it works in Chrome It must should be working in Firefox, of course And so I looked into the console and saw response has an unsupported mind type. Hmm. I was like does it though? Does it though because Chrome is clearly accepting it. It can't be that wrong and so I've been using HP server for a while now I've been using it here as well for development. Oh, this is the node the one. That's my PM, right? Yeah Yeah, okay. Yeah, use that did a curl and I found this content. I was like Ah, I mean, it doesn't really make sense to define Squeaky chair This is the specific but I'm aring it. I don't know. I'm looking at this and go like Okay, so I mean a character set doesn't really make sense for a binary exactly But also the the core part is that it's defined as application wasm And so it is intended to be WebAssembly and should be interpreted as WebAssembly. The chair said yeah, it doesn't really matter and I guess Firefox was being pedantic and Chrome was just being like And just accepting it and I was like well firefox Yes, and so I was opening an issue and then one of the firefox engineers came back and said like well Actually actually well the spec is quite explicit and says Extra parameters are not allowed. Did they do that? They wagged the finger like on every single line Yeah, they went into a video call like karaoke Okay, so even if you just have a semi-colon at the end which is a separator between the mime type and the extra parameters That is already not allowed. And so really what we discovered is that firefox was technically correct While chrome was being lenient but not spec compliant This is not unusual is that the amount of times I found something like oh firefox And then you look at it.  Yeah, okay, and I was like Why why isn't the tests? Why you know this these are kind of things that I would because it's so explicit in the spec It's well and this seems very strict. I don't know why there's a strict But if you are the strict put it in a test well oh There are tests and sadly you can't read what these tests are about because they kind of clip it away And only if you hovered with your mouse they should hide the results to show you the rule whatever, okay? We're basically these tests test if you reject invalid mime types Okay, and chrome okay, it's not fair not passing those tests because it is not rejecting those invalid mime types Firefox on the other hand well done them. They are completely test compliant Safari well safari is a fail because they don't implement the streaming API's at all Okay, they don't even have the phone call, so they're not getting the mime type wrong. They just don't have the API exactly fair enough What we see here is that tests have been failing and Nobody cared about it and I was very surprised by that and it turns out as I learned later on that usually whenever you Accept that a test is not failing you have to attach a bug to it inside chrome Okay, this test not failing. He's a bug that we're working on it. Yeah, these tests didn't have that what I don't know Why my hunch is that these tests were added after we already implemented the streaming API's and that? Some oversight happened a long way. This is still not good. I'm just saying like it's probably This state Nobody was aware of this is the current set of things that's that's fair because sometimes when like the first browser You know a browser will implement something first the other one will come and start working on it And they'll find like an inconsistency in the spec and they'll add tests to that and so that's probably something like that We'll have happened here, and so I was thinking That's an intro problem. Yes, it is because something works in Chrome and not in Firefox and That is where you usually end up in issues in production. It was like, you know what? Can I fix it? Is this something that is simple enough that I can do it myself and so I did I had two engineers Andrea's house and Clements who is not on here and who helped me through this entire process of learning how to actually clone Chrome compile Chrome Which how to find the files I need to patch? Yes, it's a big old project Chrome, isn't it? Yeah, yeah, if I didn't have a computer farm full of compilers that Google gave me it probably takes a good amount of hours Yes, it does in this case. It only takes 20 minutes Yes, but in the end it turned down to be a one-line change So basically the the function Streaming instantiate and streaming compile do this mime type check with it like is the mime type not application web assembly Right. Yeah, now the fun fact is that mime type is a helper function by us That strips away extra parameters from the content type header. So anything after the semi colon Right. Yes. No, it's what I'm doing instead. I now call content type Which gives me the literal content type header turn to lowercase because you know if you write application uppercase a lowercase or something like that doesn't matter That's the change. That's the entire change and now ship it. Yeah, and I did I opened the CL as you saw and Then someone came along and said well well I applaud your effort because this is interop nice try, but but we are doing the worst kind of interop