0:13

So combining all these observations that we have been making so

far, we have obtained that the variation of the action,

of the Einstein-Hilbert Action, under the variations of the metric,

is proportional to the following quantity.

1:03

Multiplied by the following thing.

R mu, let me actually make sure that whether there is a square root of g,

yeah there is a square root of g, so it's correct.

1:19

So, multiplied by square root

of g times R, times g mu and

times the full-length,

thing, one-half g mu R

minus half g mu lambda,

minus 8 pi kappa T mu.

So this has to be equal to 0 for

any small delta g mu, for any value of this guy.

For this to be true,

we obtained the following

equation R mu times one half

Rg mu minus one half lambda g

mu equals to 8 pi kappa T mu.

2:31

And it does relate geometry to the matter, to the energy of matter.

Well the attribution of this term is arguable.

Whether we should attribute to the geometry,

because it corresponds to the volume factor or to the matter, because it

corresponds to the cosmological constant whose origin can be falling from matter.

So, it's arguable, but in any case, this equation relates geometry and

energy, and that is the essential point.

So, our goal during the following lectures

is to study this equation and define the solution of this equation.

3:11

But before we go further and

to finish this lecture, let us describe a few properties of this equation, and

a few properties of the energy momentum tensor for the matter.

First thing, let us assume that we are discussing the things

in the absence of matter, so no matter and the cosmological constant is 0.

Then this equation reduces to the following one.

R mu- one-half Rg mu = 0.

If we multiply both sides of this equation by g mu,

and using the fact that, g mu

times g mu = 4 number of dimensions,

because is just delta mu mu,

the trace of unit matrix 4 by 4 unit matrix.

So as a result of the multiplication we obtain that R = 0,

and because R = 0 as a consequence of this equation,

we obtain that result, cosmological constant and matter.

Einstein equations reduced to the condition of so-called reach of flatness.

That reach is vanishing.

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Let me stress that reach of flatness doesn't imply that Riemann tensor is zero.

Riemann tensor is not zero, this is Riemann flatness.

This, the condition that Riemann tensor vanishes

is the fact that the spacetime is flat.

Of course, flat spacetime solves this equation.

But the resolutions of this equation, which do not have

vanishing Reimann tensor, so there are non flat space times,

which solve Einstein equation without matter and

cosmological constant, this is important thing to have in mind.

5:15

And another point that we want to say, let us take covariant

derivative of the, now we are discussing this equation together with this and this,

let us take the covariant derivative of both sides of this equation.

Have in mind that metric tensor is

covariantly constant for any even result contraction of indices this is zero.

So having this in mind, we obtain from this equation the following relation.

That R nu mu covariant derivative for

nu minus one-half multiplied by

d mu R is equal to 8 pi kappa D mu times T mu.

So, but what is standing on the right on the left hand side,

we have observed to be equal to zero due to Bianchi identity.

At the very end of previous lecture we saw that

Bianchi identity implies that this is zero.

As a result, even if we didn't assume from the very beginning

that energy momentum tensor is conserved,

the conservation of energy momentum tensor is just the consequence of this equation.

This equation implies that energy momentum tensor is conserved.

This has some important consequences that we're going to discuss now.

So we have obtained Einstein equations of motion.

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G mu nu.

Now we're discussing the properties of this equation before going on with

its solution, we're discussing various forms of etc, etc.

So but we, what we have observed, that if we apply

7:37

This is somewhat similar to the sedation one encounters in electrodynamics,

because there the cup to equations of motion, of Maxwell.

Maxwell equation of motion have the following form

in electrodynamics where this is a fore current, and

this is an electromagnetic tensor, which is d mu A nu minus d mu A mu.

So if we apply to the both sides of this equation the ordinary derivative,

this is in flat space, so in Minkowski space, so

if we apply d nu to both sides of this equation.

As a consequence of the fact that F mu is nu mu and if it is

differentiable as a result the left hand side vanishes,

and we obtain conservation of electric charge,

electric current condition.

