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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%"*%."#$;%
x
¯
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/"0',12%
3+",(%"*%45+,% 6578%"*%45+,% 3+",(%"*%."#$% 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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