## Sunday, August 28, 2011

### Solutions to Special Relativity Problems (5)

Recall the last set of problems:

1) Assume that the lifetime of quasar 3c-9 is 1 million years as measured in its own rest frame. Over what total time span (in Earth-measured time) would its radiation be received at the Earth? (Assume 3c-9's velocity relative to the Earth remains constant)

2) Suppose that you happen to be moving at a velocity of 3c/4 past a remote observer who picks up a stopwatch and then sets it down. Using a high power telescope you observe he held the watch for 9 secs. How long would HE think that he held it?

3) An astronaut orbits the Earth at a distance of 7 x 10^6 m from its center for a week. How much younger than his twin on Earth is he when he lands? Assume standard orbital speed of 18,000 mph and neglect the rotation of the Earth.

4) Consider 3 galaxies: A, B and C. An observer in A measures the velocities of B and C and finds they are moving in opposite directions - each with a speed of 0.7c relative to him., i.e.

(0.7c) <--------(B)----(A)-----(C)-------->(0.7c)

What is the speed of A observed by someone in B?

What is the speed of C observed by someone in B?

The observer in A thinks that the two other galaxies are receding from him at a rate 1.4c. Show him how this is wrong, by providing the correct result.

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solutions

(1) Let t = 10^6 yrs be the quasar's lifetime in its own rest frame. Then the total Earth-based time that its radiation will be received is:

t' = t [1 - v^2/c^2]^ ½

where v = 0.8c

so:

t' = (10^6 yrs.) [1 - (0.8c)^2/c^2]^ ½

t' = (10^6 yrs.) [1 - 0.64]^ ½ = 10^6 yrs(0.36)^ ½

t' = 0.6 (10^6 yrs.) = 6 x 10^5 yrs.

(2) Let delta t' = t2' - t1' = 9 secs

t2' = (t2 - 3c/4)(1 - 9/16))^-½

t1' = (t1 - 3c/4)(1 - 9/16)^-½

or:

t2' = (t2 - 3c/4) (7/16)^-½

t1' = (t1 - 3c/4) (7/16)^-½

whence:

(t2' - t1') = [(t2 - 3c/4) - (t1 - 3c/4)]/0.661

or: 9 sec = (t2 - t1)/0.661 and

delta t = 5.95 sec

(3) The speed of the astronaut is given by:

v = (2 gr)^½

and r = 7 x 10^6 m, g = 10 ms^-2

v = 11, 832 m/s = 3.94 x 10^-5 (c)

In the astronaut's frame, 1 week - 86,400 s x (7) = 604, 800 secs

For the twin on Earth:

t = t'/ [1 - v^2/c^2]^ ½

t = 604, 800.0002s

and so the time difference = 0.0002s, or the twin in orbit will be (t' - t) or 0.0002 s younger on his return.

(4) The speed of A observed in B = 0.7c, exactly equal to the speed of B observed in A, by principle of relative velocities.

To find the speed of C observed in B, we use relativistic addition of velocities or:

u = (u' + v)/ 1 + u'v/c^2

u = (0.7c + 0.7c)/ [1 + (0.7c)(0.7c)/c^2]

u = (1.4c)/ 1 + 0.49 = 1.4c/1.49 = 0.94c