 really sorry because at 12.10 I have to go for something urgent we're gonna have an extra makeup class tomorrow at 11.30 in the mathematics seminar room okay so just for those who came late I'm today's gonna be very short class like half an hour class because I have to go for something urgent at 12.10 I'm sorry about this okay so but we're having a makeup class tomorrow in the mathematics seminar room okay so now let's let's move quickly okay so in the last class we derived Einstein's equations and this class we want to start using Einstein's equations okay so we want to start using Einstein's equations but before we start using it indeed these equations in detail which will be much of this class the first thing we're going to try to do is try to understand them the non-relativistic or the Newtonian limit of the Einstein equations okay so you remember that in the previous class we had checked that provided g00 is equal one plus two phi okay provided g00 is one plus two phi test particles in a background metric will in the appropriate limit test particles in a background metric will in the appropriate limit behave as if they are subjected to Newtonian potential phi okay and the appropriate limit was let me say it was of course that g is equal to one plus plus epsilon and h some degree of smallness in epsilon and we had also that velocities velocity squares were like one plus excellent so velocity squares were of the same order of smallness okay so velocity is like a measure of x distance by time distance okay so velocity squares was of the same order of smallness as the deviation of the metric from flat space and the way to see that those had to be of the same order was to note that in the Newtonian theory phi and v squared are often exchanged in ordinary motion so a planet that has only phi ball up here that has only phi at the top has v squared at the bottom but v squared and phi of the same order right so v squared and phi were of the same order this was the the Newtonian limit under under consideration okay so what we're interested now is in looking at what Einstein's equations become in this limit so we look at the limit in which g squared is one plus epsilon of epsilon times h and these squares are small now how do we precise the statement these squares okay the kind of thing that we want to say is that now v this is sort of like delta x by delta t the whole thing squared is okay so what what we're saying is that in the configurations that we're looking at time derivatives one of the things we're saying is that time derivatives are of order square root epsilon times space derivatives because d by dx is like one over a spatial distance d by dt is like one over a time distance so this will give us delta x by delta t is over epsilon squared it we get this okay so we're interested in the limit in which things are varying much faster in right but a derivative is one over delta x okay so this is one by delta t is over the square root epsilon times one by delta x so delta x by delta t whole thing squared is of order square root epsilon okay so basically this tells us the time derivatives are much smaller than space derivatives by a factor of square root epsilon and we want to examine Einstein's equations in this limit okay really clear what we want to see is that Newton's law has come out of Einstein's equations okay now as we've seen and in order to get Newton's laws all that's important is that g 0 0 takes the form 1 plus 2 phi where phi is the Newtonian potential we got agreement with behave with motion in a Newtonian potential provided that all g's were 1 plus order epsilon and g 0 0 was of this form it didn't matter to us what g i j was do you remember that okay so we have to demonstrate is that g 0 0 when written as 1 plus something is that that something becomes a Newtonian potential the g i j's can be whatever they want it doesn't affect the Newtonian limit the strict Newtonian limit so what we want to do is to find an equation for g 0 0 sorry what we want to do is to find an equation for the the deviation from g 0 0 so we write g 0 0 is equal to 1 plus h 0 0 and try to find one of Einstein's equations that in this limit of smallness gives us an autonomous equation just for h 0 0 so the equation we look at is this one the equation we look at is that okay now there's something else that I should have said this the something else that I should have said is so far we've been looking at Einstein's equations to the answer Einstein part of the equations but there's also the matter but we're looking at the non-relativistic limit in matter so imagine that the matter has made up of a bunch of non-relativistic objects moving around okay in this limit the rest energy the stress tensor of this this this system is dominated by the rest energy of the system right so t 0 0 is going to be much much larger than okay it's going to be of order t 0 i will be of order v i times t 0 0 is this clear and and t i j will be will be down even even smaller t i j will have a v i v j in it would be of order v square compared to t 0 0 so the dominant term on the right hand side is t 0 0 and t 0 0 is larger than t 0 i by a factor of square root epsilon and t 0 i is factor by as larger than t i j by another factor of square root epsilon okay so that's the other thing that we're going to use because the matter itself is assumed to be non-relativistic so given all these assumptions we want to check that we get back newtons equations okay so now let's look at this in this equation r 0 0 what was Einstein's equations Einstein's equation was that r mu nu is equal to 8 pi k we're dropping the C's t mu nu minus half delta mu nu t so keeping things just to leading order on the right hand side okay what we're going to do is to reduce everything here just to the to the t 0 0 parts but this should be an r you think okay good so that's something that I forgot to tell you about last time let's let's let's move back to our so our mu nu we had our mu nu minus half r by 2 delta mu nu is equal to how was it 8 pi k tends to be okay now there's another way of writing this equation that we can get by just taking