 So I can see that he was sleeping during one of my lectures because he thinks that the second one. So let me remind you where we ended yesterday. So there's something called SCE, Scholar Constraint Equation, which says that the reti-scaler of a Riemannian metric, we're looking at initial data. So we have a surface sigma which has an induced metric and an extrinsic curvature tensor. So the reti-scaler of this metric is equal 16 pi rho, rho being defined through the energy momentum tensor in this way. So this is the geometric way of saying this is the energy density coming from matter fields on this surface. So the N mu is the normal vector to this, to the surface. If you have a space time, well if you don't then this is the starting point to get one. SCE is this famous extrinsic curvature tensor and one way of writing it is as follows. It's a symmetric, obviously, tensor. It uses this normal vector N mu and it's just some kind of space derivatives of this vector. So this is the norm square of this vector. This is the trace and perhaps a cosmological constant if you have one. So this is the scalar constraint equation and VCE being the vector constraint equation. Divergence of a specific tensor here is this one. So I got confused by Rick's lecture today because he wrote a plus here. You can get away with a plus or minus by changing the orientation of this N vector so he didn't write anything wrong. He didn't just tell you what is the orientation he's using. But this is a very painful sign if you're writing a paper and you want to have your conventions correct and then you just end up with something which is minus what you expected to be. So to the best of my knowledge, this is the right sign if N is future directed. Of course if you change the sign of the normal then sign changes here. Of course in what Rick was saying the dominant energy condition is that the length of this vector is smaller than this number then the sign doesn't matter and that's the vector constraint equation for me anyway. We kind of derived this formula or at least I showed you how you get it from the scalar curvature that this is essentially you collect the second derivatives of the metric which are given this expression, write them as a divergence and you call this the mass period. So it looks completely arbitrary but the point was that if you assume that the metric has a asymptotic so that it goes to a flat one at rate so odd to some power and it's derivative go a little bit faster well a little bit mean one more and if the integral of the scalar curvature is finite and this rate is alpha one half then this is well defined. And I also showed you that you allow one half here, hell breaks loose. And so this is a metric all of you know this is the Schwarzschild metric I didn't dare writing it last time I didn't want to offend you but now I'm writing it to make sure that we agree that this number here is actually what you're going to get if you take this metric which plug it in here and obviously so we're looking only at the space part of the metric so it's sitting here and if I look at this part of the metric so I can make a Taylor expansion of this as one well if you expand one minus x you get one plus x so one plus two m over r plus lower order terms so if you have a metric written in this form then you immediately know what the mass is by looking at this one so one has to be a little careful there's a change of sign right if you have it in an emulator you put it up you have a change of sign so let's start with a simple exercise in the same spirit as the ones we've been doing before playing with this exponent alpha one half so we go to Minkowski and we introduce a new time tau which is t plus a square root of r where a is a constant so again this exponent one half is showing up now it's positive so let's rewrite the Minkowski metric so we go from t r at r theta Phi to coordinates tau r theta Phi so the Minkowski metric I like to write it as eta so minus dt square plus dr square plus r square d omega square now we have to substitute for t this expression so this is minus d tau plus well let me just do it slowly it's a beginning of the lecture I have to warm up as well so t minus a square root of r the whole square plus whatever follows good so let's do this calculation so this is minus dt very virtue of square root yes something wrong good thanks what did I do wrong again now this is not torture this is come on this is a two I mean you can come here I agree that's not the greatest to I've ever written that but look this is the same to us here right good thanks good so so that's a detail okay now you differentiate square root I'm going to get well a minus a over 2 1 over square root dr square plus dr square plus r square d omega square good so minus d tau square that's easy this double product with a minus becomes a plus and the two vanishes plus a d tau dr over square root of r and then this guy square root is squared becomes one over r so when I collect the dr terms I get one from here and the square of this gives me a square over 4 is a minus over all minus a square over 4 r dr square plus r square d omega square so this is my Minkowski metric in this coordinates I mean nobody tells me to use this coordinates is a stupid idea but nobody would prevent me to use them if I want to right so now you look at this metric and you say well it's a bit of a funny metric but it's still if I go to infinity this goes to zero right so this goes to zero this goes to zero so in this goes to a Minkowski metric like large distances there's no question about it and now if I look at the surface tau equals zero the surface tau equals zero means I drop this term means I drop this term so this is my induced metric on the surface tau equals zero right well what is the mass this is my formula here right if you have a metric which is one plus something over r plus lower order then the something over r is actually 12 to mass right so the mass is minus a square of eight or something like that so does it make any sense I mean so so this is really a surface