Timeline for Equivalence of speed and time flow
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18 at 18:56 | comment | added | WillO | "The photons don't experience time. They can't because they always travel perpendicular to time." The path of a photon is perpendicular only to itself. | |
| Feb 18 at 15:52 | answer | added | KDP | timeline score: 1 | |
| Feb 18 at 13:43 | vote | accept | ElmoVanKielmo | ||
| Feb 18 at 0:54 | answer | added | Albertus Magnus | timeline score: 2 | |
| Feb 18 at 0:08 | comment | added | ElmoVanKielmo | @Community I want to understand if movement through time and space can be represented uniformely using a vector with constant value but varying direction. | |
| Feb 18 at 0:05 | comment | added | ElmoVanKielmo | @josephh that's more or less my point. The photons don't experience time. They can't because they always travel perpendicular to time. | |
| Feb 18 at 0:01 | comment | added | CommunityBot | Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. | |
| Feb 17 at 23:33 | comment | added | joseph h | then photons travelling at light speed would effectively be frozen in time. Light/photons do not experience time. They follow what’s called “light-like” or “null” geodesics. That is, the distance it travels from one point in spacetime to another is “zero” and it has no proper time (and this is true for all massless objects). The question about light and time has been asked here many times, so you may want to try the site search tool above for detailed information. | |
| S Feb 17 at 23:04 | review | First questions | |||
| Feb 18 at 0:01 | |||||
| S Feb 17 at 23:04 | history | asked | ElmoVanKielmo | CC BY-SA 4.0 |