the trace of both sides of this equation so if I take the trace on both sides of this equation what I get I get r by 2 wait so let me be careful here the trace of this guy is just r the trace of this guy is 4 times r but there's a by 2 here so that's a minus 2 times r so this is minus r okay is equal to 8 pi k times t let me plug this in here so now what do I get I get our mu nu plus 4 pi k times t delta mu nu is equal to 8 pi k mu nu which implies that our mu nu is equal to 8 pi k in a T mu nu minus half minus t by 2 delta mu nu so this is an equivalent way of writing Einstein's equations this and this are the same equation this is the form that I'm going to find more convenient for whatever don't know is this clear okay so the next thing I'm going to do is to try to understand I'm going to take this equation so I'm sorry I'm extra chaotic today we're going okay but I'm going to write here so now what I'm going to do is now take the indices mu and nu to be 0 and 0 so we get the equation r 0 0 is equal to 8 pi k then there's a t 0 0 okay and we get minus delta 0 0 so that's 1 times half of t since we're only keeping leading a leaving on on the right-hand side we can replace that what was t t was t 0 0 plus t 1 1 plus t 2 2 plus t 3 3 1 index lower 1 index higher 1 they suffer and doubt and we're only going to keep the leading order term so this was half of t 0 0 keeping just the leading order okay so we've got all 0 0 is equal to 4 pi k times t 0 0 in the approximation of which we're keeping things this is what one of Einstein's equations being because okay now to further process this this so remember I'll keep this on the board because this is we want to we're going to write g 0 0 0 1 plus 2 phi and find an equation for phi okay but let's go there slowly so we've got we've got this equation now and I want to process this r 0 0 so firstly let's remember what let's remember the formula for r a b cd okay so the formula for r a b cd was what it was gamma a b we had c comma d minus gamma a b d comma c plus terms that are quadratic in gamma now if we write g as 1 plus order epsilon the Christopher symbols are obviously for the epsilon because they involve derivatives of the metric and one has no derivative so Christopher symbol square is obviously for the epsilon square and we're gonna only good only going to be keeping this equation to order epsilon the minus sign is wrong the first one is a minus sign the second ones plus thank you okay thank you okay good now that's right. That's it thank you thank you, thank you thank you thank you. Thank you thank you thank you, think let me write it as plus okay good yes and thank you yes okay so we only need to keep these two terms okay more over every term that we are keeping has at least one factor of the small metric as we said so this automatically of order epsilon so any further smallness is we kill it what does that mean that means that all derivatives that we are taking should be spaced derivatives okay so now let's actually let's process the formula for our so firstly now what was our we want to contract the first and third indices together so we get a so gamma a a and now we want okay and then of course we this is going to give us our zero zero with both indices lower but the difference between upper and lower is a factor of the inverse which differs from delta and one higher order of smallness so we ignore that so r zero zero with both uppers is the same as r zero zero with both lowest because we're using the signature no minus sign because we're using the signature plus minus minus minus okay so to the same order what we're interested in is okay let me be more systematic r zero zero which is the same as r zero zero to the order which we're working is equal to gamma a a zero zero minus gamma first and third first and third first and third sorry first and third a and c gamma a a zero zero minus gamma a a zero zero now we decided that because all these are already order epsilon any further smallnesses for instance by differentiating with respect to zero should be ignored so we ignore the stuff is this clear so all that remains is this guy so we've got gamma a zero zero comma a now this a runs over all values but we'll only allow it to run over space because we don't want a time derivative so gamma i zero zero i so now all we need to do is to compute this gamma i zero zero zero i but we know how to compute that that is equal to minus half g zero zero comma i plus terms that it will involve time derivatives which we're going to throw we've concluded from here that this equation in the in the Newtonian limit that we're considering simply boils down to minus half g zero zero comma i is equal to eight pi four pi k t zero zero okay now we substitute g zero zero this is basically we've met our objective we've got an equation that autonomously determines g zero zero forgetting about all the others okay and now we just now we just oh sorry and there was another eye so this is del square spatial del square okay so now we substitute g zero zero is this so we get minus del square phi is equal to four pi k now this we can call mass density energy density it doesn't matter because we're setting c equals one we work it non relativistic limit okay and this is of course Newton's equation once you identify well in some convention once you identify the gravitational constant Newton's constant as how yeah case G but without four pies okay I think this is case what yeah case G right so K is equal to G it's Newton's constant once we make this identification we've got Newton's law Newton's law of gravitation so in the appropriate limit neglecting many things that are small inappropriate circumstances these are the appropriate circumstances include motion of the sun for instance okay we've recovered Newton's equations so we see that Einstein's equations why much much much bigger than Newton's equations Newton's equations gravity much more beautiful much more complicated having much more structure reduced to it in the right limit which is of course a necessary condition for it to be the right equation of gravity since Newton's equations have been