 to know was that if you thought of gamma a b mu is that in the way that you might think of a mu. So, these two indices are thought of as matrix indices and this has a space back. And then you found the same combination that you would get what does that mean that should be del mu, gamma a b mu minus del mu, gamma a b mu plus gamma a theta mu, gamma theta mu, gamma theta mu, gamma theta b. This is first to make sure that the sine of the last two terms, right? That is the first to the next matrix of that is always the top one. So, this is the expression for R 18 union. So, just structurally there is a close analogy between potentials, amuse and gammas and field strikes and couches. This is something that is going to be given tonight, at least help you to remember the formula and help you to remember various other things. The Bianchi identity that we proved also can be remembered in this language. So, Bianchi identity for gauge bearing was del alpha f e to the gamma cyclically cement trice to the 0. What would this way of thinking of that suggest? This would suggest that the Bianchi identity was gamma a b mu mu comma comma theta psi mu mu theta is equal to 0. And that is precisely the identity we proved. Now, but this was not a code incident. I am going to make it in the fact that this is not an ordinary code. So, this is the first time it was used. The fact that it is behind this is that e square is actually a differential equation that proved the first one. The gauge case. That e square is 0. Yes, yes, yes. In the second one or is it the same or is it? We proved it last class. In some sense, yes. In some sense, yes. Okay. So, just keep this analogy in mind a little deeper as we go along when we talk about the tetraformalism of tetraformality. But it is useful at least to remember formulas. Now, the next thing we talked about was okay, now that we have curvature completely now and we were able to write down the understanding. So, we wrote down the action as minus 1 by 16 pi integral r plus integral square root lambda. But this part was talking to me. Because, guys, we spent some time motivating while this could be, by this is an instant action, just to repeat the motivation and the problem behind the motivation. The motivation was that suppose you got some unknown, you know, we don't understand gravity well yet. We don't understand the underlying theory or anything. Suppose you got some unknown gravitational theory at blank scale. Well, the first thing we noted is that the scale here had dimension length square. And we gave this length a name. It's called lp square, blank length square. In the real world, blank length was density minus 33 centimeters. It's very small. The next thing we noted was that if you had some unknown, some theory of gravity at a particular scale, okay, we made the following assumption. We made the assumption that there was only one length scale. So, suppose you had some complicated quantum theory of gravity here, that length scale, and energy is much lower than that, you would get some effective classical description. And we wrote down possible corrections to this action. It's one of the corrections we learned out of square, and that goes on to the form. Just by dimension analysis, this term at one over length squared is r and 2 divided by 2. But d4x is 4x. The same reason I gave you that this term was dimensionless, and this term was lp before, and so on. And then a crude estimate of how these terms contributed to dynamics on length scale l gave you that this contributed like l by lp squared. This contributed like 1, and this contributed like lp squared by lp squared. So, length scale is much larger than lp. Clearly, this is turbulence dynamics. So, at least to the extent that we understand gravity today, we expect that there is nothing particularly sacrosanct, particularly sacrosanct above the unsancted term. All terms exist in gravitational action. It's just that we are probing in a length scale is much larger than the length scale. And so, all physics is governed by this term. This sounds very nice and very satisfying except for one problem. The problem is that if you take this logic to its conclusion, there is one more term in the action that you can write down that is even more important than Einstein's term. This was the term lp to the 4, that was d4x squared of g, the so-called cosmological constant. And it was at one minute. And this term occurred, it would be more important than this term than the Einstein term by a factor of l squared over lp squared. I think it is one of the central puzzles of theoretical physics today. So, just experimentally, it's, there's lots of evidence that this is right for your gravity. But this term is either absent or is present with a coefficient that is not 1 over lp squared, but 1 over l Hubble squared, the size of the universe. Now, there's a big scale difference between 10 to the minus 33 and the size of the universe. And I think that it's one of the key outstanding problems of fundamental physics. To explain why this is the case, any nine estimates for such a term would have been the same order as this term. And that would have resulted in our universe incurred the lens scales of all the 10 to the power minus 33 centimeters, not being huge like it is. Okay? So, there's a key quality to think about our universe we don't understand. Why this term is absent or is so small? Okay? Module of that, if we hide our ignorance of the spin of why this is the case, then this seems like the reason for the next term. For a long time, what people have thought was that some unknown, as a unknown theoretical mechanism that senses this term you see. You know, in the theory of supersymmetry, for instance, I have a supersymmetric theory such term is formed. The world around us is now supersymmetric. So, this is now a special explanation. Well, that's just an example of theoretical realism. So, a certain symmetry could sense such a term to see. So, for a long time, it was assumed that there was some unknown reason this term was to see. Okay? But now that there's increasing experimental evidence that it's not zero, but it's some ridiculously small number, except by the size of the universe. And that sounds really strange. I mean, there are no, there's no understanding for why any reasonable theory of gravity would set to zero. There's even less understanding