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Next two lectures we're going to devote to the study of the gradational radiation.

To define what means gradational radiation,

we have to specify what means energy flux in the presence of gravity.

That is a problem we are going to address in this lecture.

Well to understand what is the problem.

Let us start with the standard situation of Minkowski spacetime.

In Minkowski spacetime,

the energy momentum conservation condition is expressed like this.

With a short derivative rather than than covariant long derivative.

And because of this, if we integrate both sides of this equation

over a surface sigma and take 0 component.

For nu equals to 0, we obtain the following equation.

That integral

over dp0/ dt +

integral over sigma over boundary of this sigma.

D 2 write as three dimensional.

Because it's three dimensional.

D 2 sigma i T i 0 equals to 0.

Where P mu is the falling full vector.

When it's a function of time, it's an interval over sigma T mu 0 t,

x d3 sigma.

So this is three dimensional Kushi surface.

0 is a long time.

So we have a Kushi surface sigma.

And we have a normal direction to this Kushi surface,

which is designated by t or 0 component.

So this is what we obtain.

And P0 as one can see, P0 is nothing but the energy.

Because it's integral over the energy density over the three dimensional

hypersurface.

And this is volume element, area element at the boundary of this surface sigma.

So this law establishes conservation, actual conservation of this quantity.

There is no energy flux through the boundary.

This is conserved in this region.

If there is energy flux going out, energy is reduced.

If in, energy is increased.

But in case of the gravity.

What we have been calling as the energy momentum conservation

doesn't have this form.

It does have the following form.

It has a form like this with the covariant long derivative.

And as a result we have the following situation.

So 0 is equal to, we just write explicitly this covariant derivative.

It's T mu nu we express it in this form, sorry.

It was one up and this lower index, and this is upper index.

Plus gamma, mu beta,

mu T, beta nu.

Minus gamma beta nu mu.

T mu beta.

And as a result, this is equivalent 1 over square root of the modulus of

the determinant of the metric, d mu acting on T mu nu.

Square root of the determinant

of the metric, minus 1/2

of d mu g mu alpha T mu alpha.

To obtain this expression, we have used explicit form.

Explicit form for gamma mu nu alpha through the metric.

It's symmetry properties.

And also the fact that energy momentum tensor is symmetric.

T mu nu, T nu mu and finally we have

used that gamma, this gamma.

The trace of gamma, gamma mu nu mu with two indices.

And some summation of the indices is 1 over square root

of g d nu square root of modulus of g.

This is, it follows from the definition of the gamma matrices.

So now what we have is that there is no such conservation.

Well, we have something similar to that expression here, but

we have additional term.

This conservation law, although we call it conservation law,

it doesn't express any explicit conservation of some quantity.

And it should be expected, on general grounds,

that if we are dealing with a theory containing gravity,

the energy momentum tensor of matter shouldn't be simply conserved.

Because there can be energy transmission from matter to gravity and

from gravity to matter.

So the question is what is the quantity which specifies what

means energy, momentum for the gravity.

How to find it and that's the issue we going to address now.

So now we're going to specify what we mean by the energy momentum for the gravity.

And what is the total energy momentum of gravity plus matter?

So let me first write this relation that we have ended with.

So this is a relation like this.

G mu nu square root of modulus of

the determinant of the metric minus

1/2 d mu g mu alpha T mu alpha.

And so let us fix the following reference frame.

Set your reference frame around

arbitrary point x0 such that in this

frame d alpha g mu nu of x0 is equal to 0.

It doesn't mean that g mu nu is equal

to eta mu nu, it's not equal.

It's constant but at this point it's derivative of 0.

But it doesn't mean that it is equal to this.

But it is important that at this point d mu of square root of modulus

of g at the point x0 is equal to 0.

It's important.

As a result from this equation because

this is true we obtain that at this

point d mu T mu nu of x0 is equal to 0.

So in this particular frame, in this particular point, we have this relation.

So now let us see how this relation changes in another reference system.

So let us use Einstein equation to express energy

momentum tensor through the Ricci tensor.

So from Einstein equations, we have this relation

g mu nu R.

So in this frame, we have that

because, derivative gi = 0.

We also have that Christoffel symbols, mu nu alpha(X0) = 0.

But, d beta gamma mu nu alpha

(X0) is not equal to 0.

As a result Ricci tensor at this point and

in this frame is as follows.

It's one-half g

mu alpha g nu gamma g

sigma delta [d alpha

d delta g sigma gamma

+ d sigma d gamma g

alpha delta- d

alpha d gamma g sigma

delta- d sigma d delta

g sigma gamma].

So this is a form Ricci tensor has in this frame.

As a result, putting this together,

we obtain from this equation and

this equation,

after straightforward calculation,

that T mu nu (X0) = d (alpha)

{1/16 pi kappa

1/modulus of g d beta {

modulus of g, (g mu,

g alpha beta- g mu alpha

g mu beta)]}.

And finally closing this.

And now, we going to define this as follows,

we going to designate this as follows.

It's 1/ modulus of g d alpha tensor eta mu nu alpha.

And this eta mu nu alpha

= 1/16 pi kappa d

beta acting on modulus

of g (determinant of

metric g mu nu g alpha beta-

g mu alpha g nu beta)]

To obtain this we have extracted this modulus of the determinant

from the derivative because recall that in this frame we have this fact.

