0:39

So where this guy box is the d'Alembert operator, so it's d mu d mu.

Â Which is the same as dt squared- Laplacian.

Â Which is the same as dt squared- dx

Â squared- dy squared- dz squared.

Â 1:21

Then the solution of this equation can be found.

Â Before saying that, let me show what happens here.

Â This equation is invariant under the shift of x.

Â If I simultaneously shift x and y this equation doesn't change.

Â Then it means that G is necessarily a function of the function

Â of difference, x- y.

Â 1:43

So if we know this function then the solution of this equation,

Â phi(x), is easy to find.

Â It's just integral over d4 y G(x- y)j(y).

Â Indeed, why do we know it?

Â Because if we act by d'Alembertian on both sides of this guy,

Â then d'Alembertian on the right-hand side acts only on this guy.

Â So then it means box of phi(x) Is nothing but

Â d4y box acting on x G(x-y)j(y).

Â But this guy is nothing but the delta function.

Â So continue this equality,

Â we have d4y delta 4 (x-y).

Â 3:02

So, we have product of four delta functions and four integrals,

Â integral of y0, y1, y2, etc.

Â So taking all these four integrals with four delta functions, we obtain j(x).

Â So we obtain this equation.

Â So indeed, this phi(x) solves this equation.

Â Now we want to find the solution of this equation.

Â This is so-called Green function for the d'Alembert operator.

Â 3:31

How do we do that?

Â Well we do Fourier transform of both sides.

Â First Fourier transform of the delta function.

Â x- y is just d4 p/2

Â pi to the fourth exponent

Â of ik mu (x-y) mu

Â resulting in division.

Â So this is the Fourier transform of the delta function.

Â Which is according to the rule that delta(x) is just dk d pi over 2 pi.

Â Sorry, here it's p.

Â dp e ipx.

Â So product of four delta functions is product of four integrals.

Â And the product of four exponents is just this guy.

Â 4:31

So this is three real.

Â This is a Fourier transform.

Â Now we want to consider Fourier transform of x-y.

Â We define it as d4p/2 pi

Â to the fourth sum G tilde of

Â P exponent of ip mu (x-y) mu.

Â Okay.

Â Now, this is just by definition.

Â So we take Fourier transform.

Â Now plugging this here.

Â Sorry, this here and this here.

Â We obtain the following situation.

Â The d'Alembertian acting on

Â integral over d4 p/2 pi to

Â the fourth G tilde(p) e ip mu

Â (x-y) mu = the integral over

Â d4p/2 pi to the fourth e ip mu (x- y) mu.

Â d'Alembertian acts only on this guy because this is the only guy who

Â depends on x.

Â But it's not hard to check that d mu d mu which is

Â d'Alembertian acting on exponent of e ip mu (x- y) mu.

Â Is nothing but

Â -p mu p mu e ip mu (x- y) mu.

Â 6:17

Then from here what do we obtain?

Â We obtain that combining everything on one side we obtain the following situation.

Â d4p/2 pi to the fourth

Â e ip mu (x-y) mu acting on p squared.

Â p squared is just this guy, p squared.

Â p squared G twiddle (p) + 1 and this is all = 0.

Â Because this is the expansion over the complete basis of functions,

Â this = 0 only if this = 0.

Â As a result we obtain that G twiddle

Â of p is equal to -1/p squared.

Â p squared is p mu times p mu.

Â 7:14

So that was actually the reason to perform Fourier transform.

Â Because Fourier transform allowed us to convert differential equation into

Â algebraic equation.

Â And the solution of algebraic equation is very easy to find.

Â What remains to be done is to calculate this integral

Â with the substitution there of 1/p squared.

Â If we do that, we are done.

Â 7:41

Actually there is one simple command.

Â This is a solution of this up to the addition of

Â some function C(p) which when multiplied 2 p squared.

Â Sorry, C(p) times delta p squared.

Â The B squared for any smooth function of p,

Â because delta p of b square multiple by b square is just zero.

Â Additional, this sides,

Â into here, gives us something which is specified.

Â Well you see the solution of this guy is always specified up to addition of

Â the solution of homogenous equation with the right hand side equal to zero.

Â And this term is Important in this discussion.

Â We will clarify this in a moment.

Â Okay, so we are looking for the solution of this equation.

Â 8:50

And already have found That the solution

Â of this equation can be represented as

Â d-four-k over two-pi to the fourths.

Â G-k-i-k-mu (x-y) mu and

Â with all that G tilde k is just 1 minus 1 over k squared.

Â Where k squared is k mu times k mu,

Â it's a full vector k mu, plus some smooth

Â function of k times delta k squared.

Â 10:07

Okay, where G zero is the same quantity where this guy stands here.

Â It is not hard to check which is my good exercise,

Â that this G0 of x- y solves homogeneous equation,

Â exactly due to the presence of this quantity.

