 So let me again remind you where we stopped. So last thing we've been doing, we've been considering Lagrangian theory for some fields. And the Lagrangian was supposed to be first order. Then I defined something which I called another current. So Bob has been telling you that this is the wrong notation. And then you should write j mu here. I don't like j mu because j mu is like an electric current or something. So I'm not going to use this either. So you tell me what should I do. Well, so let's maybe settle for h mu. Because this has to do with Hamiltonians or something like that. I never dared to use this dirty word in my lecture, but since Bob did it, then I can live with writing h mu. Then to do generativity or a theory of a metric, we introduced, well, this is generativity since I'm taking this Lagrangian. So we take a Lagrangian, which is 1 over 16 pi r, which we all now give normal equations for general activity, and remove a divergence so that this becomes a function of first derivative of the metric. And because it's ugly, if we just do it with partial derivatives, we just introduce a second metric so that everything becomes geometric and covariant. By the way, I've forgotten a square root here, the g. And so I have a scalar density, first order, at the price of introducing some extra structure. By the way, in my previous lectures, this g was a space metric. And there was this gothic g, which I don't write as an 8, but it's a gothic g. So for this formula, this is the gothic g of my previous lectures. So this g here is actually a space metric now. And if, moreover, you assume that Lxg bar is 0, then I can calculate these other charges. And well, you can always calculate them. But if, moreover, this is 0, then the integral over the surface and the field equations are satisfied, which in this case say our vacuum, the result is true with matter fields, then you get a formula like that. Now Bob told you this morning that the nether current doesn't give you energy. Yes and no. So if you apply this procedure and you start with this Lagrangian, in which case it's this higher order theory, and you calculate the nether current, you're not going to get the same thing as I did because you're starting from a different Lagrangian and you get something else. And then he has to correct his boundary term to get the energy. If I do this the way I'm doing here, I don't have to correct anything. I get immediately the right thing. There's nothing mysterious or better or worse in what he's doing or what I'm doing because at the end what is important is the boundary conditions. And if you start with what Bob was doing, so start with this second order, calculate another nether current, which is appropriate in this case, then the boundary conditions for the variations will not be satisfied unless you correct this term to what I have directly. So it turns out that for these boundary conditions the approach here works better. But if you were working at a finite boundary, neither his approach nor my approach will give you directly the right formula. You have to work much more with the boundary term to understand what you have to do. So there's nothing better with this except probably, in fact, the calculations are easier in Bob's formulation because you don't have to remove all these divergences. So maybe there is an advantage of going his way. But then you have to do something with the boundary term. So if you go this way, you have the result directly. You don't have to think why it's better. Good. So this is a boundary term. And let me just mention that I wrote it, I think, like that last time. But I changed the order of the indices here. There's nothing wrong with doing this as long as I don't tell you where your alpha beta is. But if I want to be consistent with the nodes, it's just an issue question of convention. If I want to be consistent with my lecture nodes, you have to change the order here. So yesterday I wrote d alpha wedge d beta wedge this. And you can do this, but then you have to change the sign here. I didn't write anything wrong. But the formula here is going to consistent with this convention. And if I do this convention, that's the convention with my nodes. So maybe it's easier for you just to use this one. Good. So I never wrote you the formula for this thing. And it's pretty terrible. It has two parts. It has one part, which is linear. Well, so there are two indices, right? So in this tensor, alpha beta is alpha beta. By the way, of course, one way of thinking about this is that this is because you're integrating over a sphere at infinity. That's what I'm interested in. So in this case, you only have, I think, with this convention that's going to be ui0 dsi, where the dsi's will be your space matrix. So one of these indices immediately becomes a 0. And then you don't have to worry why there are two indices here. Then this is the usual formula if you want to with a normal vector, right? So ui0 times ni d2s. Good. So there is one part. Everything