0:13

To move further, let me define so-called locally

Â Minkowskian reference system, or reference frame.

Â In such a reference frame, in the vicinity of point x0,

Â which has the following property.

Â That we choose a coordinate system such that in the vicinity of this guy,

Â g mu nu (X0) is at the mu nu + corrections.

Â 0:44

Some corrections which I'm going to specify a bit later.

Â And gamma mu mu alpha,

Â around that point,

Â = 0 + some corrections.

Â It doesn't necessarily mean that the derivatives of g or

Â gamma are also vanishing at this point.

Â But the idea of this reference frame, reference system, is very simple.

Â Suppose we have any curved space, like any curved space.

Â But in the vicinity of any point,

Â when the vicinity is very small the space looks almost as flat.

Â Almost means that there are corrections, but it looks almost as flat.

Â And then, in that flat fraction of this space,

Â we can specify some coordinate system because it's almost flat.

Â We can choose locally Minkowskian reference system,

Â where the metric is like that, and this guy is 0.

Â So this is just a simple idea lying behind this thing.

Â But let me explain mathematically how one can do this fixing.

Â Suppose we have two reference systems in the vicinity of this point,

Â the regional, k, and another one, k bar.

Â 2:13

Let me choose for both of these reference system, the origin at X0.

Â And introduce the coordinate xi, which is X- X0,

Â xi bar, which is X bar- X0.

Â 2:30

Then I'd do a coordinate transformation from xi to xi bar.

Â It means that xi bar is a function of xi.

Â And because I am the vicinity, small vicinity, of this point,

Â I can do Taylor expansion of this guy.

Â Which means that xi mu,

Â as a function of xi,

Â is A mu nu times xi nu + 1/2 B

Â mu nu alpha xi nu xi alpha

Â + of order of xi cubed.

Â Now using the transformation rule for the metric,

Â it is not hard to see that g bar alpha beta as

Â a function of x(0) transforms as follows.

Â A mu alpha, A mu beta,

Â g mu nu (X0) + of order (xi).

Â Well this just follows from the transformation rule of the matrix,

Â when I care only about terms which are, and

Â neglect the terms which are higher powers in xi.

Â So keep only the later term.

Â The transformation rule for gamma is as follows.

Â Gamma alpha beta gamma

Â (X0) = A alpha mu,

Â A mu beta, A sigma gamma,

Â Gamma mu nu sigma (X0)- B.

Â So it's minus.

Â This is a continuation.

Â B alpha mu nu, A mu beta,

Â A nu gamma + of order (xi).

Â 4:47

Notice that from this fact, B is symmetric.

Â B mu alpha is equal to B mu alpha nu.

Â That one has to bear in mind.

Â Now it is not hard to see that this guy,

Â this matrix, is a four by four matrix.

Â So it has 16 components.

Â A has 16 components.

Â This guy is symmetric, four by four matrix, so g has ten components.

Â So of course using this 16 guys, we can fix ten

Â components of this to be equal eta alpha beta.

Â That's exactly what is said here.

Â So we transformed the regional reference system, where this guy

Â was complicated function, to such guy up to, of course, corrections.

Â 5:54

So what are the remaining six transformations?

Â It's very simple.

Â This matrix has isometries, has such transformations which do not change it.

Â This Lorentz boost, rotations, etc., etc.

Â And there are six of them which keep this form of the matrix,

Â so this is perfectly clear.

Â So there are three Lorentz boosts, three rotations, which give this guy.

Â And also, if we choose B alpha

Â mu nu to be equal to A alpha gamma,

Â Gamma gamma mu nu (X0).

Â We also can set, up to this terms, we also can set this quantity to be 0.

Â 6:46

So when it is possible to do that,

Â it is possible to do that when gamma is symmetric in this indices.

Â It is possible to do that because this guy is symmetric in it's lower case indices.

Â So let us look more closer to what is going on.

Â Consider so-called torsion.

Â Torsion.

Â Torsion is just the difference of two gammas.

Â 7:24

Despite the fact that this guy and this guy do not transform as tensors.

Â Remember, in the transformation rule,

Â there is a part which makes it different from a tensor.

Â But the difference between these two guys does transform as a tensor.

Â One can check it by explicitly looking at the expressions

Â how these guys are transforming.

