 And then a few classes later we'll start the discussion on that. Okay, let's complete the discussion that we were having near the last class. I'll correct some algebra. So the discussion that we were having was a problem. We were trying to solve Einstein's equations in a systematic expansion in weakness of the forces. Or we put the gene velocities in the fields. These two things are good because it causes strong accelerations in time. Okay? And you'll remember that Newtonians, you'll see that's five of Newtonian's potentials and all of these. Okay? And really, not really well done, we had the equation that ds squared is equal to... Now, ds squared is equal to... Now, 1 minus 2 pi pi out, which is 1 plus 2 pi. ds squared. Minus 1 minus 2 pi dx, i dx. We could have think by solving the equations we wrote down for the metric. And thought we should have solved. And we actually knew that. It works for the first order. But we can also think of the way we have taken last class, just to remind you. Whereas one by doing that complicated procedure, it works by noting that we have the exact solution for a given mass. For a given mass at rest. And that exact solution was the torsion solution. Now, at the leading order, the equation for the linearization is linearized. So if we know what the fluctuation metric is for one mass, and the leading order is equal to one mass. Okay? And you might have checked that, look, we only knew what the fluctuation metric was for... That was already all in this book. And we love to be able to introduce one higher order of smoothness into that. So if you boost that. That's one higher order of smoothness, that's how we meet you. So the leading order, which is all in this book, we don't need any higher order of boost. Okay? And so, how do you get this book precisely? You remember, we started from the negative torsion network, which was one, which was one plus two five in an empty square. And it was minus one minus two five in an empty odd square. Plus one plus one minus odd square into the omega plus two. Now, this metric is not quite the same. Because this time, I think that it's not just the auspender, it's also the auspender of the order of the equation. It's possible to go between these two metrics, and work with each other. I mentioned this last time, but let me just show that to you. Let me show that to you. Transformation. Where is zeta alpha? So we're looking at a linearization. Zeta alpha is equal to some number which we fixed, too, like x alpha divided by r. Let's see what changes the matrix. This kind of works with transformation. So we know the formula for changing the matrix. We recall that. The formula for changing the matrix is delta h alpha lambda. So, I will do it only in the i matrix. How does space flow? Say, how do I do it? And i is not on the spatial manager. No time. So delta h, delta h is one of the zero-permissions. And delta h ij is equal to delta i zeta j plus delta j is the i. And depending on, there may be a minus sign here. I'll just skip this one. Okay. Let's evaluate this particular one. So this particular one, the j is going to be delta i xj divided by r plus delta j of x i divided by r. What do you think that it actually is? That gives you delta ij of 2 divided by r. That is your minus sign. Minus, and now let's think of r as equal to this value of x i xj. So we get minus 1 by r cube xj xi. That gives you the same sign. 2 divided by delta ij by r of j hat by r. Then we can get the function for this. This is the function. This is the r i hat r j hat d x i d x j. Obviously, because r i hat projects under the r hat. Actually, this is the definition. If I add this coordinate change to the correct coefficient, it will cancel this term in the short term. But as instead, the corresponding, with r, with the d r squared here, this number, we will have to choose the small deviation in the small deviation in the Schwarzschild metric. That's basically it. This is my fault that we need a starting point for our expansion in the Schwarzschild metric. Because it doesn't make a particular choice, it doesn't make a singular or a particular choice of function. It doesn't make the rotational isometric space. We can get this from the Schwarzschild metric after a point of change. So, to leave all that out, we have to leave the Schwarzschild equation. Because d x squared is equal to 1 plus 2 to the power 5 in a d t squared minus 1 minus 2 to the power 5 in the d x i d x j. Then the next question is about how many people are in my higher order? We determined to work to automate the set. Now, the single particle Lagrangian to automate the set. In order to do that, you remember we argued that we need h 3 to 0 to automate the set. h 0 i to automate the set. And h i j to automate the set. Because that's how it enters into the square into the match. Is this fair? So then we're going to proceed to compute. Again, the idea is that you don't need equation. n squared pi is equal to 4 pi g. So, finding the basis of this equation, so we've already obtained the number of