However recall from problem 4 which used nothing but

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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 the form ∆t 2 − ∆z 2 = ∆t 2 − ∆z 2 , (25) 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...
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This document was uploaded on 02/28/2014.

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