Timeline for Is it possible to use two different metrics for one observer in distinct times (related to twin paradox)
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2019 at 4:00 | comment | added | JEB | If you go back to the 2 lightning flashes that are at the front and back to the train car (simultaneous for the guy at the station), if said guy suddenly jumps on the train when the lightning strikes, the front strike moves to his new past, and the rear strike moves to his new future. If he then jumps onto an opposite moving train, the front flash is now in his future and the rear flash is in his past. It's all outside the light cone, so like Mermin said: the simultaneity of distant events has no inherent meaning. Basically you can tilt "now" ($t=0$) as close to light cone as you want. | |
| Jun 30, 2019 at 16:57 | vote | accept | Paradoxy | ||
| Jun 27, 2019 at 5:58 | comment | added | Paradoxy | I am convinced,your answer also solve my main problem,i should just use proper values for initial conditions,using 2 different metrics doesn't matter.However my last question is, as i said above assume that S' is in the middle of the way where he retrieves a light signal signal from S. He will see S is too old now (because S is sent to the past by factor vx/c^2, where x is the distance between two planets), if S' suddenly stops, what will happen to S age?(He can't be too old by then)You said S' by continuing his motion can send S to the past more,but i don't get it from Lorentz transformation. | |
| Jun 27, 2019 at 5:37 | comment | added | JEB | @Paradoxy It's very odd. For any event outside the light cone, say one that happens in 2021 at alpha centauri...there is a reference frame right here on Earth, right now, where it is in the past. | |
| Jun 25, 2019 at 17:02 | comment | added | Paradoxy | So in other words, just as S' starts its movement, he will consider $t \neq t'$ at $t'=0$. But this arises another question, what he will see from his telescope after a year or so (which light had enough time to cross distance between S and S')? I mean, S' will see S ages a lot suddenly in the middle of way? Isn't it kind of you know, odd? | |
| Jun 25, 2019 at 15:22 | comment | added | JEB | $t = t' - vx'$. So when $v$ goes from zero to near $c$, events in the forward direction get a new time coordinate that is before where they used to be. All this is happening outside the light cone, so the relative timing really is meaningless. | |
| Jun 24, 2019 at 17:17 | comment | added | Paradoxy | Ok your answer is a lot better now, but can you prove "only because his definition of "now" at $s_i$ jumped into the future, relative to the initial conditions." ?It does make sense, but how can we understand this from formulas? check my other question here: physics.stackexchange.com/questions/487692/… it's 100% related! it's as if position of $s_i$ doesn't matter. | |
| Jun 24, 2019 at 14:14 | comment | added | Paradoxy | I am not sure how you are going to draw diagram in S' frame (it can be done in S frame easily though) | |
| Jun 24, 2019 at 14:00 | comment | added | JEB | If you draw a Minkowski diagram, the "now" line (also called the hyperplane of simultaneity) tilts (it's always a straight line), but the angle changes depending on velocity. For instance, our definition of "now" at the Andromeda galaxy cycles about 17 hours as we rotate around the globe each day. Events in the morning's past can be in the evening's future. | |
| Jun 24, 2019 at 13:55 | history | edited | JEB | CC BY-SA 4.0 |
added 1362 characters in body
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| Jun 24, 2019 at 13:41 | comment | added | Paradoxy | Can you prove this line "but just by moving, he sends it back to the past" by math? | |
| Jun 24, 2019 at 13:18 | comment | added | JEB | That's why I said last week (and got down voted) that the Andromeda Paradox is at the heart of the Twin Paradox: if you change your velocity, "now" at distance locations can change bigly. The S,S' synchronization events (s, s') are space-like separated, so when S' moves towards S, "s" is now in the distant past. If he then stops, it is only 0.07s in the past, but just by moving, he sends it back to the past. If he turns around, it is then in his future. | |
| Jun 24, 2019 at 8:57 | comment | added | Paradoxy | At the end of the acceleration S is not much older according to S', that's the problem. He will be older like 0.07sec(see my calculation), and this is the core of the problem where S' sees S ages slowly, he will deduce that in their meeting S will be younger. Also note that my question is the title itself, not that example | |
| Jun 24, 2019 at 2:41 | history | answered | JEB | CC BY-SA 4.0 |