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10. The Riemann Curvature Tensor
First, we need to know how to translate a vector along a curve C. Let X_{j} be a vector field. We have seen that a parallel vector field of constant length on M must satisfy
dt 
=  0  ...... (I) 
for any path C in M.
Definition 9.1 The vector field X^{j} is parallel along the curve C if it satisfies
for the specific curve C. 
If X^{j} is parallel along C, which has parametrization with domain [a, b] and corresponding points and _{} on M, then, since
dt 
=  _{i}^{j}_{h }X^{i }  dt 
......... (I) 
we can integrate to obtain
X^{j}(_{})  =  X^{j}()    b a 
_{i}^{j}_{h }X^{i }  dt 
......... (II) 
Question Given a fixed vector X^{j}() at the point M, and a curve C originating at , it is possible to define a vector field along C by transporting the vector along C in a parallel fashion?
Answer Yes. Notice that the formula (II) is no good for this, since the integral already requires X^{j} to be defined along the curve before we start. But we can go back to (I), which is a system of first order linear differential equations. Such a system always has a unique solution with given initial conditions specified by X^{j}(). Note however that it gives X^{j} as a function of the parameter t, and not necessarily as a welldefined function of position on M. If it does not, then we have a parallelizable manifold.
Definition 10.2 If X^{j}() is any vector at the point M, and if C is any path from to _{} in M, then the parallel transport of X^{j}() along C is the vector X^{j}(_{}) given by the solution to the system (I) with initial conditions given by X^{j}(). 
Examples 10.3
(a) If C is a geodesic in M given by x^{i} = x^{i}(s), where we are using arclength s as the parameter (see Exercise Set 8 #1) then the vector field dx^{i}/ds is parallel along C. (Note that this field is only defined along C, but (I) still makes sense.) Why? because
Ds 
=  ds^{2} 
+  _{i}^{j}_{h }  ds 
ds 
, 
which must be zero for a geodesic.
(b) Proper Coordinates in Relativity Along Geodesics
According to relativity, we live in a Riemannian 4manifold M, but not the flat Minkowski space. Further, the metric in M has signature (1, 1, 1, 1). Suppose C is a geodesic in M given by x^{i} = x^{i}(t), satisfying the property
dt 
,  dt 
<  0. 
Recall that we refer to such a geodesic as timelike. Looking at the discussion before Definition 7.1, we see that this corresponds, in Minkowski space, to a particle traveling at sublight speed. It follows that we can choose an orthonormal basis of vectors {V(1), V(2), V(3), V(4)} of the tangent space at m with the property given in the proof of 9.2, with V(4) = dx^{i}/dt. We think of V(4) as the unit vector in the direction of time, and V(1), V(2) and V(3) as the spatial basis vectors. Using parallel translation, we obtain a similar set of vectors at each point along the path. (The fact that the curve is a geodesic guarantees that parallel translation of the time axis will remain parallel to the curve.) Finally, we can use the construction in 9.2 to flesh these frames out to full coordinate systems defined along the path. (Just having a set of orthogonal vectors in a manifold does not give a unique coordinate system, so we choose the unique local inertial one there, because in the eyes of the observer, spacetime should be flat.)
Question Does parallel transport preserve the relationship of these vectors to the curve. That is, does the vector V(4) remain parallel, and do the vectors {V(1), V(2), V(3), V(4)} remain orthogonal in the sense of 8.2?
Answer If X and Y are vector fields, then
dt 
X, Y  =  dt 
,  Y  +  X ,  dt 
, 
where the big D's denote covariant differentiation. (Exercise Set 8 #9). But, since the terms on the right vanish for fields that have been parallel transported, we see that X, Y is independent of t, which means that orthogonal vectors remain orthogonal and that all the directions and magnitudes are preserved, as claimed.
Note At each point on the curve, we have a different coordinate system! All this means is that we have a huge collection of charts in our atlas; one corresponding to each point on the path. This (moving) coordinate system is called the momentary comoving frame of reference and corresponds to the "real life" coordinate systems.
(c) Proper Coordinates in Relativity Along NonGeodesics
If the curve is not a geodesic, then parallel transport of a tangent vector need no longer be tangent. Thus, we cannot simply parallel translate the coordinate axes along the world line to obtain new ones, since the resulting frame may not be Lorentz. We shall see in Section 11 how to correct for that when we construct our comoving reference frames
Question Under what conditions is parallel transport independent of the path? If this were the case, then we could use formula (I) to create a whole parallel vector field of constant length on M, since then DX^{j}/dt = 0.
Answer To answer this question, let us experiment a little with a fixed vector V = X^{j}(a) by parallel translating it around a little rectangle consisting of four little paths. To simplify notation, let the first two coordinates of the starting point of the path (in some coordinates) be given by
Then, choose r and s so small that the following paths are within the coordinate neighborhood in question:
C_{1}: x^{j}(t)  = 