for why reasonable theory of gravity would set to something that all this here would aggregate. Okay? It's a key outstanding issue in my opinion. And you know, understanding of the basic structure of the universe. You know, it's a very basic issue. Why are we living in a universe that is big, rather than curled up on time scale science? As you attempt that. Okay? And understanding this is a key question. Okay. So, we talked about that, you know, model of this universe, which one of you will hopefully have resolved. Okay? Why do you know this, this issue? It seems like a reasonable postulate to make the business of the, the action of gravity. As you did, it turns out to be correct. Okay. Let me turn to the company of complex action, Matt. Okay? So, we started talking about Einstein's equations. So, we've got a Lagrangian and we want to find the equation of motion that's called E. So, we decided to look at the Lagrangian, take action, and look at the variations. We look at the term that was delta g mu nu times, times square root g times e, whatever it is, some equation of motion, which I call E. Now, the E means you get contributions from two terms, from the matter part and from the outside. The, we computed last time the variation from the matter part. We computed it for the special example where I can get something. And then we discussed in general what, what its properties are. Okay? So, we discussed in general that I should take the matter part that's got, square root g lambda x. You take the delta of this, we get delta of g mu nu times square root g times, and then by definition whatever came, we'd call the stress tensor of the factor of two, perhaps a sign, take that. About the action, we defined to be this object, t mu nu. Okay? And then, in the last class, we went through this little exercise to show that by this definition, t mu nu was always conserved. Okay? And let me remind you of the simple demonstration of that fact. The demonstration went, went, went as follows. It went as follows. So, I said, look, suppose you take any general equivalent action and barely all feels by a quantum transmission, then the variation in the action is clear. That's clear. The feels that you have to barely have the matter feels and the metric for this to be zero. If you're very high in the matter, but keep metric fixed to higher in the metric, keep matter fixed to against it. But the whole thing is zero. Now, let's, this is true for variation, for ordinary. But now, and so far, what I said was kind matters. It's true for arbitrary variations. And it's true for off-shell without inputting equations. However, on-shell, there is something else that we know. That is, any variation of the matter feels on-shell. First order gives you managing action. That's the principle that you use to derive the equations of motion. So, if you put these two things together, all the solutions to the equation of motion, a coordinate variation of the metric does not change the action. Of the metric, the limit matter, the matter fixed does not change the action. This is not a kind of matter. It's true only when the matter feels obeyed. But we have, and I'm going to please note that just to show that, but that if you take its mu plus zeta, this is an arbitrary vector field, the first order, the change in the metric was equal to up to plus minus sign, but I don't show that. Then mu zeta, mu plus mu zeta. Okay? This is something, but okay, just as an aside, coordinate transformations that leave the form of the metric unchanged, are called iso entries. And infinitesimal coordinate transformations that leave the form of the metric unchanged. A parameterized, called infinitesimal transformations are parameterized by vector field, as you can see. And zeta is a parameterized internal coordinate transformation that leaves the metric unchanged, are called Killing Equations. Okay? This is, set this equal to zero is called the Killing Equation. This is the equation that determines where you see that. It's K. Okay? Well, that's one of the considerations. We have this delta G mu nu, it's given by this. Then, putting into this definition, and using the fact that G mu nu is symmetric because it multiplies the symmetric variation, we conclude that that must be that delta S, which was equal to integral del mu zeta nu, square root G mu nu, was equal to zero. For arbitrary zeta, coordinate transformations zeta. Now, we decided to do this for a coordinate transformation that vanished in the boundary on space. So, we can take this equation integrated by parts. Okay? And so, conclude that square root G zeta mu del square root G of, yeah, zeta mu del mu. And again, some of you, why was that integration by parts justified? Have you reminded of this? Why? Because? Because it's Gauss's theorem. Gauss's theorem. Gauss's theorem applies to ordinary Gauss's theorem. Because this, because of the formula that E mu is equal to one by square root G del mu square root G. And so, the term that, this and this differ by a term of this form. And this square root G cancels that, and that becomes a one. Then we can also say that if the integration measure is recognized square root G, g, d, and d4x, then even the coordinate derivative is, you know, as long as it's acting on, as long as it's acting on something with no other indices. If there were other indices in the game, then, and there were not all indices in the game. Then a formula like this would not work. There were three indices. So, this is specific to either having no free indices or having only anti-selection free indices. Good, yeah. But with that carry out, it's just true. Okay? And then we can kill it that therefore, by setting the coefficient of z down here, you know the stress density. Okay? And of course, this stress density will form part of Einstein's equations. Okay, now we, now we're going to, within the right or many part of Einstein's equations, what you get by taking Einstein's actually, the square root of g, r, r, y, and varying the strength of the equation. Any other questions? Any questions or comments? So, when you write this total action, whatever it is, when you say I mean, you vary with the strength of the model of the equation, and that implies that when you vary with the strength of the equation, that variable part would be zero. I didn't understand that. So, I mean, you say something