On 12/10/2024 11:44 PM, Alan Grayson wrote:
On Tuesday, December 10, 2024 at 11:40:10 PM UTC-7 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. I can't tell if you're drawing plots to satisfy
your biases, 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 *
*Brent; your numbers check out. The car fits with ease from the pov of
the garage frame, but not from the pov of the car frame. But this
bothers me since we know that all frames are equivalent in SR. How
then can two, so-called equivalent frames, gives different results?
Using the LT, measurements in different frames generally differ, but
here something more fundamental seems to be happening; namely, that
the car fits and doesn't fit, depending on the frame being analyzed. AG
*
"All frames are equivalent", doesn't mean they agree on frame dependent
things like simultaneity. They agree on physical events. That's why I
plotted the simultaneous endpoints of the car from the frame where it
wasn't moving into the frame where it was moving. It illustrates that
the car doesn't shrink, rather the identification of simultaneous
endpoints changes. The simultaneous endpoints of the car in the second
diagram, in which the car is moving, are different and closer together
in space because of the change in simultaneity, not because the car is
shorter.
Brent
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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