C_{2}: x^{j}(t)  = 


C_{3}: x^{j}(t)  = 


C_{4}: x^{j}(t)  = 

These paths are shown in the following diagram.
Now, if we parallel transport X^{j}(a) along C_{1}, we must have, by (II),
X^{j}(b)  = 

(since t goes from 0 to 1 in the path C_{1})  
= 

(see the definition of C_{1} above; only x_{1} changes...) 
The integrand term _{i}^{j}_{1 }X^{i} is not constant, and must be evaluated as a function of t using the path C_{1}. However, if the path is a small one, then the integrand is approximately equal to its value at the midpoint of the path segment:
X^{j}(b) 



where the partial derivative is evaluated at the point a. Similarly,
X^{j}(c)  = 





where all partial derivatives are evaluated at the point a. (This makes sense because the field is defined where we need it.)
X^{j}(d)  = 





and the vector arrives back at the point a according to
X*^{j}(a)  = 





To get the total change in the vector, you substitute back a few times and cancel lots of terms (including the ones with 0.5 in front), being left with
X*^{j}(a)  X^{j}(a)  =  X^{j}  x^{2} 
(_{i}^{j}_{1 }X^{i})    x^{1} 
(_{i}^{j}_{2 }X^{i})  rs 
To analyze the partial derivatives in there, we first use the product rule, getting
X^{j}  X^{i}  x^{2} 
_{i}^{j}_{1}  +  _{i}^{j}_{1}  x^{2} 
X^{i}    X^{i}  x^{1} 
_{i}^{j}_{2}    _{i}^{j}_{2}  x^{1} 
X^{i}  rs  ......... (III) 
Next, we recall the "chain rule" formula
dt 
=  X^{j}_{h}  dt 
in the homework. Since the term on the right must be zero along each of the path segments we see that (I) is equivalent to saying that the partial derivatives
for every index p and k (and along the relevant path segment; notice that we are taking partial derivatives in the direction of the path, so that they do make sense for this curious field that is only defined along the square path!) since the terms dx^{h}/dt are nonzero. By definition of the partial derivatives, this means that
x^{h} 
+  _{i}^{j}_{h}X^{i}  =  0, 
so that
x^{h} 
=   _{i}^{j}_{h}X^{i}. 
We now substitute these expressions in (III) to obtain
X^{j}  X^{i}  x^{2} 
_{i}^{j}_{1}    _{i}^{j}_{1}_{p}^{i}_{2}X^{p}    X^{i}  x^{1} 
_{i}^{j}_{2}  +  _{i}^{j}_{2}_{p}^{i}_{1}X^{p}  rs 
where everything in the square brackets is evaluated at a. Now change the dummy indices in the first and third terms and obtain
X^{j}  x^{2} 
_{p}^{j}_{1}    _{i}^{j}_{1}_{p}^{i}_{2}    x^{1} 
_{p}^{j}_{2}  +  _{i}^{j}_{2}_{p}^{i}_{1}  X^{p}rs 
This formula has the form
X^{j}  R_{p}^{j}_{12}X^{p} rs  ............ (IV) 
(indices borrowed from the Christoffel symbol in the first term, with the extra index from the x in the denominator) where the quantity R_{p}^{j}_{12} is known as the curvature tensor.
Curvature Tensor

The terms are rearranged (and the Christoffel symbols switched) so you can see the index pattern, and also that the curvature is antisymmetric in the last two covariant indices.
The fact that it is a tensor follows from the homework.
It now follows from a grid argument, that if C is any (possibly) large planar closed path within a coordinate neighborhood, then, if X is parallel transported around the loop, it arrives back to the starting point with change given by a sum of contributions of the form (IV). If the loop is not planar, we choose a coordinate system that makes it planar, and if the loop is too large for a single coordinate chart, then we can break it into a grid so that each piece falls within a coordinate neighborhood. Thus we see the following.
Proposition 10.4 (Curvature and Parallel Transport)
Assume M is simply connected. A necessary and sufficient condition that parallel transport be independent of the path is that the curvature tensor vanishes. 
Definition 10.5 A manifold with zero curvature is called flat. 
Properties of the Curvature Tensor
We first obtain a more explicit description of R_{b}^{a}_{cd} in terms of the partial derivatives of the g_{ij}. First, we have the notation
g_{ij,k}  =  x^{k} 
for partial derivatives, and remember that these are not tensors. Then, the Christoffel symbols and curvature tensor are given in the convenient form

