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Â So in the previous lecture we have considered free gravitational waves and

Â the energy carried by them, energy flux carried by them.

Â In this lecture, we continue the consideration of the gravitational waves.

Â But we concretely consider waves that are created by

Â 1:16

and in this expression of course there are terms which are.

Â Of our order nh, normally your terms that I, in particular,

Â contain in the gravitational cell dame tensor.

Â And in here we contain only linear.

Â 1:35

nh terms.

Â And one comment is in order here,

Â that the gage condition here is in the agreement with the.

Â Conservation in linearized approximation, conservation of this energy

Â momentum tensor.

Â So, to solve such an equation one has to use so called green function.

Â Green function for the box operator.

Â This is down version right, probably I have to define that.

Â Boxes d t squared minus so

Â to solve this equation we have to use retarded green function for the operator.

Â 2:25

Retarded Green function.

Â So I assume that all the listeners of my lectures are familiar with the standard

Â course of special theory of relativity and classical electrodynamics,

Â where people study the standard electromagnetic radiation.

Â And hence they are familiar with such a notion as a green function and

Â how to find solution of such equations with the use of the green function.

Â So the green function solve for the box operator,

Â for the d'Alembertian operator solves this equation for

Â pi delta 4 (x- y).

Â Where x bar is just for vector, and y bar also.

Â And so (t and x vector), for vector.

Â So this is a product of 4 delta function,

Â delta (4) (x) is a product of four delta function for each component of x and y.

Â And solution of this equation should be known to those who are familiar with

Â electromagnetic radiation retired solution of this equation is as follows.

Â 3:37

That the function of tx- ty times

Â delta tx minus ty.

Â Minus x minus y.

Â Model is divided by x minus y.

Â Models.

Â So tech to function is a heavy site tech to function so tech of delta t.

Â Is equal to one when delta t is greater than 0 and

Â 0 when delta t is less than 0 and

Â before moving farther,

Â let me clarify the physical meaning of the returning green function.

Â So as one can see, from this delta function.

Â Retarded green function in fact, so

Â we have a point y for point y, so

Â this is space time p x1 x2.

Â So this is point y which is a source here.

Â And the green function basically gives you

Â the value of the field, which is sourced at the point y.

Â So there is a light cone which

Â has this point y as the, as a apex.

Â And we have, due to this these at a function this

Â 5:28

So basically if we have this function which solves this equation.

Â Then the solution of this equation can be represented as false.

Â So it's sin nu of T and

Â X is approximately equal to 4 pie

Â 4 copra integral over d3y.

Â T mu nu Y, t-

Â x- y divided

Â by x- y modulus.

Â So how did I obtain this?

Â Well basically T mu nu is equal, that solves this equation,

Â is equal to minus 16 pi kappa

Â integral over D four Y.

Â G retarded X minus Y,

Â T function of light.

Â And then I took one of the integrals of this.

Â There are four integrals here.

Â I took the integral over tx ty, one of the integrals.

Â With the use of this delta function.

Â 7:06

and we're going to use it.

Â So actually it physically describes the full insertation?

Â There is basically a source which is just some collection of bodies moving somehow.

Â So this is just a bunch of bodies which are responsible for

Â this energy momentum tensor.

Â And at each point x which,

Â there is an active creation of the radiation at every point.

Â 7:36

So there is a right corn starting from any point where the source is and

Â we can observe ration at every point

Â X created somewhere by

Â the a bunch of bodies that are responsible for this energy momentum tensor.

Â So this is a physical meaning of the obtained expression.

Â Well anyway, everybody who has studied Electro magnetic radiation

Â should know this, so, I just give a small command.

Â So now, we assuming the following situation that

Â 8:25

So now, I assumed that the region where the body where the energy

Â momentum tensor is not 0, it's compact.

Â So I assumed there is some volume in space.

Â Some volume in space where the body is not moving.

Â 8:42

There is motion, complicated motion of bodies.

Â So this is a region V.

Â And we want to observe the creation of radiation in a very far date, so

Â the region where Y takes values, Y is here.

Â And we want to observe the radiation that is created by this

Â particles very far away.

Â So size that models of fix,

Â so as an origin of the coordinate system somewhere here.

Â So we want to consider such that all that is a fix much greater than Y.

Â So we are in the so called wave zone.

Â Then, one can immediately delaying approximation,

Â neglect y in the denominator and

Â the citation is such that to neglect Y here,

Â Y here in numerator, one has to consider A so called non relativistic

Â motion of the bodies, if the motion is non relativistic one can neglect y here also,

Â then in such a situation we are basically considering

Â the following solution, let me write, well before writing it I should come.

Â 9:57

State that if we are very far from the source of the radiation

Â in that region TMU is 0.

Â So basically our side source homogenous equation.

Â The same kind of equation as we have been considering at the end of the previous

Â lecture.

Â And hence that solution has similar

Â properties to those which we haven't counted in the previous lecture.

Â In particular, one of the properties is such that it's traceless.

Â So, it means that H R for alpha is 0.

Â Free gravitational waves as we remember, are traceless.

Â As a result H R for alpha is 0.

Â 10:57

As we remember from the previous lecture,

Â as a result taking into account all these statements that I have made,

Â the solution that we are going to work with is as follows.

Â H I J is approximately equal to minus 4.