about you varying with the strength of the model, if the equation for the model is zero. And then when you write in this reading, when you substitute the equation for the model, then when you vary with the action, that is when you understand. When you vary, okay. So, let me just say that first, about you. You see, the first manipulation we did, so the full action was Einstein's term plus an average. So, s is equal to minus 1 by 65, plus 16, k, 65 k, square root g, r, r plus integral square root g. The first manipulation that we discussed had to do just with this. We take that term, vary it, by itself is generally called. We vary it with respect to we vary it with respect to the nature and the matter field, that obviously vanishes. But the variation, this is the only part that contains that. So, the variation with respect to the matter field also gives us something vanishing at first order by equations of motion. First state of dynamism, second state of dynamism. Therefore, one of the solutions to the equation of motion, the variation of waste term by itself, with respect to the electric field, must vanish, provides that the variation of the metric is the variation caused by co-eventure. This is an example of a kind of a almost suspiciously slick kind of raising. There is often use with quantum field theory in great effect. It's good to get used to. It's a very beautiful idea in my opinion. Therefore, this vanishes. Now, you take the matter problem, that's the first part, second part is the definition. The definition is take the matter problem to the ground, vary it with respect to the metric. The co-evation of that variation, apart from a factor of square root g, is the definition of stress. Put these two terms together. Now, put these two facts together. That's the definition of this one. For an arbitrary c-thumb. Then the integration of that power is the definition of this T union, that has a covariant derivative of the stress, that's the definition. Now, what we're going to do when we try to find Einstein's equations, we're going to vary the metric in the whole part of the equation, in everything. And the variation of the metric will not be restricted to that variation that comes from the equation. We're going to look at arbitrary variation with respect to the metric. The whole action to derive an axiom. So, what is the action that is needed? One of the equations will be. So, delta s is equal to, one part we've seen, is equal to half delta g mu mu square root g d media. That was our definition. Okay. What remains? What remains from here? Oh, and we learnt last class how to take every time I have a symmetric variation that should be square root minus g. Sorry. Okay, last class we learnt how to take the variation of the determinant of the metric. Okay, we have this nice formula which was delta of square root minus g is equal to minus g mu mu minus delta g mu mu minus delta g mu mu times square root mu. So, this is just at the side delta s, the formula for delta s. So, we get minus half delta g mu mu g mu mu times r square root minus g. That came from bearing the square root minus g. And then we remember that r itself r mu nu is plus delta g mu nu r mu nu square root minus g. Finally, we have plus delta r mu nu. Let's express the terms of delta of the metric. Let me just connect the terms of the equations putting these terms together. I am going to ignore it because as I have now proved to you that variation vanishes up to the end. Okay, so let's just see what we get from the equation, ignore it that for a moment. What we get is r mu nu the coefficient of delta g mu, so r mu nu minus half g mu r half g mu nu. This part came with the 1 by 16 phi k is equal to 0. Except that we had a minus sign in this part of the action. So, this whole thing is 1 by 10 to the minus. Okay, and therefore we have the equation r mu nu minus g mu nu by 2 r is equal to 8 by k This is our next equation. Equation of the determinant how the metric field is done with this equation. The stress. But before we've clearly derived minus sign equations, I have to show you that this term is a total level. So that's what we're going to turn to now unless there are questions. So let's conclude that the equation of minus sign equations I want to show you minus g times delta of r mu. To do this, I'm going to use because this is a total level. To do this, what I'm going to do is to use the kind of arguments that we started using last class. Slicker arguments than just doing the algebra completely mindlessly, just to ease things. And the style of the argumentation is always one of the reasons. If you're trying to compute a tensor and go to a coordinate system, which makes your i easy, use the fact that it's a tensor to then take the answer from that easy coordinate system and apply it to g. Now you see, as you know there's a formula expressing r as a function of Christopher's and Mu's and his derivatives. So the variation of r can be expressed in terms of the variation of the Christophas symbol. A Christophas symbol by itself is not a tensor. But the difference between two Christophas symbols is a tensor. Now how do we know that? Well, we know that first the algebra is here. When we derive the transformation properties of the Christophas you remember we had a vergingy as far as and an inner vergingy as far as that just depended on derivatives of x by x by x. This was true but that transformation property was true in respect of what the matrix was. Which is an absolute different transformation. So suppose you have a Christophas symbol built out of one matrix and a Christophas symbol built out of another matrix. We subtracted those two Christophas symbols. The inner vergingy just far cancels in the transformation. So the difference between two Christophas symbols behaves like a tensor. This is the algebraic statement of the fact and you can give a physical interpretation of this. Christophas symbol tells you how to parallely transport something here. Two different ways of parallely transporting gives you two different vectors at the same point. The difference between two vectors at the same point. Christophas symbol itself is not a tensor. The difference between two Christophas symbols is this. So the small variation of the Christophas symbol is a tensor. Why is that important? Because we're going to write the variation of our variable in