We can lower the index by defining
Substituting the first of the above (boxed) formulas into the second, and using symmetry of the second derivatives and the metric tensor, we find (exercise set)
Covariant Curvature Tensor in Terms of the Metric Tensor

(We can remember this by breaking the indices a, b, c, d into pairs other than ab, cd (we can do this two ways) the pairs with a and d together are positive, the others negative.)
Notes
1. The "new kinds" of Christoffel symbols _{ijk} are given by
2. Some symmetry properties: R_{abcd} = R_{abdc} = R_{bacd} and R_{abcd} = R_{cdab} (see the exercise set)
3. We can raise the index again by noting that
Now, let us evaluate some partial derivatives in an inertial frame (so that we can ignore the Christoffel symbols) cyclically permuting the last three indices as we go:
R_{abcd,e} + R_{abec,d} + R_{abde,c}  =  2 
(g_{ad,bce}  g_{ac,bde} + g_{bc,ade}  g_{bd,ace}
+ g_{ac,bed}  g_{ae,bcd} + g_{be,acd}  g_{bc,aed} + g_{ae,bdc}  g_{ad,bec} + g_{bd,aec}  g_{be,adc}) 
=  0. 
Now, I claim this is also true for the covariant partial derivatives:
Bianchi Identities

Indeed, let us evaluate the lefthand side at any point m M. Choose an inertial frame at m. Then the lefthand side coincides with R_{abcd,e} + R_{abec,d} + R_{abde,c}, which we have shown to be zero. Now, since a tensor which is zero is sone frame is zero in all frames, we get the result!
Definitions 10.6 The Ricci tensor is defined by
we can raise the indices of any tensor in the usual way, getting

In the exercise set, you will show that it is symmetric, and also (up to sign) is the only nonzero contraction of the curvature tensor.
We also define the Ricci scalar by
The last thing we will do in this section is play around with the Bianchi identities. Multiplying them by g^{bc}:
Since g^{ij}_{k} = 0 (see Exercise Set 8), we can slip the g^{bc} into the derivative, getting
Contracting again gives
or
or
Combining terms and switching the order now gives
R^{b}_{eb}    2 
R_{e}  =  0, 
or
R^{b}_{eb}    2 
^{be Rb}  =  0. 
Multiplying this by g^{ae}, we now get
R^{ab}_{b}    2 
g^{ab}R_{b}  =  0, 
or
where we make the following definition:
Einstein Tensor

Einstein's field equation for a vacuum states that
(as we shall see later...).
Example 10.7
Take the 2sphere of radius r with polar coordinates, where we saw that
g_{**}  = 

. 
The coordinates of the covariant curvature tensor are given by
R_{abcd}  =  2 
(g_{bc,ad}  g_{bd,ac} + g_{ad,bc}  g_{ac,bd}) + _{a}^{j}_{d}_{bjc}  _{a}^{j}_{c}_{bjd}  . 
Let us calculate R_{}. (Note: when we use Greek letters, we are referring to specific terms, so there is no summation when the indices repeat!) So, a = c = , and b = d = . (Incidentally, this is the same as R_{} by the last exercise below.)
The only nonvanishing second derivative of g_{**} is
giving
since b = d = eliminates the second term (two of these indices need to be in order for the term not to vanish.)
_{a}^{j}_{d}_{jbc}  =  _{}^{j}_{}_{j}  =  4 
sin 
(2r^{2}sin cos )  =  r^{2}cos^{2}. 
Combining all these terms gives
We now calculate
and
R_{}  =  g^{}R_{}  
=  sin^{2} 
=  1. 
All other terms vanish, since g is diagonal and R_{****} is assymetric. Click here to see an instance of this! This gives
R  = 


= 

Summary of Some Properties of Curvature Etc.
_{a}^{b}_{c} = _{c}^{b}_{a} _{abc} = _{cba} 
Exercise Set 10
1. Derive the formula for the covariant form of the curvature tensor in terms of the g_{ij}.
2. (a) Show that the curvature tensor is antisymmetric in the last pair of variables:
3. (cf. Rund, pp. 8283)
(a) Show that
X^{j}_{hk}  = 


= 

where
4. Show that R_{abcd} is antisymmetric on the pairs (a, b) and (c, d).
5. Show that R_{abcd} = R_{cdab} by first checking the identity in an inertial frame.
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