![]() When B and C change their velocities to stay in A frame, A will see a nearby C clone as younger with respect to her as C with respect to B in the previous case. If they exchange selfies, they will show a younger B compared to C. When B changes its velocity to stay in the same frame of C, he now ages at the same speed of the nearby C clone, and also all of C clones including the original. The same for A and B with respect to the C clones, and all of them at the same rate ( $\approx 1:1.14)$.īut A observes B clones passing by younger and younger at a higher rate ( $\approx 1:1.61$) because of their bigger relative speed. The relative distance is kept constant between the clones of A, B and C in theirs respective frames of reference.Ĭ observes the A and B clones passing by younger and younger than him. The same for A and for B for their frames. So, C has a clone each light-day in the both directions, and all of them in its own frame. This approach avoids the complications of the delays of exchanging signals. Of course all selfies send to C from A and B will show the same age of A and B, if C receives them at the same time.īut to compare the time dilation of each one in relation to the other, it is better to suppose 'trains of ships' instead of just one per frame of reference. That way A and B have the same magnitude of velocity with respect to C, that is the idea (I suppose). I have also chosen nicer numbers so that many of the calculations are rational numbers (with $v_ = 0,785c$$ Chrono is the only inertial observer that is at the separation event O and reunion event Z. With the same outgoing and incoming speeds. Where Alice and Bob move in opposite directions (but I'm sure that you can change it to your own scenario) I've drawn a different situation in timekeeper-Chrono's frame This should hopefully give some intuition that you might not get from just applying formulas, especially if you want to consider piecewise-inertial trips. Here's a graphical method (a spacetime diagram on "rotated graph paper") that you could use to analyze various situations you might want to dream up. This is just my attempt to wrap my head around time dilation involving 3 parties rather than two. PS - I know this might feel like a homework problem but I assure everyone it's not. How does this apparent paradox get resolved? I'm guessing there's an error in my assumptions about what everyone's relative velocities and time dilation factors are, and I should be using the addition of velocities formula somewhere, but I'm not sure where exactly the error lies. In other words, when Bob and Chrono decelerate down to Alice's frame, it seems they should disagree on how much Alice has aged, which is of course absurd. But because of the time dilation caused while v ab = 0.97c, Alice should have aged 40 years compared to Bob's 10. Because of the time dilation caused while v ac = 0.48c, Alice should have aged 12 years while Chrono aged 11. Because the time dilation between Bob and Chrono was relatively small (1.1:1), while Bob aged 10 years, Chrono aged 11 in that same time.īut now let's say Chrono and Bob both immediately decelerate back down to Alice's frame, so v ab = v ac = v bc = 0. ![]() Suppose after 10 years from his perspective, Bob decelerates to the same frame as Chrono, so v ab = v ac = 0.48c, while v bc = 0. Assuming yes to both, can the time dilation between the twins be physically detected in ANY way prior to one of them decelerating, or is it only happening "on paper" at this point?.Do the two twins each perceive the same amount of time between their receipt of consecutive selfies relayed by Chrono?.Will the twins appear to be aging at the same rate when they compare selfies? (Put differently, do they agree on how much time has passed between the receipt of consecutive signals from Chrono?).When they get Chrono's signal, they each snap a selfie and send it to Chrono, who then immediately relays it to the other twin. Let's also pretend that everyone can accelerate and decelerate effectively instantaneously so that the amount of time they're all travelling at anything other than their target speeds is negligible.Ĭhrono, at a regular interval from his perspective, sends a signal to both Alice and Bob. This yields a time dilation factor of 1.1:1. Suppose that Chrono is going to act as a third-party time keeper, and takes off right behind Bob in the same direction but at half the speed, such that v ac and v bc are both 0.48c. Specifically, let's say Alice is on earth and Bob is an astronaut zooming away from Alice at a velocity ( v ab) of 0.97c which gives a time dilation factor of roughly 4:1. To try to understand the twin paradox better I thought of a variant involving a third party time keeper that both twins can agree on, and am trying to understand what each party would actually observe while the astronaut is moving as well as when he decelerates back to earth speed and when both return home.
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