From: g.scholten@nospam.gmx.de
Luigi Fortunati wrote:
> There is an unbroken row of bowling pins (wide ) that are
> approached to each other and there is a ball (diameter=l) that
> throws down only one bowling pins if it hits it full or two
> consecutive bowling pins in other cases.
This won't work. If the ball is moving e.g. in x-direction with constant
velocity and the pins set up a row in x-direction, then the ball with
throw down the first pin, continues moving, hits the second pin, throws
it down, again continues moving, hits the third pin, throws it down, and
so on.
Alternatively, instead of considering what happens in total, you can
consider what happens until some moment in time, e.g. until the back of
the ball crosses the position of the first pin. Let's assume this
hereinafter.
> This happens if the pitcher is at rest with the bowling pins.
>
> But if the bowling pins are on a train that moves to gamma=4 and
> the ball is fired from the station, the bowling pins are contracted
> to 1/4 of their width and then the ball hits 4 or 5 bowling pins.
According to our upper assumptions, this means that if x1 is the
position of the first pin and t1 the moment when the back of the ball
crosses this position, the front of the ball crosses the position x4 of
the 4th pin at that moment in the rest frame S of the pitcher.
> However, in the train's reference frame the ball is to be contracted
> at 1/4 of its width while the bowling pins do not and then fall
> down a single bowling pin or at most two.
>
> So how many bowling pins fall down?
Let's again consider the event that the back of the ball crosses the
position of the first pin. In the train's rest frame S', the position of
the first pin is x1', the moment when the ball crosses this position is
t1'. In the frame S', the front of the ball does not cross the position
x4' of the 4th pin at the moment t1', but is mutch closer the position
x1' at this moment. This is due to the relativity of simultaneity: Let P
denote the event when the back of the ball crosses the position of the
1st pin and Q the event when the front of the ball crosses the position
of the 4th pin, then P und Q are simultaneous in the frame S, but not in
the frame S'. In S', the event Q is later than the event P. Instead of
Q, there is a third event, let's denote it R, simultaneous to P, namely
the event that the front of the ball crosses a position between the
first and the second pin.
In frame S, the event P hat the coordinates (x,t) = (x1,t1), and the
event Q the coordinates (x4,t1), indicating the both event are
simultaneous. In frame S', the event P has the coordinates (x',t') =
(x1',t1'), and the event Q the coordinates (x4',t4') with t4' > t1',
indicating that both events are not simultaneous, but Q is later than P.
So, the answer to your question is: in frame S', only one pin has fallen
down at time t1', but at the later time t4', four pins have fallen down.
In frame S, on the other hand, all four pins have fallen down at time
t1. The different situations in S und S' are due to the relativity of
simultaneity.
For more on the relativity of simultaneity see here:
https://en.wikipedia.org/wiki/Relativity_of_simultaneity
This is a very interesting animation:
https://en.wikipedia.org/wiki/Relativity_of_simultaneity#/media/
ile:Relativity_of_Simultaneity_Animation.gif
Also see:
https://www.youtube.com/watch?v=Xrqj88zQZJg&t=7s
especially between 2:50 and 4:45. As they say there at 3:50:
"From Adam's point of view, the doors did not close at the same time".
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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