Re: TOM ROBERTS - Dono is confused, please help him out (was SR cannot determine Contraction)

Dono wrote:
On Feb 25, 8:39 pm, Tom Roberts <tjroberts...@xxxxxxxxxxxxx> wrote:
Assume the following description of the pole in the barn paradox:
-The pole is 50 feet long and a barn is 30 feet.
-What would happen if you closed both barn doors simultaneously (in
the
barn frame) after 1 foot of the pole made it all the way to the other
side of the barn, thru the exit door.
-Assume that the pole is moving at .8c and that the door closure takes
10^-21 seconds.

OK. So v/c=\beta=0.800, \gamma=1/sqrt(1-\beta^2)=1.67. Unprimed coordinates are in the barn frame, primed coordinates are in the pole frame. I use units with c=1 and distance measured in feet, with +x along the motion and with x=0 at the exit door; x'=0 is at the front of the pole. At time t=0 the front of the pole is at the exit door (the x and x' origins coincide at t=t'=0), so the time in the barn frame that 1 foot extends past the exit door is:
t = 1/0.800 = 1.25 (feet) [@]
We need to compute where the rear of the pole is at t=1.25 (feet); we know it is located at x'=-50 (feet), so we solve the Lorenz transform for the known value of x' in terms of x [#]:
x' = -50 = \gamma(x - \beta*t) = 1.67*(x - 0.800*1.25)
Solving for x:
x = -50/1.67 + 0.800*1.25 = -28.94 (feet)
The entry door is at x=-30 (feet). So at t=1.25 (feet) in the barn frame (when the front of the pole is 1 foot past the exit door), the rear end of the pole is 1.06 foot inside the entrance door.

[@} Yes, time is measured in feet -- I did not select feet as
the unit. But, of course, any unit of distance will do.

[#] We do it this way because the simultaneity is in the barn
(unprimed) frame: we know x' and t, and want x; this is the
only LT equation relating those 3 quantities.


-Will the entry door clip the pole or not?

No. Even if door closure is instantaneous (which is not possible). The rear end of the pole is already 1.06 feet past the entry door when it needs to close.


Tom Roberts
.

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