Where we go from something that works to now that is something that doesn't work Right, which is correct because if there's a site in the wild that does this header with char set blah, blah, blah at the end It worked so far and now it stops working and if you break a large site for a lot of users And they go what did you do and you say haha? I think you'll find it's now Wow, that's okay then great Yeah, we'll find that millions of users couldn't access our products So they said the code we were was fine like is this the correct code? I mean it didn't arrive here instantaneously. I had code reviews and they caught some bugs and blah But they said we probably either want to send out the deprecation notice first Mm-hmm, or make sure we are not actually breaking a significant amount of sites Right, and so it turns out in Chrome. We often we gather some user metrics where you know What kind of files are even being loaded is a certain API even being used how often this is this code path happening Yeah, this kind of thing is not tracked So it's not like we track the content type header which with a certain type is delivered over the network That is something it's like you could add that but it takes six weeks to roll out Yeah, you have to gather the data. That's a long old process. We've done it with service worker stuff Yeah, yeah what we have instead and which apparently some Chrome engineers and think of what I find quite interesting is we have a GP archive Yes, we do. HGP archive crawls Five million pages every month. I think roughly. Yes, and it's it's slanted towards the like a set of top sites Yeah, most popular sites are crawls and like the amount of pages increase But they add you know from the top of a list they keep adding Yeah, and all the results basically all the fetches all the assets are put into a big old database and then you can use big query With some pseudo sequel syntax. Yeah, run some analysis on this is what I did So what I did is I took the summary requests all the requests of our sense during one of these crawls and I got the content type column and Beta just counted for all the files at end and wasm Yep, what that content type header was and apparently this is about 60 gigabytes of data, which is you know, because it's just text It's quite a bit and this is not just something that's like internal Google thing No, this is public anyone can do this. Yeah, open source database. Yep and So yeah, I did this query and I turned it into a Spreadsheet. Yay spreadsheet. So out of these five million pages that are being crawls Roughly two thousand even use WebAssembly at all. That's awesome. It's popular, isn't it? Yeah It's pretty new. Yeah And out of these two thousand pages or actually out of these five million pages exactly one page Has a WebAssembly file that would go from working to not working Right because it has it even then and that's I think it's very important I say that it doesn't mean that this site would break because it would only break if it uses the streaming API Which it probably does not because Safari doesn't implement it Safari doesn't implement it And this would be failing a firefox and it will be fine So let's I so that HTTP server thing it definitely says on the MPM package like this is for development Yeah, not for life That's the same server. Yeah, it could be so that means overall point zero zero zero zero two percent of pages are Potentially affected by my chain change So it would actually be given like the amount of time you have to have a deprecation notice in Chrome It would actually be a better use of your time to go and hand deliver a letter to the one person Yeah, runs that site, but let's go round to the house and just I know you can fix things remotely, but it would still be quicker to go round Yes, and say excuse me. Hi, can I just can I just get to your laptop and so when so they say Something will be closely Investigated something at this like a deprecation or a breaking change if you are on point zero three percent or more So we're way below that here. So in this case, they said, you know, it's fine You don't need to hand out a deprecation notice or anything like that. We can go ahead with this And so just now press press the merge button literally just now So this this is a change to the other two takes of this episode We've done because it hadn't gone through yet. No, just literally of the delay The bots finally finished a test all the tests went green. I could press the merge button Which means did you engineer the delays? Did you have you been tampering with the equipment? Just so you could have been found out So yeah by tomorrow or day after tomorrow, it should be in Chrome canary Which is sometime in the past when this episode goes out. Just just it's incanary now. Yeah at home Brilliant, check it out now. And if it's not incanary now It means the patch has been reverted because some are broke every time I put the I put the committed the CL I do Go check. Yeah, you can go check it out and make fun of me What did I do? Bye. Bye. Bye. See you happy next time Click the subscribe button Press the bell if you want to get notifications click the bell push the bell the bell I have no idea what you're on about to do all of those things Bell bell and this is the end. So it's the bell end You


