[SOUND] [MUSIC] So in the previous lecture we have considered free gravitational waves and the energy carried by them, energy flux carried by them. In this lecture, we continue the consideration of the gravitational waves. But we concretely consider waves that are created by massive bodies, moving massive bodies. We have to solve the following equation. Box Si mu nu equals 2 16 approximately equals 2 16 pi kappa t mu nu and d mu c si mu nu equals 0. I remind you that si mu nu. As h mu nu minus half traceless part of this is h and h is actually alpha alpha trace part of h and in this expression of course there are terms which are. Of our order nh, normally your terms that I, in particular, contain in the gravitational cell dame tensor. And in here we contain only linear. nh terms. And one comment is in order here, that the gage condition here is in the agreement with the. Conservation in linearized approximation, conservation of this energy momentum tensor. So, to solve such an equation one has to use so called green function. Green function for the box operator. This is down version right, probably I have to define that. Boxes d t squared minus so to solve this equation we have to use retarded green function for the operator. Retarded Green function. So I assume that all the listeners of my lectures are familiar with the standard course of special theory of relativity and classical electrodynamics, where people study the standard electromagnetic radiation. And hence they are familiar with such a notion as a green function and how to find solution of such equations with the use of the green function. So the green function solve for the box operator, for the d'Alembertian operator solves this equation for pi delta 4 (x- y). Where x bar is just for vector, and y bar also. And so (t and x vector), for vector. So this is a product of 4 delta function, delta (4) (x) is a product of four delta function for each component of x and y. And solution of this equation should be known to those who are familiar with electromagnetic radiation retired solution of this equation is as follows. That the function of tx- ty times delta tx minus ty. Minus x minus y. Model is divided by x minus y. Models. So tech to function is a heavy site tech to function so tech of delta t. Is equal to one when delta t is greater than 0 and 0 when delta t is less than 0 and before moving farther, let me clarify the physical meaning of the returning green function. So as one can see, from this delta function. Retarded green function in fact, so we have a point y for point y, so this is space time p x1 x2. So this is point y which is a source here. And the green function basically gives you the value of the field, which is sourced at the point y. So there is a light cone which has this point y as the, as a apex. And we have, due to this these at a function this function is not 0 on the light cone, on the future light cone, starting from y from the point Y, the event Y. So this is at any point X laying on this light cone this function is not zero. So basically if we have this function which solves this equation. Then the solution of this equation can be represented as false. So it's sin nu of T and X is approximately equal to 4 pie 4 copra integral over d3y. T mu nu Y, t- x- y divided by x- y modulus. So how did I obtain this? Well basically T mu nu is equal, that solves this equation, is equal to minus 16 pi kappa integral over D four Y. G retarded X minus Y, T function of light. And then I took one of the integrals of this. There are four integrals here. I took the integral over tx ty, one of the integrals. With the use of this delta function. And. That's how I obtained this expression. So now we have this solution at our disposal and we're going to use it. So actually it physically describes the full insertation? There is basically a source which is just some collection of bodies moving somehow. So this is just a bunch of bodies which are responsible for this energy momentum tensor. And at each point x which, there is an active creation of the radiation at every point. So there is a right corn starting from any point where the source is and we can observe ration at every point X created somewhere by the a bunch of bodies that are responsible for this energy momentum tensor. So this is a physical meaning of the obtained expression. Well anyway, everybody who has studied Electro magnetic radiation should know this, so, I just give a small command. So now, we assuming the following situation that this is a physical meaning of this formula after the integration of a T, the physical meaning is not that obvious A bit less over this. So now, I assumed that the region where the body where the energy momentum tensor is not 0, it's compact. So I assumed there is some volume in space. Some volume in space where the body is not moving. There is motion, complicated motion of bodies. So this is a region V. And we want to observe the creation of radiation in a very far date, so the region where Y takes values, Y is here. And we want to observe the radiation that is created by this particles very far away. So size that models of fix, so as an origin of the coordinate system somewhere here. So we want to consider such that all that is a fix much greater than Y. So we are in the so called wave zone. Then, one can immediately delaying approximation, neglect y in the denominator and the citation is such that to neglect Y here, Y here in numerator, one has to consider A so called non relativistic motion of the bodies, if the motion is non relativistic one can neglect y here also, then in such a situation we are basically considering the following solution, let me write, well before writing it I should come. State that if we are very far from the source of the radiation in that region TMU is 0. So basically our side source homogenous equation. The same kind of equation as we have been considering at the end of the previous lecture. And hence that solution has similar properties to those which we haven't counted in the previous lecture. In particular, one of the properties is such that it's traceless. So, it means that H R for alpha is 0. Free gravitational waves as we remember, are traceless. As a result H R for alpha is 0. And hence we have such as situation that sign and use is basically H MU. And also there we have only spatial directed components. So only zero, components are H I J. As we remember from the previous lecture, as a result taking into account all these statements that I have made, the solution that we are going to work with is as follows. H I J is approximately equal to minus 4. Kappa divided by molars of x that is due to the denominator times the integral over the complex volume d 3 y times T I J which is a function of T. The same this