depends upon x, right? So you have to choose your vector x. And depending upon the vector, we're going to get various things. There's one part, which is linear in the vector field. And there is another one, which is linear in the derivatives in the vector field. So I have this d bar alpha x lambda here where this d bar derivative is a covariant derivative of the background. So this is the part which is depends upon the derivatives. And if you're looking at translations, translations are in Minkowski spacetime are co-variantly constant. Then this term drops out. So you get immediately a simplification when you look at translations. Then you can forget this term. You only have to worry about this one. But then this one is pretty terrible anyway. So it's written here. And I'm not going to check this evening if you remember the formula. So that's a little terrible. It's a density as it should because there's a determinant of the background divided by a square root. So this behaves like a square root. Here you have two determinants. So this is a scalar. And this covariant derivative, again, it's a covariant derivative of the background. What's wrong here? That was it. Your eight tensor appears as a product here. But exactly. So this would be two gothic g's here. Except that my gothic g was only the metric where your gothic g had a. So we should synchronize notations better next school. Good. So before doing anything with these equations, let me write you some simpler formula for this. So now this works with any metric g bar. So as I already told you, you can use this to do asymptotically anti-deceto space times. And in the unlikely event, I have some time left at the end of this lecture. I will tell you a little more about them. But let's just do asymptotically flat. And let's just look at the energy momentum. So this is going to be my paragraph 2-3, I believe. This would be asymptotically flat. So energy momentum in asymptotically flat space times. AF being asymptotically flat. So let me start with some alternative formulae. So that everyone can read what I'm doing. Just move here. So this is pretty terrible, but there are nicer ways to rewrite it. So now assume that we have a translational keeling vector for the background. Of course, it's not a keeling vector. Yes, what did I do wrong? There's something mismatched here, right? Terrible, terrible. Let me just check. What did I do? There's a g beta gamma floating around. Good, thanks. So the g beta gamma missing. So you have to differentiate this and this index. Thanks a lot. We assume this. So I'm not going to do the calculation, but there is a much nicer expression. So let me write it like that. First, this term is gone. So I get something which is linear in x. So the numbers which come with it are going to call p mu x mu. And so this is the same as the Hamiltonian or this another charge here. And I mean three dimensions. You can do it in any dimensions, but let me just bother about the expression. The analogous expressions. Higher dimensions. So this is always the spacetime metric now. So you're going to tell me, why is this any better? I did like that. It's still terrible. Let me just see if I have the indices. So if you just want to think about p mu, so just forget this. And then you have the formula for p mu. So integral of S infinity takes a sphere past to the limit. This is the epsilon tensor with the square root of g factor in. So that's a density. So lambda is contracted here. Gamma is not contracted, so this doesn't look good. Yeah, one of them should be beta, right? So it's probably a beta. You just, yeah, that's a gamma here. So now everything looks to be contracted. Whether it's correct is another issue, but at least. OK, so is it better? It's a little better. Somehow, well, there are some anti-symmetrizations and stuff like that. These anti-symmetrizations are, of course, taken by putting an epsilon here. And then you can replace this by gammas. What I have gained is an integration by part cute formula, which I think was first noticed by Ashtekar. You can integrate things by parts here. And you're going to get x mu, x, x mu, x mu, r mu, mu, alpha beta, square root of g. And I'm back with my forms, s alpha beta here. But this one is much nicer. So this is an integration by parts. This is still my vector x, which was constant in asymptotic of scan coordinates. And here I have introduced the position vector. So the position vector is something which is completely not geometric. It comes with the coordinates. Well, we're not shocked because this was not geometric. So that's not astonishing. However, this looks already much more geometric than the previous one because it doesn't use the Christophels, which transform in a terrible way, but a Riemann tensor, which transform is in a nice way. So this is a much cooler formula. This is the Riemann tensor. This is the Riemann tensor, the curvature tensor. And so the way you get this formula is you write your Riemann tensor in terms of