connections of what is the first number we're going to automate. And we don't need money. We've done it to zero. Well, if you go further for h 0 0, if you find h 0 0, you've got to stop on it. If you have money, you have everything. Stupid. Reminded you of that. From the stress test. Now we want to solve my equation. We only want to do it on the right-hand side of the set. And we have a nice equation for the stress test because the x mu by the cloud, the x mu by the cloud, delta of x mu delta 4 by square root g of x mu minus the x mu of n times n. Okay, sum it all up. But this is x 80. But the degree, which I'm going to push, this is the standard. My name is Expansion and you can play this is the four eventual data functions. This whole thing is divided. It's integrated over the world line of each article. By doing the same integral, we equivalently wrote this as c. So this was t mu nu. We couldn't write t mu nu. As t mu nu is equal to the x mu by the cloud, the x mu by the t. And the three managers provide the best value. Does he remember? This is what we are going to divide in line dimensions. It's possible it's written this long and it's basically correct. But from here, that's all. Okay. So t is 0 to 0 and we just go with that. So t is 0 to 0. I'm going to point these below us. Okay. So we should just write this as g 0 0 squared of t 0 0. Because we are only interested in t 0 0 over up to order v squared. Okay. And the terms with t 0 i times h 0 i are at least the order. The t 0 i itself is an order. So we can ignore. And then this was written in a nice form. It was written as g 0 0 by squared by square root g. And then we wrote this as x mu dot, x mu dot divided by the square root factor. But the square root factor was g 0 0 minus as full direct limit v s by d t. But I'm from K to the level of 0. Okay. So if we expand this. Okay. So in doing this we got a square root g is like an additional factor of g 0 0 minus. That's all. So what's this? So square root. If you see that square root, you're going to take 1 plus 2 5 expand it in here. So I'll take that square root that this part to the order which is effectively like g 0 0 by v 2. Because this is 1 plus 2 5 of all things, square root g 0 0 square. But then square root g was 1 minus 2 5. 1 minus 2 5. But 1 minus 2 5. Now they say it's 1 plus 2 5 in here. Okay. We do this all. So that was like a g 0 0 q but none of that was effectively g 0 0 minus v square. So g 0 0 is 1 plus 2 5. Expand the square root in that. Taking both 5 and the square root in small. And we found that this was 1 by plus v square by 2. Okay. So g 0 0 was equal to 5 by 2. This is the effect of stress energy and self-expression of particles. The energy particles. For t 0 i, we're only going to be interested to remember it's 0 i to order v q. t 0 i is a simple contribution that's already in the end area. Okay. The next connection to this is order v q and then by diverting the equation to get an additional this is all that we put on purposes. These are the coordinates. We want to make it correct. So we try to solve nice and equations. Okay. Just go to this formula. The answer is 0 is equal to h 0 alpha del h 0 alpha by del x alpha minus half del h alpha alpha. Let the equations use this alpha space and i for space time. We've got the lambda using also the space time and i for space. Let's think about the spatial del square of h 0 0 plus half h i j del square del 0 by del psi del x j minus 1 by 4 del h 0 0 by del x i del x square minus 1 by 4 del h 0 0 by del x i del 2 h i j by del x i del h i i del 2 I don't know how to put a lower this is in the highs or I don't know how to say this but I don't know. Let's move. Ok So we have a square path adopting a gauge and the gauge equation we have is the distance. We don't have to answer any equations. But before we do that we don't have a formula for constant. So there is one formula that is unknown which is del square the order between the four parts of the basis in h is at least the order v square. So the remaining terms are already of order v to the fourth just by drawing the order v square so the rational equation getting out of the remaining terms should be far less known. It is del square of the order v to the fourth b to the fourth b to the fourth. Here I have a v square path of h. It is the next step. So then I am going to give you this part. So what we do is that here we have to get hold so remember h 0 is equal to 3 I am j. It is also equal to 3 and h ij is equal to minus it is minus i. So now with this in mind I will tell you that with this term. So this guy here is simply 2 5 plus plus sign of the square path times half times del square so we have this guy here is del 2 5 the whole x square with a minus from here. Minus plus i for 2 5 del i this was into 2 into ij. So effectively del i of 2 5 but with the minus because and this first with the plus because the straight is with the 3 so plus v into del i of let's put all this together this stuff is del square 5 so here the 4 caps is due to 2 there is plus 3 minus 2 but