Consequences of Special Relativity: Problems (Serway et al. p.45; Ph311p45.doc)

6. interval of improper time = (gamma)(interval of proper time)

Here, 2T = (gamma)(T) where T is any time interval

Gamma = 2 = 1/(1-v^2/c^2)^(1/2), so (1-v^2/c^2) = 1/4 = 0.25

(v/c)^2 = 0.75, v = 0.866 c = 2.60E8 m/s

8. Suppose the clock ticks exactly once every second when stationary,

but once every Ts seconds when moving at a relative speed v.

Then (N - 1)(Ts) = (N)(1s) where N = number of seconds in one day.

1/(1-v^2/c^2)^(1/2) = gamma = (improper time)/(proper time) = Ts/1 = N/(N-1)

Therefore (1-v^2/c^2)^(1/2) = (N-1)/N = 1 - 1/N

But (1-v^2/c^2)^(1/2) = 1 + (1/2) (-v^2/c^2) approximately, by binomial theorem

So (1/2)(v^2/c^2) = 1/N , v^2 = (2/N) (c^2) = (2)(3.00E8)^2 / (24)(60)(60)

= 18E16/(86400) = 2.08E12, giving v = 1.44E6 m/s

Note that we subtracted 1 from each side in the algebra, NOT in an electronic calculator (which may have insufficient accuracy when v^2/c^2 is much less than unity).

10. Improper length = (1/gamma) (proper length) = L/gamma = L (1-v^2/c^2)^(1/2)

= L [1-(0.9c)^2/c^2]^1/2 = L [(1 - 0.81)^1/2] = L [(0.19)^1/2] = 0.436 L

12. Proper time = 3600s since clock is stationary relative to the observer.

Let time measured on ground = 3600 + T = improper time

3600 + T = gamma (3600s) = (3600s)(1-v^2/c^2)^-1/2

= (3600s) [1 + (-1/2)(-v^2/c^2)] = 3600s + (3600s)(1/2)(v^2/c^2)

T = (1800s)(400)^2 / (3.00E8)^2 = 3.20E-9 s = 3.2 ns

The fact that T comes out positive means that the grounded clock measures a longer time.

14. (a) Observed lifetime = (gamma)(torr) = [(1-.95^2)^-1/2](2.2 microsec)

= (0.0975^-1/2 )(torr) =(3.20)(2.2 microseconds) = 7.05 microseconds

(b) Journey time t = distance / speed = 3000m / [(0.95)(3.00E8 m/s)] = 1.053E-5 s

N = (N0) exp(-t/torr) = (5.0E4) / exp(1.053E-5 / 6.05E-6) = (5.04E4)/1.49

Remaining number N = 33.6 particles, on average.