Solutions To Simpler Relativity Problems

 1) Given L = 11m and l = 5.9 x 10^-7 m

The limit for lowest resolution: 2 D d = 0.005 fringe

Then:

d = (0.005 fringe)/ 2 = 0.0025 fringe

and since:

D d = Lv²/ (c²)

v = {(D d)( l) c²)/ L}^½

v = {(0.0025)( 5.9 x 10^-7 m)(3 x 10⁸ ms^-1)/11 m}^½

= 3.5 km/s

2) We have: x' = 60m, t' = 8 x 10^-8 s and y' = y, z' = z

v = 0.6c = 1.8 x 10⁸ m/s

Then:

x = [60m + (1.8 x 10⁸ m/s)( 8 x 10^-8 s)]/ (0.64)^½

x = [60m + 14.4m]/ 0.8 = 74.4m/0.8 = 93m

and t =

 [(8 x 10^-8 s)+ (1.8 x 10⁸ ms^-1)(60m)/(3 x 10⁸ ms^-1)²/0.8

t = 2.5 x 10^-7 s/ 0.8 = 2.33 x 10^-7 s

The space time coordinates are: (93 m, 2.33 x 10^-7 s)


3) The problem requires zero relative motion, i.e. defined specifically in the x-direction so the equations:

t = t' + x'v/c²/(1 - v²/c²)^½

and

t' = t - xv/c²/(1 - v²/c²)^½

are immediately simplified by the terms in x becoming zero, so:

t = t'/(1 - v²/c²)^½

and

t' = t /(1 - v²/c²)^½

Here: t = time passage on Earth clock

and t' = time passage on astronaut's clock

For t = 1 Earth year = 365 ¼ days:

t' = (365 ¼ days)/ [1 - (0.9c)²/c²]^½

= (365 ¼ days)/(1 - 0.81)^½

t' = (365 ¼ days)/0.436 = 837.7 days 

This is the time elapsed on the astronaut's clock when the Earth has made one revolution equal to 365 ¼ days. In other words, each of his days is roughly equal to 2.29 Earth days. Hence, his clock is obviously running slower than the Earth clock.