derivative of Christophels, integrate by parts. And by integration by parts, this axis will disappear. You'll get this. This is capital X. This is the same vector as here. And this is the position vector. So this is the coordinate. So x mu is really t x, y, z. So this is not geometric at all. Putting an x is not a geometric thing to do. Good, but so that's the formula. You can replace it by the vial tensor if you want to because I've assumed vacuum. And if you don't assume vacuum, you assume that the matter fields decay fast. And therefore, any contribution from the matter fields from the Ricci tensor here will have to be zero. Otherwise, these integrals don't converge. So I'm going to come back to the question of convergence of these integrals and stuff a little later. But let me just first show you the formal formula. So this one is from a direct calculation. This one is a lot of juggling with gammas and derivatives and stuff. This is integration by part. There is a cute formula in terms of the Cauchy problem. So because if I'm working on a surface t equals zero, I can decompose the Riemann tensor. So the R of the spacetime metric, spacetime metric, sorry, I need more space, R nu mu alpha beta of the spacetime metric. And if I just take the spacetime metric, it's equal the Riemann tensor of the, and I don't want to use a different symbol, I could, but of the spacetime metric plus k square terms. Well, if you want to, I can even write what they are. Maybe, but I know roughly up to sine what they are. So if you're happy with up to sine, then it's for a i k a j a minus a k i l k j k. And it's a little chance maybe it's even correct. I think it's correct. So you're integrating at t equals zero. So at t equals zero, this would be zero. So only the space part here will enter in this formula if you just work it out. And moreover, if you think about what we've been assuming before, that k i j is O of R minus 1 minus alpha, which alpha is larger than 1 half, then this k square terms are O of R minus 2 minus 2 alpha, which is slower than R to minus 3. And if you look at this formula, you have a position vector, which behaves like R. You have an area coming from the integration, which is R square. So this gives you an R. Area gives you an R square. This is a constant. So you get a R3, multiplying things which are a little over i minus 3. So this will give no contribution. So no contribution. So you can go ahead and put in, rather than putting the spacetime, curvature tensor, you can put in the space curvature tensor. Now in three dimensions, the space curvature tensor can be expressed by the Ricci tensor of the space metric. So I'm not going to do the calculation, but if you continue, if you do this, so insert this in there and use the fact that Riemann is rich in three dimension. You get another cute formula, which is minus 1 h over pi integral s infinity, the trace free part of the space reach, x i ds xj ds. So I think the coolest formula I know for the ADM mass, and that's the one I think everybody should use. So rather than writing this terrible formula in terms of partial derivatives, they have a nice formula in terms of the Ricci tensor of the space metric with a non-geometric object, which is this vector. Well, you can make it kind of geometric by saying that this vector is a conformal killing vector of the background. So the xj corresponds to a dilation. It generates dilations. It's a conformal transformation of the flat metric. So you could rather writing a component, you can write it as a zj vector zs xj dj in asymptotically flat coordinates. This is a dilation vector in the background. So that's good. So these are the formula that you get. And now the question is, in any case, does all this converge? And what about conservation laws now? So conservation will come to Lorentz invariance later. I will discuss maybe, again, at the end, angular momentum and center of mass. But I think it's more important that in my lectures, I tell you something about the Troutman-Bondimas, then keep discussing ADM. So I'll talk about conservation now very quickly. And then go to Troutman-Bondimas. And maybe, at the end, come back to angular momentum center of mass if we have time. So conservation of energy momentum. Well, before we talk about conservation, you should ask, is it well-defined at all? So first question well-defined. By the way, so I'm finally able to answer a question that somebody asked my first lecture. You did. Is this mass? Is this energy? What's the business here? Well, so now here I have a four vector. So this is my energy momentum vector. This is the P mu. And the mass is actually the zero component of it. So mass is not really mass. It's energy. So if you've done elementary special relativity, the zero component of the energy momentum vector is not the mass. The mass is the square. So the mass is actually, so this is a very bad notation. I mean, well, actually the mass should be square root of minus P mu P mu. This is the invariant mass coming