Timeline for Relativity, Lorentz Transforms and the time it takes to perceive
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5 at 12:58 | comment | added | WillO | @TKoL : Yes, definitely. | |
| Feb 5 at 12:46 | comment | added | TKoL | @WillO thank you - and the output of the transormation is the synced time of the new frame as well, so if I wanted to then calculate when someone in that other frame would see the same event, I'd have to do extra work to calculate that. | |
| Feb 5 at 12:41 | comment | added | WillO | @TKoL : If you are just now, at 11lpm, seeing an event that happened eight minutes ago on the sun, then the time to input into the Lorentz transformation for that event is 10:52. | |
| Feb 5 at 12:35 | comment | added | WillO | @TKoL ; Yes, what you are saying is exactly right. (Responding to your first comment.) | |
| Feb 5 at 12:35 | comment | added | TKoL | For just a bit of extra clarify, what I would LIKE, ideally, is to be able to take synchronized time/place of events in one frame, and convert them into synchronized time/place of events in another frame, and I thought that's what I was doing with Lorentz transforms @WillO -- am I using transforms the wrong way if I input sync time and treat the output as sync time? | |
| Feb 5 at 12:33 | comment | added | WillO | I believe the OP was asking about trying to transform the coordinates of $A$ by inputting the coordinates of $C$, which is a third alternative that the OP correctly expected to be wrong. | |
| Feb 5 at 12:32 | comment | added | TKoL | but when I do that for C and D, I'm finding out when C and D happen to me still, just in another reference frame right? When I transform C and D, I'm not finding out what an observer in that other reference frame experienced, I'm finding out what they think I experienced? (or rather, when they think I experienced it) | |
| Feb 5 at 12:16 | history | answered | Professor Sushing | CC BY-SA 4.0 |