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qr code reader c# .net R sin in VS .NET
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(R) (R sin ) This example has shown us how to carry out the rst step needed to compute curvature, getting the Christoffel symbols. Usually, we will carry our calculations further in the local frame using the Ricci rotation coef cients. The procedure used to do so is explored in the next section. Computing Curvature
Working in a coordinate basis to nd all the components of the af ne connection and then calculating an endless series of derivatives to get the Riemann tensor is a hairy mess that one would like to avoid. Thankfully, a method was developed by Cartan that is a bit more sophisticated but saves a great deal of tedium. Cartan s Structure Equations
To get the curvature and ultimately the Einstein tensor so that we can learn about gravitational elds, we need the Riemann tensor and the quantities that can be derived from it. Previously we learned that R a bcd = c a bd
d This is a daunting formula that seems to ooze airs of tedious calculation. Only a person who was completely insane would enjoy calculating such a beast. Luckily, Cartan has saved us with a more compact equation that is a bit easier on the mind once you get the hang of it. The key insight is to notice that we obtain the Riemann tensor from the Christoffel symbols by differentiating them. In the previous section, we calculated a set of curvature one forms, which contain the Ricci rotation coef cients as their components in the local frame. So it makes perfect sense that we should differentiate these things to get the curvature tensor. We do so in a way applicable to forms and de ne a new set of quantities called the curvature two forms which we label by a b in this book. They are given by a b
(5.27) It turns out that they are related to the Riemann tensor in the following way: a b
1 a R bcd c d 2
(5.28) Now, notice that the Riemann tensor in this equation is expressed in the orthonormal basis (there are hats on the indices). That means we need to transform to the coordinate basis if we need or want to work there. This is done by using the following handy transformation formula: R a bcd = 1 a e e R f g h f b g c h d
(5.29) Let s do a few calculations to see how to use these quantities. We start with a very simple case and then consider one that s a bit more complicated. EXAMPLE 52 Perhaps the simplest metric for which we have nonzero curvature is the unit sphere, where ds 2 = d 2 + sin2 d 2 Find the Ricci scalar using Cartan s structure equations. Cartan s Structure Equations
SOLUTION 52 This is a very simple metric. The basis one forms (for the orthonormal basis) are given by = d = sin d
In the quiz, you will show that the nonzero Ricci rotation coef cient for this metric is given by
= cot
In this case, (5.27) becomes
= cot
=d =d =d
= d cot = d (cos d ) cos sin d sin
Moving from the rst to the second line, we used we obtain
= 0. Moving on, = d (cos d ) = sin d d
The next step is to rewrite the differentials in terms of the nonholonomic basis one forms. Looking at our de nitions = d and = sin d , we see that 1 we can write d = sin , and so we obtain = sin d d = =
(5.30) This result can be used to obtain the components of the Riemann tensor via (5.28). Since there are only two dimensions, this equation will be very simple, because the only nonzero terms are those that contain a b where a = b. This gives 1 1 1 R cd c d = R + R 2 2 2
Cartan s Structure Equations
Now, we use = to rewrite this as
1 R R 2
To get this into a form where we can read off the components of the Riemann tensor from (5.30), we need to use the symmetries of the Riemann tensor. First we need to lower some indices. Since the metric is ds 2 = d 2 + sin2 d 2 , a b is nothing but the identity matrix: a b = 1 0 0 1 Now recall the symmetries of the Riemann tensor: Rabcd = Rbacd = Rabdc These symmetries still apply in the local frame where we are using the nonholonomic basis. Therefore, we can write R = R = R = R = R = R This allows us to rewrite the following: 1 1 R R = R + R = R 2 2 With this change, we get = R

