Timeline for Special relativity change of frame confusion
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| May 22, 2021 at 20:52 | comment | added | Nakshatra Gangopadhay | @Iti I'm sure you are experienced in lorentz and inverse lorentz transformations. In lorentz transformation, S is at rest, S' is moving. In the inverse, S' is at rest, S is moving. What I did was just an ordinary inverse transform, but I changed the name of S' to S, and S to S', if you get what I mean. That created all the confusion. This means del x in case 2, is del x' of case 1 which is obviously 0. | |
| May 22, 2021 at 20:49 | comment | added | Nakshatra Gangopadhay | @Iti absolutely. Basically the way I defined the events in case 2 is wrong, and different from case 1. Remember, in case 1, the appearance of asteroid is event 1 and collision is event 2, and these happen at x=14.4 and 0. In the second case, the event must not change, but since the asteroid is the frame, think the events in its perspective. Event 1: where is the asteroid visible to itself. The answer is, at its own origin, since it is at rest in this frame and the world passing by. Event 2: where does the collision take place? Again at x=0. This is why, in Case 2, del x = 0. | |
| May 22, 2021 at 16:43 | comment | added | Iti | @NakshatraGangopadhay, as you have got the answer, can you please post answer to your problem, I want to see the solution. I myself get very confused | |
| May 22, 2021 at 15:22 | vote | accept | Nakshatra Gangopadhay | ||
| May 22, 2021 at 15:04 | answer | added | Professor Sushing | timeline score: 1 | |
| May 22, 2021 at 15:01 | comment | added | Nakshatra Gangopadhay | But didn't know inverse lorentz transformation means exchange of frames, so I created case 2, which turns out to be a new problem. If you see, if I exchange rocket and asteroid in Case 1, I get case 2. Then the time should have been 60 min, and I would have got time on rocket to be 58 min. By misunderstanding inverse lorentz transformation, I seemed to have created a new question, different from case 1. | |
| May 22, 2021 at 15:01 | comment | added | Nakshatra Gangopadhay | I think I found the problem. In the first case, I considered rocket to be S and asteroid to be S'. If I wanted to make the asteroid at rest and rocket to be moving, I could just take the inverse lorentz transformation, which you've said. | |
| May 22, 2021 at 14:26 | comment | added | Iti | I think in the case of with respect to asteroid frame, the time interval $58.25min$ is proper time interval. If we are sitting on particle, then everything moves but particle remains at rest. So I think $\Delta x=0$. Thus , we get $\Delta t'=\gamma\Delta t=60min\; (approx)$. Then we get consistent result. | |
| May 22, 2021 at 13:29 | comment | added | Nakshatra Gangopadhay | I've added my lorentz transformations, and you please check it ? Thank you. | |
| May 22, 2021 at 13:28 | history | edited | Nakshatra Gangopadhay | CC BY-SA 4.0 |
added 624 characters in body
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| May 22, 2021 at 12:47 | comment | added | silverrahul | If you show the your lorentz calculations in detail, then people will be able to point out, where you made any mistakes | |
| May 22, 2021 at 12:01 | comment | added | Adrian Howard | I may be confused here, but since we are only given a closing speed wouldn't symmetry give the same time to close for both rocket FOR and asteroid FOR? | |
| May 22, 2021 at 10:41 | review | First posts | |||
| May 22, 2021 at 11:19 | |||||
| May 22, 2021 at 10:38 | history | asked | Nakshatra Gangopadhay | CC BY-SA 4.0 |