# qr code reader c# .net A Null Tetrad in .NET Encode QR in .NET A Null Tetrad

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Now that we ve reminded ourselves about null vectors in Minkowski space, let s move on to the topic of this chapter. We begin by introducing a tetrad or set of four basis vectors that are labeled by l, n, m, and m. In a departure from what you re used to in de ning basis vectors, we allow two of these vectors to be complex. Speci cally, l, n are real m, m are complex
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(9.1)
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As the notation makes obvious, m and m are complex conjugates of one another. The second important observation is that these basis vectors are null vectors with respect to the Minkowski metric. Based on our quick review in the last section, this means that l l = n n = m m = m m = 0 (9.2)
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or for example we can write ab l a l b = 0. In addition, these vectors satisfy a set of orthogonality relations that hold for inner products between a real and a complex vector: l m =l m =n m =n m =0 (9.3)
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We need two more pieces of information to characterize the tetrad. The rst is we specify that the two real vectors satisfy l n =1 and nally, the complex vectors satisfy m m = 1 (9.5) (9.4)
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It turns out that a null tetrad can be constructed rather easily with what we already know. We can do it using the basis of orthonormal one forms a . First we state some relationships and then write down a matrix that will transform the a into a null tetrad. Suppose that for a given spacetime we have de ned an orthonormal tetrad as follows: va timelike vector i a , j a , k a spacelike vectors We can then construct the null tetrad using a simple recipe. The two real vectors are given by la = va + ia 2 and na = va ia 2 (9.6)
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while the two complex vectors can be constructed using ma = j a + ik a 2 and m a = j a ik a 2 (9.7)
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EXAMPLE 9-2 Show that the de nition given in (9.6) leads to abl a l b = 0 and abl a n b = 1. SOLUTION 9-2 The vectors v a , i a , j a , and k a form an orthonormal tetrad. Since v a is timelike, we have ab v a v b = +1, while i a being spacelike satis es ab i a i b = 1. Orthonormality also tells us that v i = 0. Proceeding with this knowledge, we have abl a l b = ab va + ia 2 vb + ib 2 1 = ab v a v b + v a i b + i a v b + i a i b 2 = In the second case, we have abl a n b = ab = va + ia 2 vb ib 2 1 = ab v a v b v a i b + i a v b i a i b 2 1 (1 + 0 + 0 1) = 0 2
1 (1 + 0 + 0 + 1) = 1 2