1)
We have: t = t’/ [1 - v2/c2] ½
But the proper time is defined such that:
t' = t/2 or t'/ t = ½
Then:
[1 - v2/c2] = (t'/ t)2
and:
v2/c2 = 1 - (t'/ t)2
v2 = c2[1 - (t'/ t)2]
so:
v = c[1 - (t'/ t)2]½ = c[1 - 0.52]½ = c[0.75]½ = 0.866c
2) The proper time t' = 3600 s
Since v = 300 m/s = (10-6) c and hence v/c << 1 we need the form:
But the proper time is defined such that:
t' = t/2 or t'/ t = ½
Then:
[1 - v2/c2] = (t'/ t)2
and:
v2/c2 = 1 - (t'/ t)2
v2 = c2[1 - (t'/ t)2]
so:
v = c[1 - (t'/ t)2]½ = c[1 - 0.52]½ = c[0.75]½ = 0.866c
2) The proper time t' = 3600 s
Since v = 300 m/s = (10-6) c and hence v/c << 1 we need the form:
t = t’/ [1 + v2/2c2] t = 3600s/ [1 + (10 -12) c2/2c2]
Since the numerator is only slightly larger than 1, the time t will be:
3600
s/(1.000000000001)= 3600.0000000018
=
3600 + 1.8 x 10-9 s or slightly longer than one hour.
3) (a) The proper time t' applies to the muon's reference frame.
So: t = t’/ [1 - v2/c2] ½ and t' = t [1 - v2/c2] ½
where v = 0.99 c and v2 = (0.99c)2 = 0.98c2
Then: t' = t [1 - 0.98c2/c2]½ = t [0.02] ½ = t(0.14)
recall distance travelled = 4.6 km = 4600 m
To get t' we need to find t first, e.g.
t = 4600 m/ (2.97 x 108 m/s) = 1.55 x 10-5 s
Then: t' = (1.55 x 10-5 s) (0.14) = 2.1 x 10-6 s
b) The distance traveled by the muon as measured in its frame is just the
proper length, L' so:
L' = 4600 m [1 - v2/c2]½ = 4600 m (0.02)½
L' = 4600 m (0.14) = 644 m
4) The proper time t' = 2.6 x 10-8 s
t = t’/ [1 - v2/c2] ½ and v = 0.95c
so:
t = (2.6 x 10-8 s)/ [1 - (0.95c)2/c2] ½
t = (2.6 x 10-8 s)/ [0.0975]½ = (2.6 x 10-8 s)/ 0.312
t = 8.3 x 10-8 s
L' = 4600 m [1 - v2/c2]½ = 4600 m (0.02)½
L' = 4600 m (0.14) = 644 m
4) The proper time t' = 2.6 x 10-8 s
t = t’/ [1 - v2/c2] ½ and v = 0.95c
so:
t = (2.6 x 10-8 s)/ [1 - (0.95c)2/c2] ½
t = (2.6 x 10-8 s)/ [0.0975]½ = (2.6 x 10-8 s)/ 0.312
t = 8.3 x 10-8 s
Which is the mean lifetime of the muon as measured by an observer on Earth.