there is a lower or minus sign so it is plus 4 caps is for minus 1 so minus 2 del i of 5 and that if you take it is equal to half del square 8 0 0 2 5 del square 5 minus 2 del i of 5 del x square it gives some access equations so access equations were having 0 0 0 which was r 0 0 is made by k times t 0 0 minus half t 0 0 and here we made a mistake that actually somebody pointed out to me at the end somebody pointed out that is too late when we did last time the mistake I was making was now including the lower of the raising of g 0 0 in this so we had 8 by k 8 times t 0 0 now it is left by this this is equal to minus half of g 0 0 0 t 0 0 0 t 0 0 0 minus t 0 i g 0 g 0 i and so on but those are not the other equations these are all the game pieces of the other equations now in the g 0 0 notice that this g 0 0 gets it so this is equal to 8 by k 8 0 and here when we work on last time we will take this g 0 0 and ignore this one which is the important of course do the same thing last one but that we don't care about it is not important we are only interested in right hand side problems here is already a problem but is this it does g 0 0 equals 1 plus 2 so it is equal to 1 that is what is this g 0 0 we add somewhere there so that is equal to 8 by k into by 2 by 2 then plus this square by 2 simply plus this rather than this so by 2 by 2 we find this square 8 0 0 is equal to minus 2 5 del square 5 plus 2 del 5 square 8 by k by 2 0 1 plus 5 5 plus 3 square no problem no problem what do we do with this del square by 0 0 it is just this 8 by k you know 1 plus this m e del x minus x x a and some other way k this equation del square by 5 is equal to 4 by k looking at this we can do that first order h is to 8 0 0 is to 5 that is consistent we need to have to do this so we said 8 0 0 0 is equal to 2 5 plus all of e to the 4 and then we have to do all this we take it down by 2 5 so if we add 8 0 0 we have got 10 square by 8 0 0 4 is equal to minus 4 5 del square by 5 plus 4 del 5 0 to the square plus 8 by k in a 5 5 plus 3 square by 2 and 8 del tau x minus x a some other way what do we do with this what do we do with this what do we do with this what do we do with this what do we do what should we do with this what do we do what do we do in this what do we have so for many in 8 this thing is what should we do that is let's look at this this is what we have lookухd then you write this as a square of 5 x square, the terms that are 2 derivative of the same as 5 or terms which have a 1 derivative given in shape, and when you add them in the same 5, you have a choice of which is bicep, so this is active of 2, so that we must 4 x square 5, so the terms are 2 derivative of that in the same 5 axis, so the terms where 1 derivative acts on each line, again this is active of 2, because 2 derivative of the 2 5 is the derivative of 2 x, that is this, you could just work it out, you have this look like, it is so simple, simple, and you are always in it, so a, a, a, 4 theta percentage is 5, and therefore a is actually 4rd omega beta prime, so now that is published to be as you write, what, in many of them have written as, as in do not add it in the breaks, like before the Edward this week at definitely have a helped us then okay, at least how much you have okay, so as you know in not important in scores, what are important is using almost always a some particular 80004 is equal to 5L squared of 5 and now we are going to mistake this term which was from here, I take it to the left hand side so I take this 10 squared of 8000 minus 5 minus 2 5 this will be coming in this term and this term 8 plus 8 sum over 8 5 plus 3 squared by 2 m8 divided by 8 minus exactly is equal to 4 by 8 squared by 5 is equal to 4 by 8 and squared by 5 is equal to 4 by 8 yes yes so that is 4 by 8 times sum over 8 so now I am going to make 5k out here so that will be equal to the minus 4 so that cancels out these 5 fights that we cancels out so I am taking this is 8 by 8 and this is sum over 8 5 plus 2 m8 this is the thing that we do when we solve it so that will cool up and blow up our problems we use the green function we use the fact that 10 squared of 5 by 8 is equal to 4 by 8 times sum over 8 minus 2 by 8 times sum over 8 so therefore 10 squared by 4 by minus 4 by 8 is equal to 0 by 4 minus 2 by 5 squared is equal to minus whatever we have there 8 by 8 8 5 plus 3 d squared by 2 we take it like that sum over 8 5 anything what is the distance between A and the point which you are completing 8 is equal to 0 so 8 is equal to 0 by 4 at r is equal to 1 here that means that if you take this formula because 5 is the potential at this is at r potential at r is equal to 5 at r I know this because there is an r minus r a sum over r d including r and now the reason for this divergence is that we actually try to deal with this point I am looking at a distributed connection of each point we would not have got any such divergence and the end result will be the whole thing correctly I must have observed this before that is just to place this 5 r in another 5 