Solutions To Simpler Problems In Special Relativity (March 17 post)

Solutions:

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 very 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 = 4600m/ (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 in its frame is just the proper length, L'  so:

L' = 4600 m [1 - v²/c²]^½ = 4600m (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