0:13
To move further, let me define so-called locally
Minkowskian reference system, or reference frame.
In such a reference frame, in the vicinity of point x0,
which has the following property.
That we choose a coordinate system such that in the vicinity of this guy,
g mu nu (X0) is at the mu nu + corrections.
0:44
Some corrections which I'm going to specify a bit later.
And gamma mu mu alpha,
around that point,
= 0 + some corrections.
It doesn't necessarily mean that the derivatives of g or
gamma are also vanishing at this point.
But the idea of this reference frame, reference system, is very simple.
Suppose we have any curved space, like any curved space.
But in the vicinity of any point,
when the vicinity is very small the space looks almost as flat.
Almost means that there are corrections, but it looks almost as flat.
And then, in that flat fraction of this space,
we can specify some coordinate system because it's almost flat.
We can choose locally Minkowskian reference system,
where the metric is like that, and this guy is 0.
So this is just a simple idea lying behind this thing.
But let me explain mathematically how one can do this fixing.
Suppose we have two reference systems in the vicinity of this point,
the regional, k, and another one, k bar.
2:13
Let me choose for both of these reference system, the origin at X0.
And introduce the coordinate xi, which is X- X0,
xi bar, which is X bar- X0.
2:30
Then I'd do a coordinate transformation from xi to xi bar.
It means that xi bar is a function of xi.
And because I am the vicinity, small vicinity, of this point,
I can do Taylor expansion of this guy.
Which means that xi mu,
as a function of xi,
is A mu nu times xi nu + 1/2 B
mu nu alpha xi nu xi alpha
+ of order of xi cubed.
Now using the transformation rule for the metric,
it is not hard to see that g bar alpha beta as
a function of x(0) transforms as follows.
A mu alpha, A mu beta,
g mu nu (X0) + of order (xi).
Well this just follows from the transformation rule of the matrix,
when I care only about terms which are, and
neglect the terms which are higher powers in xi.
So keep only the later term.
The transformation rule for gamma is as follows.
Gamma alpha beta gamma
(X0) = A alpha mu,
A mu beta, A sigma gamma,
Gamma mu nu sigma (X0)- B.
So it's minus.
This is a continuation.
B alpha mu nu, A mu beta,
A nu gamma + of order (xi).
4:47
Notice that from this fact, B is symmetric.
B mu alpha is equal to B mu alpha nu.
That one has to bear in mind.
Now it is not hard to see that this guy,
this matrix, is a four by four matrix.
So it has 16 components.
A has 16 components.
This guy is symmetric, four by four matrix, so g has ten components.
So of course using this 16 guys, we can fix ten
components of this to be equal eta alpha beta.
That's exactly what is said here.
So we transformed the regional reference system, where this guy
was complicated function, to such guy up to, of course, corrections.
5:54
So what are the remaining six transformations?
It's very simple.
This matrix has isometries, has such transformations which do not change it.
This Lorentz boost, rotations, etc., etc.
And there are six of them which keep this form of the matrix,
so this is perfectly clear.
So there are three Lorentz boosts, three rotations, which give this guy.
And also, if we choose B alpha
mu nu to be equal to A alpha gamma,
Gamma gamma mu nu (X0).
We also can set, up to this terms, we also can set this quantity to be 0.
6:46
So when it is possible to do that,
it is possible to do that when gamma is symmetric in this indices.
It is possible to do that because this guy is symmetric in it's lower case indices.
So let us look more closer to what is going on.
Consider so-called torsion.
Torsion.
Torsion is just the difference of two gammas.
7:24
Despite the fact that this guy and this guy do not transform as tensors.
Remember, in the transformation rule,
there is a part which makes it different from a tensor.
But the difference between these two guys does transform as a tensor.
One can check it by explicitly looking at the expressions
how these guys are transforming.
So this guy does transform as a tensor.
7:44
So now if we can fix a reference frame where
gamma is 0, then this guy is 0 in that
locally Minkowskian reference system.
But because this is a tensor, this is a tensor,
it means that it transforms multiplactively.
So it multiplies by something.
So if it is 0 in one reference frame, in one reference system,
then it is 0 in any other.
8:18
But now one can see that if it is 0,
it means that Gamma mu nu alpha is symmetric.
Which is in accordance with what I have just said here.
So the manifolds where the torsion is 0,
where gamma is symmetric, and where one can
use locally Minkowskian reference system.
8:57
related to curved spacetimes to Riemannian geometry.
So let me stress that metric tensor, g mu nu, and
curvature, Gamma mu nu alpha, are not independent.
They are interrelated, for Riemannian manifolds at least.
So let us see how it happens.
Consider the following covariant derivative of differential of vector A mu.
9:25
So one can write it like this, D alpha g
mu nu A nu.
So using just the definition of this guy.
And this, if we use Leibniz rule,
this is just (D alpha g mu nu)
A nu + g mu nu D alpha A nu.
But to have an agreement between different definitions, and
to have in mind this is a tensor quantity, we have to have that this is 0.
10:41
This is very important relation which establishes relation between metric and
connection.
Let me write this equation like this.
D alpha, well let us just use the definition of the covariant derivative.
D alpha g mu nu- Gamma
mu nu alpha- Gamma nu mu alpha.
11:12
This is just a vertical line, slash.
Here I just used the definition of the covariant derivative.
And I used the multiplication of gamma to the metric tensor such that I
lowered some of the indices.
Now we can write this relation in different manner by just
reshuffling the indices.
So it means that I can write
another equation,
which is d nu g alpha mu- Gamma
alpha mu nu- Gamma mu / alpha nu.
This is another way of writing the same equation and there is third way.
Let me write it here.
0 = d mu g nu
alpha- Gamma
alpha nu mu- Gamma
nu alpha mu.
So we get three equations.
Actually, on three quantities, we have to use the symmetricity property of this guy.
12:37
Is symmetric.
So then we have algebraic equation, algebraic equation on Gamma.
So we can solve these three equations with respect to Gamma to
obtain the following equation for Gamma.
Gamma alpha mu nu is nothing but
1/2 (d nu g alpha mu + d
mu g nu alpha- d alpha g mu nu).
13:14
Or Gamma, with uppercase index alpha,
mu nu + 1/2 g alpha beta
(d nu g beta mu + d mu g nu
beta- d beta g mu nu).
Now we see that this connection Gamma is nothing but the Christoffel symbol that
have appeared in the previous lecture, at the very end of the previous lecture.
Where we found it appearing in the equation for
the geodesic, for the extreme world line and space time.
So this is just nothing but Christoffel symbol.
Moreover, as by-product,
we obtain another interpretation of the geodesic equation.
Just to remind you, geodesic equation looks like this.
It's just d/ds of
A mu + Gamma mu nu alpha
a U nu U alpha = 0.
But this thing can be written
as Zed dot nu d/d Zed nu.
I mean this derivative can be written like this,
but this is nothing but U nu d/d Zed nu.
But then this whole equation can be written, as a result,
in the following form.
It's U nu (S) D nu
of U mu (S) = 0.
So this is a geodesic equation.
This is another form of the geodesic equation.
And this just means that we have the following thing.
We have a geodesic.
We have a tangent vector to it.
We take the covariant derivative, this, and then project on itself,
on the tangent vector to the geodesic.
So geodesic equation just has the following geometric meaning.
That tangent vector to the geodesic is covariantly constant along the geodesic.
That's the meaning of the geodesic equation that we have now.
[MUSIC]
Emil AkhmedovAssociate Professor