is T and X. Function of T and X. This is the function and T and Y. So remember we have neglected the Y here in this expression and let's hope we get this solution. That's it. So we're going to work with this expression. So, we have obtained the following expression for the radiation field, rotational radiation field. I have to correct the last formula, which I have just obtained, a little bit. The correction will be obvious. So we have models of X here, integral over D3Y over the volume where the energy momentum tends there is no 0. And argument of T is actually T minus at model of X Y. So this is observation moment and this is a radiation movement. So let me draw the picture again, there is a radiation moment defined by Y if the bodies are moving non relativistically. So we neglect here the argument of Y. And the observation moment is somewhere here. And the moment of radiation is light, light distance away from here in this space time. Okay, anyway, so this the expression where we obtain. Again, this expression is obtained in non-relativistic approximation, and very far away from the region where the energy momentum tensor is not 0. Very far away from in the wave zone. So to study in greater details there are limits of validity of these formulas, and the way one obtains these formulas, one should consult the standard electromagnetic radiation, consideration is very similar. One can say, read that in Landau Lifschitz second volume. Anyway, so here the problem with this formula is T I J here depends on the details of motion of radiating massive bodies and we want, so it doesn't just depend on the mass distribution. Of these kind of things. So we want to do some manipulation to express this quantity through the details of the distribution, not the details of the motion. So we will do the following trick. Let us consider the conservation of this S partial part of energy conservation. So this conservation law. Multiply it by ij and integrate over dy. So as a result we obtain the following that Z We have the following expression equal to 0. So if we multiply by ij and integrate, we get 0. So it's yj d mu T mu y. And rewriting this as d 3 y y j (d t T 0 i- d k T k i. Now in this expression we can make the integration by parts and drop off the boundary terms, well as usual we drop off in this lecture of course we ignore the discussion of the boundary terms which is a. Separate subject. But we assume that everything is vanishing, all the fields and masses, flexes, not the flexes. All the currents are vanishing at infinity. So when we integrate bi-parts we obtain boundary terms, which is at partial infinity, we drop it off. And the remaining volume term gives us the following expression. So, continuing that equality we obtain the following situation that dt, well here we can take dt take out from the integral. So d c y y j d t, sorry, not d t, no d t here d t is taken away. Ti zero plus after integration by part d three y t i j. So, after integration by part the derivative acts on this we get chronicle symbol which makes k equal to j. So this is actually the expression that we have here. And we have that it is equal to this and then one can, because t i j is symmetric on the exchange of indices, as a result we have the following equality as well. That the integral over d three y of t i j is equal to minus 1/2 times d t of the following expression, d three y. Y j t zero i plus y i t zero j, well, this is easy. This is times. Okay. So, let us move further. Let us multiply now.. FInal, so we have also apart from this, we have also t nu zero of y equal to zero. Let us multiply this by y, k, and y l and again integrate. Take the integral, as a result we get the following expression that two zero is equal the following expression, d three y, y k y l, d mu t mu zero. Again doing the same here, as we did here. So, expanding this, we expanded this expression like this. So, we can expand this expression like this. In similar manner,as a result we get d t acting on d three y y k y l t zero zero plus integral over d three y y k t zero l plus y l t zero k After integration by path. After integration by path in the second term. So this is exactly the same expression as we get here. As the result taking, combining these two formulas together We obtained the following expression that integral over d3yTij is inus 1/2 double derivative over t of integral over d3. Y t zero, zero. Now, using this in here. We obtain that h i j is approximately equal to minus two kappa over modulus of fix times d t squared over d three y y i y j t zero zero. This expression already depends only on mass. T zero zero is nothing but the mass density. Now, having in mind that h i j is traceless we can subtract from this expression it's trace. And as a result, we get the following expression, final expression that we're going to use in the following. Final expression for the radiation field. Two kappa over three modulus of x times q i j where q i j is a quadrapole mass moment, quadrupole moment because it is equal to the following quantity d three y times ro times three y i y j minus delta i j y squared. Basically, we multiplied this by three, divided by three, and then subtracted from this expression its traceless path. So here is double derivative. So this is a final expression that we get. Now it can be explained on general grounds that, why should we expect such an answer. Remember that for the extra magnetic radiation we have a vector field. It's partial component is a first derivative of a dipole moment. So it's not so unexpected that for tensor field we have a second derivative of a tensor quantity. The other way to observe it, of course it's a very heuristic and vague explanation. As a second reason to expect something proportional to quadrupple moment in case of gravity is the falling situation. In case of electromagnetic radiation we know that If the gyromagnetic ratio, e over m, is equal to one, sorry, for such a system where this gyromagnetic ratio, in the electromagnetic case, is the same for all charges. Contributing to the radiation field, if it is the same for all charges radiation field then dipole radiation and magnetic dipole radiation are vanishing and the only contribution to the radiation process in the electromagnetic case is quadrupole contribution. But in case of gravity we have basically charge proportional to the mass because what is charge for gravity it is mass, so this ratio for the gravity is also always equal to one and the same for all sorts of radiating bodies. And as a result, it's not such an unexpected fact that radiation field in gravity is proportional to the quadrupole moment. Now we going to use this expression to calculate intensity of the gravitational radiation that is created by moving bodies. [MUSIC]