Special Relativity 1

Special Relativity 1

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Unformatted text preview: ) L 1 ,1 ( β ) ￿￿ ct￿ −ct￿ ￿ from which we obtain. . . L 0 ,0 ( β ) − L 0 ,1 ( β ) = − L 1 ,0 ( β ) + L 1 ,1 ( β ) If we put the two results together, we come up with: L 1 ,1 ( β ) = L 0 ,0 ( β ) L 1 ,0 ( β ) = L 0 ,1 ( β ) Moving on to the third constraint, let’s write the coordinates of O￿ in each frame: ￿￿ ￿ ￿￿ ct ct ￿ r= r= β ct 0 The two descriptions are related by. . . ￿￿￿ ￿ ￿ ￿￿ ct L 0 ,0 ( β ) L 0 ,1 ( β ) ct = β ct L 0 ,1 ( β ) L 0 ,0 ( β ) 0 so. . . ct = L0,0 (β )ct￿ β ct = L0,1 (β )ct￿ 8 Divide the second equation by the first, and you get. . . L 0 ,1 ( β ) = β L 0 ,0 ( β ) Now, let’s wrap up by returning to that second constraint. What it means for us is that, if we design a transformation operator to take descriptions of events in the rocket into descriptions in the lab, we ought to be able to use that same operator to take descriptions in the lab back over to the rocket by reversing the sign of β . That is, if r = L( β ) r ￿ then r ￿ = L( − β ) r Substitute that first equation into the second. It should be obvious that L( − β ) L( β ) = I At this point, we can write ￿ L ( β ) = L 0 ,0 ( β ) 1 β β 1 so that second constraint boils down to ￿ ￿￿ 1 −β 1 L 0 ,0 ( − β ) L 0 ,0 ( β ) −β 1 β L 0 ,0 ( − β ) L 0 ,0 ( β ) ￿ 1 − β2 0 ￿ β 1 0 1 − β2 ￿ ￿ = = ￿ ￿ 1 0 1 0 which leads us to the inevitable conclusion L 0 ,0 ( β ) = ￿ 1 1 − β2 We’ll make use of the standard definiti...
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This note was uploaded on 12/03/2013 for the course PHYSICS 105A taught by Professor Corbin during the Winter '13 term at UCLA.

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