s1131 - PROBLEM 31 The Matrix L is defined by L= l00 l10...

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Unformatted text preview: PROBLEM 31 The Matrix L is defined by: L= l00 l10 l20 l30 l01 l11 l21 l31 l02 l12 l22 l32 l03 l13 l23 l33 . (a)–Calculate −gL. (b)–Write down the transpose matrix L. (c)–CalculateLg . (d)–Compare (a) and (c) to find the general form of L (i.e. use Lg = −gL). (a)–The matrix g is defined by: g= 10 0 0 0 −1 0 0 0 0 −1 0 00 0 −1 . –By matrix multiplication: −gL = −l00 −l01 −l02 −l03 l13 l12 l11 l10 2 2 2 l0 l1 l2 l23 l30 l31 l32 l33 . (b)–The transpose of L is: L= l00 l01 l02 l03 l10 l11 l12 l13 l20 l21 l22 l23 l30 l31 l32 l33 . (c)–By matrix multiplication: Lg = l00 l01 l02 l03 −l10 −l11 −l12 −l13 −l20 −l21 −l22 −l23 −l30 −l31 −l32 −l33 . (d)–Set −gL = Lg and compare the components: −l00 = l00 −l01 = −l10 −l02 = −l20 −l03 = −l30 l10 = l01 l11 = −l11 l12 = −l21 l13 = −l31 l20 = l02 l21 = −l12 l22 = −l22 l23 = −l32 l30 = l03 l31 = −l13 l32 = −l23 l33 = −l33 . (1) –The resulting matrix is: L= l02 l03 0 l01 0 0 l12 l13 l1 l02 −l12 0 l23 l03 −l13 −l23 0 . (2) (e)–Obtain the same result by discussing the elements of g αβ ˜βγ gγδ = −lαδ . Perl forming the contractions we get ˜α = −lα . lδ δ Using the definition ˜α = l α lδ δ we have the equations lδ α = −lαδ , which are identical with the equations (1), namely (no summations, a = 0, 1, 2, 3, i = 1, 2, 3 and j = 1, 2, 3): laa = +laa ⇒ laa = −laa = 0 , l0i = −l0i ⇒ l0i = +li 0 , lj i = +lji ⇒ lji = −li j and the resulting matrix is again (2). ...
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This note was uploaded on 11/10/2011 for the course PHY 4241 taught by Professor Berg during the Spring '11 term at University of Florida.

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s1131 - PROBLEM 31 The Matrix L is defined by L= l00 l10...

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