0:39

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

Which is the same as dt squared- Laplacian.

Which is the same as dt squared- dx

squared- dy squared- dz squared.

1:21

Then the solution of this equation can be found.

Before saying that, let me show what happens here.

This equation is invariant under the shift of x.

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

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

of difference, x- y.

1:43

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

phi(x), is easy to find.

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

Indeed, why do we know it?

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

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

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

d4y box acting on x G(x-y)j(y).

But this guy is nothing but the delta function.

So continue this equality,

we have d4y delta 4 (x-y).

3:02

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

integral of y0, y1, y2, etc.

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

So we obtain this equation.

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

Now we want to find the solution of this equation.

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

3:31

How do we do that?

Well we do Fourier transform of both sides.

First Fourier transform of the delta function.

x- y is just d4 p/2

pi to the fourth exponent

of ik mu (x-y) mu

resulting in division.

So this is the Fourier transform of the delta function.

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

Sorry, here it's p.

dp e ipx.

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

And the product of four exponents is just this guy.

4:31

So this is three real.

This is a Fourier transform.

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

We define it as d4p/2 pi

to the fourth sum G tilde of

P exponent of ip mu (x-y) mu.

Okay.

Now, this is just by definition.

So we take Fourier transform.

Now plugging this here.

Sorry, this here and this here.

We obtain the following situation.

The d'Alembertian acting on

integral over d4 p/2 pi to

the fourth G tilde(p) e ip mu

(x-y) mu = the integral over

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

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

depends on x.

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

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

Is nothing but

-p mu p mu e ip mu (x- y) mu.

6:17

Then from here what do we obtain?

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

d4p/2 pi to the fourth

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

p squared is just this guy, p squared.

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

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

this = 0 only if this = 0.

As a result we obtain that G twiddle

of p is equal to -1/p squared.

p squared is p mu times p mu.

7:14

So that was actually the reason to perform Fourier transform.

Because Fourier transform allowed us to convert differential equation into

algebraic equation.

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

What remains to be done is to calculate this integral

with the substitution there of 1/p squared.

If we do that, we are done.

7:41

Actually there is one simple command.

This is a solution of this up to the addition of

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

Sorry, C(p) times delta p squared.

The B squared for any smooth function of p,

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

Additional, this sides,

into here, gives us something which is specified.

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

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

And this term is Important in this discussion.

We will clarify this in a moment.

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

8:50

And already have found That the solution

of this equation can be represented as

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

G-k-i-k-mu (x-y) mu and

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

Where k squared is k mu times k mu,

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

function of k times delta k squared.

10:07

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

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

that this G0 of x- y solves homogeneous equation,

exactly due to the presence of this quantity.

But the solution of this equation always is defined up to

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

Because we will always drop off these kind of terms,

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

Let me stress one thing, one thing.

Which appears here for this quantity.

We drop this off for this quantity.

This integral G(x-y),

integral of d k zero over two pi,

11:06

integral of d three k over two pi cubed i,

k0, x minus y0,

minus ik, x minus y,

divided by k0 squared, minus k vector squared.

It is not have to see that this

integral from the mathematical point of view is not defined.

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

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

The contour of integration goes along the k zero axis.

And there are two poles on this axis.

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

plus models of k.

12:08

So, there is a problem.

We cannot find this integral.

What should we do?

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

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

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

We expect, what is the meaning of quantity?

What is the meaning of this?

To understand the meaning of this quantity.

Let us, for a second,

look at the following X minus y.

12:56

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

And is.

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

the field that it creates at the position x.

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

We have a subset y And

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

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

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

This is a source at position in space.

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

So we have space-time.

13:40

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

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

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

y And zero everywhere else.

13:59

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

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

light because this is a wave equation,

describing voice traveling with the speed of light.

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

which has apex at the point y.

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

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

What do we expect?

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

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

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

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

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

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

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

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

15:12

So, let me say that this statements

specifies how we define the contour of integration.

The contour if integration we define according to this rule,

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

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

the contour of integration over k0, like this,

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

That it is 0 If X zero

is before than Y zero.

It is zero But

it is not 0 if x0

is after y0.

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

And if we would choose a different contour,

that would correspond to the solution of this equation,

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

something like solution of homogeneous equation.

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

which is minus dk0 over 2 pi

integral over d3k over 2 pi cubed,

exponent of ik0(x0-y0) minus

i k vector (x- y) divided by k0

squared minus k vector squared.

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

So we have k0 plane.

17:14

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

And it has poles here, -k and k,

where k is just modulus of this k.

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

that it goes around the pole like this.

So our contour is like this.

17:39

Now, why?

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

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

The reason for that is following.

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

which means that this quantity is negative.

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

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

Why?

Because on this circle, the integrand is always 0,

because when this negative on this circle,

the imaginary part of k0 is minus infinity.

As a result, the exponent is 0.

And I add nothing to this integral over this contour,

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

The integrand is 0.

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

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

It is a holomorphic function of k0 inside this contour.

So according to the Cauchy theorem, for this case,

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

Now, if x0 > y0,

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

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

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

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

19:41

According to the Cauchy theorem, this guy, well,

it's a home exercise, along this contour,

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

20:00

I'm not going to give too many details.

This is a home exercise to take this integral.

It is equal to minus exponent ik,

this k, x0-y0,

minus exponent i,

divided by 2 ik.

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

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

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

Delta x0 is just x0-y0.

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

Integral over d3k, so plugging this into here,

what remains to be done is to take this integral

21:25

Ik(x0-y0) minus

minus ik(x0-y0)

divided by 2ik.

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

To take this integral, we do the following.

We use instead of k1, k2, k3,

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

to k, modulus of k, theta and phi.

Where they're angles in the spherical coordinates.

And direct delta x along the third direction such that k

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

22:21

So what do we obtain then, then this integral.

Continuing it, let me continue it.

This integral acquires the following form.

So we have theta function of delta x0,

we have 2 pi cube here, integral from

0 to plus infinity, dk times k squared,

and we have integral over d phi.

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

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

we have sine theta times d theta.

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

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

The integrand is just -ik,

modulus of delta x times cosine theta times

this exponent ik(x0-y0),

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

While taking the integral over the dk,

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

we obtain the following theta function

of delta x0 divided by 2pi squared.

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

24:09

What remains is a falling integral of DK.

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

Exponent of -ik.

Modulus of delta x-

ik modulus of delta x

multiplied by the exponent.

Well Exponent of ik delta x,

0 minus ik delta x 0,

d ivided by 2i.

So, k cancel here.

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

So what remains is just the product of 2 brackets.

25:17

If we expand the brackets, we attain 4 terms,

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

But then, we can reshuffle these two terms.

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

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

an integral from minas infinity two plus infinity DK.

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

into the integral from zero to the plus infinity.

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

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

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

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

So the result is the following situation,

ik delta X0 + models of delta

X- ik delta X0 + delta X models.

26:38

Okay, but these integrals are very simple.

They just lead to delta function.

So what do we attain?

Theta of delta X0 divided by 4 pi

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

27:08

minus delta function delta X0 minus delta

X Sorry vice versa,

here is minus and here is a plus sign.

But the argument of delta function for

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

So this can be dropped off.

27:34

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

use in the calculations in our lectures.

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

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

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

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

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

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