 Okay, can I start? Okay, great. So, okay, so this is my last lecture on the relationship between black hole dynamics in large D and the dynamics of a membrane. And let me briefly remind you what we have so far learned. What we have learned is the following. If you look at the motion of black holes in a large number of dimensions, and you concentrate on motions that happen on time scales and length scales, order one rather than order one over D, then all solutions of this black hole motion are in one to one correspondence with these solutions of an auxiliary problem. The auxiliary problem is a problem of the motion of a membrane, a co-dimension one membrane in flat space. The variables of that problem are the shape of the membrane and a divergence-free velocity field that lives on that membrane. Okay, the membrane is subject to the following equations of motion. The degrees of freedom of the membrane are subject to the following equations of motion, project of how null to u times this is equal to zero. Okay, it's a very simple equation for the motion of the membrane. And we saw last time that there were as many equations as variables. I emphasize that the membrane lives in flat space. Okay, there's no general relativity and the dual version of the problem. How did that come about in our construction? It came about from the fact that we constructed black hole space times using quantities that were defined in flat space. The function rho was a function in flat space. It's normal field was a normal field in flat space. It's norm was one in flat space. The velocity field, the velocity one-form field was velocity one-form field in flat space. It's norm was one in flat space. This fact, very conveniently, and this was the key to using these Kershiel coordinates, this fact very conveniently turned the problem of general relativity into a problem of flat space because we've got some auxiliary constraints on the data used in our construction and the data was all data of one-forms and functions in flat space. So we had this quote-unquote ADS CFT, baby ADS CFT like thing, a gravitational problem related to a flat space problem. Okay, it's like the fluid gravity correspondence except in a way structurally more interesting because fluid gravity lived in flat space but in an unmoving flat space because it lived at the boundary of ADS which was just where the field theory lives, some static flat space. Here, once again, like fluid gravity we have the reduction of dimension. Our effective non-gravitational system lives in one dimension less than the gravitational theory. However, where it lives is itself a dynamical thing. It's the shape of the membrane. It's the membrane and the membrane's shape is part of dynamics, okay? So in a sense, there's a little more structure here than fluid gravity, okay? The place where this holographic theory lives is itself dynamical, okay? This equation of motion is one of the central results of the lectures. Any questions or comments about what we've already seen before we proceed? Excellent. Now in order to keep us disciplined I made a time schedule here. So that we go through everything that is important to say. So let us move on. We've done our review. Okay. Now everything I said was about uncharged black holes. It was about the equation r mu nu is equal to zero. But there is a very natural extension to the max where the vacuum Einstein's equation. It is the extension to studying the Lagrangian square of gr plus f mu nu, f m n, f m n by four, the Einstein Maxwell system, okay? And you can ask, what about this system? Can you generalize your large D analysis to the study of black holes in the system? And the answer is yes, and the generalization while technically requires some work conceptually is trivial. What's the difference? The difference is that instead of Schwarzschild black holes we have charged Schwarzschild black holes otherwise known as Reisen and Nordstrom black holes. Our little patches of event horizon are labeled by the size of the black hole, the position of the black hole, the velocity of the black hole and the charge of the black hole. So there's one additional degree of freedom labeling little patches. So our membrane should have one more degree of freedom, namely the charge of the little patch of black hole that we're making the membrane out of at that point, okay? How you have to implement this technically is just like before. You write down the Reisen and Nordstrom black hole in Kerr-Schild coordinates and then you write down a generalized ansatz for including this rho function, this u function and an arbitrary q function. The charge of this black hole is also allowed to vary in an arbitrary manner. Sometimes we've got one additional function. Then you roll the same machinery again, okay? Now if everything's working nicely, I mean this is a test to see how general this procedure is. Because if everything's working nicely, we have one more degree of freedom. So we need to have one more equation. Otherwise it would not be a closed dynamical system. Indeed when you play the game, you find that this equation of motion is generalized and generalizes to the following equation, p times. So that's this equation gets generalized to this in a manner that clearly reduces to that equation if you set the charge equals zero. Charge is a new variable. That's great. So generalizes the old equation. But as we said, we need a new equation that turns out that there is a new equation. And the new equation is del square q by kappa minus u dot del q minus q. So we have a new degree of freedom. This equation here, as we discussed before, has d minus two components. Because it was a vector equation. This equation is a scalar equation. How many equations? There's one. One additional equation for one additional degree of freedom. We continue to have a good initial value problem. At least counting points, okay? This equation, of course, like this equation, it reminds those of you who are familiar with the equations of hydrodynamics, of the equations of hydrodynamics, this equation should remind you of the equation of charge transport. For instance, this is a diffusive term. The diffusive term for charge, okay? And we'll see some of this a little more in detail as we look at charge components and stress distances. Ah, I'm sorry. So I should, in my paper, we use curly k for this, which I'm very bad at drawing. This is the scalar, whereas this is the extrinsic curvature itself. Okay, so this is, I'll write this out explicitly. It's uab, ua, ub, kab. That's the only place where you've got the tensor k. Everywhere else is the scalar k. Okay, so I think this is all I want to say about the generalization to charge at this first point in our discussion. Any questions or comments about this before we proceed? Please. You see, magnetic charges are point-like charges in four dimensions, but in d dimensions are some sort