Â So this guy does transform as a tensor.

Â 7:44

So now if we can fix a reference frame where

Â gamma is 0, then this guy is 0 in that

Â locally Minkowskian reference system.

Â But because this is a tensor, this is a tensor,

Â it means that it transforms multiplactively.

Â So it multiplies by something.

Â So if it is 0 in one reference frame, in one reference system,

Â then it is 0 in any other.

Â 8:18

But now one can see that if it is 0,

Â it means that Gamma mu nu alpha is symmetric.

Â Which is in accordance with what I have just said here.

Â So the manifolds where the torsion is 0,

Â where gamma is symmetric, and where one can

Â use locally Minkowskian reference system.

Â 8:57

related to curved spacetimes to Riemannian geometry.

Â So let me stress that metric tensor, g mu nu, and

Â curvature, Gamma mu nu alpha, are not independent.

Â They are interrelated, for Riemannian manifolds at least.

Â So let us see how it happens.

Â Consider the following covariant derivative of differential of vector A mu.

Â 9:25

So one can write it like this, D alpha g

Â mu nu A nu.

Â So using just the definition of this guy.

Â And this, if we use Leibniz rule,

Â this is just (D alpha g mu nu)

Â A nu + g mu nu D alpha A nu.

Â But to have an agreement between different definitions, and

Â to have in mind this is a tensor quantity, we have to have that this is 0.

Â 10:41

This is very important relation which establishes relation between metric and

Â connection.

Â Let me write this equation like this.

Â D alpha, well let us just use the definition of the covariant derivative.

Â D alpha g mu nu- Gamma

Â mu nu alpha- Gamma nu mu alpha.

Â 11:12

This is just a vertical line, slash.

Â Here I just used the definition of the covariant derivative.

Â And I used the multiplication of gamma to the metric tensor such that I

Â lowered some of the indices.

Â Now we can write this relation in different manner by just

Â reshuffling the indices.

Â So it means that I can write

Â another equation,

Â which is d nu g alpha mu- Gamma

Â alpha mu nu- Gamma mu / alpha nu.

Â This is another way of writing the same equation and there is third way.

Â Let me write it here.

Â 0 = d mu g nu

Â alpha- Gamma

Â alpha nu mu- Gamma

Â nu alpha mu.

Â So we get three equations.

Â Actually, on three quantities, we have to use the symmetricity property of this guy.

Â 12:37

Is symmetric.

Â So then we have algebraic equation, algebraic equation on Gamma.

Â So we can solve these three equations with respect to Gamma to

Â obtain the following equation for Gamma.

Â Gamma alpha mu nu is nothing but

Â 1/2 (d nu g alpha mu + d

Â mu g nu alpha- d alpha g mu nu).

Â 13:14

Or Gamma, with uppercase index alpha,

Â mu nu + 1/2 g alpha beta

Â (d nu g beta mu + d mu g nu

Â beta- d beta g mu nu).

Â Now we see that this connection Gamma is nothing but the Christoffel symbol that

Â have appeared in the previous lecture, at the very end of the previous lecture.

Â Where we found it appearing in the equation for

Â the geodesic, for the extreme world line and space time.

Â So this is just nothing but Christoffel symbol.

Â Moreover, as by-product,

Â we obtain another interpretation of the geodesic equation.

Â Just to remind you, geodesic equation looks like this.

Â It's just d/ds of

Â A mu + Gamma mu nu alpha

Â a U nu U alpha = 0.

Â But this thing can be written

Â as Zed dot nu d/d Zed nu.

Â I mean this derivative can be written like this,

Â but this is nothing but U nu d/d Zed nu.

Â But then this whole equation can be written, as a result,

Â in the following form.

Â It's U nu (S) D nu

Â of U mu (S) = 0.

Â So this is a geodesic equation.

Â This is another form of the geodesic equation.

Â And this just means that we have the following thing.

Â We have a geodesic.

Â We have a tangent vector to it.

Â We take the covariant derivative, this, and then project on itself,

Â on the tangent vector to the geodesic.

Â So geodesic equation just has the following geometric meaning.

Â That tangent vector to the geodesic is covariantly constant along the geodesic.

Â That's the meaning of the geodesic equation that we have now.

Â [MUSIC]

Â