Twins who are 19.0 years of age leave the earth and travel to a distant planet 12.0 light-years away. Assume that the planet and earth are at rest with respect to each other. The twins depart at the same time on different spaceships. One twin travels at a speed of 0.854c, and the other twin travels at 0.513c.
(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...
twin #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?
Lo = 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)
To understand how the answer of 4.3 years is derived, let's break down the problem step by step.
1. Start by calculating the time it takes for each twin to reach the distant planet. We can use the formula for time dilation in special relativity:
t' = t / sqrt(1 - v^2/c^2)
where:
t' is the time experienced by the traveling twin,
t is the time experienced by the stationary twin (on Earth),
v is the velocity of the traveling twin's spaceship, and
c is the speed of light.
For the twin traveling at 0.854c:
t'1 = 19.0 / sqrt(1 - 0.854^2)
For the twin traveling at 0.513c:
t'2 = 19.0 / sqrt(1 - 0.513^2)
2. Calculate the time it takes for the traveling twin to return to Earth. Since both twins leave at the same time, the traveling twin's time to reach the planet and return will be twice the time it took to reach the planet:
T1 = 2 * t'1
T2 = 2 * t'2
3. Determine the time difference between the twins when they meet again. This can be done by subtracting their respective travel times from their initial age difference:
Δt = 19.0 - (T1 + T2)
Plug in the calculated values of T1 and T2 to find the time difference.
By following these steps, you should be able to find the time difference between the twins when they meet again.