of form charges, okay? So that would be very different since we're looking at largely. Other questions, comments? Excellent, yeah, fine. So, now, let us, you see, we've got these nice equations. We've got these nice equations, yeah, please. What's, I could, I could, but you see then the objects that would be charged under them would be extended objects. And if you had nice, stable, extended objects in, I think one could. It's just that I haven't done it, okay? I was looking for the generalization of point-like charges. Charges like, got it by black holes. You know, yeah, it's mainly that if you're motivated by eventually using it for four dimensions, okay? You would, these are the most natural things to study, but certainly one could extend. Suppose one was interested in looking at 10-dimensional dynamics, that would be very natural thing to do. That's a good, good suggestion. I just haven't done that. Okay, other questions or comments? Particular, you know, one could, yeah. Your question makes, makes contact with this thing that Impran and collaborators called the Blackfold Program. You know, they've been looking at stationary configurations of extended objects, often that carry charge. And they have some effective membrane-like picture of those objects, but only for stationary solutions. The problem with that whole program is that if you start vibrating the things, the radiation from here will go there and use non-local interactions in a finite number of dimensions. The great thing about d-dimension, large d, however, is that all that gets decoupled. So you'll find effective local dynamics for those blackfolds in d-dimension. So your suggestion is more or less equivalent to asking for how the blackfold program extends to dynamics in large d. It's an interesting question. And one that would be worth studying. Okay, other questions or comments? Excellent. So we've looked at the extension to charge. Now, these equations look sort of unfamiliar and I wanna quickly tell you about, I wanna quickly tell you about some simple solutions of these equations. Okay? This came up in the discussion session yesterday. I'll be very brief about this. I'll repeat some of what I said in the discussion session. The first thing to note is that, let's look at the uncharged case just for simplicity. Everything I say about the uncharged case is generalizations to charge, but it's a bit more complicated to say, so I won't tell you about it. We're writing a paper on this and the radiation, which should come out in, this is not published yet, but we're writing a paper about this and the radiation should come out in the next month, so you can look it up there. Yeah, so let's first look here. As was pointed out in the question session yesterday, suppose we look at very simple stationary configurations. So static configuration. The configuration in review was equal to 1, 0, 0, 0, and so on. Okay, then this is so simple, and we look for membrane shapes that are time-independent. Okay, membrane shapes that are some surface in space, and then just time. For such configurations, you can easily convince yourself that this is 0 and this is 0. So for the equation of motion to be satisfied, we need del kappa is equal to 0, which means kappa must be a constant. Okay? So static configurations of the system are systems of constant mean curvature. Okay? There's one very simple solution of constant mean curvature, which is a sphere. As was pointed out in the question session yesterday, that's not the only, there is quite a rich class of solutions even of constant mean curvature. There are so bubble-like configurations that are sort of like strings that wiggle and so on. Already there's an interesting question of how to characterize the most general class of static solutions. Okay, that's a question we will not explore. Now, I should point out that the special case of membrane dynamics restricted to static and stationary configurations has been studied in examples in quite a lot of detail by Emperor Suzuki and Tanabe, a little bit after our first, parallel to an after our first paper on the subject. And for a while it was not so clear what the connection between what they were doing, what the connection between what they were doing and what we were doing was, the connection is simply that, as we've just seen for static, and I'm going to tell you about stationary, if you take our general equations and plug in the conditions of staticness or stationaritiveness, you get their equations. Okay? So the equations they've written are the things that they've done restricted to static and stationary configurations, special cases of the general equation, you can derive their equations from us. Okay? So this equation, the equation that, for static configurations, you have constant mean curve, which are first appeared in Emperor Suzuki and Tanabe. Okay? Very good. So stationary configurations have this solution. What about, if you're interested in something like a rotating black hole, if you're interested in something like a rotating black hole, you don't expect the velocity vector just to be 1, 0, 0, 0, because then there's no rotation. Okay? So what should you expect? Now see, suppose you've got a black hole under you, it's in some general motion. Since our membrane dynamics is highly dissipative. Okay? Since our membrane dynamics is highly dissipative, you expect everything to settle down into some equilibrium configuration. What are the most general equilibrium configurations? As I tried to say in the question answer session yesterday, what you'd expect is that there is one equilibrium configuration or discreetly many equilibrium configurations for every conserved charge you can add to the problem. The conserved charges you can add to the problem here, once you go into the rest frame, are basically angular momentum. So for every angular momentum, now, the commuting angular momentum, angular momentum corresponding to rotations in orthogonal two planes. So for every orthogonal two plane, you can add an angular momentum. That should cause the membrane to rotate in that two plane. When things settle down, what do you expect? The thing that you expect intuitively is that the membrane should settle down into a configuration of rigid rotation. In the study of hydrodynamics as a theorem, that the only equilibrium configurations of fluid motions are rigid rotations. And you might expect the same thing to be true here. We haven't proved it as a theorem from our equations, but it sounds like somebody might be able to prove this. We might be able to prove it, but we haven't tried. But it certainly sounds like the first thing you would guess. What does that mean? What do rigid rotations mean? It means that our velocity field is equal to 1. And now I'm choosing a particular set of coordinates. What I'm doing is choosing the metric of flat space to be minus dt squared plus dr1 squared plus r1 squared d theta1 squared plus dr2 squared plus r2 squared d theta2 squared, and so on. Breaking up space into a bunch of orthogonal two planes, such that on each two plane, r1 is the radius vector and theta1 is the angle. r2 is the radius vector, theta2 is the angle. Are my coordinates clear? It's a simple generalization of polar coordinates to products of two-dimensional polar coordinates. In these coordinates, what you mean by a rigid rotation is very clear, very simple. In these coordinates, rigid rotation is the statement that u mu is equal to 1, 0, because there's no motion in the radial direction. All motion is rotational. So my coordinates are t, r1, then theta1. And theta1 has a constant angular speed. And that's what rigid rotation is, so omega1. 