qr code reader c# .net Copyright 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use. in VS .NET

Make QR in VS .NET Copyright 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

Copyright 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.
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Tensor Calculus
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where is a scalar and is invariant for all vectors V a , then a is a one form. We can carry this process further. If Tab V a = Ub such that Ub is a one form, then Tab is a (0, 2) tensor. Another test on Tab is Tab V a W b = where V a and W b are vectors and is an invariant scalar, then Tab is a (0, 2) tensor.
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The Importance of Tensor Equations
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In physics we seek invariance, i.e., we seek laws of physics written in an invariant form that is true for all observers. Tensors are a key ingredient in this recipe, because if a tensor equation is true in one coordinate system then it is true in all coordinate systems. This can greatly simplify analysis because we can often transform to a coordinate system where the mathematics will be easier. There we can nd the form of a result we need and then express it in another coordinate system if desired. A simple example of this is provided by the vacuum eld equations. We will see that these can be expressed in the form of a (0, 2) tensor called the Ricci tensor, where Rab = 0 It is immediately obvious that this equation is true in any coordinate system. Let s transform to a different coordinate system using the [Lambda] matrix discussed Rab =
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c a d b Rc d =
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x c x d Rc d x a x b
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On the right-hand side, the transformation of 0 is of course just 0, so we have xc xd Rc d = 0 x a x b
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xc xd xd xb
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We can divide both sides by , which gives
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Tensor Calculus
Rc d = 0
The Covariant Derivative
Consider the problem of taking the derivative of a vector. In ordinary cartesian coordinates, this is not really that complicated of an issue because the basis vectors are constant. So we can get the derivative of a vector by differentiating its components. However, in general, things are not so straightforward. In curved spaces the basis vectors themselves may vary from point to point. This means that when we take a derivative of a vector, we will have to differentiate the basis vectors as well. Consider some vector A. To compute the derivative, we expand the vector in a basis and then apply the Leibniz rule ( fg) = f g + g f to obtain A Ab = a A b eb = e + Ab a (eb ) a a b x x x x In cartesian coordinates, we could just throw away the second term. But this isn t true in general. To see how this works in practice, we turn to a speci c example. We start with something familiar the spherical polar coordinates. Cartesian coordinates are related to spherical coordinates in the following way: x = r sin cos y = r sin sin z = r cos
(4.1)
The rst step in this exercise will be to work out the basis vectors in spherical coordinates in terms of the cartesian basis vectors. We ll do this using the procedures outlined in 2. This means we will need the transformation matrices that allow us to move back and forth between the two coordinate systems. Again, we denote the transformation matrix by the symbol a b , where an element of this matrix is given by xa xb
Tensor Calculus
We will consider the unprimed coordinates to be (x, y, z) and the primed coordinates to be (r, , ). In that case, the transformation matrix assumes the form x
=
r x x
y r y y
z r z z
(4.2)
Using the relationship among the coordinates described by (4.1), we obtain sin cos = r cos cos r sin sin sin sin r cos sin r sin cos cos r sin 0
(4.3)
Now we have the machinery we need to work out the form of the basis vectors. We have already seen many times that the basis vectors transform as eb = ab ea . So let s write down the basis vectors for spherical coordinates as expansions in terms of the cartesian basis using (4.3). We get er = e = e =
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