On Wednesday, December 11, 2024 at 12:03:06 PM UTC-7 Brent Meeker wrote:
On 12/10/2024 10:40 PM, Alan Grayson wrote:
On Tuesday, December 10, 2024 at 11:15:16 PM UTC-7 Brent Meeker wrote:
Do I not only have provide a diagram I also have to explain it in detail
just to end this silly thread??
*Yes you do. Providing plots without the numerical values in the LT, is
useless. *
There are numerical values on the plots.
*Only on the axes. And in your earlier plots there were several typos. AG*
*I can't tell if you're drawing plots to satisfy your biases, *
You still can't tell since you seem unwilling or incapable of doing the
simplest calculations.
*True, sometimes I'm lazy, but in this case I figured you'd use v = .8c so
the sqrt would calculate easily. I'm considering a new model using an
optical scanner on the the garage end points instead of doors, which given
the length of the car and its speed, we can determine if its degree or type
of enclosure. With this observation, the car can reveal only one type of
enclosure, even though several exist (exact fit, etc.) from the car's
frame, and the garage's frame. In this model, the need for simultaneity is
minimized of possibly eliminated. I am aware that the equivalence of
inertial frames allows for frame dependent measurements, keeping the laws
of physics unchanged, but this case seems different, where there's an
underlying reality of a specific outcome which can be observed. AG *
*or if the numbers support the case you're making. Lesson learned; always
do a real proof, which means supplying the arguments, or STFU. AG *
First note by comparing the two diagrams that the car is longer than the
garage, 12' vs 10'. So the car doesn't fit at small relative speed. What
does "fit" mean? It means that the event of the front of the car
coinciding with the right-hand end of the garage is after or at the same
time as the rear of the car coinciding with the left-had end of the
garage. In both diagrams the car is moving to the right at 0.8c so
\gamma=sqrt{1-0.8^2}=0.6. Consequently, in the car's reference frame, the
garage is contracted to 6' length and when the rear of the car is just
entering the garage, the front is *simultaneously*, in the car's reference
frame, already 6' beyond the right-hand end of the garage.
Then in the garage's reference frame the car's length is contracted to
0.6*12'=7.2' so at the moment the front of the car coincides with the right
end of the garage, the rear of the car will simultaneously, in the garage
reference system, be 2.8' inside the garage as shown below.
Note that in the above diagram I have marked two simultaneous events with
small \delta's. The diagram below is just the Lorentz transform of the one
above. The two simultaneous \delta's are also in the diagram below. You
can confirm they are the same events by referring to the time blips along
the world lines, which are also just the Lorentz transforms of those
above. But clearly the events marking the simultaneous locations of the
rear and front of the car above are NOT simultaneous in the garage frame
below. Conversely, the front and rear simultaneous locations of the car
below are not simultaneous in the above diagram, as the reader is invited
to confirm by plotting them. Simultaneity is frame dependent.
Incidentally, when I was in graduate school this was still know as the
"Tank Trap Paradox". The idea was that if one dug a tank trap shorter than
the enemy tank, then the tank would just bridge the hole, UNLESS the tank
were going very fast in which its contracted length would allow it to fall
into the trap. This was being explained to me by Jurgen Ehlers, whom you
may correctly infer from his name was a German professor recently hired at
Univ Texas. I said, "What is it with you Germans, illustrating things with
tank traps and cats in boxes with poison gas?" Jurgen who was too young to
have fought in the war didn't realize I was pulling his leg and he was
struck speechless.
Brent
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