## Lecture 10: The Riemann Curvature Tensor

10. The Riemann Curvature Tensor

First, we need to know how to translate a vector along a curve C. Let Xj be a vector field. We have seen that a parallel vector field of constant length on M must satisfy

 DXjdt = 0 ......     (I)

for any path C in M.

Definition 9.1 The vector field Xj is parallel along the curve C if it satisfies

 DXjdt = dXjdt + ijh Xi dxhdt = 0,

for the specific curve C.

If Xj is parallel along C, which has parametrization with domain [a, b] and corresponding points and on M, then, since

 dXjdt = -ijh Xi dxhdt .........     (I)

we can integrate to obtain

 Xj() = Xj() - ba ijh Xi dxhdt .........     (II)

Question Given a fixed vector Xj() 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 Xj 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 Xj(). Note however that it gives Xj as a function of the parameter t, and not necessarily as a well-defined function of position on M. If it does not, then we have a parallelizable manifold.

 Definition 10.2 If Xj() is any vector at the point M, and if C is any path from to in M, then the parallel transport of Xj() along C is the vector Xj() given by the solution to the system (I) with initial conditions given by Xj().

Examples 10.3
(a)
If C is a geodesic in M given by xi = xi(s), where we are using arc-length s as the parameter (see Exercise Set 8 #1) then the vector field dxi/ds is parallel along C. (Note that this field is only defined along C, but (I) still makes sense.) Why? because

 D(dxj/ds)Ds = d2xjds2 + ijh dxids dxhds ,

which must be zero for a geodesic.

(b) Proper Coordinates in Relativity Along Geodesics
According to relativity, we live in a Riemannian 4-manifold 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 xi = xi(t), satisfying the property

 dxidt , dxidt < 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 sub-light 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) = dxi/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

 ddt X, Y = DXdt , Y + X , DYdt ,

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 Non-Geodesics

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 DXj/dt = 0.
Answer To answer this question, let us experiment a little with a fixed vector V = Xj(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

x1(a) = r, x2(a) = s.

Then, choose r and s so small that the following paths are within the coordinate neighborhood in question:

C1: xj(t)=  xi(a) if i 1 or 2 r+tr if i = 1 s if i = 2
C2: xj(t)=  xi(a) if i 1 or 2 r+r if i = 1 s+ts if i = 2
C3: xj(t)=  xi(a) if i 1 or 2 r+(1-t)r if i = 1 s+s if i = 2
C4: xj(t)=  xi(a) if i 1 or 2 r if i = 1 s+(1-t)s if i = 2

These paths are shown in the following diagram.

Now, if we parallel transport Xj(a) along C1, we must have, by (II),

Xj(b) =  Xj(a) - 10 ijh Xi dxhdt dt
(since t goes from 0 to 1 in the path C1)
=  Xj(a) - 10 ij1 Xi r dt
(see the definition of C1 above; only x1 changes...)

Warning:   The integrand term ij1 Xi is not constant, and must be evaluated as a function of t using the path C1. However, if the path is a small one, then the integrand is approximately equal to its value at the midpoint of the path segment:

Xj(b)  Xj(a) - ij1 Xi(midpoint of C1)r
 Xj(a) - ij1 Xi(a) + 0.5 x1 (ij1 Xi)r r

where the partial derivative is evaluated at the point a. Similarly,

Xj(c) =  Xj(b) - 10 ij2 Xi s dt
 Xj(b) - ij2 Xi(midpoint of C2)s
 Xj(b) - ij2 Xi(a) + x1 (ij2 Xi)r + 0.5 x2 (ij2 Xi)s s

where all partial derivatives are evaluated at the point a. (This makes sense because the field is defined where we need it.)

Xj(d) =  Xj(c) + 10 ij1 Xi r dt
 Xj(c) + ij1 Xi(midpoint of C3)r
 Xj(c) + ij1 Xi(a) + 0.5 x1 (ij1 Xi)r + x2 (ij1 Xi)s r

and the vector arrives back at the point a according to

X*j(a) =  Xj(d) + 10 ij2 Xi s dt (X*j(a) is the new vector at the point a)
 Xj(d) + ij2 Xi(midpoint of C4)r
 Xj(d) + ij2 Xi(a) + 0.5 x2 (ij2 Xi)s s

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) - Xj(a) = Xj x2 (ij1 Xi) - x1 (ij2 Xi) rs

To analyze the partial derivatives in there, we first use the product rule, getting

 Xj Xi x2 ij1 + ij1 x2 Xi - Xi x1 ij2 - ij2 x1 Xi rs .........   (III)

Next, we recall the "chain rule" formula

 DXjdt = Xj|h dxhdt

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

Xj|h = 0

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 dxh/dt are non-zero. By definition of the partial derivatives, this means that

 Xjxh + ijhXi = 0,

so that

 Xjxh = - ijhXi.

We now substitute these expressions in (III) to obtain

 Xj Xi x2 ij1 - ij1pi2Xp - Xi x1 ij2 + ij2pi1Xp rs

where everything in the square brackets is evaluated at a. Now change the dummy indices in the first and third terms and obtain

 Xj x2 pj1 - ij1pi2 - x1 pj2 + ij2pi1 Xprs

This formula has the form

 Xj Rpj12Xp 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 Rpj12 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 Rbacd in terms of the partial derivatives of the gij. First, we have the notation

 gij,k = gijxk

for partial derivatives, and remember that these are not tensors. Then, the Christoffel symbols and curvature tensor are given in the convenient form

bac =
 12 gak(gck,b + gkb,c - gbc,k)
(Notice that the indices c and d are switched in the negative terms.)