fabulously tested for instance in planetary motion okay questions or comments about this okay so now let's move on now we understand that these equations reduced to reduce to Newton's equations so that's one well this is minus sign is the minus sign minus sign is not changed so we've messed up on the minus sign but did I wait sorry I've messed up the minus sign the minus sign shouldn't be there let me let me look at that okay yeah wait and was correct oh where did this minus sorry no no no that was okay the minus sign is okay let's see let's see let's to check what is the gravitational potential of a delta function source according to this equation so according to this equation if we do the Stokes theorem kind of business I mean the Gauss theorem kind of business then integral of del phi over the surface will be equal to something negative okay so that would tell us that the gravitational potential is equal to 1 by r so that its gradient is minus r hat by r square okay not bad that's the usual answer right okay thank you okay fine okay maybe we've not messed up maybe I just was checking with Landau-Liffchitz they've got the opposite sign sorry wait I think we have messed up you know in this sorry we have messed up here because we're using this weird metric you see we have this gamma ii now I want to write this in terms of both indices lower because that's what del i squared is del x times del x okay and just to say it again that this formula that I wrote down was a formula for the gamma with all three indices lower then there was a G inverse of the upper index there was a G inverse of the upper index and in special directions G inverse in our conventions has a minus sign sorry so okay so this is what we're getting now now you've confused me into thinking this is not this should not be what we're getting what no no no I think he just just a minute and now we just have to remember Newton's laws okay how does it normally go but let's look the gravitational force should be attractive okay so the gradient of the gravitational potential should be negative so the gravitational potential should be positive and this sign gives us right one potential one by our gives us force minus our hat by our square which is correct okay so we so this is what we expect if K is equal to G no no no no no force should be attractive so one by our now differentiate forces minus gratified okay okay so the question is with this one by two five did we get T minus T minus V or T plus V in our when we looked at our square okay let's let's get the straight let's go back suppose we looked at our at the motion of a particle okay so this was square root G mu nu DX mu DX nu now this thing came with a minus M okay now in this we write minus M integral square root of 1 plus 2 phi we put a DT here so that's 1 minus DX I minus V squared squared and you're right that that what we're getting here is so the action is turning out to be MD XI by DT the whole thing squared minus M phi okay so this was our action now as you say this will give us so but when we differentiate we will get acceleration minus acceleration minus grad phi is he is equal to zero so acceleration is equal to grad phi so force as you say thank you is minus gratified correct so the force is minus gratified so minus gratified should be negative okay that's right so minus gratified negative gratifies positive therefore phi should be negative according to Landau leaf shits well you know we should be able to work consistently in writing 1 plus 2 phi we've identified it with the potential okay so now unless we flip our sign of K something is just one minute have I messed up somewhere else wait okay maybe we have got it right okay let's let's let's let's keep thinking now yeah I think it's all okay look see we wanted we wanted grad phi to be positive okay sorry about this we want to grad phi to be positive but that's consistent with del square phi is equal to this because del square phi is in integrated over volume is integral grad phi over the surface and so in grad phi is positive so that the force is negative this is actually positive so I'm saying that the solution to this will be minus 1 by r the solution to del square phi is equal to positive delta function is minus 1 by r okay sorry about this sorry about this this was embarrassing okay so okay so now we are all happy with our signs it all works we believe that Newton was correct gravity is attractive rather than repulsive as we were finding briefly and and and we're all happy okay now the next thing we want to do is to go beyond the next thing we want to do is to go beyond working this Newtonian approximation because it's great that it reproduces Newton's laws approximately but we want to see what happens exactly okay so though I'm only going to be able to teach for the next six minutes I have something urgent to do it in past 12 we'll have to continue this mainly tomorrow but let's start setting it up the thing that we want to do is to try to solve Einstein's equations in some interesting situations and the most interesting situation is like to try to understand the well and all these first thing to do is try to understand planetary motion around the sun so we want to find the grad the solution to science and equations due to the sun okay but we're going to do this in a way that is a little more general than to be so that we can use in other situations as well let us try to set up Einstein's equations let us try to set up Einstein's equations in yeah so let us try to set up Einstein's equations in in a manner that that is that you that that is correct for any spherically symmetric situation okay at the moment we won't assume that this the situation is static but we will assume spherical symmetry okay so what is spherical symmetry mean spherical symmetry means that you know things in that direction in that direction are the same around a given origin okay so the metric in our situation should have a rotational symmetry symmetry of rotational rotations around this origin point okay what does that mean that means that the metric should have a factor okay so let me write down what it means