Question

In the twin paradox example (in Section 35.2 ), Alice boards a spaceship that flies to space station 3.25 light-years away and then returns with a speed of \(0.65 c .\) a) Calculate the total distance Alice traveled during the trip, as measured by Alice. b) Using the total distance from part (a), calculate the total time duration for the trip, as measured by Alice.

Short answer

b) What is the total time duration for the trip as measured by Alice?

Step-by-step solution

Calculate the total distance of the trip in the rest frame

First, recognize that Alice travels to the space station and then returns, which means she covers a total distance of 2 times the distance to the space station in the rest frame. So, the total distance in the rest frame is: \(d = 2 \times 3.25 \text{ light-years} = 6.50 \text{ light-years}\)

Apply length contraction formula to find the distance as measured by Alice

Now, we will use the length contraction formula to find the total distance traveled by Alice as measured in Alice's frame: \(d_A = \frac{d}{\sqrt{1 - \frac{v^2}{c^2}}}\) Where \(d_A\) is the distance as measured by Alice, \(d = 6.50 \text{ light-years}\) (total distance in the rest frame), \(v = 0.65c\), and \(c\) is the speed of light. Plugging in the values, we get: \(d_A = \frac{6.50 \text{ light-years}}{\sqrt{1 - \frac{(0.65c)^2}{c^2}}}\) \(d_A = \frac{6.50 \text{ light-years}}{\sqrt{1 - 0.4225}}\) \(d_A = \frac{6.50 \text{ light-years}}{\sqrt{0.5775}}\) Now, calculating the value of \(d_A\): \(d_A \approx 8.56 \text{ light-years}\)

Calculate total time duration in the rest frame

To find the total time duration in the rest frame, we need to know how long it takes for Alice to travel the total distance. Using the formula for speed, we can find the time in the rest frame: \(t = \frac{d}{v}\) Where \(t\) is the time in the rest frame, \(d = 6.50 \text{ light-years}\), and \(v = 0.65c\). Plugging in the values, we get: \(t = \frac{6.50 \text{ light-years}}{0.65c}\) \(t \approx 10 \text{ years}\)

Apply time dilation formula to find the time duration as measured by Alice

Now, we will use the time dilation formula to find the total time duration for the trip as measured by Alice: \(t_A = t \cdot \sqrt{1 - \frac{v^2}{c^2}}\) Where \(t_A\) is the time measured by Alice, \(t = 10 \text{ years}\) (time in the rest frame), \(v = 0.65c\), and \(c\) is the speed of light. Plugging in the values, we get: \(t_A = 10 \text{ years} \cdot \sqrt{1 - \frac{(0.65c)^2}{c^2}}\) \(t_A = 10 \text{ years} \cdot \sqrt{1 - 0.4225}\) \(t_A = 10 \text{ years} \cdot \sqrt{0.5775}\) Now, calculating the value of \(t_A\): \(t_A \approx 7.43 \text{ years}\) So, the final answers are: a) The total distance Alice traveled during the trip, as measured by Alice, is approximately 8.56 light-years. b) The total time duration for the trip, as measured by Alice, is approximately 7.43 years.