[SOUND]. So, on general grounds, based just on symmetry and assumptions, what are dynamical degrees of freedom, we have proposed an action for the gravitational series? Now we're going to use this action and minimal, least action principle to derive equations of motion that fall from this action. So, to do that, we have to vary the proposed action, Einstein Hilbert, Einstein Hilbert action, with respect to the metric. So, we have to vary the falling quantity one, 16 Pi, kappa Delta g, and d 4 x, square root of g, g R plus ambda plus delta g of Sm matter. So, this is just arc, and this is variation of this quantity. Now, this is equal to minus 1 over 16 pi kappa Integral over d 4 x, and here, variation of the square root of the determinant metric multiplied by R plus lambda. Well, this is just this, so this is variation of the first. Now we're going to write a variation of this, and you will see. Square root of g multiplied by delta g mu mu time r mu. So, this is variation of this plus square root of modulus of g, delta R mu nu multiplied by g mu nu. And the remaining thing delta g of S M. So now, we want to find separately each contribution like this one after the other. So the bracket is closed here after that. Let us do first this. To do this, let us do the following. Let us do the variation of the log of the determinant of an arbitrary matrix M. So, in this chain of relations that I'm going to write now, we'll keep track only of linear terms with respect to the variation of M. Only linear terms with respect to this. So, this variation is just log determinant of m plus delta m minus log determinant of M. Well, before I move farther, I just realized. So, what we do. Of course, when I wrote this, I assumed that everybody knows how to write the right field action with respect to the corresponding field, but let me just say a few words. So what we do, we consider this action, and we want to slightly deviate the metric. So, we add to the metric small deviation. And then the variation of the action, of this full action, Einstein-Hilbert action, Is just the difference of s plus g plus delta, g minus s over g. And if we are in the vicinity of the critical point of such a metric that minimizes the action, this variation is equal to 0, that's what we do. So now, coming back to what we have been doing, I am looking for the variation of this, but I do a more generic thing. The log of the determinant of an arbitrary matrix M. So this is, just by definition, the same, and let me do the following. So, this is log of the determinant of M plus delta M divided by the determinant of M. Well, this is just the property of the log So this is equal to log, ratio of the determinants is the determinant of the ratio of the matrixes, due to the property of the determinant. So this is just then m minus first m plus delta m, and then I open up the brackets. So, this is log determinant 1, so unit metrics, plus M to the minus first power times M, like this. And then by the property of the such that log determinant is nothing but the trace log. So, this is then trace log over the same quantity. Delta M, and at the very last step we're just going to keep only linear in delta M term. So to linear order this quantity Is as follows. This is just approximately. To linear order, trace times M minus first, delta M. So let us apply this expression to the variation of the log of the square root of g. Well, this is just variation of the log of the square root of the determinant of g minuet. Well, by definition. But this is minus one half variation of the log of the determinate. Sorry, this is modless of the determinate of the inverse metric, g minue with upper indices. Well, applying this formula, this is nothing but minus 1/2 gmunu delta gmunu. So as a result, as follows from these formulas, we can find that what is this, this is nothing but a variation of the square root of the modulus of g divided square root of modulus of g. So from this expression, we can find this variation. So we have found the variation of this guy that square root of, variation of the square root of g minus 1/2 of square root of g. Times g minue, delta g minue. This term is obvious, it is already proportional to delta g minue, like this one. And now we continue with this term. So, to find this term, let us fix local mean co reference system, where gμν is approximately etaμν around the point At the point x and gamma nu, alpha nu is 0 approximately, they're all the same point. Then in the vicinity of this point in this reference frame, we have that delta r minue is equal to delta, a variation of gamma alpha mu. New comma alpha minus gamma of alpha mu, mu. So this is equal to d alpha, well, by definition, delta gamma alpha mu nu minus d. Nu delta of y two delta gamma mu alpha alpha. And in the local Minkowkskian reference frame, this is equal to d. Alpha coordinate derivative of delta gamma alpha nu, minus d nu delta gamma, alpha, mu, alpha. And now, one can see this quantity has tanza properties. It is a tensors, despite the fact that these two guys are not tensors. This is a tensor because no tensor plot cancels between this contribution, so we have a relation between this tanzarial quantity and this tanzarial quantity. Although this relation was obtained in this local reference frame, because this obtains quantities and transform multiplicatively. So, if this relation is valid in this reference frame, it is valid anywhere else, so this relation is true in any reference frame. From this relation From this relation, we have the g Mu times delta R Mu, which is exactly the variation of delta R, sorry, no, no, no this is not true. So, this quantity stands here, so this quantity is equal to d mew applied to g alpha beta delta gamma mew alpha beta minus g. Alpha, Mu, times Delta, Gamma, Beta, Alpha, Beta. So, this is nothing but a co-variant derivative of some quantity, Which we can call delta mu. This quantity, we call delta minue mu. Now, we're going to use this expression in here, under the integral. So just to stress again where we are in our duration. So we are looking for the first variation of the Einstein-Hilbert action and I want to put it to be zero. So, the first variation of the Einstein-Hilbert action is as we have been observing already is sixteen pi kappa Integral over d4x, variation of the square root of g times r plus lambda plus square root of G delta g mean minue times R minue, plus square root of modules of g times delta R minue times g minue Plus variations of matter action. So far, we have found that this guy is the following. It is minus one half. Square root of g times delta g mew so this proportion to this quantity. This term is already proportional to this quantity and has appropriate form for further study. Now, we have been studying this guy and this guy we found to be g mu nu times delta R mu nu is equal to D mu of g alpha beta gamma delta. Gamma r for beta mu minus g alpha mu delta gamma beta alpha beta. And this, we denoted as D mew times delta mew u. So, this variation is a total derivative as can be observed from this formula. So under the integral over the space time manifold, over the space time manifold this quantity, so this variation, Is equal to, D4x square root of g, d mu to the total derivative. Delta u mu, and because it's a derivative of integral of the total derivative, one can apply Stokes' theorem or Ostrogradsky-Gauss theorem. In this case, in curved space time, and then this is the integral over the boundary of our space time manifold D sigma mu times delta u mu. Well, this is where d sigma mu, d sigma mu can be written. This is vectorial quantity which is a vector, whose modulus is proportional to their area or volume of their three dimensional boundary of the four dimensional space time. And the vector is directed along the normal to this element So it can be written like this. N mu normal vector square root of the determinate of the three dimensional induced metric on the boundary. Induced from the metric [INAUDIBLE] in space time, times the parameterization of the boundary of space time. So, then the study of this boundary terms is very interesting and separate subject, but at this point, we can see there, in this course, we can see that the variation. Such variations at the boundary, the variation is 0. As a result, this term doesn't contribute. So, this term is just 0 after the integration by parts, it doesn't contribute to equations of motion. But the consideration of the boundary terms is separate interesting subject which is not covered in our course here. So as a result, combining all these things together, combining, so I assume that this is zero at the boundary of the manifold. So as a result, combining this variation, this, and this, we obtained that one has to put to zero the following quantity: variation of the matter, action, minus one 16 pi kappa, integral over d4 x, square root of modulus of g times, r mean mu, minus 1 1/2 g mu new times r minus 1/2 g mu new times lambda, and multiplied delta g mu nu. So, to complete our exercise, what we need to find Is this variation. [SOUND]