0:13

So, this our goal is to

work with this metric.

dt squared minus a squared of t multiplied by

dr squared 1 minus k times r squared plus r squared d omega squared,

for the three cases of K equal to 1,0 or minus 1.

And our goal is to fix this factor from Einstein equations of motion.

Well, it is straightforward to calculate from this metric the components of

the Christoffel symbols,

the non zero components of the Christoffel symbols as follows.

It's a times a dot, a is a function of time only,

it is not function of partial coordinates.

That's this due to the fact that we have spatially homogeneous situation here.

So, dot means it's differential with respect to time.

1:41

And gamma 1, 01 is equal

to a dot over a and gamma 1,

11 is equal to cappa times

r 1 minus k r squared gamma

1 2 2 is equal to minus r

times 1 minus k r squared.

And gamma 1 3 3 is equal to

minus r 1 minus kr squared

times sine squared of theta and

gamma 2 02 is equal to a dot

over a gamma 2 1 2 is 1 over r.

And gamma 2 33 is equal to minus

sine theta times cosine theta and

gamma 3 03 is a dot over a.

And gamma 3 13 1 over r and

gamma 3 23 is equal to

cotangent of theta.

These are non zero components of Christoffel symbols amma nu,

nu, alpha for this metric tensor gb nu.

And the other non zero components of the Christoffel symbols can be

obtained from the symmetry properties of this tensor.

3:39

And otherwise if do not follow, if other components do not follow

from this by the symmetry properties of the Christoffel symbol they are zero.

We have written for you the components of the Christoffel symbols for this metric.

Now we are going to write the components of the Ricci tensor for this metric.

The Ricci tensor components for this metric like this.

a double dot, second differential over time of a,

r 11 is a times a double dot

plus 2a dot squared plus 2k.

Over 1 minus kr squared

R22 is equal to r squared

aa double dot plus 2a

dot squared plus 2k.

R33 is equal to r squared

sine squared theta a times a double

dot plus 2 a dot squared plus 2 k.

And then Ricci's scalar following from this,

riemann tensor and this metric is

minus 6 a times a double dot plus

a dot squared plus k over a squared.

To find the behavior of a of time,

to find this function from Einstein equations

we have to specify the energy momentum tensor.

The energy momentum tensor suitable for

the situation that we are discussing was already presented before.

So, it has this form of this perfect fluid.

So, it's rho of t times u mu u nu plus

p of t times u mu u nu minus g mu nu,

where g mu nu is a metric tensor written here.

And the only peculiarity of this tensor follows from the spacial homogeneity.

Because of that we assume that ro and

p are functions of time only but not of spacial coordinates.

In fact, in the frame, in the free

falling matter n this metric has a following

full velocity vector, 1, 0, 0, 0.

And as a result, this form of the energy

momentum tensor has the following matrix.

It is diagonal matrix of the form rho of t minus p of t,

minus p of t, minus p of t.

So this form of the energy momentum tensor of the matter is in perfect

agreement with the spatial homogeneity.

So, first of all, we have rotation symmetry in space.

7:19

And so all directions in space are equivalent,

and all the components of this energy momentum doesn't depend only on time.

So we are going to use this energy momentum.

Well, to move further, we will specify the components

with both lower indices, T00, as follows from here,

is equivalent to ro, T11, again as follows from that,

is p a squared, over 1, minus k times r squared.

8:12

So we're going to use this formulas, this expression and

the expression for the Ricci tensor plug in the expression for

the energy momentum tensor.

Plug them into Einstein equations and look for

the solutions which follow from the Einstein equations.

So we're going to solve Einstein equation

R mu minus one half g mu r equal to 8 pi kappa t mu.

And we so far have specified this tensor, this tensor, this tensor and this tensor.

Now we going to plug them into this equation and

find the equation for unknowns,

a of t, rho of t and p of t.

Well we have to specify the relation between these two,

which is called equation of state.

So the equations that we obtain are as follows,

it's 3 a dot squared plus k equal

to 8 pi kappa ro a squared.

This is 00 component of this equation.

And, two a a dot dot plus

a dot squared plus k equal

to minus 8 pi kappa pa squared.

This is 11 component of this equation.

Actually 22 and 33 components of this equation

give the same due to spatial homogeneity as it should be.

And the other component of diagonal components of this equation give trivial

relation that 0 equals to 0 and nothing else.

And one more thing, we have also the condition of energy momentum conservation.

10:24

This a full equation.

For mu equals to 0 we obtain

the following equation

rho dot plus 3 p plus rho,

a dot over a equals to 0.

And this equation is not independent from these two,

it follows from these two equations.

Actually these equations are referred to as Friedman equations.

So, this equation follow from this.

And the equation for mu running from 1, 2,

3 give trivial relation, nothing non trivial.

Now to solve these equations these two or this, they're not independent again.

We have to specify equation of state.

The standard equation of state in cosmology is p

equal to some parameter omega times rho,

where omega is constant, omega is constant.

11:46

This is the case of dust, and

in cosmology describes galactic matter spread over the universe.

And there other option which is encountered in cosmology

is omega equals to 1 3rd.

For omega equals to 1 3rd, we obtain the energy momentum tensor

which is traceless, as one can check straightforwardly.

And this corresponds to, this is called radiation.

So this is dust, this is radiation.

12:23

Why it is radiation, because this case we encounter in Maxwell's theory.

And that is one of the reasons why it is called radiation.

Now one can look at this equation for these two options.

First of all, for the case when W equals to 0, we as a consequence of this

equation, we obtained that rho, we already encountered this equation in fact.

Rho times a cube is equal to constant, which just tells us

that if the volume of special sections behaves like a cube, sorry.

If the volume behaves like a cube so it either expands or shrinks.

Hence the density behaves like 1 over a cube.

The density of the dust, as it should be.

13:29

Why is it so for the radiation?

It is easy to understand, because for the radiation, again, for

P equals to 1 3rd rho, we obtain this situation.

Why is it, so for the radiation can be also understood, because

13:53

the density of radiation also behaves

like this if universe expands and shrinks according to this law.

But at the same time the wavelengths of radiation behaves like this.

So this is the wavelengths,

the wavelength is scaling as a, so it increases.

The frequency hence decreases, and

as a result the energy of radiation decreases according to this law.

So the total energy density for

the radiation hence is the product of this times this.

And as a result we obtained that rho behaves like

1 over a to the 4th power for the radiation.

That is the reason we have this law for the radiation.

So these two types of behavior for

these two types of matter can be explained on general grounds.

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