[SOUND].

So we have obtained the following metric tensor

whose metrics expression matrix is as follows.

It is diagonal and its components are e

to the power of mu, minus e to the power of lambda,

minus r squared, minus r squared sine

squared theta, so this is four by four matrix.

The components are like this.

Then nonzero components of

the Christoffel symbols are like this,

1 11 lambda prime over 2,

gamma 0 10 nu prime over 2,

gamma 2 33 equals 2 minus sine

theta cosine theta then gamma 0

11 is equal to lambda dot over

2 e to the power of lambda minus nu.

Gamma 1 22 is equal to minus r e to

the power of exponent minus lambda,

gamma 1 00 Is equal to new prime

over 2 exponent new minus lambda.

And gamma 2, 12 equals to gamma 3,

13, is equal to 1 over r.

And the gamma 3 23 is

equal to ctg of theta,

then gamma 0 00 is nu

dot over 2 gamma 1 10 is

lambda dot over 2, and

finally gamma 1 33 is equal

to minus r sine squared theta theta

exponent of minus lambda.

So, all non-zero components of

the tensor gamma mu nu alpha, are written here or

obtained from these guys with the use of the symmetry of this guy.

All the rest of the components, which do not follow from these expressions

where the application of this symmetry are equal to zero.

So this is prime, this is prime.

And prime means differentiation with

respect to d partial differential d over dr,

while dot means partial differential with respect to d over dt.

So, this is dot, this is dot, this is prime, this is prime.

Here is dot, what else?

Yeah.

That's it to understand what is standing for the christople symbol.

So to move farther and for future convenience for

the future lectures, instead of writing this form of Einstein equations,

let us write it in the form as if there is non zero energy momentum tensor.

So we'll write it in this form, r nu.

Actually, we will write it in

this r nu nu minus 1 half delta

mu r equals to 4 pi kappa T mu nu.

So, we'll write in this form.

In this form, for the metric under consideration,

Einstein's equations look like,

actually, it should be probably 8 Pi.

But anyway, so, 8 Pi Kappa T11

equals 2 minus exponent of

minus lambda nu prime over r plus

1/2 plus 1 over r squared.

Sorry.

1/r squared.

Plus 1 over r squared.

Then there are two equations.

8 pi kappa t 2, 2 is equal.

This is the same, two equations of the same form.

8 pi kappa T 3 3 is equal to minus half

exponent of minus lambda mu double

prime plus mu prime squared over 2 plus

mu prime minus lambda prime over 2

minus mu prime time under primer over 2,

here is over r, sorry, over r rather than 2.

2 plus 1 half exponent of mu multiplied by

Lambda double dot plus lambda

dot squared over 2 minus lambda

dot times nu dot over 2.

And few more equations.

Then we have equation 8 pi kappa.

T 0 0 equals 2 minus exponent of minus

lambda 1 over r squared minus lambda

prime over r plus 1 over r squared.

And the last equation 8 pi

kappa t01 is equal to minus exponent of

minus lambda times lambda dot over r.

So, we have written these equations as if T mu is not 0.

The other part of Feinstein equation, for example, corresponding to T1 3 or

T1 2 or other components of mu nu for

the other values of mu nu, the other equations are trivially satisfied

meaning just that they have a relation, something like zero equals to zero,

if the corresponding components of energy-momentum tensor are zero.

So if we put the other components of energy-momentum tensor to zero,

the other components of Einstein equations are truly very satisfied like this.

So now, having written Einstein equations in this form for

the given ansatz for the metric, we continue with

the consideration of the case when t mu nu equals to zero.

So when this is 0, these equations

are reduced to exponent of minus

lambda nu prime over r plus 1 r squared

minus 1 over r squared equals 0.

So this reduces this equation.

Then we have equation exponent to minus lambda,

lambda prime divided by r minus one over r

squared plus one over r squared equals to zero.

So this form you reuses this equation, and

finally this equation gives us lambda dot equals to zero.

So this is the equations we obtain for

the metric answers for

the given form of the metric from Einstein equations.