in Minkowski space-time right Minkowski space-time should have zero mass doesn't but I've been explaining to you that a metric like that has a well-defined mass right there's no issue about this this has a well-defined mass as long as you stay within the right system of coordinates that's what you're going to get right so something unphysical is going on right so mathematically that's what I did but physically you'd say well there's something physically wrong yeah because I should get if I take a reasonable slice in Minkowski space-time I should get zero mass because zero mass should be a property of a space-time what you would hope at least and so what goes wrong well certainly the scalar curvature of this you can check goes to zero very fast because it has the same asymptotics as the Schwarzschild so so so this condition wait okay maybe this condition will not be satisfied right so if we look at the scalar constraint certainly lambda is zero certainly row is zero in Minkowski right so team union is zero something must go wrong and so so this condition maybe doesn't is not satisfied why is it not satisfied because I have produced a kaj here I have bent this surface T equals zero to another surface in Minkowski space-time you can just check your plotting skills to draw the graph of this function right so you've deformed the surface and in this introduces explicit curvature and this explicit curvature will be started I will not be in a one the integral of I will be not finite and so let's check this right so let's check this I have a formula for kaj here so I only need the radial component so this is one half well there'll be di dr and r twice and there'll be Christophels times times the n but Christophels go to zero so maybe they're not that important so let's look at this term here well the Christophels of this metric actually are easy to work out right so there'll be of order one over r square just from this right so because the derivative of this metric so whatever n does this is order of one over r square and one over r square is very good just you can integrate its square and because now this condition that integral of r is smaller than infinity because this formula means that the k square integral of k square should be finite so integral of k square should be finite so let's see what is the normal vector here well if you know a little you have some experience you can just read this from this term but if you don't let's do this calculation well so if a equals zero then the normal would be just one zero zero right so this we're Minkowski equal zero so this is the d over dt metric so for large distances everything goes to Minkowski so n should be like dt and there'll be some lower order terms well just I know what the result of the calculation is so I just let me just add this to you and plus maybe things which decay right so this doesn't have to be exactly one but it's going to be leading order d over dt plus a radial component how do I determine the normal vector where I'm requiring that g what the spacetime metric actually g so this would be in this case the Minkowski metric so eta of d over dr and of the normal vector should be zero the scalar product of the normal vector to a tangent vector should be zero right so let's calculate this with this metric so well because I have only the sketch there so this is eta r mu and mu so if I use this guess that this is going how it's going to look like then there's going to be eta r t and t what else and plus right plus eta r r and r well this is actually exact if I know that there are no more terms but this one is about one eta r t it's minus one eta r r no what nonsense eta r t well actually I should go in torque holding it so sorry about this okay this should be d tau this this is correct this is good so eta r tau this is here right and normally this comes with a factor two so this is a over two square root of r well eta r this is about one that's a leading order it's one so this is about one so and this should be zero so I can read and are from this right if I approximate and tau by one and lower order terms I'm going to get n r is minus this right so minus a two square root of r so this is the index up plus lower order terms index down will be obviously the same and I want to calculate k r r so this is if I just keep the asymptotics one half of d r of what I just calculated here minus a over two square root of r plus lower order terms okay should be able to differentiate this right the two a two it's one over eight maybe the minus would go away a over all three halves something like that yeah no I forgot okay this one is okay good so now it's okay good yes no but to leading order it is yeah that's a good point right if I'm lowering indices I have to I normally if I did the calculation correctly I would have to take into account this but this goes to zero so the leading order behavior is correct so the question was this formula is approximate because I had the index up I have to lower but lowering with the spacetime metric so there might be some other stuff coming here but good so we see that while the norm of k square is certainly going to be louder than this so whatever a constant times r to power 3 and integral of k square is going to be louder than integral of r 3 of or say r louder than r because we're only working in large distances of constant times r 3 in three dimensions this is infinite right because you have an r square from the measure when you go to spherical coordinates so you're integrating one over r from infinity to get logarithmic divergence okay so the lesson of this exercise we is that when you're defining asymptotic flatness and you are in a spacetime set up not only you have to be careful about what the metric does but also what the extrinsic curvature does and this factor one half is showing up all the time so if I didn't put a one half in my silly coordinate system but just some general alpha I will get that everything is swell if the exponent here is smaller than one half if it's larger we get infinite mass actually minus