Kyrie do you think that this season serves as a maybe a lesson for what is required to go through 82 games? The amount of commitment that an entire team or franchise needs to have amongst each other and secondly Do you plan on pursuing an extension with the Brooklyn that's this summer? man To answer your first question. Can you ask your first question again? Just oh for the 82 game season. I apologize Yeah, no, I just think it's important when You know you you can start as a team in October and and just Really build a relationships outside of the game. So when you're on the floor, it's just it becomes a trust and trusting environment You know when you're playing with so many new guys and you're trying to implement offense defense day-to-day and drill it and You know, you got guys coming off injury and stuff like that Just a realistic viewpoint of being in the NBA things happen.  So 82 games in the playoffs they definitely matter and In terms of my extension, man, I don't really plan on going anywhere. So You know, this is like I said, this added motivation for our franchise to be At the top of the league for the next few years and You know, I'm just looking forward to the summer and just building with our guys here


Alola everyone and welcome to the channel, my name is PhillyBetsu and in this video I'm going to be showing you guys an awesome cool fun trick that you can use using the move weather ball in Pokemon Sun and Moon. Now of course you're going to be equipping this with a Z crystal, the normalian Z, and then using weather ball and we're going to see exactly what happens when you do it. We got our Pokemon that we're going to use for this experiment. We got our two cast forms. One cast form is going to be used for causing the weather effects and our second cast form is going to be using the move weather ball. So we're going to be doing this live so you guys can see it. So without further ado, let's get right into this to see what is going down. Let's go in this grass and get to work. Here we go. Alright, so there is our wingle that first shows up. So I think what we're first going to do is we're going to make it hail. We're going to want to make it hail. We're going to test out the hail weather effect first and then we're going to use our weather ball afterwards. Of course, cast form is then going to change. Wingle is going to do absolutely nothing to it. And now we're going to switch out our Pokemon right over here. We're going to go into our second cast form. The second cast form has our move. And of course, cast form is going to change with the forecast. And of course, if you guys want to get a cast form, you can get it in the lush jungle in the game. Alright, so here we go.  Now, like I mentioned before, it does have a normalian Z crystal equipped to it. So I'm going to hit the Z power, right? And then we're going to go into breakneck blitz, which is using the weather ball move. We're going to see exactly what happens. Oh, that's not a break, break, break, break, break, break, neck blitz move. Check it out. Look what it's doing. Cast form is full for Z move, but breakneck blitz turned to sub zero slammer. So it's turned into an ice move because of the weather. This is a really cool thing to do. So there we go. Ice type Z move based off weather ball. Pretty cool. Pretty cool. We're going to take a look at the other ones, too. While we're at this, we go rolling in the grass. And of course, there's only a few weather types. We got hail, we got the sandstorm, we got the sunny day, and we got rain. So those are the four types that we're going to be playing with as well when we're doing this. So here we go. Cast form is going to go again. This time we're going to do the rain dance. Now also be aware it's not, this doesn't only apply to cast form. I'm going to show you guys later on the other Pokemon that can learn this beautiful weather move, the weather ball. There you go. Cast form transformed into the rain. It's rain version. All right. Now we're going to switch out our cast forms again and show you guys another weather ball using the break, break, breakneck blitz move, or the Z move. So here we go. We got a normally Z equipped, forecast enables again, right outside I ain't going to do nothing to us. We're going to Z power up and we're going to use our breakneck blitz utilizing the weather ball move. So here we go. This time we get the water Z move, guys. It's pretty cool. There we go. Hydro vortex. Just because of the weather. Boom. Get wrecked, potato. Oh man. This is awesome stuff, man. Which means you don't have to have every single Z move equipped in order to perform that Z move. There are some cool options that you can do in the game. Meaning I don't have to have my water crystal equipped just to get it. I don't have to have my ice crystal, Z crystal equipped just to get that Z move. So it's cool that there's ways of working around with these Z moves in this game when you don't have to equip the exact type Z move. All right. Here we go. This time we're going to make it sandstorm. Now this is a pretty good example for you