0, omega2. 0, omega3, and so on. This is good except that u squared is not equal to minus 1. So all we have to do to make u squared equal to minus 1 is multiply by a factor and choose that factor so that u squared is minus 1. If you just look at this metric here and compute the norm of my u squared, you'll see that my gamma has to be equal to, square root of 1 minus omega1 squared r1 squared minus omega2 squared r2 squared plus so on. This is what I mean by rigid rotation. By rigid rotation, I mean that the velocity field on the membrane is chosen to be this. So what do you intuitively think of as rigid rotation? Things that just rotate. It's like taking the membrane and viewing it at a rotating frame. Is this clear? What do you expect? OK, the first question is a mathematical question. The first question is a mathematical question because now what we've done is this. Generically speaking, we had as many equations as variables. But most of our variables, namely d minus 3 of them, were in the velocity field. Most of our equations, well, they were d minus 2 equations. If we just plug in our favorite form of the velocity field, we will have only one variable, namely the shape, and d minus 2 equations plus a few constant parameters. If your favorite guess for what your solution is is a wrong guess, you will have no solutions. Because, generically speaking, d minus 2 equations cannot be solved. You can't solve d minus 2 equations with one variable function. However, if you made the right guess, all the equations would turn out to be the same equation. That would be one equation left, it would allow you to solve. I won't go through the details. But it's not difficult to check that this is the case with these equations. And that the equation generalizes to kappa, the mean curvature, is not constant, but is equal to gamma squared times some alpha, which is constant. That's great. Now, that's great. Now, these equations, a nice equation, it's an equation that you should think of as a partial differential equation on the shape function. You can try to make this partial differential equation an actually practical partial differential equation. And maybe I should take the two minutes to do that for you, just so that you see how you can translate between these abstract-sounding equations and equations that you can actually lay your hands on. See, suppose we restrict attention to the following configurations, configurations in which only in a finite number of two planes do we have rotations. And then there are no rotations in all the remaining two planes. Actually, if we look at the logic of what we had done, we might be forced to do that, because we assumed that things aren't too exciting in more than a finite number of dimensions. This is this SOD minus P minus 3 symmetry. So we're forced to do that. So in the language of this symmetry analysis of the previous lecture, our z squared is equal to s squared. And in the z directions, there are no rotations. The rotations are all in the x-y directions. What we could try to do is to instead think of this abstractly, turn this into a differential equation for try to parametrize this function by s squared is equal to 2g of xa. s is a particular coordinate. Nothing depends on the angles in the z directions. Everything depends only on s. And if we know this function s squared is equal to 2g of xa, we know our shape. Is this clear? For instance, for a round sphere, what would this be? It would be that s squared is equal to r squared minus r1 squared minus r2 squared r, however many r's there. Let's say there are nr's, little nr's, rn squared. Because then you take the r's to the left-hand side. And then this gives you spatial radius squared is equal to constant, which would be s squared. Is this clear? So if g turned out to be this function, that would be a sphere. And we could try to find other g's. But this will not be the solution when we've got angular momentum turned on. We can try to find the solutions with angular momentum with these velocities turned on. Is what I'm trying to do clear? Now let's turn this. You see, this equation can easily be used with this parametrization of the function. We can easily compute this constant mean curvature. I'm not going to go through the actual algebra, but the basic point is once you've got an equation like this, you write it as s squared minus 2g of xa. By the way, these xa's, we know that we've got these rotations. So we'll assume that it's function only of the ra's. Nothing's going to be a function of the theta's. So we've got x squared is equal to 2g of r minus ra is equal to 0. Then you find the normal vector dual to this membrane. What's the normal vector? n is equal to 2 times s ds minus 2 times del ag dxa. But then you have to normalize it. So let's get rid of the 2 because that will vanish with normalization. And s squared plus del ag, the whole thing squared. So the n vector is simply this. But s squared was simply 2g, so we can write this as 2g. And then we briefly discussed in yesterday's lecture that in large d, this mean curvature has a very simple expression in terms of this s coordinate. kappa at leading order, and I've forgotten a factor of d. So this alpha is over d. So kappa at leading order was just d times ns by s. Let me roughly show you how this comes about. We discussed it yesterday, if you remember, by computing this k. But let me give you another quick on exceeding my time for this. Let me give you another quick derivation of this. What is kappa? It's essentially del a and a. Now let's compute this del a and a in polar coordinates in the symmetry directions. So this is del mu n mu. So if you work in polar coordinates in the symmetry directions, then you get a factor of square root g that is equal to s to the power d minus p minus 3. Is this clear? Now the formula for del a and a has this del mu square root g, g upper. The g uppers are all trivial in our case, times del mu. The important thing is the square root g. So while you get del mu n mu, so 1 by square root g, del mu square root g and mu, since this g is a function only of s, you can take the square root g out for