infinite because of this sign which is weird okay and okay so this is the lesson is the decay of kaj matters and to understand properly how this works we really need to go to the spacetime picture so I would I've been cheating and doing initial data but you really sorry need to go to the spacetime yes I don't well no actually I know because if there was a consolation integral of r would have been finite then I would get a mass which no so whatever consolation there are I mean what could happen right okay so if I was suppose the thing was positive and still cancelled so that the integral is finite then you could use rick's theorem that the mass must be finite okay so good so therefore whatever consolation there are they will not help you here so so what I would need to do and to make it completely I would need to check that the other components of kaj follow faster which is the case so this is the dominant term and then if you look at this good well in any case I get a negative mass so if it doesn't worry you I doesn't worry me and I will show you a theorem which says that if kaj we will prove a theorem if kaj falls off faster namely like minus alpha minus one with alpha larger than one half then this problem will never occur the mass will be well defined as a spacetime notion not only as a surface but in a space yes say it again so this one this one the vector dr is tangential to my surface the surface is t equal constant at all equal constant right so this the new surface is not t equal constant because t equal constant would be just the previous one the new surface is tau equal constant right and then dr is tangent to the surface and so that's the condition yes it's just a simple coordinate information in the t coordinate why so sensible the result I don't know this example was you know just meant to make you perplexed and to think that this adm mass doesn't make any sense right good and so the point is that the asymptotic is delicate the asymptotic should be faster than one half and then you'll be fine and that we're going to prove this right away any other questions yes the condition that you have with the law and in being normal so that if I write it says that in lower are is actually zero so how are you getting so let me do this again right so this would be Minkowski but now well you're still in Minkowski but the surface is if a equals zero d over d tau would be d over dt so this would be the usual normal vector in Minkowski to the slice t equals zero but because we're not then we're looking at the slice tau equals zero then there are correction terms so there'll be a n tau term here which you should believe me goes to one I don't want to do the calculation because it will take me another five minutes maybe I should have done the calculation and it would take taken me less time than questions are as a one form not as a vector here's a vector right so here I have a vector and the formula therefore if I set this to zero I get this with an index up and I have to lower it but I lower it with this metric and this metric has a one as a leading order term and a correction and that's why I wrote more or less because there'll be lower order terms which I ignore I just want to know the leading order asymptotics now it's lower order then you'll get lower order terms maybe we do this calculation together late any other questions good so you're all suitably perplexed and let me show you worse I normally I wouldn't do this because it takes a little time and we want to go on but somebody asked me can you make a time dependent ADM mass because obviously mass is conserved in physics so if it's useful it should be conserved so if I take different slices I should get the same number well I can write in minkowski a slicing in minkowski where each slice has a well-defined mass but changes in time all right so this is the next example you just take t is equal to one plus a square root of course what else let me just check that I have it right so that I don't have to repeat the calculations but I think that's the right formula yes okay let's let's calculate so still the metric is eta so the minkowski metric and well the only thing which matters is the TR sector so so it's minus dT square plus dR square and just to save chalk I'm not going to write what happens here but you welcome to do angular stuff good so now I have to do my calculation again minus so t is what actually directly here plus a tau root r plus dR square plus the angular stuff good so let's you know maybe let me just write it like that so there's a d tau term which is multiplied by 1 plus a square root of r d tau and let me just expand immediately right so I get the d tau square is just by applying d here and squaring so I get some of kind of funny asymptotics here good and then there will be a term which is a cross term from let me do it carefully because otherwise you'll get it wrong okay so d tau times 1 plus a root r plus tau times the differential of this so a over 2 square root rho rd on everyone agrees and the square so d tau times this plus tau times the differential of this which is square root gives me a one-half so thank you plus dR square good and now I'm going to look only at Toyqualk constant so let me just ignore the cross terms and the these terms so there'll be something d tau square which I don't care plus something d tau dR which I don't care plus dR square is coming with a one from here and a square from this right so 1 plus tau square a square over 4r 0 square and now let me just to make things clear this is the angular part okay so if you look at Toyqualk constant this is gone this is gone this is a perfectly fine asymptotically flat metric what is the mass well we can just read it it's twice I mean this coefficient is twice the mass right so the mass is well positive this time I think if I have my signs correct do I have my signs correct I have a minus yeah so actually it's become negative now so the mass is minus tau square a square over 8 so tau is our new time so we have a perfectly defined asymptotically flat metric right