guys here because this cast form doesn't have a sand type of change, you know, form that it changes into. So we don't have to worry about that. So now you guys are going to see that you don't necessarily have to even be the same type Pokemon to be using that move. Here we go. We got our second cast form out here. And let's see what goes down, fam. Time to use our Z crystal. Rattata's calling for help. Please, Rattata. Stop it. We're not trying to SOS chain here. All right. Now we're going to do the Z move again here and check this out, guys. Check this out. This is just proof that you don't need to only use cast form. Oh yeah. Look at that. That is so cool, man. That is so cool. Now it's going to use continental crush. And that's without transforming. Oh, man. How many ways can you kill Rattata? So Rattata, fam. So there you guys have it with the, what sandstorm does. You get to use the ground, the rock type Z move as well. So there you guys go with that. And then we're going to go in the grass one more time. Let me show you the final one. And we got a slow poke this time. We'll set out our cast form again.  Of course, cast form is going to change with this one. But like I mentioned before, the Pokemon does not have to change forms or change its type in order to perform the weather ball Z move effect, which is really nice. And there's a lot of stuff you can throw out there that people would not be expecting. So here we go. We're going to go and do our Z move. Our final Z move for this. Final Z move for this. Oh, it's a switcher cast form. Whoopsie, Daisy. So there we go switcher cast form. I got confused for a second. I thought my theory broke. I wasn't going to work. But here we go. Here's our next cast form coming out. It transformed. Slowpoke's going to use tackle, but it's not going to affect our level 71 cast form at all. Z power this up. And here we go, guys. Use the breakneck blitz. And it's going to transform its type. Look at that. Here we go. Inferno overdrive. Oh my gosh. Rip slowpoke, man. Rest in peace. Rest in peace. So there you guys pretty much have it. Those are the examples of weather effects and how that works with Z moves. And if you guys for the final test, we're just going to I'll show you just want to do it without any, any other weather effects. Don't want to do it with any sand streams. Don't want to do it with hails. Don't want to do it with rain days. Don't want to do it with sunny days.  It's not going to be anything different. This is the last one I'll show you guys. Let me just switch out the cast forms here and to our next cast form. Remember, normally in Z guys, it's what you got to have. That's going to use its water gun on us. And here we go. Breakneck blitz coming up again. And this is going to be the exact same type. It's not going to change anything. No weather change. Right. So weather changes do affect it. So if we have a rainy, rainy terrain going on, you're going to get that water move, the hydro vortex. But here we go. That's pretty funny looking at cast form do this. And there you go. That's the regular effect for it. So weather ball, weather ball can do quite a lot and you can do a lot of strats in the game with it. Now, at the same time, I want you guys to take a look at this stuff. We open up Serb, just to give you guys a look at the weather ball. Like we said, its base power is 50. Its accuracy is 100. The battle effect is that this attack varies in power and type according or depending on the weather type. Now, here's some Pokemon that can learn weather ball leveling up. So you got Lugia. We got Ho Ho. In the game, in Pokemon Sun and Moon, you can obviously get a cast form. You can get a Roserade through the island scan. And a Vinylux through it, a Volcanion. As well can learn it, but you can't get Volcanion in the game. That's going to be happening after Pokemon happens. And here are some other Pokemon that you can get in the game. Bellsprout, Weepin Bell, Victory Bell, Snow Run, Galeo. You cannot get Cherubi and Cherum in the game. You have to get that through the transfer as well. You can get Drifbloon, Drifbloon, Frostlass, Fomentis, and Laurentis. Now guys, I want you guys to think about this very carefully, alright. If you get Sunny Day on a Laurentis and you have weather ball. Where Laurentis can perform a fire type Z move. And Laurentis is not able to actually do that. Because if we take a look at Laurentis, and we go down to its usable Z moves. We don't see any fire types here. So with weather ball, we can actually go above and beyond its capabilities for a Pokemon by using maybe a fire type Z move on certain Pokemon or using a water type Z move. So that's something they want to keep in mind as well. You can apply it to a lot of these Pokemon as well. Use a Galeo, and you can also use a fire Z move, which is really awesome. So guys, I hope you enjoyed this video. If you enjoyed this, smash that like button. I hope you guys subscribe, and I'm going to bring you guys more awesome content like this. Spread the word, share what happens, try this gimmick with your friends. It's going to be a fun time. My name is Philly Beats you, and I'll be seeing you guys in the next video. I'm out.