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**Unformatted text preview: **st velocity is smaller than 1.
(b) This is almost exactly the same, as the two events are spacelike separated we have
∆x2 > ∆t2 , (18) the equation we want to solve is now
γ 1
v v
1 ∆t
∆x = 0
∆¯
x , (19) ∆t
the boost velocity is v = − ∆x , and we see by the spacelikeness condition that
|v | < 1 implying a sensible boost. 4. There are many ways to do this. Here’s my favorite: take two meter sticks. On one
of them attach two blue pens on the two ends of the sticks; on the other attach two
red pens. Put the blue one in frame F and the red one in frame F ; but also set up a
piece of paper so that as the meter sticks move the pens draw lines on the paper. In
frame F we see the red meter stick moving while the blue one is at rest; thus if there
is any contraction the red lines should be inside the blue lines on the paper. In frame
F the blue stick is moving and the red one is at rest; thus if there is any contraction
now the blue lines should be inside the red lines. These obviously can’t be true at the
same time; thus there can be no co...

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