And of course, in obtaining all these relations

we had in mind the following thing.

That around the point, so

g mu nu, around the point (X) 0 can be expressed like this.

It's g mu nu X(0).

First derivative is absent but

there are second derivatives d alpha,

d beta, g mu nu (X0),

(X- X0) alpha, (X- X0) beta.

So in expressing this like this,

we kept the terms in this expansion.

So we have obtained that at the point X0,

in the gauge d alpha g mu nu (X0) = 0.

That T mu nu, first of all,

d mu T mu nu = 0, in this frame or gauge.

But at the same time T mu nu

(X0) = 1/ modulus of

determinant of the metric,

d alpha eta mu nu alpha.

And eta mu nu alpha is the following quantity.

It's 1/16 pi kappa

d beta [modulus of g( g

mu nu g alpha beta-

g mu alpha g mu beta)].

And one can see that this tensor has the following symmetry properties that

under the exchange of the first two indices, it is symmetric.

But under the exchange of the alpha index with any of mu and

nu, it is anti-symmetric.

So if we keep the first one untouched, but exchange the third and

the second index, it changes the sign, as it's seen from this property.

And now because in this

frame d mu l g l = 0.

One can immediately see that from this property of this tensor if we apply to

this relation d mu, immediately we obtain this.

Now let us see what happens beyond this gauge.

If we change the frame.

Then of course, it's not zero.

Let us denote.

So d alpha eta mu nu alpha-

modulus of the determinant T of T mu nu is not any more zero.

As you can see because of this.

It's some quantity which we denote like this, t mu nu.

Let us find it, let us find this quantity.

This quantity one can find from the following relation.

That it is the difference between

what follows from here and

Einstein equations.

Let me see what happens.

It's 1/16 pi.

Kappa d alpha d beta

acting on modulus of

g times g mu nu g alpha

beta- g mu alpha g mu

beta- modulus of g

times 8 pi kappa (R

mu nu- one-half g mu nu R).

Well, this just follows from this,

the definition of this is this,

the definition of this is this while this follows from here.

And this follows from equation.

So in this gauge, they cancel each other and this is 0.

But beyond this gauge, we have often tedious but

straightforward calculation were find the following.

Well, it will be a long expression.

So let me start writing here.

So modulus of g times t mu nu is equal to the following quantity.

It's 1 over 16 pi kappa.

I first write the expression and then explain the indentation.

G mu nu comma alpha G alpha

beta comma beta- G mu

alpha comma alpha G nu

beta comma beta + one-half

g mu nu g alpha beta.

G alpha gamma comma sigma G

sigma beta comma gamma -,

( g mu gamma g alpha

beta G nu beta comma

sigma G alpha sigma

comma gamma + g nu

alpha g beta gamma,

G mu gamma comma sigma

G beta sigma comma alpha

) + g alpha beta.

g gamma sigma G mu

alpha comma gamma G

nu beta comma sigma

+ one-eighth.

( 2 g mu alpha g nu

beta- g mu nu g alpha

beta ) times ( 2 g

gamma sigma g delta

psi- g sigma delta

g gamma psi times G

gamma psi alpha G,

this is comma,

G sigma delta comma beta ).

So this is a tedious relation for

this quantity where G mu nu.

G mu nu, by definition,

is just square root of g times g mu nu.

And I remind you that this comma of any quantity is just,

well anyway, G mu nu comma alpha is just d alpha of G mu nu.

Well, one can immediately see that in the gauge, in the frame, where this is 0.

This just vanishes, because it contains only derivatives of the metric.

But beyond this gauge, this is not 0.

Well, it's a tedious calculation to obtain from here this quantity.

So what we have obtained is the following.

That d alpha eta mu nu alpha- modulus

of g T mu nu = modulus of g T mu nu.

And we have obtained a rather complicated expression for this quantity.

The reason for introducing this quantity, which is we did that after Landau and

Lifshitz and this quantity is referred to as pseudotensor energy momentum,

tensor for gravity.

Why it is called pseudotensor,

because it is not a tensor quantity under the coordinate transformations.

Because here from its definition, one can see that this expression contains short

derivatives rather than covariant derivatives.

And as a result, the transformation law for

this quantity is not appropriate for the tensor.

But the reason for

introducing this pseudo-energy momentum tensor was as follows.

As you can sequence of this relation and

the symmetric properties of this guy under the exchange

of index alpha with any of this, that is antisymmetric.

As you can sequence, we have the following conservation law.

T mu nu + T mu nu = 0.

So this is a real conservation law with short derivative and

this quantity is conserved.

So we can define for momentum as follows.

d3 sigma nu modulus of g T mu nu + t mu nu.

And this is for vector which is perpendicular to the surface sigma.

So we integrate, it is perpendicular to surface sigma.

And the modulus of it is equal to the elementary volume on this surface.

Elementary volume on this surface.

So in the absence of the gravitational field, this is 1, this is 0,

and we just go back to the same quantity as space.

In the presence, that was the gravitational field,

this is the quantity we have to consider.

So in the presence of the gravitational field to have a conservation of some

quantity, we have to fix a bare ground and consider.

Well, this is the issue we're going to discuss right now.

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Energy–momentum pseudo–tensor for gravity

Introduction into General Theory of Relativity

Meet the Instructors