Solutions To Special Relativity Problems

Alas, there were no submissions of solutions to the special relativity problems - for award of a  $50 book prize.  Anyway, the solutions are given below:

1) Given x' =  1/a (x - vt) and t' = 1/a (t - vx/c²),

Then: x' = x/a - vt/a and t' = t/a - vx/ac²

and: x' + vt/a = x/a and t' + vx/ ac²  = t/a

so:

x = a(x' + vt/a) and t' = a(t' + vx/ ac² )

finally: x = a(x' + vt) and t = a(t' + vx/c²)


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 no relative motion 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.

    A more intuitive way to look at the result would be in terms of the time transformation: 

t = t'/ (1 - v²/c²)^½  

-  or asking how much time elapses on an Earth clock for each year elapsed on the astronaut's? The result will be found to be 2.29 years, or in other words his Earth counterparts are aging 2.29x faster.