the mu coordinates. That's the x, sorry, the a coordinates, the x a coordinates. For the s coordinate, this becomes del s. 1 by s to the power d minus p minus 3. s to the power d minus p minus 3 times ns. And now you have two options. You can act this del s in the ns. Or you can act the del s of the s. But this guy dominates everything else because it picks out a factor of d. So you see this dominant term is ns by s times d, which is this formula. You can also get it from the formulas we had in the last lecture. This is a quick way of getting it. And ns by s divided by d is simply just s divided by square root of 2g plus del a g, the whole thing squared. So this quantity, which was some abstract quantity, in terms of this parametrization, is this very particular parametrization? Is this clear? Is this clear? Oh, sorry, I needed to get rid of the s because I had ns by s. So this left-hand side is d times, so kappa is equal to this quantity. Now in a very similar way, it's very easy to write down a formula for this gamma. It's basically the formula we wrote down before, which is equal to 1 over square root of 1 minus r1 squared omega 1 squared minus r2 squared omega 2 squared and so on. And we just square this equation times a constant. We square this equation and equate the two sides. This gives you a nice partial differential equation for g. Did I write? Sorry, this is one factor of gamma. I'm sorry. So if I square this equation and equate the two sides, I get the equation 2g plus del a g, the whole thing squared, is equal to up to a constant, some constant, times 1 minus r1 squared omega 1 squared and so on. Now notice that this equation has the following nice property. The right-hand side is a quadratic function of the r's. If you plug in a quadratic ansatz into g, the left-hand side is also for quadratic function of the r's because g is quadratic and del a g squared is also quadratic. So quadratic ansatz works perfectly. So all you have to do is take this equation, plug in a quadratic ansatz, and solve, OK? And I'll give you the solution, the solution. I can find it. The solution turns out to be s squared plus sum over n rn squared into 1 plus minus 1 minus 4 r squared omega n squared by 2 is equal to r squared. What can we say about the solution? Firstly, suppose we said all omega n is to 0. This is s squared plus all the r squared is equal to capital R squared. That's a spherical solution. If you don't set the omega as equal to 0, this basically rescales, effectively rescales the rn's. So it makes it like an ellipsoid. What's the interpretation of that? The interpretation is simply centrifugal force. If you're rotating, those directions are pushed out. It's not a sphere anymore. It's an ellipsoid. Exactly the same thing would happen with a soap bubble. It takes soap bubble. You rotate it, it'll become like an ellipsoid. That's what's happening here. You can solve this exactly, use the fluid gravity, use the membrane map to find the metric of the solution, and check with non-metrics. Kerr solutions are known in all dimensions exactly. You can do that, and the match is perfect. What I wanted to show you is how simple it is from this point. It's so simple. It's almost trivial, partly because you have so much intuition. Finding the Kerr black hole solutions in a large number of dimensions was not a sneeze. This is a sneeze. You can do similar things with charged black hole solutions and various such things. You could play a big game of trying to generate lots of solutions. We've only touched the surface of that, but there's lots one can do. My five minutes went a little over here. Questions or comments on this? OK, let's move. I have 25 minutes to go, is that right? Excellent. We'll have to speed up a bit. Now, the next thing that I want to do is to quickly tell you about the quasi-normal modes. I've been planning to do the calculation for you, but if I did that, I think we will not end up talking about all the other things that we have to do. 10 minutes might be a realistic estimation for doing that calculation, but I don't want to spend that 10 minutes. I really wanted to do the calculation, not because it's very illuminating, but just so that I could show you how fast you can compute the quasi-normal modes using this point of view. On the other hand, if you wanted to compute the quasi-normal modes actually in gravity, that's a big exercise. I don't know if any of you have gone through these computations of quasi-normal modes. Finding these linearized fluctuations, finding the gauge invariant fluctuations, computing the spectrum, it's a big exercise. It's many papers. One guy diagonalizes the fluctuations by these gauge invariant. Somebody else fixes a mistake. Somebody else does the calculation. It's a big exercise. I could really do it for you in 10 minutes. And all you have to do is the following. All you have to do is the following. You've got the spherical sub-bubble. Computing the quasi-normal modes above the Schwarzschild black hole would not be difficult to do it all about these curved black holes if you wanted. But I'm just, yes. Perhaps I could do it in the discussion session. Yes. Okay, excellent. Yes. But let me quickly just tell you what we do here. What we do is we've got the spherical black hole, which is a sphere. And so for this problem, it's useful to use the spherical symmetry. So let's work with the coordinates such that the metric of flat space is equal to dr squared plus r squared d omega d minus two squared minus dt squared. In this problem, the background black hole solutions r is equal to one. This is r equals one. And, or you can take r equals any constant. But because of the scale in variance-veinstein's equations, you don't lose any generality in taking r equals one. Okay? And then what you do is to, for instance, this is what we did. On the membrane, you use the angular coordinates, theta a, as coordinates on this membrane. Okay? And then let delta r be a function, r of theta a. And let u mu be u mu of theta a. And of course of time. Now, you have to be a bit careful of using these equations in the right metric. Because the metric on which you have to solve these equations are the metric on the membrane. That's not quite the round metric. It's a round metric plus some small perturbations because you've got this delta r fluctuation. Okay? When you put the, but you just plug in this configuration into these equations. And you get some differential equations. Because of the symmetry of the problem, delta r can be expanded in scalar spherical harmonics. u can be expanded in vector spherical harmonics. The equations of motion tend to eigenvalue equations relating Laplacian's on spherical harmonics to omega turns into a algebraic equation, never need to solve a differential equation. Okay? And that algebraic equation gives you the answer. And you get this thing that we've been talking about. About a