Â Kappa divided by molars of x that is due to the denominator times

Â the integral over the complex volume d 3 y

Â 12:05

So, we have obtained the following expression for

Â the radiation field, rotational radiation field.

Â I have to correct the last formula, which I have just obtained, a little bit.

Â 12:48

So let me draw the picture again, there is

Â a radiation moment defined by Y if the bodies are moving non relativistically.

Â So we neglect here the argument of Y.

Â And the observation moment is somewhere here.

Â 13:22

Again, this expression is obtained in non-relativistic approximation,

Â and very far away from the region where the energy momentum tensor is not 0.

Â Very far away from in the wave zone.

Â So to study in greater details there are limits of validity of these formulas,

Â and the way one obtains these formulas, one should consult the standard

Â electromagnetic radiation, consideration is very similar.

Â One can say, read that in Landau Lifschitz second volume.

Â 14:01

Anyway, so here the problem with this formula is T I J here

Â depends on the details of motion of radiating massive bodies and

Â we want, so it doesn't just depend on the mass distribution.

Â Of these kind of things.

Â So we want to do some manipulation to express this quantity

Â through the details of the distribution, not the details of the motion.

Â 15:31

k T k i.

Â Now in this expression we can make the integration by parts and

Â drop off the boundary terms, well as usual we drop off in this

Â lecture of course we ignore the discussion of the boundary terms which is a.

Â Separate subject.

Â But we assume that everything is vanishing, all the fields and

Â masses, flexes, not the flexes.

Â All the currents are vanishing at infinity.

Â So when we integrate bi-parts we obtain boundary terms,

Â which is at partial infinity, we drop it off.

Â And the remaining volume term gives us the following expression.

Â 16:14

So, continuing that equality we obtain the following situation that dt,

Â well here we can take dt take out from the integral.

Â So d c y y j d t, sorry, not d t,

Â no d t here d t is taken away.

Â Ti zero plus after integration

Â by part d three y t i j.

Â So, after integration

Â by part the derivative acts on this we get chronicle symbol which makes k equal to j.

Â 17:03

And we have that it is equal to this and then one can,

Â because t i j is symmetric on the exchange

Â of indices, as a result we have the following equality as well.

Â That the integral over d three

Â y of t i j is equal to minus 1/2

Â times d t of the following

Â expression, d three y.

Â Y j t zero i plus y i t zero j,

Â well, this is easy.

Â This is times.

Â Okay. So, let us move further.

Â Let us multiply now..

Â FInal, so we have also apart from this,

Â we have also t nu zero of y equal to zero.

Â Let us multiply this by y, k, and

Â y l and again integrate.

Â Take the integral, as a result we get

Â the following expression that two zero

Â is equal the following expression,

Â d three y, y k y l, d mu t mu zero.

Â Again doing the same here, as we did here.

Â So, expanding this, we expanded this expression like this.

Â So, we can expand this expression like this.

Â In similar manner,as

Â a result we get d t acting

Â on d three y y k y l t zero zero

Â plus integral over d three

Â y y k t zero l plus y l t zero

Â k After integration by path.

Â After integration by path in the second term.

Â So this is exactly the same expression as we get here.

Â 20:07

We obtain that h i j is

Â approximately equal to minus

Â two kappa over modulus of

Â fix times d t squared over d

Â three y y i y j t zero zero.

Â This expression already depends only on mass.

Â T zero zero is nothing but the mass density.

Â Now, having in mind that h i j is traceless we can

Â subtract from this expression it's trace.

Â 20:53

And as a result, we get the following expression,

Â final expression that we're going to use in the following.

Â Final expression for the radiation field.

Â Two kappa over three modulus of x

Â times q i j where q i j is a quadrapole mass moment,

Â quadrupole moment because it

Â is equal to the following quantity

Â d three y times ro times three y

Â i y j minus delta i j y squared.

Â 21:39

Basically, we multiplied this by three, divided by three, and

Â then subtracted from this expression its traceless path.

Â So here is double derivative.

Â So this is a final expression that we get.

Â 21:56

Now it can be explained on general grounds

Â that, why should we expect such an answer.

Â Remember that for the extra magnetic radiation we have a vector field.

Â It's partial component is a first derivative of a dipole moment.

Â 22:17

So it's not so unexpected that for

Â tensor field we have a second derivative of a tensor quantity.

Â The other way to observe it, of course it's a very heuristic and

Â vague explanation.

Â As a second reason to expect something proportional to quadrupple

Â moment in case of gravity is the falling situation.

Â In case of electromagnetic radiation we know that If the gyromagnetic ratio,

Â e over m, is equal to one, sorry, for such a system where this gyromagnetic

Â ratio, in the electromagnetic case, is the same for all charges.

Â Contributing to the radiation field, if it is the same for

Â all charges radiation field then dipole radiation and

Â magnetic dipole radiation are vanishing and the only contribution to the radiation

Â process in the electromagnetic case is quadrupole contribution.

Â But in case of gravity we have basically charge proportional to the mass because

Â what is charge for gravity it is mass, so this ratio for the gravity is also

Â always equal to one and the same for all sorts of radiating bodies.

Â And as a result, it's not such an unexpected fact that

Â radiation field in gravity is proportional to the quadrupole moment.

Â Now we going to use this expression to calculate intensity of

Â the gravitational radiation that is created by moving bodies.

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Â