## Saturday, August 13, 2011

### Solutions to Special Relativity Problems (2)

Recalling the last problem set:

1- Given that x' = 1/a (x - vt) and t' = 1/a (t - vx/c^2), derive similar equations for x and t in terms of x' and t'.
(Recall: 1/a = (1 - v^2/c^2)^½)

2- An event in space-time occurs at x' = 60 m, t = 8 x 10^-8 s, in a frame S' (y' = 0, z' = 0). The frame S' has a velocity of 0.6c along the x-direction with respect to a frame S. The origins O and O' coincide at time t = t' = 0. Find the space-time coordinates of the event in S.

3- Suppose an astronaut is traveling at 0.9c in a space ship with respect to the Earth. How long a time interval will his clock indicate when the Earth has revolved once around the Sun? (Take the duration of one standard revolution of Earth around the Sun to be 365 ¼ days.)

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Solutions

1) Given x' = = 1/a (x - vt) and t' = 1/a (t - vx/c^2),

Then: x' = x/a - vt/a and t' = t/a - vx/ac^2

and: x' + vt/a = x/a and t' + vx/ac^2 = t/a

so:

x = a(x' + vt/a) and t' = a(t' + vx/ac^2

finally: x = a(x' + vt) and t = a(t' + vx/c^2)

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

v = 0.6c = 1.8 x 10^8 m/s

Then:

x = [60m + (1.8 x 10^8 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 10^8 m/s)(60m)/(3 x 10^8 m/s)^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/c^2/ (1 - v^2/c^2)^½

and

t' = t - xv/c^2/(1 - v^2/c^2)^½

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

t = t'/ (1 - v^2/c^2)^½

and

t' = t /(1 - v^2/c^2)^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/c^2]^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 other time transformation: t = t'/ (1 - v^2/c^2)^½ and ask how much time elapses on an Earth clock for 1 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.