just looks like Schwarzschild up to lower order terms at every tall line but the mass changes in time good so now you should be as I said perplexed square and now let's therefore think that this idea of defining mass using only the Rittice scalar was stupid we need to do something better so let's go to the spacetime picture and let's see so what do I raise don't think I need this doesn't hurt but we're starting chapter two of this course which is the spacetime picture good so I'm going to erase so let's make a tabula rasa here while I'm raising can stretch relax go for a cigarette and try to summarize what we've done so far right so we introduced the notion of mass based on Koshi data made perfect sense but starts being a little dangerous if you start thinking about spacetime and since the general activity is not only about initial data sets but also about spacetimes we need really to go back to square one and figure out how to define something which has a feeling of the mass in general activity so let's go even one step back and let's ask well what about field theory in general so we'll talk about something know that calls that Bob world calls another charges and I think he will explain you tomorrow that whatever I told you was wrong and he has a better way of understanding that which is certainly fine with me but so the question is right so is there some conserved quantity associated with spacetime flows so this would be the argument would be as follows there's something called energy in mechanics and if you've done another theorem seen classical mechanics you know the conservation of energy is either a calculation or a complicated argument saying that there is invariance under translation in time so that you from this you get the result right so in other words the energy in mechanics has something to do with invariance under changes and a motion in time and well here we are in a field theory on a manifold we don't really know what means what it means to move in time so the best thing we can do is just we give yourselves a vector field on the manifold and we start moving along this vector field okay so we take a x a vector field on a manifold so now this on spacetime right so on spacetime forget all the nonsense with initial data and so the equivalent of moving in time in classical mechanics will be flowing along this vector field so certainly we have introduced something like that and I suppose we have a Lagrangian field theory where we have a Lagrangian which depends and for simplicity I'm going to get to talk about Lagrangians which depend upon only upon field and its first derivatives and you're going to say okay good go ahead but we are clever we already know that general activity you can get it from the Hilbert Lagrangian depends upon two derivatives so don't bore us with this well you can repeat what I'm saying with smaller derivatives but it's more complicated and I just make a quick and dirty trick to get Einstein theory in this form so Lagrangian field theory and I'm going to require that this is by the way this should be a density that's not only function but the density so density so you should think of this as square root of that G over R if you could over 16 pi maybe so the density part is here and in fact you can write Einstein theory like that if you use the the metric and the connection as independent fields so that would be one way of doing this right so so if you look at this as a Lagrangian of which depends upon the metric well in this case no derivatives but who cares in principle and gamma and D gamma then sets of this form right so I'm going to ask for geometric invariance the Lagrangian takes the same form no matter what coordinates you use right so this is an example right so this one no matter what coordinate system you use it's going to be the same algebraic expression of G DG in this case no DG of gamma and D gamma right you write it in local coordinates that's an explicit form no matter what coordinates use an example which doesn't work would be a so so example is there's the equation one today right and non-example would be so the Maxwell Lagrangian for example which would be probably one over four pi or something like that I never know the coefficients but so F no G alpha beta you knew F alpha mu F beta new well this is the Maxwell field so you're saying well of course this looks the same in every coordinate system well if you think of this as being a function of G and F yes if you think of this as being a function of F only no because in Minkowski coordinates suppose you're in Minkowski space time in Kowski's coordinates you had ones and minus ones and think like that here but if you oops but if you go to spherical coordinates you're going to get r square is all over the place right all from the inverse matrix and sine square theta's and minus one over sine square theta right so this is not geometric if you look at this as a function of a and da well obviously F alpha beta is the alpha beta minus d beta alpha it's going to become geometric if you couple it twice in an equation and then you have a gradually depends but the metric is derivative and so forth right so this is the geometric invariance and there is a elementary fact that if you say define something which Bob calls the nether current theta mu is defined and depends upon a vector field X is okay so D L over D phi alpha over mu the lead derivative of the fields phi a so I didn't tell you what the indices are but they can be anything right so if you just do the Palatini so the fee I would be G mu news and the Christophels right so this is a collective index which describes all indices if you're throwing the F then the fee phi a will also have the ace good and so this is minus the Lagrangian times x mu then the divergence is zero so this is like my claim that if you calculate this then the divergence of this nether current so a lot of you know here that you have a vector field zero divergence it's useful because it gives you you can use stocks theorem and obtain some kind of conservation though so conservation