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# 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 &gt; ∆x2 3 (16) t ¯ t95,:%4578%&quot;*%.&quot;#\$;% x ¯ !&quot;#\$%&amp;'()%*+&quot;,(-%)(&quot;.)% /&quot;0',12% 3+&quot;,(%&quot;*%45+,% 6578%&quot;*%45+,% 3+&quot;,(%&quot;*%.&quot;#\$% 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 | &gt; 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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