A first we assume that the two events are timelike

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Unformatted text preview: ion. This can always be done, and after the rotation I have the unprimed ∆X = (∆t, |∆x|, 0, 0) (15) Now define ∆x ≡ |∆x|. Note that now we are doing a two-dimensional problem involving only (∆t, ∆x). (a) First, we assume that the two events are timelike separated; this means that ∆t2 > ∆x2 3 (16) t ¯ t95,:%4578%"*%."#$;% x ¯ !"#$%&'()%*+",(-%)(".)% /"0',12% 3+",(%"*%45+,% 6578%"*%45+,% 3+",(%"*%."#$% 4 x Now we seek a frame in which the two events happen at the same spatial point; ¯ i.e. we seek to transform to a (barred) frame where ∆X points entirely in the time direction. Thus recalling the explicit form of the Lorentz boost, we seek a v that satisfies the following equation: γ 1 v v 1 ∆t ∆x = ¯ ∆t 0 (17) ¯ The right hand side of this formula is the components of ∆X , i.e. the four-vector ∆x in the barred frame. Clearly v = − ∆t will do the job; but wait, you cry, perhaps we will find |v | > 1 implying a boost faster than the speed of light, which makes little sense. Of course however we don’t, because the condition (16) ensures that the boo...
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