Let's have some fun. Here's a basic special relativity problem from John R. Taylor's Classical Mechanics . 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…
The time dilation formula tells us that the time experienced by a moving clock will appear slower to a stationary observer by a factor of $\gamma$, where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ and $v$ is the speed of the moving clock and $c$ is the speed of light. In this case, the speed of the space explorer's rocket is 0.95c, so $\gamma = \frac{1}{\sqrt{1 - 0.95^2}} = \frac{1}{\sqrt{0.0025}} \approx 7.071$.
To determine how long the space explorer's clocks say that he was gone, we need to apply the time dilation formula to each half of the journey. Since the speed of the rocket is the same on the way out as it is on the way back, the value of $\gamma$ will be the same in both cases.
First, let's consider the outbound leg of the journey. From the perspective of an observer on Earth, this journey takes 40 years. However, from the perspective of the space explorer, the time experienced on the outbound leg of the journey will be slower by a factor of $\gamma$, so the time experienced by the space explorer on this leg of the journey will be 40 years / $\gamma$ = $\frac{40}{7.071} \approx 5.648$ years.
Now let's consider the inbound leg of the journey. From the perspective of an observer on Earth, this journey also takes 40 years. However, from the perspective of the space explorer, the time experienced on the inbound leg of the journey will also be slower by a factor of $\gamma$, so the time experienced by the space explorer on this leg of the journey will be 40 years / $\gamma$ = $\frac{40}{7.071} \approx 5.648$ years.
To determine the total time experienced by the space explorer, we need to add the time experienced on the outbound leg of the journey to the time experienced on the inbound leg of the journey. This gives us a total time experienced by the space explorer of 5.648 years + 5.648 years = 11.296 years.
To determine the difference in aging between the space explorer and their twin who stayed on Earth, we need to multiply the time experienced by the space explorer by $\gamma$. This gives us a difference in aging of 11.296 years * $\gamma$ = 11.296 years * 7.071 = 79.8 years. This means that the space explorer has aged 79.8 years less than their twin who stayed on Earth.