Fig. 3-2. A torus is an example of a manifold. in .NET framework

Maker QR Code JIS X 0510 in .NET framework Fig. 3-2. A torus is an example of a manifold.

Fig. 3-2. A torus is an example of a manifold.
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Fig. 3-3. A crude abstract illustration of a manifold. The manifold cannot be covered by
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a single coordinate system, but we cover it with a set of coordinate patches that we designate by Ui . These patches may overlap. The points in each patch can be mapped to at Euclidean space. Here we illustrate a mapping for a coordinate patch we ve called U1 . A point p that belongs to the manifold M (say in U1 ) is mapped to a coordinate that we designate x a (p ). (Houghston and Todd, 1992)
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The notion of a curve as the plot of some function is a familiar one. In the context of relativity, however, it is more useful to think of curves in terms of parameterization. In this view, a point that moves through space traces out the curve. The parameter of the curve, which we denote by , is a real number. We describe a curve by a set of parametric equations that give the coordinates along the curve for a given value of : x a = x a ( ) (3.1)
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In n dimensions, since there are n coordinates x a , there will be n such equations. EXAMPLE 3-1 Consider the curve traced out by the unit circle in the plane. Describe a parametric equation for the curve. SOLUTION 3-1 We call the parameter . Since we are in two dimensions, we need two functions that will determine (x, y) for a given value of . Since the equation of a circle is given by x 2 + y 2 = r 2 , in the case of the unit circle the equation that describes
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Fig. 3-4. A parabola with x 0.
the curve is x 2 + y2 = 1 This is an easy equation to describe parametrically. Since we know that cos2 + sin2 = 1, we choose the parametric representation to be x = cos and y = sin
EXAMPLE 3-2 Find a parameterization for the curve y = x 2 such that x 0. SOLUTION 3-2 The curve is shown in Fig. 3-4. To parameterize this curve with the restriction that x 0, we can take x to be some positive number . To ensure that the number is positive we square it; i.e., x = 2 Since y = x 2 , it follows that y = 4 will work.
Tangent Vectors and One Forms, Again
A curved space or spacetime is one that is going to change from place to place. As such, in a curved spacetime, one cannot speak of a vector that stretches from point to point. Instead, we must de ne all quantities like vectors and one forms locally. In the previous chapter, we basically tossed out the idea that a basis vector is de ned as a partial derivative along some coordinate direction. Now
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let s explore that notion a bit more carefully. We can do so with parameterized curves. Let x a ( ) be a parameterized curve. Then the components of the tangent vector to the curve are given by dx a d EXAMPLE 3-3 Describe the tangent vectors to the curves in Examples 3-1 and 3-2. SOLUTION 3-3 In Example 3-1, we parameterized the curve traced out by the unit circle with (cos , sin ). Using (3.2), we see that the components of the tangent vector to the unit circle are given by ( sin , cos ). Calling the tangent vector and writing the basis vectors in cartesian coordinates as ex and e y , we have = sin ex + cos e y Now we consider Example 3-2, where the given curve was y = x 2 . We found that this curve could parameterize the curve with 2 , 4 . Using (3.2), we obtain the tangent vector v = 2 , 4 3 . Let s invert the relation used to describe the curve. We have = Therefore, we can write v as v = 2 xex + 4x 3/2 e y Now that we ve found out how to make a vector that is tangent to a parameterized curve, the next step is to nd a basis. Let s expand this idea by considering some arbitrary continuous and differentiable function f . We can compute the d derivative of f in the direction of the curve by d f . Therefore, d is a vector that d maps f to a real number that is given by d f . The chain rule allows us to write d this as d f xa df dx a f = = d d x a d x a x (3.2)
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