law is actually a very dangerous and overused term in this context because what you're going to get I'm going to do the calculation it's for sorry for Bob boring those of you who already done this calculation hundred times then you get some kind of flux you get a balance law between energy at one time and compared with another one and some fluxes this is the right thing and that doesn't have to be a conservation involved because things can leak so talking about thinking immediately that this implies conservation of anything is actually a stupid thinking don't do this for example this will give us the law of energy loss by gravitation radiation which is obviously not a conservation law and saying that the mass is not conserved it's actually decreasing so so saying that this is a conservation law is a bit well it's not a great idea any case we have this formula and the proof is just two lines I can't resist the pleasure of doing this so they just calculate this right now if you're calculated in a brute force way it's actually not two lines then this becomes annoying especially since I didn't tell you what this lead derivative does but well the idea is first to use a coordinate system where the vector field x is as simple as possible namely say the pointing along one direction with this coefficient one so so you'd ask me why did I use one and not zero well it just to confuse you but also not to think I don't want you to think that this is necessarily energy because if I put d over the x zero then it would be time and I will prove you something about energy but this will not apply to anything else well this applies also when this is a translation so one would be the one core first coordinate along but this could be also rotation right so we have a rotation you call this coordinate one and the argument I'm doing here applies good so well can I always do this I let you think about this answer is almost always and you can try to figure out why almost always and what's the almost here but almost is good enough because if you can do it almost always by continuity you can do it always so it's true it's true okay so anyone can tell me when can you not do this if the vector field is a zero right so a vector have a zero no matter how hard you try you're not going to find a decent coordinate which does this right so good so you prove this for any vector fields which have no zeros and then by continuity if this is an identity you can just approximate a vector field which has zero by vectors which don't no no the index theorems right now but what you can do you can do something very silly if you have a vector field which has a zero you can write it as sum of two vector fields neither of them having zeros this is this you can always do locally right but if you can do it locally this expression is linear in my vector field so if it's true for any of them good anyway so we take a vector field which look like that calculates this divergence right so d mu of dL over d phi a over mu okay so if x is d over d1 in any recent decent field theory the derivative is just the partial derivative so let's accept this and then we have so d phi a over dx1 here minus the Lagrangian times delta mu 1 good so this something that mr. Leibniz invented I think this created a first Brexit at least scientific one where the Brits stopped collaborating scientifically with Europe because Newton said I invented this so we don't talk to any Germans anymore and this was 400 years or so so so in any case there's a slight me through which says that we have a term like this time and this and then we have this d mu acting on mu dx1 let me continue so this is the first term and the second one is minus well derivative of the Lagrangian in the direction one and that's it right so well let me just work on this term and that's where my geometric invariant condition happens L has the same form no matter what coordinates you use right so if I had a stupid theory I changed coordinates the Lagrangian will start depending upon the coordinates and the hypothesis here tells me that in this coordinates the Lagrangian still depends only upon the fields and their derivatives okay so that's what happens here so so let me call this star and star is then well I have to differentiate L with respect phi a d phi a over dx1 plus derivative of L over d phi a of mu d2 phi a over dx1 dx mu so there's something important I should have written if the field equations are satisfied okay so because I have a variational theory so this is equal that if the field equations are satisfied right and there's a minus here so this cancels with this one and this is obviously the same as this one so that's as easy as that nothing as I said if you worked in any coordinates throw down this lead derivative in a terrible way with derivatives of x everything you'll get there but would be a little more work good so what's the bottom line of this so before I draw the bottom line for some of you who are not that familiar with integration on space time and stuff like that so let me recall you the stock serum the way I taught it that I was told they used to teach it in Warsaw some many years ago which says that if you have a region omega then integral over omega mu teta mu is equal integral boundary of teta mu times some funny forms and these forms are well actually I should write here d say d okay whatever the dimension is right so dx0 times dxn if you don't like integrating forms and I'm sure that you don't see what I wrote here which is good then you think of this is just whatever you used to integrate so this is this and I have to tell you what this yes muse are so well I it's kind of the thing that you learn in calculus in high dimensions but it becomes confusing when you have a silly signature like Minkowski space time because this is you can think of this as teta mu times the normal within this is down okay and some measure d mu of d omega whatever this is so this is what it is this is the normal but with the indices at the right place so you have to be careful with a normal in a conomal because