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(c) r 2 cos cos2 sin (d) r 2 cos2 cos sin2
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Tensor Calculus
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For Questions 7 and 8, let ds 2 = 2 du dv (1 u)2 dx 2 (1 + u)2 dy 2 . 7.
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is 1+u 1 + u 1 v 1 u
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The Ricci scalar is (a) 1 (b) 0 (c) u 2 (d) v 2
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yy is given by (a) 0 (b) cos2 x 2 (c) 1+tan yy 2 tan
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(d) 10.
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The Ricci scalar is (a) 0 2 2 2 2 (b) y sin x+y xsin2 x tanx y+x ycos x tan y 2 y sin2 tan2 (c) (d) y sin x + y sin x cot y + x cos x tan2 y
x 2 sin2 x+y sin2 x tan2 y+x cos x tan2 y y 2 sin2 x tan2 y 2 2 2
CHAPTER
Cartan s Structure Equations
Introduction
A coordinate basis is one for which the basis vectors are given by ea = / x a . With the exception of cartesian coordinates, a coordinate basis is not orthonormal. While it is possible to calculate using a coordinate basis, it is often not the best or easiest way to approach a problem. An alternative exists and this is to construct an orthonormal or nonholonomic basis. Physically, this means working in an observer s local frame. To express results in the global coordinates, we use a transformation that can be constructed easily by looking at the metric. In this chapter we will begin by introducing this concept and then show how to transform between the two. Once we have this concept in place, we will develop a new set of equations that can be used to nd the Riemann tensor for a
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Cartan s Structure Equations
given metric. This procedure appears a bit daunting at rst but is actually much nicer to use than the methods we have discussed so far.
Holonomic (Coordinate) Bases
The most natural choice of a basis that comes to mind is to de ne the basis vectors directly in terms of derivatives of coordinates. Given coordinates x a , we de ne basis vectors and basis one forms in the following way: ea = a = xa and a = dx a (5.1)
When a basis set is de ned exclusively in terms of derivatives with respect to the coordinates, we call it a holonomic basis or a coordinate basis. A given vector V can be expanded in this basis as V = V a ea As an example, consider spherical polar coordinates ds 2 = dr 2 + r 2 d 2 + r 2 sin2 d 2 The coordinate basis vectors are er = r = , r e = = , e = =
The important thing to notice here is that not all of these basis vectors are of unit length. In addition, they do not have the same dimensions. These considerations will lead us to a different basis that we will examine below. In a coordinate basis, the basis vectors satisfy the following relationship: ea eb = gab Furthermore, we can write v w = gab v a w b (5.3) (5.2)
Above, we brie y mentioned a problem with a coordinate basis. If we choose a coordinate basis, it may not be orthonormal. We can see this in the case of
Cartan s Structure Equations
spherical polar coordinates using (5.2). Looking at the line element ds 2 = dr 2 + r 2 d 2 + r 2 sin2 d 2 we see that the components of the metric are grr = 1, g = r 2 , and g = r 2 sin2 . Now let s compute the lengths of the basis vectors. Using (5.2), we obtain er er = 1 |er | = grr = 1 e e = r 2 |e | = g = r e e = r 2 sin2 e = g = r sin Since two of these basis vectors do not have unit length, this set which has been de ned in terms of derivatives with respect to the coordinates is not orthonormal. To choose a basis that is orthonormal, we construct it such that the inner product of the basis vectors satis es g (ea , eb ) = ab This is actually easy to do, and as we will see, it makes the entire machinery of relativity much easier to deal with. A basis de ned in this way is known as a nonholonomic or noncoordinate basis.
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