## Sunday, October 14, 2018

### Solutions To Special Relativity Problems (2)

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:

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

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

Which is the mean lifetime of the muon as measured by an observer on Earth.