it's down is in differential geometry if you don't want to think you just say that DS mu is the contraction with of the coordinate form with the vector DS mu in other words yes zero would be dx1 times dxn dx1 because this is an anti-symmetric operation then you're going to get a minus sign here dx0 times dx2 this contraction eats up the one index and so forth okay so this is a formal formula no not not here not here and my thank you for pointing this out because so what how many saying that well if you're integrating on a manifold this looks like a very stupid formula because I don't have globally defined objects so how is it possible that this makes sense and the reason why this makes sense is that I assumed that my Lagrangian is not a function but a density so this is a density and let me just show you the formula if I was working with functions right so if I was working with functions so if teta mu is square root of g times a vector field then how does this formula look like integral over d omega y mu and maybe an absolute value if I'm in space time at g y mu ds mu and then this is the the forms you were thinking about so maybe you want to call this your surface form if you want to but I use this once then this is the divergence of teta right so d mu of square root at g y mu and I can divide it by square root of the g and put in a square root of the g here and dx0 wedge dxn and now this formula makes perfect sense if you remember that this is the same as the covariant derivative of the vector field y mu okay so sorry for those of you who know this very well but if you don't then this partial derivatives formula when teta is a density is the same as this geometric formula where now these are things which you can integrate globally because they have good transformation properties good so now that I convinced you that this is a geometric because that's actually the main aim of what I'm doing is let's see what we get out of this well so start thinking about a region which looks like that they were surface sigma one a surface sigma two and in fact let me just do it in three dimensions right away sigma one sigma two as this is the interior is the omega and this boundary is something I call tau and you can think of this as being time and this is space but you don't have to this could be angles my vector can be just rotation or a boost or or something else so just think of a region like that and in fact this region has nothing to do with the vector x anyway so a priori you can choose it adapted but you don't have to and so there's divergence theorem is telling you that if you integrate well integral over omega of the divergence of mu well this is obviously zero then this is integral of the d omega of t time u ds mu so if I call the integral so let me just write it into four pieces integral of a sigma two t time u ds mu plus integral over sigma one ds mu plus integral over tau ds mu so now let's see just for fun let's call this one h of x and sigma two so this is by definition h Hamiltonian obviously or energy or something like that so this global thing this integral and depends upon the surface and depends upon the vector field x right so I didn't write the vector field explicitly but theta depend upon x right so this is 8 now here there's a question of orientation because if I really think about a space-time domain and that's for the moment you do this then this the normal is pointing this way and here when you do this talk serum you always have to say in which direction you have the orientation so then this one will be if I decide that this is plus and this one will come with a minus so this one is h of x sigma one with a minus because of the orientation and this one is some kind of flux so this is we call this phi of x and tau I could call it still the Hamiltonian because it's the same but it has more flavor of a flux than energy so good so this is what what the divergence theorem tells me that I have if I think about moving my initial data surface in time then I have some kind of energy balance whatever energy I have here is equal the one I've put in minus whatever I've lost with this flux here right so energy of x sigma 2 is energy of x sigma 1 minus the flux so is it they have it right yeah I think so minus the flux so this was an exercise which you must have done in your elementary field theory minkowski space time you do this for a scalar field in minkowski space time and get conservation of the standard energy and conservation you'll obtain conservation when you go with this surface to infinity and your fields go to zero right because this theta which has vanished somewhere now is it yeah I just erased it I'm going to write it again so theta mu is dl over d phi a mu leader evative of phi a minus x mu l right this depends upon your fields so if you take Maxwell fields at large distances like granted will be zero this will be zero there'll be no flux and you'll get conservation of energy and indeed some cases you get conservation loss out of that not always you get you're going to get the trout man bondy flux formula if we do it in general activity later in this lecture or maybe tomorrow so good so where what do we stand now so this is quite general and let's do an application to general activity so this would be paragraph 2.2 the general activity with a background metric okay so we already saw that well you can do general activity variational principle Hilbert Lagrangian second order doesn't apply here we could do it general activity Palatini Lagrangian first order we could do this here but it's kind of more complicated so there's one way of doing general activity first order as a field the fields being just the metric and so the question is what is a Lagrangian which depends upon g and dg and the answer is simple well if you have if you were working in a manifold with a single coordinate part then you just say well the calculation we did yesterday or outlined yesterday the reticular times that g is the divergence of something plus a q which was quadratic in dg so if you want to do weak field