Â But the solution of this equation always is defined up to

Â the addition of this thing, so in the following, we'll drop it off.

Â Because we will always drop off these kind of terms,

Â because again the solution of this equation is specified up to this thing.

Â Let me stress one thing, one thing.

Â Which appears here for this quantity.

Â We drop this off for this quantity.

Â This integral G(x-y),

Â integral of d k zero over two pi,

Â 11:06

integral of d three k over two pi cubed i,

Â k0, x minus y0,

Â minus ik, x minus y,

Â divided by k0 squared, minus k vector squared.

Â It is not have to see that this

Â integral from the mathematical point of view is not defined.

Â In complex K zero plane, in complex K zero plane, where this

Â is a real part of K zero complex K zero complex [INAUDIBLE] of K zero.

Â The contour of integration goes along the k zero axis.

Â And there are two poles on this axis.

Â It's minus models of k, of this k, and

Â plus models of k.

Â 12:08

So, there is a problem.

Â We cannot find this integral.

Â What should we do?

Â Well, this, in this case we have to use physical intuition, and

Â the fact that the solution of this equation is defined up to the addition

Â of this .Well, what do we expect to get?

Â We expect, what is the meaning of quantity?

Â What is the meaning of this?

Â To understand the meaning of this quantity.

Â Let us, for a second,

Â look at the following X minus y.

Â 12:56

And this guy is nothing but the Coulomb's law In physical terms.

Â And is.

Â We have a charge placed at the position y, and if we look further,

Â the field that it creates at the position x.

Â So it's like two point correlation functions so to say.

Â We have a subset y And

Â we look at the correlation that it creates at the point x.

Â So this is guy is similar, is a generalization of the citation.

Â So we have a source which is instantaneous in space and time.

Â This is a source at position in space.

Â This is a spatial problem, and here we have a space-time.

Â So we have space-time.

Â 13:40

So this is x0, which is t, x1 x2.

Â And we have a point y, and at this point we have a source, an instantaneous source.

Â So delta function can be assumed as if it is not zero at one point,

Â y And zero everywhere else.

Â 13:59

So this source creates something outside of this, the right hand side is zero.

Â So the solution of this equation describes something which travels with the speed of

Â light because this is a wave equation,

Â describing voice traveling with the speed of light.

Â So, the solution of this equation should be not zero on the light cone,

Â which has apex at the point y.

Â So it should be none zero only on the light cone, because it describes some

Â excitation which travels with the speed of light, and passing through the point y.

Â What do we expect?

Â On physical grounds we expect it to be nonzero only in the future light cone.

Â So only for the x's which are sitting here.

Â For the point x mu, which sits on the future light cone,

Â it should be nonzero, but it should be zero here.

Â Because You see the [INAUDIBLE] created an excitation, which travels with the speed

Â of light in future, towards the future but not towards the past.

Â So we want to look for solution of this equation, so-called green function,

Â which is not zero on the future light cone from the point Y.

Â 15:12

So, let me say that this statements

Â specifies how we define the contour of integration.

Â The contour if integration we define according to this rule,

Â we will see now that if we go around these pause according to this rule.

Â Then, this guy is 0, so if we define the contour here,

Â the contour of integration over k0, like this,

Â then we obtain the entire green function, which obeys the following property.

Â That it is 0 If X zero

Â is before than Y zero.

Â It is zero But

Â it is not 0 if x0

Â is after y0.

Â We will see that the specification of the contour has to do with this property.

Â And if we would choose a different contour,

Â that would correspond to the solution of this equation,

Â which is different from the one which we are considering by the addition of

Â something like solution of homogeneous equation.

Â So we have to calculate this integral G(x-y),

Â which is minus dk0 over 2 pi

Â integral over d3k over 2 pi cubed,

Â exponent of ik0(x0-y0) minus

Â i k vector (x- y) divided by k0

Â squared minus k vector squared.

Â And the contour of integration in the k0 plane, k0 plane, let me draw it.

Â So we have k0 plane.

Â 17:14

The contour of integration in this plane goes along the real line.

Â And it has poles here, -k and k,

Â where k is just modulus of this k.

Â And I told you that we have to specify the contour like this,

Â that it goes around the pole like this.

Â So our contour is like this.

Â 17:39

Now, why?

Â It's just because if the contour, let me call it C, goes like this,

Â this guy obeys the property of the retarded Green function, retarded.

Â The reason for that is following.

Â Now, consider that situation that x0 is before than y0,

Â which means that this quantity is negative.

Â If this quantity is negative, I can close the contour in the lower

Â plane by infinite hemicircle, half circle, according to this rule.

Â Why?

Â Because on this circle, the integrand is always 0,

Â because when this negative on this circle,

Â the imaginary part of k0 is minus infinity.

Â As a result, the exponent is 0.

Â And I add nothing to this integral over this contour,

Â I add nothing along this contour for this guy being negative.

Â The integrand is 0.

