On Thursday, December 12, 2024 at 12:16:24 PM UTC-7 Jesse Mazer wrote:
On Thu, Dec 12, 2024 at 5:09 AM Alan Grayson <[email protected]> wrote: On Thursday, December 12, 2024 at 1:55:07 AM UTC-7 Alan Grayson wrote: On Thursday, December 12, 2024 at 1:43:53 AM UTC-7 Quentin Anciaux wrote: Il n'y a pire sourd que celui qui ne veut pas entendre. You're afraid to answer a simple question. Will you tell the car observer that he was suffering an illusion, or what? AG The answer is the car will crash into the end wall of the garage because the frame of the car supercedes the frame of the garage, similar to why in the Twin Paradox the traveling twin ages more slowly than the Earth-bound twin due to the asymmetry inherent in its path which is identified by its acceleration. AG If the car continues to move inertially right up to the front end hitting the back wall, then both frames predict the car crashes, the garage frame just differs in saying that the back end of the car had already entered the garage prior to the crash (so the car briefly 'fit' prior to the crash). Earlier weren't you willing to consider the simpler scenario where the garage is more like a covered bridge with both ends open, so that inertial motion can continue after the front of the car passes the back of the bridge/garage? Jesse I solved the problem using the insight of what a passenger in the car would experience, as well as changing the model so that the front door of garage is open, and back door closed. Then, using Brent's parameters, since the car's length is greater than the garage's length in the car's frame, the car will crash into the back door. I claimed that the car either fits or doesn't fit, and Brent firmly rejected that claim. And by fit, I mean the car's length equals the garage's length, but this can't happen using Brent's parameters because the car is definitely longer in length than the garage. So what will the passenger experience? -- definitely. and only, a crash at the back door of garage. Could the passenger also experience a non-crash at the back door because of what the garage frame implies, that the car's length is SMALLER than the garage's length? Definitely not! It makes absolutely zero sense that the passenger would physically experience a crash and a non-crash at the back door of garage. So what the hell is going on? It then occurred to me that this situation is somewhat analogous to the Twin Paradox (TP), where the two frames seem identical, yielding an age contradiction when the twins meet. But the solution to the TP is the recognition that the frames are NOT equivalent due to the accelerations of only the traveling twin whose clock can be shown, with SR or GR (although they likely give different numerical values), that the traveling twin's clock ticks at a SLOWER rate than the clock of the Earth-bound twin, accounting for the age difference when they meet. So, how to apply the lesson of the TP to the issue at hand? How is the garage frame different from the car frame? The answer is ACCELERATION! Specifically, in the problem at hand, these frames can only be equivalent if they have *equivalent* *histories.* But they don't. To get the car's velocity v = .8c, it must be accelerated, but the garage is never physically accelerated. So the frames in the problem we've been discussing are not equivalent in similar manner as the TP frames are not equivalent. And for this reason the implications of the garage frame cannot be put on an equal footing with the car's frame, and explains why the passenger just experiences a crash at the end door of the garage, but no non-crash (which makes no sense anyway, even absent an analysis of frames). AG Le jeu. 12 déc. 2024, 09:35, Alan Grayson <[email protected]> a écrit : On Thursday, December 12, 2024 at 12:23:54 AM UTC-7 Alan Grayson wrote: On Thursday, December 12, 2024 at 12:14:07 AM UTC-7 Alan Grayson wrote: On Wednesday, December 11, 2024 at 11:44:13 PM UTC-7 Brent Meeker wrote: On 12/11/2024 10:13 PM, Alan Grayson wrote: > The observer in the car's frame denies the cat fits in the garage, > whereas the observer in the garage's frame affirms the car fits in the > garage. But what does the observer riding in the car observe? TY, AG He would obviously be observing the car's frame. I think you could have figured that out yourself. Brent I did, but IIRC, I didn't ask you. AG I was speculating that the observer riding in the car, might be in the best position to determine the reality of what's happening. AG So if the observer in the car reported that the car crashed into the back wall, would you claim he was mistaken, or say had the wrong point of view? AG -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/8e4135ce-ed10-48db-9a1c-5c0847a34bb1n%40googlegroups.com.
