On Thursday, December 12, 2024 at 1:28:01 PM UTC-7 Jesse Mazer wrote:

On Thu, Dec 12, 2024 at 3:08 PM Alan Grayson <[email protected]> wrote:

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.


There is no scenario in which the two frames would disagree about whether a 
crash happens, different inertial frames always agree in their predictions 
about localized events. Do you disagree with my point that if we assume the 
car is moving inertially from before it enters the garage right up until 
the moment the front of the car reaches the back of the garage, then in 
this case both the garage frame and the car frame will both predict a crash?

 
I disagree. Since the car's length is greater than the garage's length from 
the pov of the car's frame, the car's frame predicts a crash, but ISTM the 
garage frame predicts the car will fit inside the garage, so no crash is 
predicted, provided of course, that the car stops when fully inside the 
garage. Note that the garage's length remains 10' in garage's frame, but 
the length of car is Lorentz contracted to 7.2'.  AG 

 

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.*


Acceleration is only relevant to the twin paradox if it happens between the 
initial and final events you are analyzing, specifically between the moment 
the two twins depart from a common location and when they reunite at a 
common location. Any acceleration done by either twin *before* the 
departure moment would be completely irrelevant to predicting their ages on 
reuniting, if you know their ages at departure and subsequent paths between 
departure and reuniting, that's sufficient to get a prediction of their 
ages on reuniting that doesn't depend in any way on what happened prior to 
them departing from one another. Similarly, in the car/garage paradox we 
can assume some initial conditions where the front of the car has not yet 
entered the garage and both the car and the garage is moving 
inertially--what happened *before* those initial conditions will be 
irrelevant to the analysis of what happens after, for example it makes no 
difference if the car accelerated before that moment, or if we replace the 
car with space rock with the same rest length that has been moving 
inertially for billions of years, while the garage is mounted to a rocket 
and accelerated towards the space rock shortly before the initial 
conditions. 


I'm not sure. It does seem there's a difference between the frames, but you 
could be correct here. Nonetheless, both frames don't predict a crash at 
back door of garage. AG

Jesse
 

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 

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