Timeline for Twin paradox - observers counter orbiting Earth
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 10, 2012 at 12:56 | comment | added | Leos Ondra | OK, don't worry. Your current answer is already very detailed :-) | |
| Apr 10, 2012 at 2:42 | comment | added | David Z | @LeosOndra actually this turns out to be more complicated than I thought. It may not actually be the Kerr metric that is the appropriate choice for this situation; I'm not sure if some coordinate-transformed version of the Schwarzschild metric would be appropriate instead... someone who's more familiar with how these things work would have to explain this one. If I can find such a person I'll see if I can ask them about it. | |
| Apr 7, 2012 at 22:09 | comment | added | David Z | @Leos I will do that when I have a chance but it might have to wait for a day or so. | |
| Apr 7, 2012 at 20:23 | vote | accept | Leos Ondra | ||
| Apr 7, 2012 at 20:23 | comment | added | Leos Ondra | Thanks for the additional comments. If you have time later to add the Kerr metric solution I would much appreciate it. | |
| Apr 7, 2012 at 19:03 | comment | added | David Z | Now, if you wanted to calculate the rate at which time passes for C as observed by B, you would have to do a more complicated calculation using the Kerr metric, because B observes the Earth to be rotating relative to its own inertial frame. I could try to add that in (when I have time) if you want. | |
| Apr 7, 2012 at 18:56 | comment | added | David Z | All I've calculated here is the difference between the various observers' clocks that accumulates over one orbit. Since all the observers meet at the same position after the orbit, this quantity is the same no matter who is observing it - in other words, there is no distant observer actually involved. The calculation tells you what A,B,C themselves would actually measure. Of course a distant observer would see the same thing as well. | |
| Apr 7, 2012 at 10:22 | comment | added | Leos Ondra | Thanks for detailed answer. How would be the entire situation seen or calculated by one of the orbiting observers (rather than Schwarzschild far away from Earth)? I assume that for him/her (B) the clock of C will has different rate but that the dilatation due to relative velocity average out during the orbit? | |
| Apr 7, 2012 at 1:07 | history | answered | David Z | CC BY-SA 3.0 |