terms of the variation of the Christophas symbol. But because it's a tensor we might as well go to a and we'll get an expression that's generally called a little bit. So we might as well go to a coordinate system in which the Christophas symbol is managed. So we go to we work at a point one point at which the Christophas symbol matches. And the variation of the Christophas symbol matches. That's important. But the Christophas symbol itself that you have mentioned in that one that you must leave out. Let's see how that works. So let's write down the expression for R. So R new new but first remember the expression for R was remember the expression for R and say A B C B was equal to can somebody help me now? Without looking here don't tell me what the expression is. Then new gamma A B minus then new gamma A B new plus gamma C gamma C B new minus gamma A C new gamma C X. This was the full remark. We want the regions. We don't get that but we do this contract first in 30 years. So we just said mu is equal to A that's A A the variation of this one with respect to a small variation of the of the Christophas symbol itself which will ultimately express the variation. Okay? Now choose these terms of obviously your contract in the spectral coordinate system of interest two of these terms of obviously your contract which are these two terms the first exactly because the variation of this multiplies gamma which has a special term. Okay? So therefore what we have is that we get delta of R B new is equal to del A delta gamma A B new minus del mu delta of gamma A A excellent and now we are interested not in just delta of mu new we are interested in G mu and etc. Okay? So that's G mu and delta of mu new so that's G mu mu new that's delta of mu new that's delta of mu new and that is equal to del A gamma A B new minus del mu gamma A B A times G mu well now this in this special coordinate system I claim the following let me check it if you if you define delta of mu okay that's well is equal to G I G gamma I G L I suppose we got that A A minus G I I L gamma I K I K then claim this quantity is equal to 1 by square root of minus G D by D D by B mu square root of minus G W okay this is going to be the final claim in an arbitrary system the point of the check is that in our coordinate system the square root of minus G is just a bit of a question so in our coordinate system what we have to check is that just del mu W mu okay is equal to this quantity and let's say that's true yeah that's obvious okay the point is that the point is that this del is not active G because in our coordinate system the first del is just G okay so you just take this and where is my I'm sorry where is the B B B should be replaced by this B should be replaced by this B mu okay so now I just take this quantity here I contract B okay so this becomes okay let's say but this is del mu of gamma A A B G B mu and this was del A of gamma A B mu G B these are the same statements we've written these now as terms of derivatives G point B okay now because of the tensorial nature where is my delta this should be delta sorry delta first of all because of the tensorial nature of of this delta of gamma in this coordinate system I can just replace ordinary delta 2s by covariant delta 2s and then the expression will be true in every course okay so what we've done is take this expression the variation and rewrite it as covariant delta 2 del mu of w mu for some w mu w mu being the sum of this current as the covariant derivative of some vector field we have this lovely expression for covariant delta 2s which was square root of G del mu square G that was the square root of G on the side this cancels this this is D mu this is now an ordinary derivative and we integrate by parts so that tells us that there is no local contribution to the variation of the action from the G mu delta is this clear sorry I said it's more confusing it is clear I'm clear that anyone can go through and also steer into this negative surface you get a surface integral of omega mu omega mu will develop delta gamma mu and in obtaining the equations of motion we always assume that variations vanish at the moment that's how the principle of this action on the street you look at variations that vanish at a boundary and then the action should so this just does not contribute to the action as to the variation of the action at all with the right so this term does not contribute to a term in Einstein's equations and therefore the equation may not down on the board but it's a full Einstein is this clear? ok so this is a significant minus term of the course we have finally written down the correct equation that comes the equation that comes in questions and comments about this immediate comments are in Einstein's equations what would remain plus plus plus the equation that we wrote down was r mu minus half r t mu independently derived the t mu mu that independently derived the t mu mu is covariately observed so if we take this equation in the contract that are both sides then mu we get z on the right so this is equal to it must be true in the left hand side of the equation the left hand side vanishing come down from a new equation of motion because if we already have all the equations of motion none of the equations of motion are third order so basically the only way this could work is if it was the derivative of a bold equation or if it was an idea the way it works in Einstein's equations is that it's an identity this in fact is obtained by contracting the Bianchi identity as we explained last class okay so the fact that it's this combination rather than just r mu mu is something that you could have deduced just by knowing that it's this combination that is covariately observed what you're saying here is that once you know another thing to say is that once you know that Einstein's equations take this form then you don't have to separately assert the conservation of the stress energy tensor it follows from the equations itself I take the covariate derivative of both sides that the thing on the right hand side which we are interpreting as the matter stress tensor is covariately conserved now this is an interesting statement because conservation of the stress tensor in the normal field theory results out of the equation of the motion once you know the form of the stress tensor of the matter of the conservation