with energy momentum vector. So this component is energy. So what people call mass is actually energy. Of course, if you're doing only the Riemannian problem, so positive scalar curvature and you want a number, it seems silly to talk about something else in the mass. But at the end, this M is actually a component of a four vector. And it's not the mass. It's the energy. And the whole thing, if you get to mass, you have to know that this vector is time like. Otherwise, you'll get a complex number. And that's part of the positive energy theorem. So you have to work a little. So this vector is time like. So minus P mu P mu is actually positive. And its square root is what is the mass. Good. So well defined. Well, to do this, we'll do the usual assumption. Well, a similar assumption to what I did before. But not on the space metric, but on the whole spacetime metric. So we will end the time derivatives, right? But now all derivatives, not only the space, but also time derivatives have to go to 0 faster than, right. So then the d sigma of T mu nu is going to be O of r minus alpha minus 1. Well, how do I check convergence of this integral? Well, I'm going to use the Gauss theorem again. Because this u was coming from this h mu. Actually, h mu is probably with my conventions here. It's probably the divergence of. And h mu is here, right? So this is a quadratic expression in the derivatives. And when the field equations are satisfied. So let us think about vacuum. Field equations are satisfied. If you remember, in the Cauchy problem case, we had this condition that this integral should be finite. But this had two parts. One was coming from the scalar constrained equation. It's telling you that r is 16 pi matter density plus k square minus trace k square. Don't ask me where lambda is, lambda is gone. This is lambda equals 0. So k square will have finite integral. Because our alpha will be, of course, louder than 1 half. So this condition, since I'm in vacuum, I don't have to worry about what this does, right. So if I had some matter fields, I would need to assume something about matter fields. But let's just do vacuum, just not to be bothered. So therefore, integral of at infinity of u mu alpha, say ds mu alpha, is probably some factors involved like 2. That is an integral of a sigma of an expression quadratic in the derivatives. And now here, in all derivatives, I don't make any difference between time derivative or space. Everything was kind of completely symmetric on those. So derivatives, so this should be quadratic in derivatives. But derivatives fall off like 1 alpha minus 1. It's exactly the same calculation that I was doing before d3x. And so this is finite if and only if alpha is louder than 1 half. So this 1 half is just showing up, again, exactly in the same manner. These integrals exist. But I get a little more across from this calculation. Namely, if I take a domain like this, sigma is t equals t1. And here sigma is t equals 0, or maybe t equals t1. Then I can use the divergence theorem between this surface and this surface. So I have that integral of u alpha beta ds alpha beta on d sigma 1. And this is d sigma 0 plus the divergence. And therefore, integral over this tau surface of something which is quadratic in dg. Quadratic dg. Well, what does my tau surface do? Well, it has a radius. And so for the integral over this, we'll have a contribution r squared from the radius. And an integral in time. But the integral in time is finite. So I get the quadratic thing in dg gives me r to minus 2 minus 2 alpha. And if I have a plus 2 coming from the radius, so this is r minus 2 alpha. Obviously, it goes to 0 as r goes to infinity. And you think, well, maybe that's so. So I'm wasting a lot of powers here, right? So because alpha was, in fact, this integral here would go to 0. Even if alpha was anything positive, right? But of course, you cannot have alpha smaller than 1 half, because otherwise the integrals wouldn't make sense. So you need this. But you also see that there's a lot of room here. And what can you do with this room? You can boost your surface. So in this calculation, when you pass with the radius to infinity, you're going to get that the Hamiltonian of sigma 1 is equal to the same as the Hamiltonian of x at sigma 0. So that's how you get conservation of energy and momentum by this formula. But because you have a lot of room here, you have a power of r for free. What you can do is you can boost your initial data surface. So let me try to do this in a 3D way. So this is your initial surface. And you boost it in spacetime, the t. And you make a boost in this direction. So if you boost, you're going to get a surface which looks like that. So this is your sigma 1 now. This is your sigma 0. And you can use the divergence theorem on this surface. So this is now your tau. So if you're integrating over tau now, what has changed? Well, we're still going to infinity with the radius. So you're still going to get an r square from the area of the