lot, omega is equal to plus minus iL minus one. Sorry, omega is iL, yeah. No, iL minus one plus minus, plus minus square root of L minus one. This is the real part of the frequency imaginary part for the scalars and omega is equal to iL minus one for the vectors. Since we interpreted the low lying modes in yesterday's discussion section, I won't go through that again. The, what I wanna emphasize about the spectrum is that the spectrum is entirely dissipative apart from the zero modes. Now, this, I wanna emphasize again because it's physically important. This is a difference from fluid gravity or from fluid dynamics. You see fluid dynamics, the leading order equation of fluid dynamics are those of perfect fluid dynamics. They are the leading equations in the extreme long wavelength limit. And perfect fluid dynamics is non-dissipative. You see that from the spectrum of small linearized oscillations. Okay? In perfect fluid dynamics, the four goldstone modes are k by square root three, the real, and omega is equal to zero. These are the sound modes and the shear modes. Once you add in dissipation, you pick up the plus i k squared by t times number and this becomes zero plus i k squared by t times number prime. First order corrections to perfect fluid dynamics lead to dissipation, lead to imaginary parts and damping. Okay? In the extreme long wavelength limit, you have no dissipation. In our problem, our problem is more dissipative than ordinary hydrodynamics. There is no limit as far as I can see in which you have non-trivial motion, but not dissipation. Okay? I don't completely understand that qualitatively yet. Must have to do with the floppiness of the surface. That must be a very dissipative action. I'm not terribly sure why that's the case. I need to understand that better. Okay? So this is more dissipative than hydrodynamics made precise by the statement that there's no limit, there's no clean limit. Notice that we're already at first order with dissipation, already leading order. Hydrodynamics, you have to work first in some leading order to see dissipation. This is more dissipative than hydrodynamics, please. It's, yes. This becomes by d minus one. What I'm not terribly sure is how this coefficient scales with me. It could be that it doesn't scale. Yes. It could be that fluid dynamics also becomes sort of non-dissipative in large number of dimensions. That's a good point. That's a good point. I should be able to estimate that because we know how the viscosity scales like area, but I won't try to do it online. Okay. It's a good point. It could be that the lack of dissipation is connected basically with being in large number of dimensions. Okay, excellent. The, not lack of dissipation, the ubiquity of dissipation. Yeah. Okay. Fine. So we thought, by the way, we've generalized the calculation of quasi-normal modes, the quasi-normal modes about the spherical, about charge black holes. Now, when we first, when we did this in our paper, we thought we will generalize this and then do the gravity computation, but we quickly found that the gravity computation was beyond us, you know, to, I mean, it would have been two months to do that gravity computation. So we just worked out our quasi-normal modes and left it as a prediction, but that prediction has since been verified. Okay. I just want to say this again because it took us maybe an hour to do the calculation and took the guy who wrote that other paper who was Tanabe, I'm sure took him months. Okay. So just to show that it's useful, at least for simple things like this. Okay. That's it for quasi-normal modes. Now I want to turn to the discussion of radiation. Yes. Yes. Quasi-normal modes. And in fact, in the case of uncharged black holes, the quasi-normal modes have been worked out by direct gravity analysis, two first sub leading order one by D and so these equations have been found, they've computed the one by D corrections and checked that it matches the right gravity answer. It's not at all increased because always what you're doing is just plugging in spherical harmonics. Once you plug that in, you get some del squares acting on spherical harmony. You're solving an algebraic equation. Okay. That algebraic equation changes order by order because you get some del square, you get some del to the fourth. Fine. But you know how to act del squares and spherical harmonics. Instead of half an hour might take 45 minutes. I mean, okay. The complexity is knowing what the one by D correction to the equations are. That takes a lot of time. Yeah. Also what? You see in four dimensions, it's, yeah, I think so. I think so. I mean, in the sense that the imaginary and real parts are comparable to each other. One thing that you can see here, I haven't looked at, you see in four dimensions all you have is numerics. I haven't looked carefully at that in all details. But one thing you see here is that you might have thought, you might have thought that perhaps you could lose dissipation by going to large area. But actually the reverse happens. The square root of L minus one, which is the oscillations, is subdominant compared to the dissipation at large area. I'm not sure that survives in four dimensions. Probably. Yeah. Comparable, comparable. Yeah. In fact, for the first few angular momenta, we've looked at comparing what the quasi-norm, in fact, Emporan and company also looked at this. Comparing what the quasi-normal mode frequencies are at D equals four just from numerics. And comparing that with what you get from the large D expansion, if you just take the large D expansion and plug D equals four in. And I haven't compared very exactly, but within five or 10% they agree with each other. I mean, if you plot them, they're not very near to each other. So the large D expansion does pretty well for D equals four just at the level of quasi-normal modes. That's a very weak test. It's just linearized analysis, but it's not. Okay. Any other questions or comments about quasi-normal modes? Yes. We've not considered it. Yes. It would be, it would be suddenly possible to do. It would suddenly be much easier to do than doing the gravity problem. Any other questions or comments? Okay, excellent. Yeah. So now let's move on to the discussion of radiation. You see, I started these lectures by talking about black hole collisions and a large D version of the LIGO experimentalist sitting on some planet looking at these black holes. And so far what we've got is this nice correspondence between black hole dynamics and membrane dynamics. But there's no mention of the LIGO observer. Okay, where's the LIGO observer? Where is he seeing? Because the LIGO observer sitting far away. But what is he seeing? If you think about this, you might find yourself quite puzzled because it was a very important part of our whole program that our solutions very rapidly decay to flat space. And so you might think the LIGO observer is