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March 6, 2015

March 6, 2015

Posted by **Anonymous** on Friday, April 6, 2012 at 4:32pm.

(a) According to the theory of special relativity, what is the difference between their ages when they meet again at the earliest possible time?

The answer is 4.3 years, but I'm not sure how they arrived at that...

- Physics Question -
**Steve**, Friday, April 6, 2012 at 5:06pmtwin #1:

speed .854c

travel time: 12c/.854c = 14.05 yr

twin #2:

speed: .513c

travel time: 12c/.513c = 23.39 yr

so, twin #1 travels for 14.05 yr and waits 9.34 yr for twin #2 to arrive.

However, time dilation slowed aging for the twins:

twin #1: t' = .520t -- sqrt(1-.854^2)

twin #2: t' = .858t -- sqrt(1-.513^2)

so,

twin #1 aged 14.05*.520 = 7.31 yrs in transit

twin #2 aged 23.29*.858 = 20.07 yrs in transit

twin #1 age: 7.31+9.34 = 16.65 yrs

twin #2 age: 20.07 yrs

difference: 3.4 yrs

I also did not get 4.3 years, though maybe I'm dyslexic.

Anyone see where I went astray?

- Physics Question -
**Elena**, Friday, April 6, 2012 at 5:52pmLo = 12 light years (ly)

β1 =0.854,

β2 = 0.513

Twin1: L1 =Lo•sqrt(1- (β1)^2) = 12• Lo•sqrt(1- (0.854)^2) = 6.24 ly,

Twin2: L2 =Lo•sqrt(1- (β2)^2) = 12• Lo•sqrt(1- (0.513.)^2) = 10.3 ly.

The time to reach the planet:

Twin1: t1 = L1/v1 = 6.24 ly/0.854c = 7.307 years,

Twin2: t2 = L2/v2 = 10.3 ly/0.513c = 20.08 years.

The age of each twin when each arrives at the planet

Twin1: 19 + 7.307 =26.307 years,

Twin2: 19 + 20.08 =39.08 years.

Twin1 has to wait for twin 2 to arrive. As seen from Earth

Δt1 = (12 ly)/0.854 c = 14.05 years,

Δt2 = (12 ly)/0.513 c = 23.39 years.

Twin1 must wait another 23.39 – 14.05 = 9.34 years for Twin2 to arrive.

When Twin2 arrives, twin A has an age of

24.8 + 9.34 =34.14 years.

The difference between their ages when they meet is

39.08 -34.14 = 4.94 years.

(The answer 4.3 years may be obtained if you take β1 =0.900, β2 = 0.500)

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