 Welcome to Relativity 5c, Twin Paradox, Redux. This video expands on video 5a, and it's assumed you've watched that. Link in the video description. Recall that in the twin paradox scenario, twins are born on Earth. One stays on Earth while the other speeds off, turns around, and returns to Earth. When the twins are reunited, Earth Twin is older than Space Twin. In the case we consider, Space Twin travels at 60% the speed of light. This causes her clock to run slower by a factor of 0.8. As a result, while Earth Twin ages 10 years, Space Twin ages only 8 years. To personalize things a bit, let's name Earth Twin Earl and Space Twin Stacey. Now imagine you are present at the start and the end of the experiment. We'll represent age by the height of the figures. When the twins are born, they are each placed in a space pod. The pods are sealed. Inside, each twin is isolated from the outside world. They can only experience local phenomena inside their pods. You go away. Years later, you return and find the pods in their original places. The pods open, and Earl is 10 years old while Stacey is 8 years old. You want to explain this difference. The only information you have are the twins' accounts of their time in the pods. You ask Earl what happened. He says, When I was born I was put in this pod and the door was closed. Nothing happened for 10 years. The door opened, and here we are. You ask Stacey what happened. She says, When we were born I was put in this pod and the door was closed. Immediately I was thrown against the wall. Then for 4 years, nothing happened. Then I was thrown against the other wall. For the next 4 years, nothing happened. Then I was thrown against the first wall once more. The door opened, and here we are. Now, what do you conclude? Based on the twins' experiences, you might generalize that if twins are put in pods, in one experience's accelerations, then that twin will end up younger than his or her sibling. And you'd be in good company. Here's Richard Feynman's view. From his classic lectures on physics. He first states the apparent paradox as, by symmetry, the only possible result is that both should be the same age when they meet. Then he explains, But in order for them to come back together and make the comparison, one must either stop at the end of the trip and make a comparison of the clocks, or more simply, he has to come back. And the one who comes back must be the man who is moving. And he knows this because he had to turn around. When he turned around, all kinds of unusual things happened in his spaceship. The rockets went off, things jammed against one wall, and so on. So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls and so on, is the one who would be the younger. That is the difference between them in an absolute sense, and it is certainly correct. Note that Feynman doesn't say that the acceleration caused the age difference, and video 5a doesn't make that claim either. However, there is no way to implement the twin paradox without accelerating one of the twins, and we are guaranteed that the accelerated twin will end up the younger. In this sense, the acceleration at the midway point is the key event that breaks the symmetry between Earl and Stacey. An interesting way to look at this is in terms of Minkowski's concept of proper time, which is central to the general theory of relativity. It's also an easy way to get numerical values for twin paradox scenarios. Identify two space-time events with coordinates T1, X1, and T2, X2. If a clock moves uniformly between these events, a straight line on the space-time diagram, it will measure an elapsed time delta s equal to the square root of the difference of the square of the change in time coordinate and the square of the change in space coordinate. This proper time will be obtained regardless of the reference frame used to describe the events. In our twin scenario, Earl stays at X equals 0 for 10 years. His velocity is 0, and the proper time of his space-time path is 10 years. On her outbound segment, Stacey moves 3 light-years in 5 years. Her velocity is 3 over 5, and the proper time of her path is 4. On her inbound segment, she again moves 3 light-years in 5 years. Her velocity is minus 3 over 5, and she experiences another 4 years of proper time. Thus, she ages a total of 8 years to Earl's 10. We can clearly see Feynman's rule. Earl followed a straight path through space-time. He felt no acceleration. Stacey followed a path with a kink, a change of velocity, a change of reference frames, and acceleration. She got thrown against the wall of her pod. She ends up the younger twin. Suppose Stacey doesn't accelerate, but moves uniformly to a distance of 6 light-years after 10 years. She will age the elapsed proper time of 8 years. Suppose Earl stays on Earth for 3.5 years. Then he accelerates to a velocity that allows him to catch up to Stacey. The proper time of this second segment is 2.5 years. So Earl ages a total of 6 years to Stacey's 8. Now he is the younger because he is the one who took the kinked path. He is the one who experienced acceleration. To see what's going on in more detail, we need to consider how, at least in principle, a coordinate system can be constructed. Let an observer's position define the origin of a coordinate system, x equals 0. The observer can use an accelerometer to verify that he is at rest in an inertial frame. A second observer, located 1 light-year away, can establish her x equals 1 coordinate by bouncing a light beam off the origin and using a clock to measure the 2-year round-trip time. By continually verifying the 2-year round-trip time, she could demonstrate that she is at rest with respect to the origin. A third observer at x equals 2 can likewise verify his distance from the x equals 1 observer and his state of rest in the coordinate system using light beams and a clock. We can continue this process to place observers at any desired x-coordinate positions. To synchronize the observer's clocks, the origin observer can transmit the ticks of his master clock. After 1 year, the x equals 1 observer will receive the t equals 0 tick. After 2 years, the x equals 1 observer will receive the t equals 1 tick and the x equals 2 observer will receive the t equals 0 tick. To synchronize their clocks, each observer adds their distance from the origin to the received master clock tick. The x equals 2 observer adds 2 plus 0 to get 2. The x equals 1 observer adds 1 plus 1 to get 2 and so on. By synchronizing their local clocks with these