seeing nothing. And you might fairly ask me, where is the radiation? If some black holes are doing something funny, somebody sitting far away can only detect it from the radiation, where is the radiation? What is the radiation doing? Okay, so that is what the next half an hour, the last part of these lectures are gonna be about. About the radiation that is emitted by these black hole motions. Now, the first thing I wanna explain to you, the first thing I wanna explain to you is that there is a certain kind of mathematical fact about large dimensions. That has nothing to do with gravity per se, just a fact about wave equations in a large number of dimensions. That tells you that radiation is highly suppressed. Okay, reasonable sources, reasonable sources which have order one strength produce very low radiation. I wanna explain that to you. And the best way to say this is by understanding the structure of the Green's function of the equation del squared g. You know, the equation del squared g is equal to delta of r. Okay? So suppose we're looking at a massless minimally coupled scalar. We're not, we're actually looking at a graviton and the gauge field, but it'll make very little difference to what we're actually gonna say. It's much easier to do the analysis because we don't have to worry about indices. Okay? So suppose I'm interested in this Green's function. Del squared g is equal to delta of r. Now I'm in Lorentzian space. So to specify the Green's function, I have to specify some boundary conditions. I have to specify which Green's functions I'm interested. What's of interest to me? The interest, you see, we're looking at configurations where a black hole was at rest to start with. Then let's say for instance, two of them banged on. You hit it in some way, oscillates and radiates out. The boundary conditions we should be looking for are Green's functions, okay? Are Green's functions that have no incoming radiation, but only outgoing radiation. Note, it's the same boundary conditions as quasi-normal modes, it's not a coincidence. Same problem, okay? So I'm gonna be looking at the Green's function with boundary condition only outgoing radiation at infinity. What is my boundary condition at the event horizon? That's a non-question because there's no event horizon. I'm looking at del squared G is equal to zero in flat space. Okay? Smart simpler problem with black holes and so on, so I'm looking at Green's functions in flat space, Minkowski space. Now, I'm going to do something that turns out to be very useful, though it mitigates against your, I mean, it feels like an ugly thing to do at first, which is I'm going to break the manifest Lorentz and Merian's of this problem and treat space and time separately. I'm gonna work in Fourier space and time, but position space and space. And you will see why this is, it turns out to be a useful thing. Okay? So let G equals e to the power i omega t times phi, which is a function of r. Okay? Integrated over some a of omega. Okay? I'm now gonna find a differential equation for this, well, that's better to say, it's phi omega of r. Okay? Integrated over r. I'm now gonna find a differential equation, turn this Green's function problem into a differential equation for this function phi omega of r. Please. In this problem we do since we're in flat space. Okay? I'm just looking at the Green's function in flat space. Okay? So great. So my del square was minus del t squared plus spatial del squared. Okay? Minus del t squared plus spatial del squared on e to the power i omega t phi omega of r is equal to zero. Now minus del t squared acting on this just becomes plus omega squared. That's clear. So we get omega squared. And then we have plus del squared. But now del squared is del squared on a function that is clearly radially symmetric. Okay? Sorry, I've written zero, I really mean delta function. But I'm gonna be looking everywhere except r equals zero and then I'll fix the condition at r equals zero presently. So let me stay away from r equals zero. Okay? That's one of the nice things about working in position space in r. You can treat r equals zero as a special point to be dealt with later. Okay? What is del square? We've assumed that this thing's a function only of radial position in r. Okay? So now we can use polar coordinates. We can use polar coordinates to figure out what del squared is. Okay? So we're in d minus one spatial dimensions. We've got a d minus two sphere. So square root g factor is r to the power d minus two. And so this is r to the power d minus two del r to the power d minus two phi omega phi omega is equal to zero. That's clear. For instance, in four dimensions that r to the power d minus two would be r squared. As is often the case with problems like this, it's useful to convert this into a Schrodinger type problem. It's useful for intuition to convert this to a Schrodinger like problem. What's a Schrodinger like problem? Schrodinger like problem is a linear differential equation with two derivatives in r, a two derivative term but no one derivative term. That's what characterizes a Schrodinger like problem. Okay? Oh, I missed a del r. Sorry. Yeah. Now see, suppose I just used phi. There would be one derivative term coming from del r on this r and this del r on this. Okay? There would be one derivative in phi. So however, if I define phi omega is equal to psi omega times r divided by r to the power d minus two by two. There are two one derivative terms you could get by acting with this del r or this del r on this guy. Each of them would produce a factor of d minus two by two with a minus sign. So minus d minus two. This guy would produce a d minus two that would cancel. So in terms of the function psi omega, there's no one derivative term. Yes. Excellent. So now we go to the variable psi omega and work out the equation. So we've got omega squared, psi omega, okay? Then there is of course the two derivative term which is del r squared, one by r squared, psi omega. And then there's a constant term. That's the term which comes by acting with this del r on r to the power d minus two by two. Okay? That makes it, that gives us a factor of minus d minus two by two. And then we get what's left over as an r to the power d minus two by two minus one. From that rate, d minus two by two. Yeah, d minus two by two minus one. I think that's right. So that's d minus. And so we will get d minus four by two on psi omega is equal to zero. The minus sign plus sign sounds good. What? I missed a minus sign because I differentiated here. That gave me the minus sign. That's why I got the minus sign. And then the second derivative gave me a plus sign. d minus one by two plus one. I believe you, let me take the large d limit so I can ignore the plus, plus nine, so it's okay. Okay, I'm sorry, we should do that more carefully. But we get omega squared psi omega plus del r squared psi omega by r