just by writing down the Einstein's equations you have the certain part of the equation of the motion for the matter then with that part that is sufficient that part that is needed for conservation of the stress tensor you are not searching the full equation of motion because there is more to the equations of the matter field than the part of the motion of the stress tensor we asserted some part of the equation of the motion however there is one form of matter for which the only equation of motion is the conservation of the stress tensor and that form matters of fluid and the fundamental level Einstein's equations are never wrong the thing on the right hand side is the stress tensor of some field and conservation of the stress tensor for no field if conservation of stress tensor is the full content of the equation at the fundamental level but macroscopic if you know you have macroscopic objects of fluid the equations of fluid dynamics are simply the statement of energy momentum conservation supplemented by an equation of state in the form of the stress tensor as a function of a tableon idea so for a phenomenon for fluidity because for fluidity right give you a new example epsilon plus pressure now we are using minus dels of deluses plus pressure times u sorry of right there we have selected u plus pressure times we are using space is negative so minus g and we think time is positive so plus you write down the stress tensor for a fluid for instance in this form which in the rest frame tells you that in the time component you have the energy tensor and in the space directions you have got an isotropic pressure this is the projection of space removing the time part ok so this is the now this is not complete unless you attend your body as long as it tells you some sort of equation of state so you need to know some information about fluidity but once you have the only dynamics of fluid dynamics is the assertion of oscillation of this stress so for the particular case of general relativity opposite to a fluid all you need is Einstein's equations supplemented by constant of this form for the fluid that the constant relationship tells you about pressure in terms of energy or what energy and pressure both are in terms of temperature ok and then these just Einstein's equations gives you all the dynamic equations you don't separately have to impose conservation stress principle is not right ok so in that case some constant relationship of this form gives you the full equation of motion of the system ok this is a very special case but it's an important special case because one we will use a lot of cosmology deals with the evolution of the universal large scales and then you don't care what the fundamental feels you treat all those things as fluids ok so this is this is the this is the structure of Einstein's equations in this special case that we need to include in general of course you need to supplement Einstein's equations for the equations that matter so for instance if you are dealing with Einstein's equations you have Einstein's equations that you do not get in terms of equations which we wrote down in two or three classes and these two equations together form the full set of equations ok so this was the first comment the second comment is about the structural nature of Einstein's equations ok and just to lead up to this comment I'm going to tell you about similar things in the studies just you and me so for the next five minutes we're going to retreat to a class of electronics and to study the structure of the equations of Maxwell's equations so let's remember the structure of the equations of Maxwell's equations the equations of Maxwell's equations which are constants that we have about this so what are these? in the case of fluid yes then like when you query with respect to the model fields and get the equations for the model fields shouldn't we always think of the fluid fields yeah you see the fluid dynamics don't follow from varying action not in a known way this is an interesting question that I'm actually working on it must be something they have a simple answer to your question that's not the way how you get the equations you know it's an effective cost grade equation they really come by cost grading the equations of the fundamental matter fields that do follow from varying action yeah it's an excellent question let's let's retreat to the study of Maxwell's equations to dynamism these are the equations of Maxwell's equations okay looking at these equations you might think now the question I'm going to ask is what is the initial data if I wanted let's suppose for a moment let's suppose for a moment that J nu was a given specified field I'm asking the question how much initial data do I need to completely specify the evolution of the of the electrodynamic field and all times of the future the naiveest case might be suppose you just have a minimum of a scalar just a scalar field okay and you ask how much data do I need to specify just to set the normalization I'm going to call that two pieces of data because you have to specify two functions and initial specifications namely the value of the scalar field and the value of its first time that is the data that specifies the future of the future in this case but there are two pieces two and infinity two functions okay now if you're completely naive you'll look at this and say well there are four scalar fields so presumably I have to specify eight pieces of data okay in order to to specify the future evolution of this this is what does Dr. Lee know okay now namely you might say well well well we know of these four functions one of them one of them is gauge so maybe the answer is I have to specify three pieces of data I'm sure this is a secret work the gauge in which he said one one three zero and the remaining three okay the answer is even less than three actually we answer two so okay but the answer is even less than three as we're going to see now by the following device look of this equation so what I'm going to do you know by by data I will function in space so by that statement okay because I've chosen my slice to make a lot of t equal to constant okay so there is one special equation there is an equation in which mu takes the value zero let's start with that equation that equation is del mu f mu is zero it's easy first thing to notice is that