r in which means that the formula of 5 don't include the set range I use the potential to do all other objects on this without doing anything taking into account that it is 5 times sum over 8 so that is the subtlety here that is what I said up to the subtlety we have got 9 times let me just do this very carefully minus 2k sum over 8 plus 3v a squared by 2 m a over r and sum over 8 for 8 0 to 0 to all the way to 0 sum comes in 0 it is whatever it is well solve it for 8 0 i you follow the similar procedure I said solve for 8 0 i you follow the similar procedure once again what we need is formula for r 0 i once again I am just going to serve that to you but not here it gives it to you so the two things you require over r that has to all be q r 0 i is equal to half del a j by del x j del t ok plus half del squared h j 0 del x i del x j plus h m m del t del x i that would be r 0 i you see this term here 8 0 i ok so we are interested in 8 0 i up to q b 1 ok so this is del squared of r before we proceed let's use our gauge condition remember we have a gauge condition let's just write it like this gauge condition 0 i exactly so the gauge condition was del m h m 0 h i and the important point is that every time we have del m h m 0 we can replace it by h m m and that's what we have here we have a del i no sorry del j h j so that del j j is going to be replaced by half del t of h m this thing is half del t so this thing by our gauge condition becomes 1 by 4 del by del x i del by del x d of h m and we already have at least that's minus half del that's very so that becomes minus 1 by 4 our gauge condition is r 0 i is equal to half of del squared h j i by del x i del d plus 1 by 4 del squared by del t del x i h m m plus half del squared h m h i j and we know that there will be squared and we can be there it's a new factor to be that new factor to be if you don't know what it is you can use that to solve it so we've got the equation half del squared by del z i that's not what I think it is 2 so del squared by del z i is equal to minus del squared h by y so that's probably what it is that h j i x i was so remember this guy was minus 2 psi so we get by 2 so that we have del squared by 2 by del on the other hand to the other side of the glass but it's not a minus because h m m has minus 2 psi the two factors, all the factors are two answers so we get glass that's it it comes to the superclass but it has minus 2 minus del squared by del x i up but there's a trace so that's a trick minus del squared by del x i del squared by del x j of 5 16 by 10 of v i a i and so we've got a right perfect so we get h j i is equal to minus del squared by del x i del squared by del x j of 5 16 by k v a del tau it is to now that remains it is to solve this equation that's very easy so let's remember that del squared x i so so that's we just solved the equation before so h 0 i so we divide it through by 1 by 4 by and minus and then put the other examples h so let's get the pattern of the equation first so that is equal to minus 4 they have del squared it's like with a minus their equation was del squared and they have minus 65 k in the letter functions t 0 i is t 0 i in the uppers is n a b i but in this funny here signature in each way we say t 0 i in the lowest is minus so so this side I will solve and so the first place first now let's also get the you see at first I consider something because all the sources were localized in particular positions this source looks weird so it looks like a theory you know a integral instead of some some of things so that's not quite right because find itself very special it comes from I mean source you know any things let's write this funny so let's see the first thing I do is say that this thing here is it's arising produce a thing like that which is 1 by 4 by delta x i delta d of f this square f is equal to d by d inside d by d outside this motion so if I can solve if I can solve the equation with this source I can solve the equation with the difference in the source just by taking the result from the difference in that source difference d by r so that's where f is equal to 5 but find set is equal to minus 4 all the present 5 is equal to 5 r that's a potential to solve solve the equation solve the equation delta square f is equal to minus sum over 8 k and a r by sum over 8 I think it was you know continuity very function some sort of mess however it's much easier it's much easier to follow that because you can just guess let's just remember that there is a square of square root of straight that will be del square of so when I write del i del i of square root of x i x i let's say first derivative that's equal to half but then there are two two x's that they act on so that's x i y square root x i x i x i an extra second derivative k and then what do I get here I get two terms this guy this guy into this guy gives me a factor of 3 pi square by x and this guy into this guy gives a factor of so now this is minus square by x q so that's equal to two x's so we know what gives you the right answer