surfaces. But if you boost, you get a linear graph. Therefore, this distance will be growing up like r. So the integral here, it's still the same formula, quadratic in dg. So the quadratic in dg gives you minus 2, minus 2 alpha. The area of the spheres gives you plus 2. But you get an extra plus 1. Because this goes far in the future, so you get a plus 1. But you're still good because you still goes to 0 because my alpha is 1 1 by 1 1 by 1 1 half. So this alpha 1 half magic actually guarantees that this thing which is just a bunch of numbers at this state is actually a 4 vector under boost. It's a Lorentz 4 vector under boost of your initial data surface. So the conclusion is x sub p mu behaves like a boost, behaves like a Lorentz vector or covector under boosts. Good. So this is basic, yes? Wait, wait, wait. Can you derive also being greater than half just by demanding that energy is conserved? Energy would be conserved with alpha positive. You don't need one half, but it wouldn't be well defined. So I have an example which is conserved but not defined. Because we needed alpha out of one half for these integrals to converge, right? So that the integrals might not exist. And we know that this is sharp. I said something before which I'm going to, last time but I'm going to repeat. So p zero is actually the ADM mass, right? So it's the same as ADM. If you just do the calculation, if you assume all these fall off rates, right? Assuming alpha louder than one half. And pi is the something called the ADM momentum. And there is a formula for this which is integral of the mass. Over, yes, one over eight pi integral at the sphere of infinity using this Kij tensor. So K delta ij minus Kij Esj. So this is the ADM momentum. And we know that it exists by just what I said before. Good. So as I said, maybe, just maybe we'll go back, we'll come back to mass and center of mass later. But at this stage, let's go to the bondi form, Troutman bondi mass. So I just showed you that the ADM mass is conserved but then you should be worried because we believe that there is gravitational radiation. So how is this compatible? And the argument is trivial, right? I can go as far as I want in the future, three times the lifetime of the universe if the universe was asymptotically flat. You'll still have the same total mass, right? In this argument, there's no room for problems. So what is the why, how possibly can energy be radiated? It has to do with this condition here. This condition is a condition on the asymptotics of the metric if I go to infinity along space-like slices. Now, if you know anything about wave equations, then there is a standard fact that you have in Minkowski space time box, phi is equal zero, then at large distances, so this is Minkowski, large distances, and large times, the field, suppose you start with compactly supported data. Of course, if you start with stupid data, then everything can happen, but suppose that you start with initial data which are supported in a compact set, so both phi and the initial derivative, then this fields propagate. There's no issue about the behavior in the space directions, just by finite speed of propagation, if you started with compactly supported initial data, then the field will be zero outside this domain of influence of this region here. So, if you go to infinity at constant slices, no matter how late, the field will be zero at some stage, so this kind of behavior at in time-like, in space-like directions will be obvious. It's going to be zero, so certainly faster than any power you want if you put a scalar field here, but if you go to infinity in these directions, this will not be true anymore, and you can actually prove something like that, that this is going to be a function of u where u is t minus r and angles divided by r plus lower order times, and in fact, r minus two, I think, with the setup here if I start with smooth data, right? So, if I go to infinity in space directions, I can assume whatever I want for the decay, but if I go to infinity along null directions, I got a decay which is one over r, which is good, that's the kind of thing I wanted. If you take the angular derivative, so of course d theta of phi or dx a, so let me write x a is theta phi, then d a phi will be, well, this function d p a psi over r plus o of r minus two again, and du phi will be, or dt phi will be dt psi over r, or du psi, right, over plus o of r minus two, and dr phi, well, if you differentiate with respect to r this term, you get one power more, but here r enters also here, so you get minus du psi over r plus o of r minus two, right? So, the derivatives don't want to go faster to zero than the other ones. That's a fact of life, and because Einstein equations have a wave character, that's why I was trying to explain you at some stage, then this is the behavior we expect to have in these metrics, and therefore these naive calculations that derivatives follow faster just don't work anymore. I mean, the formula are still there, they're