We can lower the index by defining

Rabcd = gbiRaicd

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

 Rabcd = 12 (gbc,ad - gbd,ac + gad,bc - gac,bd) + ajdbjc - ajcbjd

(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

ijk = gpjipk.

2. Some symmetry properties: Rabcd = -Rabdc = -Rbacd and Rabcd = Rcdab (see the exercise set)

3. We can raise the index again by noting that

gbiRaicd = gbigijRajcd = bjRajcd = Rabcb.

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:

 Rabcd,e + Rabec,d + Rabde,c = 12 (gad,bce - gac,bde + gbc,ade - gbd,ace     + gac,bed - gae,bcd + gbe,acd - gbc,aed       + gae,bdc - gad,bec + gbd,aec - gbe,adc) = 0

Now, I claim this is also true for the covariant partial derivatives:

 Bianchi Identities Rabcd|e + Rabec|d + Rabde|c = 0

Indeed, let us evaluate the left-hand side at any point m M. Choose an inertial frame at m. Then the left-hand side coincides with Rabcd,e + Rabec,d + Rabde,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 Rab = Raibi = gijRajbi we can raise the indices of any tensor in the usual way, getting Rab = gaigbjRij.

In the exercise set, you will show that it is symmetric, and also (up to sign) is the only non-zero contraction of the curvature tensor.

We also define the Ricci scalar by

R = gabRab = gabgcdRacbd

The last thing we will do in this section is play around with the Bianchi identities. Multiplying them by gbc:

gbc[Rabcd|e + Rabec|d + Rabde|c] = 0

Since gij|k = 0 (see Exercise Set 8), we can slip the gbc into the derivative, getting

-Rad|e + Rae|d + Racde|c = 0.

Contracting again gives

or

-R|e + Rde|d + Rdcde|c = 0,

or

-R|e + Rde|d + Rce|c = 0.

Combining terms and switching the order now gives

 Rbe|b - 12 R|e = 0,

or

 Rbe|b - 12 be R|b = 0

Multiplying this by gae, we now get

 Rab|b - 12 gabR|b = 0,

or

Gab|b = 0,

where we make the following definition:

Einstein Tensor

 Gab = Rab - 12 gabR

Einstein's field equation for a vacuum states that

Gab = 0

(as we shall see later...).

Example 10.7
Take the 2-sphere of radius r with polar coordinates, where we saw that

g**=  r2sin2 0 0 r2
.

The coordinates of the covariant curvature tensor are given by

 Rabcd = 12 (gbc,ad - gbd,ac + gad,bc - gac,bd) + ajdbjc - ajcbjd .

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 non-vanishing second derivative of g** is

g, = 2r2(cos2 - sin2),

giving

ajcjbd = jj = 0,

since b = d = eliminates the second term (two of these indices need to be in order for the term not to vanish.)

 ajdjbc = jj = 14 2 cos sin (-2r2sin cos ) = -r2cos2.

Combining all these terms gives

R = r2(sin2 - cos2) + r2cos2 = r2sin2.

We now calculate

Rab = gcdRacbd R = gR = sin2

and

 R = gR = sin2sin2 = 1.

All other terms vanish, since g is diagonal and R**** is assymetric. Click here to see an instance of this! This gives

R =  gabRab = gR + gR
=  1 r2sin2 (sin2) + 1r2 = 2r2 .

 Summary of Some Properties of Curvature Etc. abc = cba abc = cba Rabcd = Rabdc Rabcd = -Rbacd Rabcd = -Rabdc Rabcd = Rcdab       (Note that a,b and c,d always go together.) Rab = Raibi = gijRajbi Rab = Rba R = gabRab = gacgbdRabcd Rab = gaiRib Rab = gaigbjRij Gab = Rab - gabR/2

Exercise Set 10

1. Derive the formula for the covariant form of the curvature tensor in terms of the gij.

2. (a) Show that the curvature tensor is antisymmetric in the last pair of variables:

(b) Use part (a) to show that the Ricci tensor is, up to sign, the only non-zero contraction of the curvature tensor.
(c) Prove that the Ricci tensor is symmetric.

3. (cf. Rund, pp. 82-83)
(a) Show that

Xj|h|k=  xk (Xj|h) + mjk(Xm|h) - hlk(Xj| l)
=  2Xjxhxk + Xl xk ljh + ljh xk)Xl + mjk xh)Xm + mjklmhXl - hlk(Xj| l)

(b) Deduce that

Xj|h|k - Xj|k|h = RljhkXl - ShlkXj| l = RljhkXl

where

Shlk = hlk - klh = 0.
(c) Now deduce that the curvature tensor is indeed a type (1, 3) tensor.

4. Show that Rabcd is antisymmetric on the pairs (a, b) and (c, d).

5. Show that Rabcd = Rcdab by first checking the identity in an inertial frame.

Last Updated: January, 2002