0:13

So combining all these observations that we have been making so

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

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

Â is proportional to the following quantity.

Â 1:03

Multiplied by the following thing.

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

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

Â 1:19

So, multiplied by square root

Â of g times R, times g mu and

Â times the full-length,

Â thing, one-half g mu R

Â minus half g mu lambda,

Â minus 8 pi kappa T mu.

Â So this has to be equal to 0 for

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

Â For this to be true,

Â we obtained the following

Â equation R mu times one half

Â Rg mu minus one half lambda g

Â mu equals to 8 pi kappa T mu.

Â 2:31

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

Â Well the attribution of this term is arguable.

Â Whether we should attribute to the geometry,

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

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

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

Â energy, and that is the essential point.

Â So, our goal during the following lectures

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

Â 3:11

But before we go further and

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

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

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

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

Â Then this equation reduces to the following one.

Â R mu- one-half Rg mu = 0.

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

Â and using the fact that, g mu

Â times g mu = 4 number of dimensions,

Â because is just delta mu mu,

Â the trace of unit matrix 4 by 4 unit matrix.

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

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

Â we obtain that result, cosmological constant and matter.

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

Â That reach is vanishing.

Â 4:33

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

Â Riemann tensor is not zero, this is Riemann flatness.

Â This, the condition that Riemann tensor vanishes

Â is the fact that the spacetime is flat.

Â Of course, flat spacetime solves this equation.

Â But the resolutions of this equation, which do not have

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

Â which solve Einstein equation without matter and

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

Â 5:15

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

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

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

Â Have in mind that metric tensor is

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

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

Â That R nu mu covariant derivative for

Â nu minus one-half multiplied by

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

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

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

Â At the very end of previous lecture we saw that

Â Bianchi identity implies that this is zero.

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

Â that energy momentum tensor is conserved,

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

Â This equation implies that energy momentum tensor is conserved.

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

Â So we have obtained Einstein equations of motion.

Â 7:08

G mu nu.

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

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

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

Â 7:37

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

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

Â Maxwell equation of motion have the following form

Â in electrodynamics where this is a fore current, and

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

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

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

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

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

Â differentiable as a result the left hand side vanishes,

Â and we obtain conservation of electric charge,

Â electric current condition.

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

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

Â bigger consequences for the following reason, because this equation in

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

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

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

Â So that what we going to observe now.

Â Let us consider more closely this equation for an example.

Â For an example where we have dust energy momentum tensor.

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

Â 9:29

which shows that for the dust energy momentum tensor,

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

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

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

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

Â Rho is a mass density, mass density and

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

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

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

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

Â It has only one non zero component, rho.

Â 10:41

Rho u mu U nu and

Â then this can be

Â equal to D mu rho U

Â mu times U nu plus

Â rho U mu D mu U nu.

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

Â we just want to use identities, identities.

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

Â both sides of this equation U nu.

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

Â that's just a property of our velocity.

Â As we know.

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

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

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

Â As the result,

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

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

Â we obtain the following equation.

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

Â The physical meaning of this equation is actually very simple.

Â This is just a covarient form of the continuity equation.

Â In flux space, in Minkowski space,

Â we have this equation, continuity equation has this form.

Â In curved space, or

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

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

Â It has a clear physical meaning.

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

Â we obtain that this equal to zero.

Â U mu times D mu U nu = 0.

Â But this is nothing but

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

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

Â 13:02

matter content of the the dust moves along the geodesics.

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

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

Â the matter from the charge conservation.

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

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

Â Well, let me use a different pencil.

Â So, Iâ€™m going to describe a few more

Â examples of the matter action.

Â Let us describe the matter action for the scalar fields.

Â 13:54

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

Â Polynomial of the scalar field.

Â Any polynomial of the scalar field.

Â Sum over n from one to n,

Â a n phi to the power of n is invariant.

Â 14:21

So to have dynamical equations of motion for

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

Â then the simplest invariant that contains derivative of

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

Â So this is the simplest invariant.

Â Having this invariant and this invariant at our disposal,

Â we can write the scalar field matter action as follows.

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

Â So this is invariant.

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

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

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

Â where this is a potential and this is kinetic part.

Â So another type of action, of the matter action,

Â can be written for the electromagnetic fields.

Â Electromagnetic tensor in curved space or in curvilinear coordinates,

Â which is by definition as this quantity.

Â 15:54

So this is the electromagnetic tensor,

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

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

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

Â 16:15

So this is again invariant and

Â now we built an invariant from electromagnetic

Â tensor, g mu alpha, g mu beta.

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

Â This one,

Â this one, and for the particle.

Â