and then you tell me if you agree with us this fixes the metric to take the following form firstly first thing we have to you know when we're dealing with Einstein's equations you might ask metric in what coordinates okay working general coordinates except for the metric on the sphere you see because we've got rotational invariance then it must be it must be possible to foliate our space time into two spheres into a two-parameter set of two spheres because what does it mean to have rotational invariance you take one point okay you rotate it you go to another point which is physically equivalent to the first point okay now if you take any point in space and rotate it in all possible ways what does it generate to generate the two-spheres okay but a two-spheres just a two-dimensional object space time is four-dimensional she must it must be possible to take space time and foliate it into a two-parameter set of two-spheres to generate all four dimensions of space time so on the two-spheres on the points that are related by symmetry by rotational symmetry to any other point we're going to use standard coordinates standard polar coordinates of the two-spheres so we'll use the angle we'll use the coordinates theta and phi this by the way is something very this procedure something very general that while while in general there's no preferred coordinate system in Einstein's equations once you're looking at a situation with certain amount of symmetry that picks out a very natural set of coordinates which is always advantages to use okay because we're looking at solutions with rotational symmetry in the two-spheres we'll use coordinates that are naturally adapted to that symmetry on part of this on part of space what about the other two coordinates so far we made no assumptions they're just two other completely random coordinates let's call them are and why should they be on a sphere because you have this rotational symmetry the other two could be anything they need not be no no they completely arbitrary coordinates there will the range will in general what the two-dimensional right and then the requirement of rotational symmetry is that the metric take the form ds squared is equal to minus dT squared into some function plus dT let's say dT dr into some other function plus dr squared into some other function a function let's not all is d because it's e times a d theta squared plus sine squared theta d phi squared where a is equal to a of rt b is equal to b of rt c is equal to c of rt and so on just to be totally clear what does it mean that this metric here has spherical symmetry what does it mean as a mathematical statement okay what it means is that there is a change of coordinates this change of coordinates is generated by rotations under which so such that in the new coordinates the metric has the same expression and the metric is the same expression as a function of new coordinates as it had as a function of old that's always where a symmetry means in physics there's some change of variables such that something doesn't change form as a function of new variables compared to function of old variables the rotation here is how those rotations act on this metric well exactly the way rotations act on a two sphere in theta phi coordinates okay and it's obvious that this metric this rotation doesn't touch the t and t and r coordinate it acts only on this part of the space and from the fact that the metric of two sphere is rotationally invariant okay the rotational invariance of this full metric follows I mean pedantic you know what I'm saying is obvious don't let it confuse you but just to make sure you understand exactly what it means okay for instance phi goes to phi plus delta phi is obviously a symmetry of this metric because in the new phi prime coordinate it takes the same form as in the fight one now what what ensure there was a symmetry of the metric first that the metric of the sphere took this form but secondly that there was no phi of theta dependence in these functions had there been five dependence in these functions then five goes to phi plus delta phi would not have been an isometry of this metric okay how do we know that this is the most general rotationally invariant function on the sphere the reason is that the only okay so it's sort of clear that this part of the metric has to be the metric of the sphere but could we have had some function of theta and phi entering here that was rotationally invariant and the answer is no there's no rotationally invariant scalar function on the sphere apart from the constant this you know from your study spherical harmonics all functions can be expanded in spherical harmonics and the only spherical harmonic the spherical harmonics transform in representation of the rotation group the 0 1 2 up to infinity representation of the rotation group the only one that's rotationally invariant is the one representation of the constant okay so the only scalar function the only combination of theta and phi that can appear here is some function of your coordinates of the sphere some scalar function that is invariant at rotations and there's no such function apart from the constant so these things could not have depended on theta and phi without sacrificing rotationally invariant okay so this is the most general metric that preserves rotational invariance is this clear greatly simplified our task it's greatly simplified our task for two reasons firstly we know a lot about the structure of the metric and theta and phi and secondly all these other functions are not functions of four variables but only functions of two variables so in a sense imposing rotational invariance has made the problem of solving Einstein's equations roughly as easy as it is to solve Einstein's equations in two dimensions which is a very easy task as you will see in some problems it's okay so so so so this now is more or less a solvable problem as we will as we will indicate okay but unfortunately not today because I have to run so we'll continue tomorrow at 11 30 math mathematics seminar over sorry about this okay I'll just run immediately since I'm doing this late anyway