Chapter 15: Problem 5
A space explorer \(A\) sets off at a steady \(0.95 c\) to a distant star. After exploring the star for a short time, he returns at the same speed and gets home after a total absence of 80 years (as measured by earth-bound observers). How long do \(A\) 's clocks say that he was gone, and by how much has he aged as compared to his twin \(B\) who stayed behind on earth? [Note: This is the famous "twin paradox." It is fairly easy to get the right answer by judicious insertion of a factor of \(\gamma\) in the right place, but to understand it, you need to recognize that it involves three inertial frames: the earth-bound frame \(\mathcal{S}\), the frame \(\mathcal{S}^{\prime}\) of the outbound rocket, and the frame \(\mathcal{S}^{\prime \prime}\) of the returning rocket. Write down the time dilation formula for the two halves of the journey and then add. Notice that the experiment is not symmetrical between the two twins: \(A\) stays at rest in the single inertial frame \(\delta\), but \(B\) occupies at least two different frames. This is what allows the result to be unsymmetrical.]
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.