mu runs over all values but the part where mu is equal to zero is zero because f is zero zero so this is actually sum over i del i now f i zero has time difference it's a one there it has of course one there del i is a purely spatial it makes so all these equations are only one time there except all these terms of this equation so this term is an equation in terms of time a single time derivative will no time derivatives there are no terms of two time derivatives moreover which fields time derivatives appear in this function does a zero time derivative appear but only as del i is del zero of a i it is so this thing depends on a i dot a zero by itself and it's spatial derivatives again and that's it and of course a i it does not have even a single time derivative this immediately tells you that there is even less data than the three pieces of data why is that? because this relation should be of any time slice it's an equation relating bits of what you might otherwise think of as independent once there is some number less than three okay now we are going to find that number even before we do that there is another little exercise with this equation the exercise I wanted to do was the following look so what we've learned is this normally in a classical system you specify data and dynamics are opposite what we've learned is that the zero equation is not an equation that actually evolved it's an equation that constrains what data you're allowed to start where in phase space you're allowed to start the i equations on the other hand have genuine time derivatives of that double derivatives everything of the ai fields those are genuine dynamic equations so the study of Maxer equations you break up the equations into two parts if you're interested in that the zero equation is called the constraint equation it's a household constraint equation and the the i equations are called the dynamic equation something that you might worry is about you might worry suppose I've set the constraint at once once once once and I use dynamics to evolve my constraint in data wouldn't these are a one-to-contradiction wouldn't be that the data that I get in future times is not to obey the constraints of that time space if so these equations will be inconsistent because that's not the case you might you see because all data is not allowed that's what the fourth equation is saying the zero equations so now if you start with allowed data dynamic should evolve you to allow data otherwise you have trouble what will ensure that's true what will ensure that's true is if del zero of the constraint was zero by the dynamic equations let's compute del zero of the constraint del zero of del mu f mu zero okay which was del zero of del i the dynamic equation the dynamic equations are del mu of f mu i is equal to zero so that now I break this up into the zero and the j part so that's del zero of f zero i plus del j f j i is equal to zero so if I've got a term that's del zero of f zero i can replace it by del j of f j i by taking the del zero through here I've got such a term so I've got del i with a minus sign of del j of f j i which is equal to zero because of the axis of f j i you know exactly what we want to prove that is that the time derivative of the constraint equation vanishes by use of the dynamic equations so that if you take the data and it obeys the constraint in one time space then it will always obey the constraint so we've learned many things we've learned that Maxwell's equations have two types of equations that are the dynamic equations of the constraint equations constraint equations constraint data and so reduce your degrees of freedom by less than three the dynamic equations evolve data but evolve it in a way that has consistent constraints now what we have to do what's left to do is hunt down how many degrees of freedom you know how to do that convenient to fix to a particular age okay I'm going to do it in one convenient gauge but in the exercises I'll make you do I'll request you to do it in seven different regions just to make sure you see that you get the same answer okay and the convenient gauge most convenient gauge that we would call the coolant gauge in this gauge let's do it we fix this gauge this you can always do by the way because if the value is not zero then you do a gauge transformation the equation you have to solve on the gauge transformation negative zero gauge for transformation is minus all the data and the Laplace is an inverted spatial so you can always go to this gauge okay so you go to this gauge and now in this gauge let's count the degrees of freedom so let's see what the constraint equation becomes in this gauge the constraint equation becomes del so what was it was del i fi zero is equal to j zero but fi zero was del zero ai of del i zero minus del zero ai you see that the del zero ai term functions because of the gauge because it's del i of del zero ai take the del zero del zero of del i ai to zero so all the convenient del i is zero so this equation the constraint equation comes del i squared a zero is equal to j zero which tells you that a zero at any spatial slice is determined by what j zero is doing in particular j zero is zero a zero would be set to zero Laplace is an inverted zero 9 degrees of freedom are there because we're completely exhausted the constraint equations okay we're completely exhausted the content of the right this the solution of this equation is unique of imposing reasonable boundary equation okay we're completely exhausted the content of the of the constraint equation the remaining two equations are dynamic equations either ground say there's no constraint it's just set it gives you second derivatives and time in terms of first other things they all vary so now how many degrees of freedom are there got rid of one degree of freedom by gauge fixing and the constraint killing the whole other is this clear is this clear it's less easy to see this in other gauges there is always a way and we'll set a couple of exercises we'll set an exercise please no problem we'll set an exercise for you asking me to do this just to get the hang of this it's a very simple linear system of equations actually this is like why does that give up a degree of freedom because now what's left what are the remaining fields there's only AI but AI doesn't stream such that AI is equal to zero this