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**Unformatted text preview: **frequency f0 for his entire proper time,
√
which we worked out above was T 1 − v 2 . Indeed the number of counts B receives
divided by the number of counts A received is γ , which makes sense as “number of
counts” is a time telling mechanism.
2. Schutz 1.17
√
For this problem γ = 1/ 1 − .82 = 5/3.
(a) The friend on the barn measures the pole to be
lpole
3
= (20 m)( ) = 12 m
γ
5 (10) (b) After the barn door is closed, in the friend’s frame, the front of the pole has 3 m
to travel before hitting the front. Thus the time it takes is
t= 3m
= 3.75 m
.8 (11) The interval between shutting the door and hitting the front is
∆s2 = −∆t2 + ∆x2 = −(3.75 m)2 + (15 m)2 = 210.938 m2 (12) which is spacelike.
(c) In the frame of the runner, the length of the pole is 20 m while the length of the
barn is Lorentz contracted to 9 m.
(d) When the front of the pole hits the barn, the barn door is still open according to
the runner and so he does not believe the pole is in the barn.
2 (e) The “paradox” here is resolved by noting t...

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