This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ∆z )
plane? Of course; the solution works out to be
2d = √
∆t 2 − ∆z 2 u = −√ ∆z
∆t 2 + ∆z 2 (24) This always has a physical solution for any timelike separated ∆t , ∆x (note the range
of u ∈ (−1, 1)) thus by building a suﬃciently large light clock and speeding it oﬀ
6 arbitrarily fast, I can show that any two timelike separated events satisfy a relation of
∆t 2 − ∆z 2 = ∆t 2 − ∆z 2 ,
where the boost relating the two is in the z direction. However recall from Problem
4, which used nothing but symmetry arguments, that boosts cannot alter the spatial
directions perpendicular to the boost. Thus we may immediately put them back and
extend this above result to say that
∆t 2 − ∆x 2 − ∆y 2 − ∆z 2 = ∆t 2 − ∆x 2 − ∆y 2 − ∆z 2 , (26) for any two timelike separated events.
Note that in class we arrived at this conclusi...
View Full Document