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...
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)
Lo= 12 ly
[twin a] La= Lo*sqrt(1-v^2/c^2)= 5.23 ly
[twin b] Lb= Lo*sqrt(1-v^2/c^2)= 10.39 ly
the time to reach the planet:
[twin a] ta=La/va= 5.23ly/0.9c= 5.81 yr.
[twin b] tb=Lb/vb= 10.39ly/0.5c= 20.78 y
The age of each twin when each arrives at the planet:
[twin a] 19 y + 5.81 y= 24.81 y
[twin b] 19 y + 20.78 y= 39.78 y
Twin1 has to wait for twin 2 to arrive. As seen from Earth
Δt1 = (12 ly)/0.9c = 13.3 years,
Δt2 = (12 ly)/0.5c = 24 years.
Twin1 must wait another 24– 13.3 = 10.7 years for Twin2 to arrive.
When Twin2 arrives, twin A has an age of
24.81 + 10.7 =35.51 years.
The difference between their ages when they meet is
39.78 -35.51 = 4.27 years or 4.3 years
To calculate the difference in ages when the twins meet again, we need to apply the principles of special relativity, particularly time dilation. Time dilation states that time runs slower for an object moving relative to another object.
In this case, we have two twins traveling to a distant planet. Let's call the twin who travels at 0.854c as Twin A and the twin who travels at 0.513c as Twin B. Here are the steps to calculate the age difference:
Step 1: Calculate the Lorentz factor (γ) for each twin using the formula:
γ = 1 / √(1 - (v^2 / c^2))
where v is the velocity and c is the speed of light.
For Twin A:
γA = 1 / √(1 - (0.854c)^2 / c^2)
γA = 1 / √(1 - 0.729) ≈ 1.64
For Twin B:
γB = 1 / √(1 - (0.513c)^2 / c^2)
γB = 1 / √(1 - 0.263) ≈ 1.19
Step 2: Calculate the time experienced by each twin during the journey. Let's assume Earth time as the reference frame (T = 0 at the start).
For Twin A:
TA = γA * T
TA = 1.64 * 0 = 0 years
For Twin B:
TB = γB * T
TB = 1.19 * 0 = 0 years
Step 3: Calculate the time taken by each twin to reach the planet using the formula:
time = distance / velocity
For Twin A:
timeA = 12.0 light-years / (0.854c) = 14.04 years
For Twin B:
timeB = 12.0 light-years / (0.513c) = 23.40 years
Step 4: Calculate the age of each twin when they meet again.
For Twin A:
AgeA = TA + timeA
AgeA = 0 years + 14.04 years = 14.04 years
For Twin B:
AgeB = TB + timeB
AgeB = 0 years + 23.40 years = 23.40 years
Step 5: Calculate the age difference between the twins.
Difference = AgeA - AgeB
Difference = 14.04 years - 23.40 years ≈ -9.36 years
Based on the calculations, the age difference between the twins when they meet again is approximately -9.36 years, not 4.3 years. Please double-check the given answer or provide more information to further investigate the discrepancy.
To calculate the time difference between the twins when they meet again, we need to apply the theory of special relativity and use the concept of time dilation. Time dilation states that time appears to pass slower for objects that are moving relative to an observer.
In this case, one twin is traveling at a speed of 0.854c (where c is the speed of light) and the other twin is traveling at a speed of 0.513c.
To find the time dilation factor (γ) for each twin, we can use the formula:
γ = 1 / √(1 - v^2/c^2)
where v is the velocity of the twin and c is the speed of light.
For the first twin traveling at 0.854c:
γ1 = 1 / √(1 - (0.854c)^2/c^2)
γ1 = 1 / √(1 - 0.728196)
γ1 = 1 / √0.271804
γ1 = 1 / 0.521596
γ1 ≈ 1.915
For the second twin traveling at 0.513c:
γ2 = 1 / √(1 - (0.513c)^2/c^2)
γ2 = 1 / √(1 - 0.263169)
γ2 = 1 / √0.736831
γ2 = 1 / 0.858048
γ2 ≈ 1.164
Now, let's assume that the first twin, traveling at 0.854c, experiences a time interval Δt1. This will be the time according to the twin who is stationary on Earth.
Similarly, the second twin, traveling at 0.513c, experiences a time interval Δt2.
According to time dilation, the relationship between the time experienced by the moving twin and the time experienced by the stationary twin is given by:
Δt2 = γ2 * Δt1
Since the twins are traveling for the same distance and meeting at the same place, we can set Δt1 = Δt2, and the equation becomes:
Δt2 = γ2 * Δt2
Simplifying the equation:
1 = γ2
Now, we can solve for Δt2:
Δt2 = 1 / γ2
Δt2 = 1 / 1.164
Δt2 ≈ 0.859 years
So, the time difference between their ages when they meet again at the earliest possible time (according to the theory of special relativity) is approximately 0.859 years.