[MUSIC] So let me again clarify what we have made. We have obtained the falling sin at the beginning of this lecture, we have introduced up sell the tanzer t mu nu, energy momentum tensor for the gray, such that we have the following conservation law. And this guy has a tedious expression through the components of the metric tanzer. At the same time after that we have considered linearized perturbations in flat space time. So we have considered flat space time plus perturbations on top of it. In this approximation Einstein equations acquire the following form. 16 pi kappa exactly the same T mu nu plus t mu nu where this guy expresses linearized perturbations, so psi mu nu Is just traceless part of h so it's just h mu nu minus one half at the mu nu times h where h is the trace of h mu nu. So, this is just linearized part, and this guy contains no linear part of the metric tanzer. So non linear part. Non linear part describes the energy flux. So now we're going to consider this equation. This equation, in a sense, is very similar to the one we have in electromagnetism, where we have the following situation. 4pi/c U r two mu in the appropriate gauge, and we want to consider free gravitational waves. Free electromagnetic waves are obtained when the right hand side is 0 here, and free gravitational waves are obtained when this part is 0. So no sources for the gravitational waves. And we first going to solve the equation with the right hand side equal to 0 and then we're going to consider this quantity for the gradational waves. That's we going to do now. Now similarly to this situation we want to look for the solution of this equation si mu nu equals so we solve this equation we want to consider solution of this equation in the following form, that h mu nu. h mu nu of x is just real part of epsilon mu nu times exponent minus iK alpha X alpha. For this, to solve this equation, here epsilon and K are some constant. They are each independent. K is real. Epsilon is complex. So, for quantity to solve this equation, K should- if we just flag it here, we obtain that K alpha times K alpha should be 0 well, this is just the consequence of the application of two derivatives on the exponent. And also, K mu epsilon mu nu, minus half K nu epsilon mu. Mu should be 0. This is a consequence. This equation is a consequence of gauge condition. I remind you that the gauge condition, the condition in which we have obtained this equation is as follows. It's d mu times psi mu nu should be 0. So from this equation we have this. If we flag this here, we obtain this equation. Now we're going to solve these two equations in some particular frame. Let us choose This equation just says that we have a light like. This K is light like, which describes waves propagating with the speed of light. And so we want to choose such a frame, call it a frame in space, such that K alpha is just K,0,0,K. If we plug this vector into the exponent we obtain the following situation, the exponent just acquires the following form, it's k z minus t, t minus z which describes a wave propagating along the cert direction with the speed of light. So we just flag it here. Now using this vector k which solves this equation, we want to solve this equation. This equation we have 4 for each mu we have equation. Nu 0, 1, 2, 3. So, for new 1 and 2, we have the following from this equation for new 1 and 2. We have the following situation, that Epsilon 01 is equal to epsilon 31 and epsilon 02 is equal to epsilon 32. At the same time, the sum of mu = 0 + mu = 3 components of this equation for this vector implies that epsilon 11 + epsilon 22 = 0. After this, using these equations already obtained relations between components of epsilon tanzer which is a polarization tanzer for the gravitational wave. Mu = 0 component of this equation implies that epsilon 0 3 is = to 1/2 epsilon 0 0 plus epsilon 3 3. Now let us perform, remember that this equation was obtained in this gauge. But there is a remaining gauge freedom that we can transform. We can, and transform h mu nu according to the law h mu nu plus d mu epsilon 0 mu such that epsilon 0. New, solves the equational box of that is equal to 0. So it's a harmonic solution of the box operator. So now, then the solution of this operator can be represented like this, at the solution of this equation can be represented like this, if S minus i Epsilon new. Well, this is a function of X but this is already constant, real path of course, over this quantity minus i. k alpha x alpha, where k alpha solves this equation. So represented like this, it does solve this equation, if k alpha obeys this quantity, obeys this relation. So under such a transformation, epsilon mu nu transforms like this. Epsilon nu minus k nu minus epsilon nu minus k nu epsilon nu. Well this is just the [INAUDIBLE] law for this epsilon. Now, if you choose that epsilon one is equal to. This epsilon 1 is equal to epsilon 1 3 divided by k of this epsilon, of epsilon mu nu of this epsilon. And epsilon 2 is equal to epsilon 2 3. Over K and Epsilon 0 is epsilon 00 over 2k and finally Epsilon 3 equal to Epsilon 33 over. 2 k. So if we fix this epsilon through this, components of this term like this. As a result we obtain, we achieve the following goal, that epsilon13 bar is