0:13

So let me again clarify what we have made.

Â We have obtained the falling sin at the beginning of this lecture,

Â we have introduced up sell the tanzer t mu nu, energy momentum tensor for

Â the gray, such that we have the following conservation law.

Â 0:38

And this guy has a tedious expression

Â through the components of the metric tanzer.

Â At the same time after that we have considered linearized perturbations

Â in flat space time.

Â So we have considered flat space time plus perturbations on top of it.

Â 1:14

exactly the same T mu nu plus t mu nu where

Â this guy expresses linearized perturbations, so psi mu nu Is

Â just traceless part of h so it's just h

Â mu nu minus one half at the mu nu

Â 2:01

This equation, in a sense, is very similar to the one we have in electromagnetism,

Â where we have the following situation.

Â 4pi/c U r two mu in the appropriate gauge,

Â and we want to consider free gravitational waves.

Â Free electromagnetic waves are obtained when the right hand side is 0 here,

Â and free gravitational waves are obtained when this part is 0.

Â So no sources for the gravitational waves.

Â And we first going to solve the equation with the right hand side equal to 0 and

Â then we're going to consider this quantity for the gradational waves.

Â That's we going to do now.

Â 3:21

For this, to solve this equation, here epsilon and K are some constant.

Â They are each independent.

Â K is real.

Â Epsilon is complex.

Â So, for quantity to solve this equation, K should- if we just flag it here,

Â we obtain that K alpha times K alpha should be 0 well, this is just

Â the consequence of the application of two derivatives on the exponent.

Â And also,

Â K mu epsilon mu nu, minus

Â half K nu epsilon mu.

Â 4:04

Mu should be 0.

Â This is a consequence.

Â This equation is a consequence of gauge condition.

Â I remind you that the gauge condition,

Â the condition in which we have obtained this equation is as follows.

Â It's d mu times psi mu nu should be 0.

Â 4:27

So from this equation we have this.

Â If we flag this here, we obtain this equation.

Â Now we're going to solve these two equations in some particular frame.

Â Let us choose This equation just says that we have a light like.

Â 4:44

This K is light like, which describes waves propagating with the speed of light.

Â And so we want to choose such a frame, call it a frame in space,

Â such that K alpha is just K,0,0,K.

Â If we plug this vector into the exponent we obtain the following situation,

Â the exponent just acquires the following form, it's k z minus t,

Â t minus z which describes

Â a wave propagating along the cert direction with the speed of light.

Â 5:27

Now using this vector k which solves this equation, we want to solve this equation.

Â This equation we have 4 for each mu we have equation.

Â Nu 0, 1, 2, 3.

Â So, for new 1 and 2, we have the following

Â from this equation for new 1 and 2.

Â We have the following situation,

Â that Epsilon 01 is equal to epsilon 31 and

Â epsilon 02 is equal to epsilon 32.

Â 6:32

using these equations already obtained relations between components of

Â epsilon tanzer which is a polarization tanzer for the gravitational wave.

Â Mu = 0 component of this equation implies that

Â epsilon 0 3 is = to 1/2 epsilon 0 0 plus epsilon 3 3.

Â 6:57

Now let us perform, remember that this equation was obtained in this gauge.

Â But there is a remaining gauge freedom that we can transform.

Â We can, and transform h mu nu according to

Â the law h mu nu plus d mu epsilon 0

Â mu such that epsilon 0.

Â 7:36

So now, then the solution of this operator can be represented like this,

Â at the solution of this equation can be represented like this,

Â if S minus i Epsilon new.

Â 7:56

real path of course, over this quantity minus i.

Â k alpha x alpha, where k alpha solves this equation.

Â So represented like this, it does solve this equation,

Â if k alpha obeys this quantity, obeys this relation.

Â 9:14

So if we fix this epsilon through this, components of this term like this.

Â As a result we obtain, we achieve the following

Â goal, that epsilon13 bar is

Â 9:32

equal to epsilon bar 2 3 is equal to bar epsilon 0 0 and

Â equal to epsilon bar 3 3 and all of them are equal to 0.

