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(5.10)
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We have introduced a new quantity, a bc , which are the Ricci rotation coef cients. They are related to the Christoffel symbols but are, in fact, different. We use them to obtain the Christoffel symbols by applying a transformation to the coordinate basis. We will illustrate this below.
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The curvature one forms satisfy certain symmetry relations that will be useful during calculation. In particular
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(5.11) =
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j i
(5.12)
Here i and j are spatial indices and i = j. To see how these relations work, we can raise and lower indices with the metric in the local frame. If a b = diag (1, 1, 1, 1), then using (5.11) we have
i j
= i i
ij
ij
ji
= j j
This works because the elements of a b are only on the diagonal, so we must have equal indices. Moreover, since we are dealing only with spatial components in this case, each instance of i j introduces a minus sign. Of course the signs used here are by convention. Try working this out with a b = diag ( 1, 1, 1, 1) to see if things work out differently. When calculating d a , it is helpful to recall the following. Let and be two arbitrary forms. Then d (d ) = 0 = (5.13)
So we recall (selecting the coordinate r only for concreteness) that d (dr ) = 0 and dr dr = 0. The Christoffel symbols can be obtained from the Ricci rotation coef cients using the transformation matrices that take us from the orthonormal basis to the coordinate basis. In particular, we have
a bc
1 a
e f
(5.14)
where a bc are the Christoffel symbols. Remember, a plain index indicates that the coordinate basis is being used, while the hatted index indicates that an orthonormal basis is being used. In this book when we write d e f , we are referring to the Ricci rotation coef cients in the orthonormal basis. The techniques used in this chapter are far less tedious than using the coordinate methods for calculating the connection. In the next section, we will carry
Cartan s Structure Equations
the method forward and see how we can calculate the Riemann tensor. The procedure used here will be to calculate both sides of (5.9) and compare to make an educated guess as to what the curvature one forms are (this is sometimes called the guess method). Once we have done that, we can use the symmetry properties in (5.12) to see if we can nd any more of the curvature one forms. Let s apply this method to an example. EXAMPLE 5-1 Consider the Tolman-Bondi-de Sitter metric, given by ds 2 = dt 2 e 2
(t,r )
dr 2 R 2 (t, r ) d 2 R 2 (t, r ) sin2 d 2
(5.15)
This metric arises, for example, in the study of spherical dust with a cosmological constant. Find the Ricci rotation coef cients for this metric. SOLUTION 5-1 First we need to examine the metric (5.15) and write down the basis one forms. Since we are asked to nd the Ricci rotation coef cients, we will work with the noncoordinate orthonormal basis. Writing (5.15) as ds 2 = t
r
we identify the noncoordinate basis one forms as t = dt, r = e
(t,r )
= R(t, r ) d ,
= R(t, r ) sin d (5.16)
As we proceed, at times we will denote differentiation with respect to time with a dot (i.e., d f /dt = f ) and differentiation with respect to r with a prime (i.e., d f /dr = f ). Now we apply (5.9) to each basis one form. The rst one does not give us much information, since d t = d (dt) = 0 Let s move on to the second one form. This one gives us d r = d e
(t,r )
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