Â So, but this contour doesn't have inside any, for

Â this contour, the integrand doesn't have inside any singularities.

Â It is a holomorphic function of k0 inside this contour.

Â So according to the Cauchy theorem, for this case,

Â we obtain that GR(x-y) = 0 if the contour is defined like this.

Â Now, if x0 > y0,

Â this quantity is positive, and we have to call the contour in the upper plane.

Â And as a result, for the same reason as we have to close here.

Â But in this case, our contour encircles some poles, and

Â the integral is not 0 and it should be taken according to the Cauchy theorem.

Â 19:41

According to the Cauchy theorem, this guy, well,

Â it's a home exercise, along this contour,

Â this guy is equal to, so I mean this integral.

Â 20:00

I'm not going to give too many details.

Â This is a home exercise to take this integral.

Â It is equal to minus exponent ik,

Â this k, x0-y0,

Â minus exponent i,

Â divided by 2 ik.

Â And we should bear in mind that this is all true for the case when x0 > y0.

Â So, as a result, we have to take the following integral.

Â GR(x-y), is minus Heaviside theta function of delta x0.

Â Delta x0 is just x0-y0.

Â So this function is 1 if x0 is greater than and 0 if x0 is less than y0.

Â Integral over d3k, so plugging this into here,

Â what remains to be done is to take this integral

Â 21:25

Ik(x0-y0) minus

Â minus ik(x0-y0)

Â divided by 2ik.

Â So, how do we take this integral, this integral?

Â To take this integral, we do the following.

Â We use instead of k1, k2, k3,

Â we do transformation to the spherical coordinates in the k-space,

Â to k, modulus of k, theta and phi.

Â Where they're angles in the spherical coordinates.

Â And direct delta x along the third direction such that k

Â times delta x is just modulus of k times delta x times cosine theta.

Â 22:21

So what do we obtain then, then this integral.

Â Continuing it, let me continue it.

Â This integral acquires the following form.

Â So we have theta function of delta x0,

Â we have 2 pi cube here, integral from

Â 0 to plus infinity, dk times k squared,

Â and we have integral over d phi.

Â Nothing depends on the integral over the phi, so integral over d phi gives us 2 pi.

Â What remains to be taken is the integral over d theta is,

Â we have sine theta times d theta.

Â So this is d cosine theta from -1 to 1.

Â This is the integral over the d theta, times the integrand.

Â The integrand is just -ik,

Â modulus of delta x times cosine theta times

Â this exponent ik(x0-y0),

Â minus exponent -ik(x0-y0) divided by 2ik.

Â While taking the integral over the dk,

Â over the d theta, d cosine theta, with this,

Â we obtain the following theta function

Â of delta x0 divided by 2pi squared.

Â Here actually one of the K's cancels with K squared.

Â 24:09

What remains is a falling integral of DK.

Â DK times K divided by I minus IK delta X modulus.

Â Exponent of -ik.

Â Modulus of delta x-

Â ik modulus of delta x

Â multiplied by the exponent.

Â Well Exponent of ik delta x,

Â 0 minus ik delta x 0,

Â d ivided by 2i.

Â So, k cancel here.

Â Delta x can be taken out 2 together with i can also be taken out.

Â So what remains is just the product of 2 brackets.

Â 25:17

If we expand the brackets, we attain 4 terms,

Â product of this to this and this, and this to this and this.

Â But then, we can reshuffle these two terms.

Â These four terms, such that we obtain the following expression.

Â Delta X0 divided by 4 Pi, divided by models of delta X,

Â an integral from minas infinity two plus infinity DK.

Â So we shuffling in the following, we combine two of the four terms,

Â into the integral from zero to the plus infinity.

Â And in the other integral we change k to minus k, and

Â convert it into the integral from minus infinity to zero of the same expression.

Â So we have take that two four terms combined into two terms of

Â the integral from minus infinity to plus infinity I hope it's clear what I mean.

Â So the result is the following situation,

Â ik delta X0 + models of delta

Â X- ik delta X0 + delta X models.

Â 26:38

Okay, but these integrals are very simple.

Â They just lead to delta function.

Â So what do we attain?

Â Theta of delta X0 divided by 4 pi

Â models of delta X times the following delta functions, delta.

Â 27:08

minus delta function delta X0 minus delta

Â X Sorry vice versa,

Â here is minus and here is a plus sign.

Â But the argument of delta function for

Â delta X0 positive, the argument of the delta function is never 0.

Â So this can be dropped off.

Â 27:34

And as a result, what remains is green function which we

Â use in the calculations in our lectures.

Â You see it is not 0 exactly on the light cone when delta

Â 0 is equal to the model of [INAUDIBLE], which is a light cone condition.

Â And in towards the future of light cone, because of the presence of these delta

Â functions and decays with the distance, as inverse models of the distance.

Â So, I hope it's suitable for the lectures.

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Â