let's say that the solution will do half k by 2 and a the solution that we get from from this part of the equation is actually actually the actually the del down instead of getting the del down we have to get an article but otherwise that's one for each now one for each part when you say this twice with respect to r one with respect to tan and the second with respect to per once here once there because we want del by del x i del by del t u so what's this way plus let's do that the other way so all the people have such an attribute it's a side plus getting through our discussion to make we have to start next class class of that so they are just they can already have that try getting this class about for sure the other means it's going to be a problem so now so let's say that del by del t where is that for function of t it's a function of t del by del t of x is what it's here so this quantity was so x was equal to so you know del by del t is equal to minus k by 2 del by del t of we are ignoring the summation r minus r is t dot r minus r r minus r okay now since we are differentiating with respect to this guy think of this r a of t minus r dot r a of t minus r okay and here now we are not going to make a mistake size so that will then just because minus k by 2 r a of t minus r so now this is differentiating so we r a of t dot r a of t minus r by r r a of minus r this quantity is called del by del t it doesn't act on r a it will act on this r okay so we want del by del x out because we will get del by del t of f that's what one term comes from here and that will give us v k by 2 and that will give us a plus k by 2 v r so that will give us the minus v dot r a minus r the whole thing write r a minus r u time there is a minus a minus here and a minus from this state plus times r a we have k v i by 2 plus v dot r a minus r of a r minus r a i I will write r a minus r and I think I am going to write some nice letters k is v i by 2 plus v i act and then I get this time right so one of these is always r a minus r u minus r a j j i by r right so that's that's what we get we take the same narrative of when we take the second narrative of f with respect to x i so so now just put it all together put it together what do we have we have yes yes if you write this double is that just goes to the same i i i k by 2 when we go to minus k by 2 into r minus r a v a i by r minus r a minus infinity times r a double j by r minus r a q i when we write that we take that for h 0 i i take this to 4 k m a i k by 2 we write v 2 into v dot r minus r a into r minus r a divided by r minus r a as well now now let's look at this minus k minus k by 2 and a of a i by this term this is what we say for this term this term this term this term this term this term this term this term 7 by 2 k m a j i by r minus r a k m a by 2 3 dot r minus r a in r minus r i it matches except that plus sign is the reference of multiplication sign we have a story of how a couple of examines in the Landau Institute a family team I don't know if it's true, I don't know if it's a story but in the Landau we're going to give the book and say find an error I'm going to tell you now our exercise is only for H0808 how do we determine the motion of art? from H0808 to H0808 because the particle couples to the particle couples to the gravitational field that we're very familiar with but the gravitational field that we're very familiar with may actually be L is equal to minus m integral yes is equal to minus m integral mu mu dx mu by mu t dx by mu t t d and then actually because it's k we should keep the terms at about 1 plus 25 plus H004 we have a factor of 2 g0 either mu becomes 0 or mu becomes 2 next so we keep 1 plus 2 h0i then we get dx0 but that's here now we messed up the sign let's check let's check the sign what we really have is g0i and v i we've got 2 g0i is a question I suppose x0i g0i simply x0i if you're really interested this is how you do it finally we're going to do that but finally we have to do it with delta ij because g ij will go up it's minus delta ij and I committed one term although we need to compute it we already had it we should keep plus h ij2 h ij2 is trivial just 2 phi delta ij but we should keep plus these pieces so this is equal to minus m integral square root of 0 we need to expect same so that's really exceptional so it's square root so 1 plus half the first order of pieces plus h 0 0 4 plus plus h 0 0 4 by 2 plus h 0 i by i minus v i2 plus h i v i2 plus h ij v i v j by 2 so this is the half term but the next term the square root is minus 1 by 8 minus 1 by 8 minus 1 by 8 and then this term and this structure so we get 2 phi that is v i2 the whole piece and total that we would use to derive the equation equation of motion for the 8 part because the 8 part is in response to all the other particles we take this as a result and derive the equation to get to where we find the work we are finally interested in the character that is we are not interested in finding the equations of motion we are interested in finding the equations of