So this variation is similar to what we have in common, but

in gR the fact that this equation falls from this has far

bigger consequences for the following reason, because this equation in

part is just some part of math equations of motion, as we are going to show.

As a result the system of math equations of motion, and

Einstein equation is over counted, so it has more equations that necessary.

So that what we going to observe now.

Let us consider more closely this equation for an example.

For an example where we have dust energy momentum tensor.

For this lecture, you will have to solve a problem,

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which shows that for the dust energy momentum tensor,

dust is just a combination of particles which do not interact with each other.

So for the dust, energy momentum tensor has a falling form,

rho of x times U mu times U nu as a functions of x.

So, this vector fuels for the velocities of the dust.

Rho is a mass density, mass density and

is a for velocity vector field, this guy U mu.

So in the reference frame where U mu is equal to (1,0,0,0),

we have that T mu nu, as a consequence of this equation,

as a matrix, has a diagonal form and more over, so it's not simply diagonal.

It has only one non zero component, rho.

10:41

Rho u mu U nu and

then this can be

equal to D mu rho U

mu times U nu plus

rho U mu D mu U nu.

Well we so far don't know anything about this so that's fine,

we just want to use identities, identities.

Let us multiply both sides of this equation by U nu,

both sides of this equation U nu.

And we going to use that U nu times U nu is equal to 1,

that's just a property of our velocity.

As we know.

And if one applies to both sides of this equation covariant derivative D mu,

one obtains the following equation as a consequence of this identity.

We have that 2 U nu D mu times U nu is equals to zero.

As the result,

after multiplication of both sides of this equation by U nu,

we get that this side is vanishing, as a consequence of this identity, and

we obtain the following equation.

That D mu applied to rho U mu is equals to 0.

The physical meaning of this equation is actually very simple.

This is just a covarient form of the continuity equation.

In flux space, in Minkowski space,

we have this equation, continuity equation has this form.

In curved space, or

in curvilinear coordinates, this equation is extended to this form.

And so it's a covarientization of the continuity equation.

It has a clear physical meaning.

But because this is equal to zero and this is equal to zero,

we obtain that this equal to zero.

U mu times D mu U nu = 0.

But this is nothing but

the geodesic equation that we have obtained in the previous lectures.

It means that as an consequence of the energy momentum conservation, we have that

13:02

matter content of the the dust moves along the geodesics.

Which means that we have obtained from this equation, equations of motion for

the matter, while no one never obtains equations of motion for

the matter from the charge conservation.

So that's important difference that we encounter in this case.

To conclude this lecture, let me describe a few more examples of the matter action.

Well, let me use a different pencil.

So, I’m going to describe a few more

examples of the matter action.

Let us describe the matter action for the scalar fields.

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The simplest in variance that we can build from a scalar field it's a polynomial.

Polynomial of the scalar field.

Any polynomial of the scalar field.

Sum over n from one to n,

a n phi to the power of n is invariant.

14:21

So to have dynamical equations of motion for

the scalar field, we need the images of the scalar field,

then the simplest invariant that contains derivative of

the scalar field can be build as follows, d mu phi d nu phi.

So this is the simplest invariant.

Having this invariant and this invariant at our disposal,

we can write the scalar field matter action as follows.

It is just integral over d4x square root of modulus of g.

So this is invariant.

And multiplied by this invariance like g mu nu d nu phi,

it's d nu phi, like this, minus V(phi).

So this is the simplest example of the scalar field action,

where this is a potential and this is kinetic part.

So another type of action, of the matter action,

can be written for the electromagnetic fields.

Electromagnetic tensor in curved space or in curvilinear coordinates,

which is by definition as this quantity.

15:54

So this is the electromagnetic tensor,

practically it doesn't change in curved space time or in curvilinear coordinate.

As a result we can straight forwardly write the following invariant general

covariant information, the following action for the electromagnetic fields now.

16:15

So this is again invariant and

now we built an invariant from electromagnetic

tensor, g mu alpha, g mu beta.

So now, at our disposal, we already have three types of matter, actions.

This one,

this one, and for the particle.