0:25

It is diagonal and its components are e

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

Â minus r squared, minus r squared sine

Â squared theta, so this is four by four matrix.

Â The components are like this.

Â Then nonzero components of

Â the Christoffel symbols are like this,

Â 1 11 lambda prime over 2,

Â gamma 0 10 nu prime over 2,

Â gamma 2 33 equals 2 minus sine

Â theta cosine theta then gamma 0

Â 11 is equal to lambda dot over

Â 2 e to the power of lambda minus nu.

Â 1:33

Gamma 1 22 is equal to minus r e to

Â the power of exponent minus lambda,

Â gamma 1 00 Is equal to new prime

Â over 2 exponent new minus lambda.

Â 2:07

And the gamma 3 23 is

Â equal to ctg of theta,

Â then gamma 0 00 is nu

Â dot over 2 gamma 1 10 is

Â lambda dot over 2, and

Â finally gamma 1 33 is equal

Â to minus r sine squared theta theta

Â exponent of minus lambda.

Â So, all non-zero components of

Â the tensor gamma mu nu alpha, are written here or

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

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

Â where the application of this symmetry are equal to zero.

Â 3:16

So this is prime, this is prime.

Â And prime means differentiation with

Â respect to d partial differential d over dr,

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

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

Â Here is dot, what else?

Â Yeah.

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

Â So to move farther and for future convenience for

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

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

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

Â Actually, we will write it in

Â this r nu nu minus 1 half delta

Â mu r equals to 4 pi kappa T mu nu.

Â So, we'll write in this form.

Â In this form, for the metric under consideration,

Â Einstein's equations look like,

Â actually, it should be probably 8 Pi.

Â But anyway, so, 8 Pi Kappa T11

Â equals 2 minus exponent of

Â minus lambda nu prime over r plus

Â 1/2 plus 1 over r squared.

Â Sorry.

Â 1/r squared.

Â 5:16

Then there are two equations.

Â 8 pi kappa t 2, 2 is equal.

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

Â 8 pi kappa T 3 3 is equal to minus half

Â exponent of minus lambda mu double

Â prime plus mu prime squared over 2 plus

Â mu prime minus lambda prime over 2

Â minus mu prime time under primer over 2,

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

Â 2 plus 1 half exponent of mu multiplied by

Â 7:25

T1 2 or other components of mu nu for

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

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

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

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

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

Â So now, having written Einstein equations in this form for

Â the given ansatz for the metric, we continue with

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

Â So when this is 0, these equations

Â are reduced to exponent of minus

Â lambda nu prime over r plus 1 r squared

Â minus 1 over r squared equals 0.

Â So this reduces this equation.

Â Then we have equation exponent to minus lambda,

Â lambda prime divided by r minus one over r

Â squared plus one over r squared equals to zero.

Â So this form you reuses this equation, and

Â finally this equation gives us lambda dot equals to zero.

Â So this is the equations we obtain for

Â the metric answers for

Â the given form of the metric from Einstein equations.

Â 9:48

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

Â in spherical coordinates for this partial section.

Â So nothing else.

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

Â So, we have obtained for the metric

Â of the form (r,t) d

Â t squared minus lambda r t d

Â r squared minus r squared d omega squared.

Â We have obtained the following equations.

Â 10:35

Nu prime divided by r plus 1 over r squared minus

Â 1 over r squared equals to 0 minus lambda,

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

Â And lambda dot equals to 0.

Â And I forgot to say that those lengths equations,

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

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

Â This equation follow from this one.

Â So this is satisfied this is also.

Â 11:45

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

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

Â But remember that we have remaining symmetry, that mu

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

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

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

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

Â 12:46

and equals to 1 plus c2 over r.

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

Â It's some integration constant whose physical meaning and

Â concrete value we will find just in a moment.

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

Â in this case, we have flat metric again.

Â But if c2 is not equal to 0,

Â then the metric under consideration

Â has a following form 1 plus c2

Â over r dt squared minus dr squared

Â divided by 1 plus c2 over r minus

Â r squared d omega squared.

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

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

Â approximately, dr square minus r squared d omega squared.

Â So, we have the following situation.

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

Â it creates spherically gravitational field.

Â And as we go very far from this gravitational center,

Â 14:24

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

Â smaller, and eventually at asymptotic infinity, spatial infinity.

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

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

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

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

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

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

Â goes beyond the scope of our course.

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

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

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

Â we consider this as an asymptotically flat space.

Â And now what we encounter is very interesting situation.

Â That if we consider a spherically

Â symmetric metric it indicates that t mu nu

Â equals to 0 and lambda equals to 0.

Â So in the case when these two quantities are 0,

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

Â Then this metric is necessarily static.

Â Static means that it doesn't change under

Â the inversion of time and under time translations.

Â 16:19

It doesn't change.

Â This metric is referred to static.

Â So this is the essence of Birkhoff's theorem.

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

Â to this property of Einstein equations.

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

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

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

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

Â So in Maxwell series we have analog of this situation.

Â But in general Theory of Relativity this Birkhoff theorem has

Â deeper consequences as we will see in the upcoming lectures.

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Â