[MUSIC] So we need to find what remains to be found to derive the Einstein equations of motion. What remains to be found is a variation of the matter action. That's what we need. Let us assume that matter action, being generally co-variant, so invariant on the general co-variant transformations. Is an integral, Of some Lagrangian density. This is Lagrangian density. For example, the action for the particle, which is proportional to the integral over ds, which is integral over square root of g mu nu z mu dot z nu dot over d tau. This guy can be represented in such a form with the use of the delta function. So it's an integral over d 4 x square root of g times, well, in this case, is g of x. An integral over d tau delta 4 of x minus z of tau, divided by square root of sub g of z. Multiplied by square root of g mu of z times z mu dot z mu dot. So this is just an example where Lagrangian density is this quantity. So we will encounter more generic matter actions a little bit later. So if this guy is like this, then the variation of the matter action with respect to the metric which is a variation of this guy. Over the metric is given as follows. It's just d4x d L over dg mu nu delta g mu nu square root of g plus L times variation of the square root of the modulus. Well this guy we already know. So using that expression, we can rewrite this as follows. As d 4 x square root of modulus of g. Dg mu nu minus 1/2 L times g mu nu times variation of g mu nu. Well, we will denote this quantity as T mu nu. Then this is, by definition, is = 1/2. Well, not this quantity. This we will call as 1/2 g mu nu. 1/2 integral over d4x square root of modulus of g. T mu nu delta g mu nu. Where t mu nu, t mu nu = 2 delta L/dg mu nu minus L g mu nu. So it is symmetric under the exchange of indices. And our goal now is to understand the physical meaning of this quantity. Who is this guy? We're going to understand the physical meaning of this quantity. So to understand the physical meaning of T mu nu, let us consider the following fact. Among the variations of the metric, there are those variations which do not change the Riemann tensor. Such variations which do not change the Riemann tensor. This is the first type of variation. And there are also variations which do change the Riemann tensor. Do change the Riemann tensor. These variations are physical, these are un-physical. Well, y physical, one physical. Physical means that these variations of the metric, they change the space time itself. They curve it, change the curvature, change the distances between different points in space time. While this variation just the corresponds to the change of the reference frames of the coordinates nets in the space time, so this variation is un-physical. It doesn't change the space time, it just corresponds to a physical changes. So let us concentrate on this type of variations. Let us consider what are those and we want to find infinitesimal form of these variations. Well, finite form of this variations is given by this equation, the transformation of the metric. The transformation of the metric is just as we know already is as follows. Well, this is not the metric, this is inverse metric tensor, so it's variation with upper indices is as follows. dx bar dx bar u divided by dx alpha dx beta. So we want to consider the following transformation that is bar of mu as just x mu plus small change of the coordinate system. This is a small vector field. Then, from this formula and this formula, we observe to the linear order in epsilon. We observe that g bar mu nu of x bar is approximately g alpha beta. So we care only about the linear order. So it's delta mu alpha plus d alpha epsilon mu delta nu beta plus This is just chronic symbols. Plus d beta epsilon mu. That just follows from applying this here. So what this is already g mu nu alpha x + d mu epsilon nu + d nu epsilon mu and this is just we denote as g mu nu of x + d mu epsilon nu symmetrized. So this is symmetrical relation of him, this is just the definition of this guy. And what remains to be done? We have to realize that this guy depends on x bar while this guy depends on x. So then g bar mu nu of x bar is just g bar mu nu of x plus epsilon, and approximately, this is just g bar mu nu of x plus d alpha g bar mu mu of x times epsilon alpha. Well then, the variation, the infinitesimal variation of the metric, under such a coordinate transformation, which is, by definition, is just the difference of g mu nu of x minus g mu nu of x. This is just approximately as we have found is just d mu epsilon nu minus d alpha g mu epsilon alpha. Which is just the co-variant derivative symmetrized of epsilon mu. So now we're going to use this variation of metric, which is un-physical. Because it follows from the coordinate change. It doesn't let, so under such a variation Riemann tensor doesn't change. So under this variation, the variation of the matter action, the variation of the matter action under such epsilon variations is exactly equals to zero. So this is just trivial relation so to say. So it doesn't let this fact, under such variations this fact doesn't lead to equations of motion. Now we're going to use this fact to draw a conclusions about the physical meaning of this quantity. So the general co-variance demands the following identity that under that infinitesimal general co-variant transformations the action, matter action, doesn't transform. Let us see what kind of conclusions does it allow to draw. So we have integral over the space time manifold over d4x square root of the modulus of g T mu and the variation of the inverse metric tensor under the infinitesimal coordinate transformation. These guys we have found, so this is nothing but 1/2 integral over d4x of the space time manifold, square root of g g mu nu times d mu epsilon nu symmeterized. Because this guy is symmetric under the exchange of n, and this guy is symmetric under the exchange of n, this is. So we have that this is equal to the integral over mu d 4 x square root T mu nu times d mu epsilon nu. Result, 1/2 and no symmetrization. Now we want to integrate by parts. Here integrate by parts. So to do that, we represent this as a difference between two integrals d 4x square root of modulus of g of the total derivative of T mu nu times epsilon nu minus integral over space time manifold d 4 x square root of g epsilon nu times d mu T mu nu. So this guy, being the total derivative, so we continue this Q, I, E, T. This guy, being the total derivative, can be reused to the integral over the boundary of the space time manifold. d, sigma, mu. T, mu nu epsilon mu. So the same kind of situation as we have encountered before in this lecture, and this remains the same. So we don't change it. d4x square root of modulus of g, epsilon new times d nu T mu nu. So it is a separate interesting storing to consider such coordinate transformation for which this guy doesn't vanish at the boundary. This is a interesting story and a separate exercise, etc., etc. But we are not going to discuss it in this course of lectures because we are just learning from the general theory of relativity from the very beginning. So we assume that the variations, coordinate transformations, vanish at the boundary. Coordinate transformations epsilon nu at the boundary is 0. So our variation do not change the boundary terms. So as a result, this is 0. And as a consequence of this singularity, for arbitrary epsilon inside the space-time manifold, to have the 0 for this to be 0 inside the space time manifold for any epsilon mu nu, we have to have that d mu T mu nu = 0. So this equation should clarify the physical meaning of T mu nu of Q mu nu. Because this is a co-variant generalization of the following law following from coordinate transformations. This law in special theory of relativity was the conservation of energy momentum tensor. So this guy is a co-variant form of the conservation of, so this is in space, and this is in arbitrary curved space, or in curvilinear coordinates, the same kind of law. So we have conservation law. So to say conservation law following from general co-variance. So this guy is nothing but energy momentum tensor of the matter under consideration. Now we're going to use this, combine all these observations to derive Einstein equations. [SOUND] [MUSIC]