Class32 - 1.6 Four-dimensional Space The Lorentz transformation treats the xi with i = 1 2 3 as equivalent variables Lets introduce time as simply

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1.6 Four-dimensional Space The Lorentz transformation treats the x i with i =1 , 2 , 3 as equivalent variables. Let’s introduce time as simply a forth coordinate: x 0 = ct (85) This four-dimensional “space” is called Minkowski space. The Lorentz transformation now becomes x ± 0 = γ ( x 0 - βx 1 ) (86) x ± 1 = γ ( x 1 - 0 ) (87) x ± 2 = x 2 (88) x ± 3 = x 3 (89) and the inverse transformation x 0 = γ ( x ± 0 + ± 1 ) (90) x 1 = γ ( x ± 1 + ± 0 ) (91) x 2 = x ± 2 (92) x 3 = x ± 3 (93) (94) which we can write as a matrix x ± 0 x ± 1 x ± 2 x ± 3 = γ - βγ 00 - γ 0 0 1 0 0 0 0 1 x 0 x 1 x 2 x 3 (95) Which we can write, letting Greek letters run from 0 to 3 as customary, as x ± μ = 3 ± ν =0 a μν x ν =0 , 1 , 2 , 3 (96) where a is the transformation matrix elements. example: Suppose that there are three inertial reference frames: S , S ± and S ± , which are in collinear motion along their respective x 1 axes. Let the velocity of S ± w.r.t. S be v 1 , and the velocity of S ± w.r.t. S ± v 2 . What is the speed u of S ± relative to S ?
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.

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Class32 - 1.6 Four-dimensional Space The Lorentz transformation treats the xi with i = 1 2 3 as equivalent variables Lets introduce time as simply

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