Timeline for Twin Paradox, Why is everyone avoiding avoiding the obvious [closed]
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2017 at 1:32 | review | Reopen votes | |||
| Mar 8, 2017 at 5:46 | |||||
| Mar 8, 2017 at 1:06 | history | edited | 2Young1s | CC BY-SA 3.0 |
added 524 characters in body
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| Mar 7, 2017 at 16:03 | history | closed |
Bill N Jon Custer Kyle Kanos Ruslan John Rennie |
Needs details or clarity | |
| Mar 7, 2017 at 2:33 | comment | added | Mark A | @Cort If the geometry of space-time is assumed to be cylindrical the curvature and acceleration can be zero everywhere. This then becomes an interesting problem. | |
| Mar 7, 2017 at 2:17 | answer | added | JMLCarter | timeline score: 1 | |
| Mar 7, 2017 at 1:27 | answer | added | WillO | timeline score: 1 | |
| Mar 7, 2017 at 1:22 | comment | added | WillO | I was going to say I'd lost you at the "circular paths with no acceleration", but unsurprisingly, I see I wouldn't be the first to say that. | |
| Mar 7, 2017 at 1:09 | answer | added | Selene Routley | timeline score: 2 | |
| Mar 7, 2017 at 0:38 | comment | added | dmckee --- ex-moderator kitten | @JMLCarter "[...] leaving us free to observe special relativisitic effects?" I think we have to assume no frame dragging as well, but then the symmetry of the situation should guarantee it. | |
| Mar 7, 2017 at 0:11 | comment | added | user107153 | It's quite clear by symmetry that the two paths between two events where the spacecraft meet have the same proper time: therefore the two clocks read the same, since proper time is what clocks measure. All instances of variants of the twin paradox can be resolved by just measuring the proper time along the trajectories of the twins. | |
| Mar 6, 2017 at 23:46 | comment | added | JMLCarter | Assumption 3) if the circumference = 10lyr, then the ships will pass every 5ly, or half way around. This is a 6.25 year interval at .8c? | |
| Mar 6, 2017 at 23:44 | comment | added | JMLCarter | Assumption 1/2) it seems fair that if A and B are in the same free fall orbit in opposite directions any effect of general relativity will be the same for both of them, leaving us free to observe special relativisitic effects? | |
| Mar 6, 2017 at 23:42 | comment | added | gj255 | @DJohnM In my mind, special relativity deals with observers moving arbitrarily in Minkowski spacetime. Only when one changes the geometry of the spacetime that our observers live in, to something other than flat Minkowski, is it necessary to use the general theory. | |
| Mar 6, 2017 at 23:29 | review | Close votes | |||
| Mar 7, 2017 at 16:03 | |||||
| Mar 6, 2017 at 23:24 | comment | added | DJohnM | @gj255 Don't the two observers need to be in uniform relative motion? | |
| Mar 6, 2017 at 23:10 | comment | added | Bill N | I don't see a question. | |
| Mar 6, 2017 at 23:06 | comment | added | gj255 | @DJohnM A problem involving circular motion, whilst more complicated than typical special relativity problems involving motion at constant velocity, is still perfectly within the realm of special relativity. | |
| Mar 6, 2017 at 22:57 | comment | added | DJohnM | Your spaceships need an centripetal acceleration of around 0.6 m/s^2 to follow that circle... | |
| Mar 6, 2017 at 22:48 | comment | added | knzhou | "These ships, fly in a giant circle [...] Assumption 1: There is no acceleration." Circular motion is accelerated. | |
| Mar 6, 2017 at 22:48 | review | First posts | |||
| Mar 6, 2017 at 23:46 | |||||
| Mar 6, 2017 at 22:47 | comment | added | DJohnM | As soon as you place both spaceships into circular motion, you move out of the realm of Special Relativity... | |
| Mar 6, 2017 at 22:46 | comment | added | Cort Ammon | If you have a path which curves, you do have an acceleration. Your assumption 1 is not valid for this particular experiment. | |
| Mar 6, 2017 at 22:43 | history | asked | 2Young1s | CC BY-SA 3.0 |