squared minus d by two squared psi omega is equal to zero. We could do it more carefully. I just don't want to take the thing. I would have to think. Okay, now where are we? You see, if I want to write this in the Schrodinger form, I want to write this like momentum square. Momentum square comes with a minus sign. Okay, so I'll write it like this with a minus. I'll take this to the other side so that becomes, something is equal to e psi. Okay, so we've got minus del r squared psi omega plus r squared, this r squared was here, I'm sorry. Just by dimensional analysis. Where did I mess that up? Where did I mess that up? Probably here, that was an r squared. So plus d by two squared by r squared is equal to omega squared. And this one, there's an r squared here, yeah, thank you. So this is equal to zero. Sorry, not is equal to zero. Okay, sorry for the incompetence. This is the final equation. Now what I want to show you is a nice fact about this equation. This equation here is the equation of a potential. Is the equation in a potential where v of r is equal to d squared by four r squared? It's the equation of a particle moving with energy omega squared but in a potential v squared, d squared by four r squared. So let's look at this equation. This potential is huge because of the d squared by four. It goes like this. Let's look at our energy. It's like this. Where does the energy cut the potential? It cuts it where d squared by four r squared is equal to omega squared. Where r is equal to d by two omega. Only at r equals order d. Okay, for r much, much less than order d, r much, much less than this value, the solutions to our equations are tumbling solutions. They're exponentially damped or exponentially growing. Okay, they're damped to growing solutions. For r much, much greater than this. Okay, these are oscillating solutions. The key point I wanted to draw your attention to is that this is the radiation. In order to solve this problem now, it's easy to solve this problem in complete detail. What we would have to do is to impose the boundary condition that the radiation is purely outgoing. That fixes this boundary condition. And then continue the solution here and fix the overall normalization so we get the delta function right here. Okay, the solution to this is some sort of Henkel function. You can just write it down. But all the qualitative features of this solution are visible from this diagram. The solution takes the form, the qualitative form. Psi is equal to a r to the power d by two plus b r, b by r to the power d by two in this region. And psi is equal to e to the power i omega r in this region. Okay, so in this region it's like this. In this region it's like this. Matching across these two regions fixes what a and b must be. Overall normalization set by getting matching the delta function. And perhaps not surprisingly what you find is that the ratio of you get some overall number here but the ratio if you choose this to be one by d this turns out to be of order d by two omega to the power d. You can compute this by just looking at the Bessel function so on checking that this is the case. Now what does this mean? Inside you've got a damped solution and a growing solution. Inside you've got a damped solution and a growing solution. However the growing solution starts off with a very small coefficient. So almost everywhere inside the damped solution is the dominant one. However because the growing solution grows and the damped solution dabs out at some point they become equal in magnitude. What point is that? That point is r to the power that point is this transition point. r is equal to d by two omega. When the damped and the growing solutions meet they merge and turn into radiation. Is this clear? So this is the structure of this Green's function. There's a scale separation in this region we've got damped and growing. This is like coulomb damping. There's some growth and in this region we've got oscillation. What that tells you is now what is the magnitude of this oscillation? The magnitude of the oscillation can be obtained by matching the value of the function here with the value of the function here. And the magnitude of the oscillating solution therefore is simply of order d by two omega to the power d by two. Yes? Oh, time. Time is over? Give me five minutes. Oh, we haven't got a radiation too. Okay, okay, okay. Okay, what is the conclusion of this? The conclusion is two-fold. Firstly, the Green's function has an oscillating piece corresponding to radiation but it's very small. It's suppressed by one over d to the power d. The second conclusion is that the Green's function here is largely insensitive to whether it's an advanced or retarded Green's function. If you work it out, that changes the details of this coefficient. It multiplies this by a factor of two or something like that. That changes from advanced to retarded but in this region, this guy is the dominant guy. All Green's functions are more or less the same when we're within a distance one of our source. Okay? They differ when we get to a distance d and then that difference turns into different kinds of radiation. The radiation is always very small. Okay? This is a simple kinematical fact about large d that we've described by looking at this equation and it's generically true. It's true of maximal radiation, true of gravitational radiation. Okay? This is the reason why radiation is so small in our problem. It's just a fact about this Green's function. The effect of source for radiation is not going to be terribly small. Okay? But the radiation itself is very small because of this kind of action. Okay? Now, in radiation two, which I don't have the time for, I'm just going to tell you the main answer. What I wanted to tell you was something very interesting I have to tell you this. It's the following. The question we started out addressing when addressing radiation is what is the radiation away from our membrane? Okay? And in principle, once we've got this analysis, you might think that we have insufficient data to answer this problem because this damped mode is the mode that you see in our solutions. This is the one over rho to the power b. That's this damped mode. But as I just explained to you, in order to figure out what the radiation is far away is, you also need to know what this mode is. Changing the magnitude of that mode in the scales of the radiation. This mode, a little bit away from the membrane, is highly subdominant, exponentially suppressed. None of our approximations are good enough to capture it. Another way of saying it is that if you take our membrane solutions and go a little bit far away, the solution is very near flat space. So we've got a linearized solution of Einstein's equations. And you might think we could just continue that linearized equation to infinity, linearized solution to infinity with one piece of data. The damped guy. And