still correct, but the behavior of all these integrants is much more delicate, so you have to sit down and work several times harder to obtain the asymptotic behavior of these things. So, the first one to attempt this in a systematic way was Andrzej Troutman in 58, who got the right formula for the mass in the graduation regime and who proved the mass loss formula in his formulas. So, one thing that he didn't have, he didn't have a nice geometric framework to describe this, and Bondi was the first one to do this. So, Bondi introduced to study this kind of behavior he said, well, because obviously the, what is important is this null directions, so this going to infinity along null directions. So, let's try to use coordinates which are adapted to null hypersurfaces, okay? So, use coordinates adapted to null hypersurfaces. And his proposal was, well, why don't you write the metric in this form, the space-time metric, so g00 du square. Well, this one, actually he had a name for this, eudr plus gabd xa r square, he's put r square habd xa plus ua du xb plus ua du. Ub du. So, these are called Bondi coordinates. So, let's see, what's special about this metric? There's something missing here. There is plus zero times dr square, right? You don't see any dr squares here. And you don't see any, what's missing? dr du xa probably, right? Plus zero dr du xa. Good, I take any metric locally, I can certainly, well, actually certainly write it like down like that, just changing some coordinates. Here, the point is that this coordinates should not be local, they should have a local collector, so you should be able to go with r, r larger than r, and so you can go with r to infinity as far as you want. Now, because this is a zero here, this means that g over dr dr is zero. So, the vector d over dr, which normally you think of it as being a nice radial vector in Minkowski spacetime normal coordinates, here is an all vector. So, this is actually, this is wrong, right? So, this is not my dr, this is dr, right? So, dr looks like that. Now, r is constant to the surfaces, so you should think of the surfaces as light cones. And an example would be Minkowski metric, where you go to u is equal to t minus r, right? So, just take eta is minus dt square plus the usual story, and I replace t, so I get minus du plus dr plus dr square plus r square d omega square. And so, the dr square cancels out, right? So, and you're left with minus du square plus minus 2 du dr plus r square d omega square. So, this is an example of bond coordinate system, right? So, g zero, zero is minus one. Beta is, let me put a minus here. Beta is zero, this metric HAB is just the wrong metric on the sphere, and ua is zero, right? So, I think, well, if this is a good model in Minkowski spacetime, and I want to go to infinity in a spacetime which was asymptotically flat. So, maybe this should be actually a good model in general, at least for large distances. And so, why don't we try? So, what can we say about the asymptotics now? So, if this is the model, then g zero, zero should go to minus one. So, this is minus one, and I'm going to anticipate here there'll be corrections terms plus little o of r minus two. All right, so this is going to be my g zero, zero. That's not obvious that it's works, and actually, this is something you have to work out. Let's see, you probably don't want to see this formula anymore. I don't think we ever wanted to see this anyway. So, what about HAB? Well, HAB, if this is supposed to be something resembling Minkowski large distances, this should go to the round metric on a sphere, which it just depends upon angles. And, maybe there is a term H one of u x a. So, this is coming from my experience with the wave equation, right? So, this would be some kind of the equivalent of this wave field, and there'll be lower order terms. What else? This function beta should go to zero. So, beta should go to zero. In fact, if you calculate it using ancient equations, let me anticipate the formula, and then I will explain you where it's coming from. Well, let me just write it like that. If you have a function of angles and time over r square plus lower order terms, and u a, well, in the model case, u a is zero. So, it should be going to zero. So, it should be, well, that's the u a one over of u and x a divided by r plus little r of r minus one. Okay, so that's the bond is ansatz. Actually, it's a little more complicated than this, yes. Yes, yeah, in this coordinate system, the vector dr is no, right? If you calculate it in this metric, it's no. Right, yeah, so here it's not, but the dr, good. This is called, Nick Woodhouse called this the fundamental confusion of analysis. When you change coordinates, vectors change directions. So, if you, yeah, if I wanted to be, yeah, what happens is vector d over dr in this case, right? So, how does it work? So, we start with coordinates dr x a, and we go to coordinates u r x a. So, if I wanted to be perfectly clean, I put me, let me put it r bar