equal to epsilon bar 2 3 is equal to bar epsilon 0 0 and equal to epsilon bar 3 3 and all of them are equal to 0. As a result of these all relations we have the following situation that polizaration tanzer. Which is this quantity. Has the only none 0 components as follows. That's epsilon 1 1 bar is equal to minus epsilon 22 bar. And epsilon bar 12 is equal to epsilon bar 21. These are the only none 0 components of this tanzer. And they have these relations. So we are looking for free gravitational waves, so we are solving this equation with psi mu nu equal to h mu nu minus traceless part of h, basically. And we have been looking for the solution of this together with the gauge condition, the gauge condition is dim sin is equal to 0. And we have been looking for the solution of this equation in the form real part of epsilon mu where this is Tanzer, complex tanzer minus i k alpha x alpha and k and absolute r constant x independent, k is real absolute is complex. So, what we have found that After fixing this gauge and using remnant gauge transformation, you do harmonic solutions of right there. We have found that this tanzer has no 0 components epsilon 1 1 we meet, we do not use bar on top of epsilon so we consider already the concrete gauge and the need of the bars. 21, as a result we have the following situation that the metric tensor the metric for this gravitational waves as the following form. So it's a linearized solution of Einstein equation. This is linearized approximation to the Einstein equations in the vacuum. (1+h11)dx^2- 2h12dxdy- (1-h11 d y squared- d z squared. So this form follows from the fact that due to this, these are the only known 0 components of this tenzor. So the only known 0 components of this guy are 1 1, 2 2, 1 2, 2 1. As a result Curvature of the metric is as follows. So we have the following picture basically so in space, in space coordinates so if this is z direction this is y and this is x. We have a gravitational wave which propagates along this direction. Along z direction. This k is directed along z I remind you, and what it does, it does affine transformations over the transversal directions. The reason for that is the following, that h11 If we plug this here and keep in mind that this guy is complex. So, h11 is equal to Modulus of absolute 11 cosine of k times Zet minus t plus pi 0 And h 1 2 is models of epsilon 1 2 times cosine of k times z minus t si 0 where this are the initial phases which are hidden in this complex. Quantities. So this sense here, so this guy solves the linearized approximation to Einstein equations, and what it does, it describes a propagation of the wave which affinely transforms, so it's like compresses one direction for appropriately chosen, if we choose phi 0. Equal to 0 and psi 0 equal to pi over 2, this is a picture we obtained. They're like half a period, like goes along this direction, compresses in one direction and expand in the other. And then compresses in this direction and expands in which was previously compressed. And so this sit he situation we have for the gravitational leaves. For the gravitational leaves. Now let us see what happens with T nu. So we want to plug this quantity obtain quantity into so we want to consider the formation on the right hand side. We want to consider T mu nu. Which is as we remember is of the order of H square. So to find it one has to use that tedious complicated expression for t mu nu through the components of the metric stanza. And I remind you that there was g mu nu, this kind of quantities. And in our linearized approximation, it's approximately Y value equal to g mu nu comma alpha which is approximately minus h mu nu comma alpha. So the only non-zero contribution to T mu new as far as Rho is relations is obtained follows from the term like this, g mu alpha one has to look for the expression for this t mu nu through the components of the terms. G mu alpha, g mu beta, g gamma. There's the g sigma psi. g gamma psi comma r for g delta sigma comma beta. And this is approximation is approximately equal to one-half. H alpha comma mu beta h beta comma nu alpha. As a result the only component of, the only contribution to t mu nu in this approximation is approximately. For the plane waves, it's approximate to three P kapa H Alpha Beta comma new H Beta Alpha comma new. And for this particular wave, which propagates along the third direction and his this kind of form, the only nonzero component of this quantity is t 03, and it is approximately equal to 16 pi Kappa times age dot. Squared 1 2 + age dot squared 1 1. So this is, dot means remember H11 and H12 are functions of this difference. Dot means just differentiation with respect to it's argument. So. What we have is that along the direction of propagation of the gravitational wave, along the said direction, the reason energy flux, that's exactly the point we obtain for the gravitational wave, gravitational wave carries energy on top of the fixed background. So, we have a background metric, flat metric and we have perturbation of this, this perturbation curved space-time, and propagates with the speed of light, and carries energy gravitational energy along this direction. That's the reason we call this as gravitational waves, and that's what they do. [MUSIC]