motion you might be tempted to think that this will just be the sum of the quantities for each piece you say what we have to do from this for the 8 part the equation of motion also involves the position x a so if we differentiate which are some of the quantities of all the particles and differentiate we get additional terms of the equation of motion that is not that you should be keeping remember the structure of the equation other particles source the matrix and then particles just move in the j o d c in that matrix the right equation of motion for the 8th particle by doing this procedure we can't just sum the individual Lagrangian this is the Lagrangian that is for 8th motion the motion of the 8th particle regarding the motion of the other particles is given so this Lagrangian is deriving the correct equation of motion for the 8th particle no more simply sum over to get everything that we are interested in take this equation of motion for the 8th particle this is completely straightforward exercise having to deal with that equation of motion check that the Lagrangian is still in 106.7 of Lagrangian I plan to go through this so very fast time excellent so having to validate this last differential problem we have now completed our task just one second what we have achieved what we have achieved is systematically connected the neutral is given by p minus v where v is the neutral connection achieved by this task in this 107 right there is that the first systematic connection to the neutral in follows a piece where in the strength of the interaction systematic interaction how we continue doing this can we go to the next to the post neutral and the answer is no the answer is no because you see this order keeps track of energies up to order v to the unfortunately now in the next class v to the 5 between the equations in the equation the neutral the only degrees of freedom of position of particles you can imagine it was a gravitation of most things of your life but that was a fake there were no degrees of freedoms and any given space spatial size if you knew the position of particles it would be rubbish if the ion speed in the array that's not pure point particles it involves positions of particles that mean to involve the degrees of freedom that will happen in order v to the 5 that it could do even the first interaction consistently then we could take this Newtonian idea if you try to do it one order higher than this that just will not be consistent okay so around you involved in just point particles unless of course you have a little better to allow yourself to be with some dissipation in our dissipation so you expect to lose energy to the equation and that will result in dissipation in the system of particles so just as an honest, conservative classically identical system the equations that we derived are almost the right in class but they are best if they knew an approximate direction and they are quite nice the question about question can just be right now but it's essentially coming up with a version of the velocities of particles it's that kind and it can't see the light and this over there it really can be done but we have seen in the equations that have just pure energy so even no particles have solutions to these equations those can be super important exactly firstly the problem that I suppose there has been a little delay because if I can I solve this for the next given the position of particles really the answer to that is not because on any solution you can superpose on a gravitational equation even in first order even in first order you can take out solutions that you don't have and I superpose to that when we try to solve the equations in the system we didn't see anything so how can we solve these equations how can we solve these equations if the particles don't move we have to solve them you have the particles can't move I can have a gravitational wave exactly what we have done is solve these equations in the observation in which time derivatives are much smaller in space space derivatives are made to be more important but time derivatives are made to be more absolute we need to solve these equations let's see in this expansion because in a gravitational wave gravitational wave because I keep k k omega t plus k x where omega is equal to k so in a gravitational wave time derivatives are the same space derivatives same order what we have been doing is to search for solutions in which all the energy issues in time is very slow but this thing is essentially inconsistent to my right if you start looking at what I have every kind of a gravitational wave there cannot be a water meter in time so it doesn't exist in time but if you qualitatively impact the next thing we see that there is no gravitational wave yes there is no gravitational wave ok so now in electric mechanism in electric mechanism once again we can ask suppose you cannot do an expansion in the process once again so let's see you know that or in A it is a square which is like v to the fourth which I can do anything beyond ah