gives one degree of freedom one less function in the middle of a course of general relativity you don't need any geniuses to guess that the same things can be something very simple let's get a bit through the mindset okay let's see how that works so that's an exercise we're going to analyze the structure of our equations equations are going to give us that's a drop of L and let me write to this it stands equations you can also divide it up into dynamic equations and constraint equations it turns out that there are four constraint equations and the remaining equations are like this the four constraint equations are any of the equations with any factor of zero at all in this way I'm writing it zero on the top so I'm clarifying that R is zero zero minus half writing it like this this is delta v zero so R is zero I by j t zero i and R is zero zero minus R by two R is equal to 8 by k t zero zero these equations are three plus one and three but I mean all in this system it's patient yes we need any choice of coordinates actually so what am I going to show you I'm going to show that actually it doesn't matter which one so I choose one particular coordinate whether it's the angle of space or something like that I'm going to show that these equations have no terms of second derivative in this coordinate no no this is g of first zero lower which is delta zero that's part of the reason I do I went to do that to understand this it's worth taking a little time to understand this so I'm going to remind you of our expression for the second derivative you remember in the last class we developed a nice and more convenient expression for many purposes for the curvature lens that part had four second derivative cases and two parts that were products of darkness do you remember this right I'm going to write down the part that was a second derivative part for convenience which is iklm iklm is equal to 3 2 by g i x okay that's the commas g i m comma k l plus g k l comma i m g i l k m comma i just using this expression I want to show that there are no terms of second derivative in time let's start with r 0 so let's remember what was r 0 y was let's first take r i i k l m like we have here then we make it an r so we contract to make k m and we take this l index alright l is a capital right where you expect this l index and where you set it to 0 I take this expression and say g k m g l z here the i index is understood to be spatial so I want to check whether there are any terms here that have two derivatives in time but clearly this guy and this guy don't because they are i n this is what remains these two values okay now the only term with two is i l i now the only term with the time derivative here is the expression in which k takes the value 0 and l takes the value 0 okay so there is such a term it's i m 0 m w 0 m g okay that term is there now let's look at the same thing here once again the only expression in which you get both 0s when both k and m are 0 so both k and m are 0 we get minus g i i l w g 0 0 g 0 l but l is the value index these two terms get so we show that this expression have second derivative I am going to leave it as an exercise video to check but it also has little terms that I can't in fact there is a sick way of arguing this the pionki identity equation the equation that pionki identity which was the equation that l mu of r mu minus of r consider that equation now this equation is l 0 of r 0 mu let's go so r 0 mu minus r del mu so what was I doing I am just expecting this so I am going to add l and this was just minus r that's del 0 mu is equal to minus and then the i part so del i r i mu l i I have just taken that equation and expanded it out this is the term with del 0 expression notice the term with del 0 is precisely the the pieces that we want I am sorry I am sorry I did this wrong so this term contributes this term exists only when nu is equal to 0 the term is exactly the del 0 of the equations of model that we wanted to focus on the constraint equations what does it tell you the del 0 of the equations of the constraint equations is del i of dynamite equations but we know that the dynamite equations are no more than two directions of any in particular no more than two dynamites we add to the del i of them that doesn't increase the number of time derivatives so the right hand side has at most two time derivatives it does that today higher at most 10 derivatives but there is an explicit time derivative because del 0 has a factor which is just easy therefore there is if this term here had an explicit time derivative this would make the third one which would contradict the fact that we know that the right hand side is no higher than second order of that this is a sleek argument for the same conclusion that the constraint Einstein equations are at most first order of that this is what I mean for this story if the del 0 is like you are prevailing on the good shoes say it again my basic question is when you do then the conservation of the secondary derivatives can it be in some sort of variation the question is is the fact that the stress tensor is conserved in the heart of people's strengths maybe there is a way of saying it's not our thinking of it at the moment there may be a way of doing it I'm not sure so ok so let's proceed so we concluded that just like the Maxwell equations the Einstein equations also can be decomposed into constraints now there are two things we can relegate to the problem one is the demonstration that we did just a few minutes ago now that's just like the Maxwell equations if the constraint equations are satisfied and one times less they continue to be satisfied at all and that this follows simply from the dynamic I mean the answer is basically it's more or less it's that narrative the constraint equation is related to very good ok the other thing that we're going to try to add so now that we've understood of the equations the dynamic let's try this the first part of that we should know that the second derivative that equation is actually 10 again 10 again what do you want to do see if you like now you're talking about the Maxwell if it's L, U, J, U, Z, U that just very much first part is say precisely what you want to say let me write down an equation that you want to prove so the relation that we have is R, A, B, C, D what does that mean? saying that we know that we know