you need, for a second order differential equation, you need two pieces of data to continue the solution to infinity. That's another way of saying it. So how do we in principle have enough information to compute the radiation field? Answer is yes. Because where mathematics fails, you use physics. Physics tells you the second boundary condition. The boundary condition is simply that there's no radiation coming in. All the radiation is going out. So if you know the damped piece, and you know that the radiation is coming out, then you can figure this out and compute the radiation. Is this clear? So we do have enough information in our problem to compute the radiation field. The principle is clear. Our solution, which is large d, which works in a distance one by d away from the membrane, overlaps with another approximation. The approximation is linearized Einstein's equations. We have only one piece of information, a couple of that with the second piece of information that we know on physical grounds. That there's no incoming, only outgoing radiation that's sufficient to compute the radiation field. I was planning to explain to you how you do it. I don't have the time. I'll just give you the answer. The answer turns out to be very beautiful. The answer turns out to be that the final, when all the dust has settled, in order to compute the radiation field, what you have to do is this. Associate with the membrane, the following stress tensor. It's like three minutes unfinished. The principles I enunciated convert into mathematics, turn out to be equivalent to the following statement. Then the membrane is wiggling. That wiggling induced, I mean, associated with that wiggle, there is a stress tensor that lives on the membrane. So this whole thing is multiplied by a delta function of rho minus one. The stress tensor lives exactly on the membrane surface and has the following indices. The radiation field is obtained by taking the stress tensor and using it as a source with a retarded green's function. Convoluting the stress tensor with a retarded green's function gives you the radiation field. For instance, in the Dondagage, where you take del h is equal to zero, you just use the simple green's function that we discussed before. Convolut the source and you get the h field outside. Okay. The radiation is very small. Why is it small? Is it small because t is small? No. This is order one. In fact, order D. It's small because of the green's function, as we just discussed. It's just a kinematical fact about logic. Now, once I've said this, there's a consistency condition. You see, I've told you that there is a stress tensor that sources radiation. Okay. I told you that there's a stress tensor that sources radiation and that the radiation energy that you get from that source is very small. Now, the consistency condition is that if this is really true and if the answer is not to depend on what gauge we use for linearized gravity, the stress tensor better be conserved because otherwise, different choices of gauge would give you different radiation. Okay. So you can check is the stress tensor conserved? I won't go through the algebra but the answer is yes. And in fact, the stress tensor, conservation of the stress tensor is exactly the membrane equation. The membrane equation, the one with the del squared u by kappa plus del kappa by kappa plus plus the u dot del u is simply a restatement of the conservation of the stress tensor. In fact, things are a bit better. You might first think this can't be true because conservation of the stress tensor is d minus 1 equations. Whereas the number of equations of motion we had was d minus 2. Surely conservation of the stress tensor has more content than this. But it turns out that if you take the conservation of this equation, the stress tensor, del a t a b is equal to 0 and dot it with u b. Then it tells you something very nice. This really reduces to the statement del dot u is equal to 0. So although in our construction, del dot u equals 0 was put in from the beginning from the point of view of the stress tensor it just comes out. One of the conservation equations for the stress tensor is the divergencelessness of the velocity. And then the remaining conservation equations are our equations of motion. In a entirely similar story, you can play with charge. You can generalize this to add charge and you can look at the charge current. The charge current convoluted with the Green's function gives you charge radiation. We're hoping to have a speculated discussion d equals 4. We don't have the time unless somebody wants to help me speculate in the discussion section. Maybe I can just end these lectures now by saying a with a summary. What have we shown? We've shown that the problem of black hole dynamics in large d, charged and uncharged is equivalent to the motion of a membrane. A core dimension one motion membrane moving in flat space. The degrees of freedom of the membrane are its shape its charge density and a velocity field, a divergence free velocity field on the membrane. The most conceptually satisfying way of stating the equations of motion are simply that gravity allows you to find constitutive relations for the stress tensor and the charge current on the membrane in terms of its degrees of freedom. The equations of motion of the stress tensor and charge current and when this membrane moves around it sources radiation in the way that any stress tensor or any charge current sources radiation. Because the radiation is so small you can get away with linearized formulas. That's the summary of what we found. These statements I've presented this at leading order in large d that are Yogesh and Shubhajit and collaborators have worked out the first sub leading order corrections in large d. First in a systematic expansion gives you a systematic expansion for black hole physics in large d. Last statement before I stop is this. Let me end here by just saying the following. It's been great fun giving these lectures and it's been great fun hearing all your questions and I want to end with an advertisement. I come from the Tata Institute of Fundamental Research in Mumbai and this post doc program I want to invite all of you to apply to that program. It's great fun to be in TIFR it's great fun to be in Bombay please come we'll have great fun together. Okay for those of you who want to come for the full two or three year period that's great but we also consider short term post doc applications for instance for six months a slot that has worked in the past for this is for people who already got another job let's say I don't know you're going to some other place in September you've graduated you've finished your PhD work in February you want to come for six months to TIFR we will consider this this is for exceptional students please apply to us please come and visit us and please come to be our post doc thank you