So, let me just stress one thing.

Immediately one can see that mu equals to zero and

lambda equals to zero solves this equation.

It's straight forwardly.

Then the metric for this case reduces to the following form.

Which is nothing but the flat metric in Minkowski space time written

in spherical coordinates for this partial section.

So nothing else.

Now, we want to look for a non trivial solution of these equations.

So, we have obtained for the metric

of the form (r,t) d

t squared minus lambda r t d

r squared minus r squared d omega squared.

We have obtained the following equations.

Nu prime divided by r plus 1 over r squared minus

1 over r squared equals to 0 minus lambda,

lambda prime divided by r minus 1 over r squared plus 1 over r squared equals to 0.

And lambda dot equals to 0.

And I forgot to say that those lengths equations,

which were corresponding to 2 2 T 2 2 and T 3 3 components,

those length equations which, were written on the previous blackboard.

This equation follow from this one.

So this is satisfied this is also.

Now we're going to solve this equation and find nontrivial solution of this equation.

And it is not hard to see that if we sum these two equations,

if we take a sum of these two equations,

we obtain that mu prime plus lambda prime is equals to zero.

It means, because this is differentiation with respect to d over d r, it means that

mu plus lambda can be function of time only, some function of time.

But remember that we have remaining symmetry, that mu

is changed to mu bar, remaining symmetry which respects this form of the metric.

That mu bar plus log of dt over dt bar squared.

So using this symmetry, one can fix this function to be 0.

Then as a result we obtain that mu is equal to minus lambda.

And finally, using this relation one can solve this equation

to find that exponent of minus lambda is equal to exponent of mu

and equals to 1 plus c2 over r.

Where c2 is unfixed, is not fixed from this equation.

It's some integration constant whose physical meaning and

concrete value we will find just in a moment.

But let me stress at this point that if c2 is equal to 0,

in this case, we have flat metric again.

But if c2 is not equal to 0,

then the metric under consideration

has a following form 1 plus c2

over r dt squared minus dr squared

divided by 1 plus c2 over r minus

r squared d omega squared.

And if we take the limit, this metric is approximated.

If we take the limit r goes to infinity, it goes, again, to the flat metric

approximately, dr square minus r squared d omega squared.

So, we have the following situation.

We have a gravitating center at r equals to 0, and

it creates spherically gravitational field.

And as we go very far from this gravitational center,

the influence of the gravitational center on the geometry becomes smaller and

smaller, and eventually at asymptotic infinity, spatial infinity.

At spacial infinity, we see flat space time, which is almost unaffected by

the gravitating center, which is situated somewhere far away from you.

So, this is physically meaningful, and this observation will help us to find C2,

but at this point, let me stress that space times which have this form,

this sort of behavior, are referred to as asymptotically flat,

although the notion of asymptotic flatness is a bit more complicated and

goes beyond the scope of our course.

So we are not going to discuss it in great detail.

So far we're just consider this as a definition.

That if a metric goes to flat spacetime and asymptotes is partial infinity,

we consider this as an asymptotically flat space.

And now what we encounter is very interesting situation.

That if we consider a spherically

symmetric metric it indicates that t mu nu

equals to 0 and lambda equals to 0.

So in the case when these two quantities are 0,

metric is spherically symmetric and goes to flat 1, at partial infinity.

Then this metric is necessarily static.

Static means that it doesn't change under

the inversion of time and under time translations.

It doesn't change.

This metric is referred to static.

So this is the essence of Birkhoff's theorem.

We didn't prove it here, but we just hinted

to this property of Einstein equations.

But we will come back to this property many times in the coming lectures.

But at this point we, let me stress that this property is very similar to the thing

that we encount in Maxwell's theory where spherically symmetric solution that

goes to zero at spatial infinity is also unique, and that is Coulomb potential.

So in Maxwell series we have analog of this situation.

But in general Theory of Relativity this Birkhoff theorem has

deeper consequences as we will see in the upcoming lectures.

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Schwarzschild solution(part 2)

Introduction into General Theory of Relativity

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