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Unformatted text preview: PROBLEM 43
The Matrix L is deﬁned by: L= l00
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 −1 0
0 0 −1 0
0 −1 . –By matrix multiplication: −gL = −l00 −l01 −l02 −l03
l33 . (b)–The transpose of L is: L= l00
l33 . (c)–By matrix multiplication: Lg = l00
−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= 0 l01
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 .
Using the deﬁnition ˜α = l α
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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- Fall '11