Timeline for How fast do you need to travel to go 35 light-years in 2 (apparent) years?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2012 at 17:35 | vote | accept | Nick T | ||
| Jun 10, 2012 at 10:07 | answer | added | N. Virgo | timeline score: 4 | |
| Jun 10, 2012 at 4:48 | history | edited | dmckee --- ex-moderator kitten |
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| Jun 10, 2012 at 4:31 | history | migrated | from math.stackexchange.com (revisions) | ||
| Jun 10, 2012 at 4:01 | answer | added | user300 | timeline score: 1 | |
| Jun 10, 2012 at 3:54 | answer | added | Nick T | timeline score: 2 | |
| Jun 10, 2012 at 3:35 | comment | added | user300 | Actually, your first formula is backwards. The proper time experienced by an observer moving with speed $v$ for time $T$ is $t' = t\sqrt{1 - v^2/c^2}$. Otherwise, you would experience more time the faster you go, in contradiction to the twin paradox. | |
| Jun 10, 2012 at 3:27 | history | asked | Nick T | CC BY-SA 3.0 |