here, and this bar I'm going to remove, but r is r bar and x a is x bar a, right? And then if you calculate d over dr in the new coordinate system, then it's d over dr bar, right? Then this is the, how does it work? d over du du over dr, no, d over d t over dr bar, plus d over dr dr over dr bar. And because d t over dr bar is non-zero, right? Because t is u plus r, the vector d over dr bar, which as every good physicist, you call r again, right? So, yeah, so this would have been the proper way to do it, but nobody does that in physics, right? So, this vector changed directions all of a sudden because you decided to change coordinates which sounds stupid because you've just changed it. Different coordinate system is going to change again, right? But yeah, that's what it does. Good, so this is this coordinate system. So you should think of you really being a null hyper surface, but it's not only a null hyper surface in Minkowski spacetime, it's really a null hyper surface, right? So it's null with respect to the real metric. And if, yes, if somebody has ever read Troutman's paper about this, this is what is missing in Troutman's paper. Troutman doesn't have an insight to really look at real null hyper surfaces. He looks at asymptotically null hyper surfaces, which works as well, but it doesn't give us such a clean picture. What is the time-like or space-like character of the u coordinate? The level set of u are null hyper surfaces. And the way to see that these are null hyper surfaces is that this famous vector, which is d over dr, not d over dr bar, is perpendicular, right? So if you calculate the scalar product of du, well, dxa, da and dr, then this is gra and this is zero, right? So dr is proportional to itself because it has zero lengths. It's proportional to the xas. So if you add a surface xa equal constant, it's proportional, it's orthogonal to all directions on this surface. So it's a null hyper, that's a definition of a null hyper surface. Good. So if I wanted to explain this to you, I would need to open a new section called the characteristic Cauchy problem, which I will not do, but let me just sketch very, very shortly and informally. I hope it's not going to collapse. How does this work? So I've already, somebody asked before, how does it work, the characteristic Cauchy problem? And my answer was you can specify freely the tensor induced by the spacetime metric on your hyper surface, okay? So here the hyper surface is u equals zero. It's a characteristic surface. It's a null surface. And the characteristic Cauchy problem is you specify in any way you want this tensor induced on the surface. Well, what is the tensor induced on the surface? You just say take this and set u equal constant, right? So u equals zero or u equal constant. So this is gone. This is gone. This is gone. This is gone. So the induced tensor is this. So these are actually the characteristic initial data here. And you don't have to copy this or note this, it's just a very loose. I just want to give you an idea of this works. Good. And then now we have the characteristic initial data but you know that Einstein equations are wave equations for all components of the metric and that's what you want to solve. So you need to out of these things calculate all components. All components being this function beta, this function g zero zero, this vector ua. And this you do using part of the, it's very similar to the constraint equations on space like hypersurfaces, except much easier because the constraint equations on null hypersurfaces are equations which are terrible. Here, if you write them out, these are ODEs along these geodesic directions. So you have something which is called constraint equations here but you can solve them by integrating along geodesics. So these are second order ODEs, whether PDEs but somehow you solve them by integrating along this direction. Therefore, once you've given this and for example like that, you're free to do this because this is completely free. So once you've given this, you solve the characteristic constraint equations and you're going to get what I wrote here. So you get this where this is a free integration function if you want, or this is determined by some global properties. So M is a global function, whatever this means, a called a mass aspect. So you get this by integrating the equations. So this form of beta you get it by integrating the equations and this form of U, you get it by integrating this equation. So once you have decided how this looks like everything else is determined. Okay, so this is a fact of life that you can admit. Let me tell you another fact of life. So Bondi was suggesting that you can use these equations to solve the characteristic problem. And no, so these equations, if you write down nice and equations for space time metric which has this form everywhere, you'll get a system of equations which doesn't have any good PDE properties. So this is similar. So there's a PDE problem, which is similar to something I