that and therefore it should be true just from that therefore it should be true just from that that which thing do you want to know use this to prove that that there are no type derivatives you know if you use that and energy exactly give you one constraint equation gauge there is a single constraint and that's equation 4 you might say that's because you would say no, no, no that's the problem gauge theory has one matrix value constraint equation that will give you 4 and 4 the path that was your matrix also has also involves derivatives it's sort of important I would have to think but it's not clear to me that you can then you can use the energy straight forward it doesn't seem to be giving the right answer perhaps there probably is a way to think about it deeply I don't see it excellent, good question fine now now let's do some naive counting and then the actual counting a way of actually understanding and the actual counting like the kind of thing that we did in the Coulomb gauge we like I'll try it later next class but firstly naive naive is the technique constraint or 10 fields or what we were calling 20 20 degrees of freedom okay gauge theory can be used to get 4 of the 10 fields so there are 6 fields or what you might want to call 12 degrees of freedom 4 constraint equations can be used to get rid of 12 of these degrees so you might think that what's left is 8 degrees 8 functions okay but this is the kind of wrong counting you could have done in gauge theory let's review how you might have done this wrong on gauge theory gauge theory might have said well there were 11 degrees of freedom okay let's let me say what I said in 10 of these 9 degrees of obtained fields get rid of 4 by gauge 6 and then constraint equations give you get rid of 4 functions so you might have thought 2 degrees of freedom so you might have thought you left with 4 degrees of freedom that's wrong the answer is you left with 2 degrees of freedom okay that's wrong in gauge theory how you might have done it wrong in gauge theory you might have argued there were 4 functions you get rid of 1 by gauge that means 3 and then you have to know half more from the constraint constraint is one equation so you get one function but when we actually did engage theory we saw that the constraint got rid of a full degree it eliminated all of these it completely solved for the easy the second thing was that I have to read your graph the constraints are sufficient to eliminate okay a full force at a degree of freedom so you lose 4 degrees of freedom by gauge and further 4 degrees of freedom full 4 degrees of freedom by constraints and you left with 2 degrees the number of degrees of freedom in gravitational field is the same as the number of degrees of freedom in the electromagnetic field and you can think of these as the 2 polarizations of the photo or the 2 polarizations of that as we will see but we study gravitational waves gravitational waves also have 2 polarizations we will see this properly in the next class and then as we go on through the course it's better okay so this is the final answer to the issue to what structure the initial learning problem of Einstein's equation is once you've done all this blah blah blah you can strain the seven dynamics to fill the words of the views of freedom associated with the 2 polarizations okay so I'll stop this general form of structural discussion in the next class we'll return to more practical matters after maybe a little more in this form and stuff but we'll return to this through the course to understand better the structure of these equations question what is the idea for the Maxwell's case let's do the nothing it's a very good question let's do the nothing in Hamilton's equation okay let's first do the nothing for Maxwell's equations in d dimensions then we'll do the nothing by Einstein's equations for Maxwell's equations in d dimensions ABU looks like it has okay so let's suppose we were in d plus 1 d space 1 so that our universe is 3 okay then the same logic that we used in the same logic that we used we answered 1 degree of freedom by gauge 1 full degree of freedom by constraint so you got d minus 1 this is the dimension of the vector of d minus 1 it's the little group of a massless particle in d plus 1 dimensions as some of you know okay now let's look at what we got now let's look at what we get from for the sine seconds so once again we have to stop now you're counting is d into d number of components okay d plus 1 and d plus 2 by 2 number of components in d plus 1 cross d plus 1 you get to know d degrees of freedom because of gauge and d plus 1 because of the strength so minus 2 in d plus 1 so this is d plus 1 number of components in the vector okay now I subtract d plus 1 degrees of freedom because of gauge freedom and now d plus 1 because of constraints so that's minus 2 okay so what we got so we got d plus 2 a whole thing by 2 so we take d plus 1 comments in d plus 1 into d plus 2 by 2 minus 2 so that's d plus 1 into d minus 2 let's first check this suppose we said d equals 3 that would be 4 by 2 so that gives you 2 that looks okay and now let's try to interpret this um you see this should be the dimension of the irreducible representation in which the gravitation transforms now the the gravitation transforms to some representation of s of d minus 1 which representation it's a symmetric traceless number of symmetric components in d d minus 1 dimensions in a d by 2 but then you have to remove the trace just check in two dimensions to see if all dimensions are correct d plus 1 d plus 1 of s of d minus 1 the little group is d minus 1 s of d minus 1 s of 2 so this is 2 in a 3 by 2 is a 3 minus 1 okay now the remaining exercise is to check these two are equal so let's work it out should be within our capabilities d squared minus d minus 2 by 2 and this is d squared minus d minus 2 symmetric places d by 2 is a particle in d every particle transforms in some use of the representation of its little group for those of you who don't know what representation of the gravitation transforms to symmetric places so this is the answer the answer to the number of degrees of freedom okay other question so in the next class we will use four of our considerations to actually solving Einstein's equations we write down the Schwarzschild metric we compute the orbits of planets we'll start having fun okay