Â As a result of these all relations

Â we have the following situation that polizaration tanzer.

Â Which is this quantity.

Â Has the only none 0 components as follows.

Â That's epsilon 1 1 bar is equal to minus epsilon 22 bar.

Â And epsilon bar 12 is equal to epsilon bar 21.

Â These are the only none 0 components of this tanzer.

Â And they have these relations.

Â 10:29

traceless part of h, basically.

Â And we have been looking for the solution of this together with the gauge condition,

Â the gauge condition is dim sin is equal to 0.

Â And we have been looking for the solution of this equation in the form

Â real part of epsilon mu where this is Tanzer,

Â complex tanzer minus i k alpha x alpha and

Â k and absolute r constant x independent, k is real absolute is complex.

Â 11:09

So, what we have found that After fixing this gauge and

Â using remnant gauge transformation,

Â you do harmonic solutions of right there.

Â We have found that this tanzer has

Â no 0 components epsilon 1 1 we meet,

Â we do not use bar on top of epsilon so

Â we consider already the concrete gauge and the need of the bars.

Â 11:48

21, as a result we have the following situation that the metric

Â tensor the metric for this gravitational waves as the following form.

Â So it's a linearized solution of Einstein equation.

Â This is linearized approximation to the Einstein equations in the vacuum.

Â 12:10

(1+h11)dx^2-

Â 2h12dxdy-

Â (1-h11 d

Â y squared- d z squared.

Â So this form follows from the fact that due to this,

Â these are the only known 0 components of this tenzor.

Â So the only known 0 components of this guy are 1 1, 2 2, 1 2, 2 1.

Â As a result Curvature of the metric is as follows.

Â So we have the following picture basically so

Â in space, in space coordinates so

Â if this is z direction this is y and this is x.

Â We have a gravitational wave which propagates along this direction.

Â 13:32

keep in mind that this guy is complex.

Â So, h11 is equal to Modulus

Â of absolute 11 cosine of k

Â times Zet minus t plus pi 0 And

Â h 1 2 is models of epsilon 1 2

Â times cosine of k times z minus t si

Â 0 where this are the initial phases

Â which are hidden in this complex.

Â Quantities.

Â 14:14

So this sense here, so

Â this guy solves the linearized approximation to Einstein equations, and

Â what it does, it describes a propagation of the wave which affinely transforms,

Â so it's like compresses one direction for appropriately chosen, if we choose phi 0.

Â Equal to 0 and psi 0 equal to pi over 2, this is a picture we obtained.

Â They're like half a period,

Â like goes along this direction, compresses in one direction and expand in the other.

Â And then compresses in this direction and

Â expands in which was previously compressed.

Â And so this sit he situation we have for the gravitational leaves.

Â For the gravitational leaves.

Â Now let us see what happens with T nu.

Â So we want to plug this quantity obtain quantity

Â into so we want to consider the formation on the right hand side.

Â We want to consider T mu nu.

Â Which is as we remember is of the order of H square.

Â So to find it one has to use that tedious

Â complicated expression for t mu nu through the components of the metric stanza.

Â And I remind you that there was g mu nu, this kind of quantities.

Â And in our linearized approximation,

Â it's approximately Y value equal to g mu nu comma alpha

Â which is approximately minus h mu nu comma alpha.

Â So the only non-zero contribution

Â to T mu new as far as Rho is relations is obtained

Â follows from the term like this, g mu alpha one has to look for

Â the expression for this t mu nu through the components of the terms.

Â G mu alpha, g mu beta, g gamma.

Â 18:03

So.

Â What we have is that along the direction of propagation of the gravitational wave,

Â along the said direction, the reason energy flux,

Â that's exactly the point we obtain for the gravitational wave,

Â gravitational wave carries energy on top of the fixed background.

Â So, we have a background metric, flat metric and we have perturbation of this,

Â this perturbation curved space-time, and propagates with the speed of light,

Â and carries energy gravitational energy along this direction.

Â That's the reason we call this as gravitational waves,

Â and that's what they do.

Â [MUSIC]

Â