already mentioned. If you try to solve the space time metric in a coordinate system which looks like that. This is not a characteristic question problem, but the normal one. Then no, you can do it for analytic data maybe, but that's it. You cannot, there's no well-posed system known to humanity at this stage of evolution of humanity, which would give you directly that this is a well-posed problem for smooth initial data, right? But you know on the other hand that you can take this as initial data, solve in harmonic coordinates. Once you have your solution harmonic coordinates, you can transform through these coordinates, right? So this is exactly the same here. You can start with initial data in Bondi form, solve in harmonic coordinates, and then transform to a form where this is everywhere true, okay? But if you want to do this on a computer, good luck. I mean, you're going to waste time and you're going to have a code which crashes if you want just to solve this. That's experience of nowadays humanity. That's what happens if you try to do Cauchy to solve these equations in this coordinate system. Good, but what does this have to do with mass and so forth? Well, so you take the metric here, you put everything here in this terrible formula and you go to infinity with R, but not in this direction, because we know already what happens in this direction, but you go to infinity along the line. So now you take integrals over fixed R and you pass to the limit with infinity and that's what you're going to get is first something called the Bondi mass, so Bondi mass, which is integral, which I'm going to go trot on Bondi mass. It is one over, oh gosh, it's probably one over four pi. Integral of this mass aspect function, theta phi over the angles, so x a, d two x a, over a sphere, so the sphere is in infinity because I've passed to the limit and I'm only keeping this factor here. You'll get the trout on Bondi, so Tb is for trout on Bondi. You get a trout on Bondi momentum, which is a similar integral. It's really funny, it's just not clear how to get this, but that's what you get. Well, this here is just the position vector, so you think of this sphere as two, as being what you think it is in R three, right? So vectors of lengths one in R three. Then this is, you integrate this mass aspect function against the function x or the function y or the function z restricted to the units here, right? So in other words, cos theta or sin theta cos phi or sin theta, sin phi. If you want to think in terms of intrinsic quantities, you integrate against the first non-trivial eigenfunction of the Laplacian on the sphere, right? Because cos theta is actually the first, you calculate Laplace of cos theta on the sphere going to get some coefficient on cos theta and so forth. So if you don't want to think about embedding, there are three functions which span the L equal one harmonics on the sphere and integrate against this. And so the, good, that's a definition. And the reason why people would have been keen to give a Nobel Prize to Bondi when they were giving a Nobel Prize to Weiss and other people if Bondi still lived, by the way, a colleague from Vienna, about a time ago. Then this is this Maslow's formula, which says that energy can only be radiated away and the formula is that dm over du, right? So m depends upon this retarded time. Here there's a coefficient which I certainly don't remember. So it's minus one over 32p and integral over s2 of this coefficient dh1 u xa. Right, so there should be indices, by the way. Right, so this is a tensor in a sphere. You take its norm. So you take this coefficient of the expansion of the metric, take its time derivative. You get a tensor of the sphere, take its square, integrate it, and this is obviously smaller than zero because of this minus sign here. So this is the Bondi mass formula. Where does it come from? Well, it just comes from taking, doing exactly what I did previously, right? You just take two surfaces which go to infinity and you make an energy balance, energy at sigma two minus energy at sigma zero is equal to this flux and this flux is this with a minus sign. So this formula was first discovered not in this notation but a slightly different notation by Troutman in 58. There's an interesting story that Troutman went to London, he gave a series of lectures on that and Bondi wrote his paper to his later, which is fine, of course, and he had this new way of doing this. It would have been fair to quote Troutman who gave lectures that he attended to, but. Good, so this is the Troutman-Bondi story. How am I doing with time? Five minutes, okay, good. So what can I tell you in five minutes? Well, I can tell you that it's been fun to lecture here. I hope you've learned new things. I certainly enjoyed lecturing to you and interacting with many of you and the weather and the swimming. And so have a nice conference. I regret I can't stay next week. I have to go back to Vienna, but it was fun being here. Thank you for listening.