general activity where the background is Minkowski you know that variational principles don't care about divergences so this term if the field equations are satisfied from this Lagrangian they will also be satisfied from this Lagrangian so but this is kind of geometrically ugly and a little better way of doing this it says well let's use a second metric on our manifold to make this divergence covariant so idea is to have two metrics two metrics g and g bar and we can write because this is a density as a divergence again a normal divergence this is the same business as before I've been working with densities you can a normal divergence is the same as a covariant one in some in a precise sense made precise plus something which and let's divide by 16 pi plus whatever remains which now depends upon g dg this background metric g bar dg bar and is quadratic in in the derivatives right in dg and dg bar or another way of saying this is just each time you have a partial derivative of the metric when you write down this expression you replace it by a covariant one to the other metric plus a tensor which is the difference and you do this calculation okay so I'm telling you I'm going to this calculation this is essentially the same calculation I outlined the previous time but then you end up with a Lagrangian which only depends upon the metric in its first derivative good so now we have that it looks swell but of course I spent a lot of time explaining you that if you have some exterior some if you take the Maxwell fields in Minkowski space time that's not an invariant Lagrangian we have you have an extra structure the Minkowski metric so here this is not an invariant Lagrangian because you have an extra structure the g bar metric right so change coordinates doesn't quite work and the main fact here is that if x is a killing vector of the background so lx of g bar is 0 then and the equations are satisfied right and the field equations are satisfied so field equations in vacuum or maybe we have some more fields than you get that this is still true in other words you can still use this but not for any vector field anymore you can do it for vector fields which are killing vector fields of your background but for us if you are in asymptotically flat spacetime that's good enough I have a metric so what is the background if I'm looking at asymptotically flat metrics I'm not going to take as a background the anti-deseter metric I mean you can try but that will not be very constructive you should just take the asymptotic Minkowski metric right so you have a flat metric you have one metric which is yours the real one and you have one which you get asymptotically so we have a g which goes to eta the Minkowski metric as r goes to infinity and you take this one as your background so this metric has as many killing vectors that you want and for every one of these you get this divergence free another current and you get this energy balance low right so now the question is is this actually a conservation low is this only a balance low or something like that is we're going to try to analyze now good so now there are two things which happen here first the divergence is zero and h of x sigma is a boundary integral so this is of some expression so let me write this geometrically first so that your that this is manifestly geometrically invariant what are these forms ds alpha beta well ds alpha beta is just d alpha contracted with d beta contracted with the volume form and and this boundary sigma is understood in an obvious way if you're looking at a bounded region and it's understood in a limiting process if you're looking at a general region so I think it's probably a good point to stop here I'm going to write down the formula for what you are for beta is now or next time next time okay so I write you the formula for you alpha beta but the bottom line of this is that these nother charges become integrals at infinity and this is a miracle I mean there's no reason that this should happen and Bob has probably a nice way of explaining why this miracle takes place but this is some kind of miracle and so these things are going to reproduce if you tweak x to be vector d over dt you get a dm mass out of this so and this is what I told you I had this Mickey Mouse way of writing what the adm mass is here's the long way right but it took me a whole lecture to get that so if you take x equal a translation you get something called the adm momentum I'm going to write you what it is if you take x to be a rotation so something like xidj minus xj di you're going to get angular momentum and there is one last interesting vector field the boost sine looks weird but it's correct this is a vector field generating boosts and there's a stupid name for this charge it's called center of mass so this is giving you the center of mass but then of course you have all the questions that we asked about the adm mass does it convert so here we already know we don't have to repeat this the the momentum it's going to convert if kij is well behaved exactly in the way we are indicated before so the mystery that kij has to be to decay is hidden here if you want to have a well-defined momentum you need extrinsic overture tensor which is what behave and these guys will require more work so so this will talk about next time and let me also mention that if you go with this boundary to infinity along a nice a simple surface you're going to get this formula will give you conservation of mass conservation of momentum conservation of angular momentum conservation of center of mass if you go with to the name at infinity along no direction this formula will tell you that there is something called the troutman bondy mass this is the mass for the gravitational wheel field in the radiation zone and rather than having a conservation you'll get a flat formula the mass bondy troutman mass is always decreasing this flux in integral would have a sign and this is going to give you the CRM that energy can only be radiated away with gravitational waves thank you and let's continue tomorrow