This is the second time this week you ask this question. Asking the same question multiple times in different topics is only going to clutter this forum.Austin0 said:
CaptainQuasar said:
But of course, my mistake, I meant to say younger not older.JesseM said:
MeJennifer said:
In this case both will be of same age. In one case A accelerated fast (instantaneously) and for B, he accelerated slowly later to catch up with A. The time gain(?) should be related to the acceleration rate. So before A accelerates to catch up, B is younger. Bu after B catches up with his own accelerations, both are of same ageJesseM said:
No, the time gain has nothing to do with the acceleration rate, only with the geometry of the two paths in spacetime. As an analogy, if two cars are driving side by side, then car A swerves and then moves straight in a different direction while B continues to move in the same direction (so they are now moving apart), then later car B swerves again so it crosses paths with A at a later point, it will always be true that B traveled a greater distance (as measured by his odometer) between the moment A separated from B and the moment they reunited, because A traveled in a straight line from the position where they began to move apart (immediately after the swerve) to the position where they reunited, and a straight line is always the shortest distance between two points on a 2D plane. The time along worldlines in 4D spacetime is pretty similar, except that here a "straight" (constant speed and direction) worldline between two points in spacetime always has a larger time as measured by a clock moving along that wordline than any non-straight wordline.mviswanathan said:
I still have a problem with this.JesseM said:
I wasn't talking about who was older, I was using an analogy that was just about the physical distance traveled by 2 cars on a 2D plane--the analogy doesn't involve relativity at all. Do you agree that if we look at two points on the plane--the point where A and B separated, and the point where A and B reunited--and we see that A traveled in a straight line from the first point to the second, while B traveled on a path with a bend in it (so that B's path would look like two sides of a triangle while A's would look like the third side), then because "the shortest distance between two points in a plane is a straight line", we can be sure that A's path will have a shorter distance? Do you agree that this is true regardless of whether it was A or B that initially swerved to cause them to begin to separate, assuming that prior to that they had been traveling next to each other along the same path? Remember, no relativity here, this is just a question about plane geometry.mviswanathan said:
I'm talking about B's entire path from the point in the plane where A and B depart from one another (because A swerves) to the point in the plane where A and B reunite. B's path is not a single straight line between these points, it's two line segments at different angles. For example, suppose A and B were both initially moving vertically (0 degrees), then A swerves so he's moving at 30 degrees to the right of the vertical, while B continues to move along a vertical path at 0 degrees. Later, B swerves so he's moving at 45 degrees to the right of the vertical, so he's able to meet with A again. In this case, A's path from the point where they depart to the point where they reunite is a single 30-degree line segment, but B's path from the point where they depart to the point where they reunite is a combination of a 0-degree line segment and a 45-degree line segment. That's why I said B's path looked like two sides of a triangle while A's path looked like the third side. Since A follows a straight line between these two points, while B follows a path that has a bend in it, we know A's path between these points will have a shorter distance that B's path between the same two points since in 2D plane geometry a straight line is always the shortest distance between two points.mviswanathan said:
Usually for introductory problems we only deal with instantaneous accelerations. The reason is simple, instantaneous accelerations make polygonal spacetime diagrams where the intervals can be summed, but gentle accelerations make smooth curves where you need to evaluate an integral to calculate the interval. But conceptually there is no difference, you are still just measuring the interval along the worldline.mviswanathan said:
I updated the diagram with Lorentz transformation. With Case -1 as the base, I corrected the coordinated for 1, b, c and d in Case-2. The case-1 is with respect to v/c as 0.25.CompuChip said:
