Question

As part of previous adventures, formerly conjoined identical twin, Sue, is now 13.4 years younger than her identical twin Lou. (The anniversaries of their birth are no longer the same day.) It is Lou's turn to travel, and he goes away and back at 0.96c. During his round trip, Lou ages 1 year. What is the the difference in the twin's ages when he returns.

Answer 1

Answer:

Difference in Twin's Ages = 12.68 years

Explanation:

Using special theory of relativity's time dilation phenomenon, we first find the time that is passed on earth during Lou's trip.

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by the person in relative motion = 1 year

t₀ = time measured by the person at rest = ?

v = speed of relative motion = 0.96 c

c = speed of light  

Therefore,

1 year = t₀/[√(1 - 0.96² c²/c²)]

1 year = t₀/[√(1 - 0.9216)]

(1 year)(0.28 year) = t₀

t₀ = 0.28 year

Let,

y = Sue's age

x = Lou's age

so,

x - y = 13.4 years

but, after this trip Lou has aged 1 year, and on earth only 0.28 years passed so, Sue has aged only 0.28 years. Therefore,

x = x + 1

y = y + 0.28

Therefore,

(x + 1 year) - (y + 0.28 year) = 13.4 years

x - y = 13.4 years - 0.72 year

x - y = 12.68 years

Difference in Twin's Ages = 12.68 years