results, all their clocks simultaneously have the same reading as the master clock. The same procedure can be used for a set of observers who are traveling with velocity v with respect to the first frame, thereby establishing a second frame of reference. We assume that the origins of all reference frames coincide at time t equals 0. We use three reference frames. Earl remains at rest in the blue frame at x blue equals 0. Blue frame observers are stationed at various x blue values and the blue clocks are synchronized at time t blue. The red frame moves right with velocity v. For her outward journey, Stacey remains at rest in the red frame at x red equals 0. The red clocks are synchronized at time t red. The green frame moves left with velocity v. For her inward journey, Stacey remains at rest in the green frame. The green clocks are synchronized at time t green. The various frames are related by the coordinate transformations of special relativity. For a blue observer with some particular values of the blue coordinates, we can use these to calculate the coordinates of the red and green observers he sees passing by at that instant. Note that the time coordinate in one frame is a combination of the space and time coordinates of the other frame. If all blue observers have the same t blue clock reading, then the red and green observers they see passing by will have t red and t green clock readings that vary throughout space. This means that if two spatially separated events happen at the same time in the blue frame, then they must happen at different times in the other frames. Events which are simultaneous in one frame will not be simultaneous in other frames. Let's visualize the experiment as it appears to blue frame observers. The blue frame observers are in the center row. All their clocks are synchronized to t blue equals 0 to begin. Some of the red frame observers they see are shown in the top row, and some of the green frame observers they see are shown in the bottom row. And we show rings around the twins locations. Earl is in the blue ring. Stacey is in the red ring. Stacey moves to the right with the red frame. Because she's moving, the blue frame observers see her clock running slow. Stacey reaches the turnaround point when a clock reads 4. The blue frame clocks, including Earl's, read 5. Stacey now jumps from the red frame to the green frame. The green frame observer she joins has a clock reading of 8.5, and that is where we start measuring a lapse time for the return path. As Stacey moves left with the green frame, the blue frame observers again see her clock running slow. When she rejoins Earl, her green frame clock has advanced from 8.5 to 12.5 for a total of 4 years. So Stacey is 8 years old while Earl's clock shows his age as 10. Here's a summary of the journey as seen by the blue frame observers. Earl is at rest during Stacey's outward trip, which takes 5 years. Therefore Earl ages 5 years. Because Stacey is moving, her clock runs slower and she only ages 4 years. After she accelerates, that is jumps frames, Earl is still at rest and her return trip takes 5 years. Earl ages 5 years. Stacey is still moving, so once again she only ages 4 years. Earl's final age is 5 plus 5 equals 10 years. Stacey's is 4 plus 4 equals 8 years. Here we plot Earl and Stacey's ages according to Earl's bookkeeping of simultaneous events. Earl's age is on the left, Stacey's on the right. The line colors correspond to the frames the twins are in. According to Earl, Stacey's rate of aging is always 80% of Earl's. Earl's explanation of the experiment is simply a summary of the blue frame observations. Stacey was always moving at a speed of 0.6, so her clock always ran slower by a factor of 0.8. Therefore during each 5 year leg of the trip she aged 4 years for a total of 8 years. That's why I'm 10 and she's 8. Now let's visualize the experiment as it appears to red frame observers. Stacey is at rest in the red frame while Earl is moving. So red frame observers see his clock running slow. After 4 years Stacey is 4 while Earl is only 3.2. Then Stacey jumps from the red to the green frame. The green frame observer she joins has a clock reading of 8.5. Now Earl is still moving but Stacey is moving even faster. So her clock runs even slower than Earl's. In the red frame the return trip takes more than twice as long as the outward trip but Stacey's clock runs so slow that she still only ages 4 years. Earl's age increases from 3.2 to 10 years. According to the red frame the first part of the experiment lasts 4 years. Stacey is at rest so she ages 4 years. Earl is moving and he only ages 3.2 years. The second part lasts 8.5 years. Earl is still moving and he only ages 6.8 years. Stacey is moving even faster and she ages only 4 years. Earl's final age is 3.2 plus 6.8 equals 10 years. Stacey's final age is 4 plus 4 equals 8 years. Finally let's visualize the experiment as it appears to green frame observers. Earl is moving and his clock runs slow. Stacey is moving even faster and her clock runs even slower. After 8.5 years Earl is 6.8 and Stacey is 4 years old. Now Stacey jumps to the green frame. Earl continues to move at the same speed but Stacey is now at rest. So Earl's clock runs slower than Stacey's. According to the green frame the first part of the experiment lasts 8.5 years. Both twins are moving. Earl ages 6.8 years and Stacey ages 4 years. After Stacey jumps to the green frame she is at rest for the second part. This lasts 4 years and Stacey ages 4 years. Earl is still moving and he ages 3.2 years. Earl's final age is 6.8 plus 3.2 equals 10 years. Stacey's final age is 4 plus 4 equals 8 years. Here we have three different descriptions of our experiment. All three agree on the final observable outcome that when Earl and Stacey are reunited Earl is 10 and Stacey is 8. But they disagree on the sequence of events leading up to this. For example in the blue and green frames Stacey initially ages more slowly than Earl. However in the red frame Earl initially ages more slowly than Stacey. We might be tempted to demand to know what quote really happened. Relativity tells us there is no unique answer to this question. What really happens depends on your frame of reference. Because there is no preferred frame of reference defining absolute space. And there is no preferred system of synchronized clocks defining absolute time. Each frame of a reference has its own description of the sequence of events. Each is equally valid and they all agree on the observable outcomes.