Paths in phase space

The previous problem makes it clear that our billiard table is deterministic, in that if you know the position and velocity of a ball at one time you can predict the motion of the ball for all time. In reality of course it is impossible to exactly specify the position and velocity of a ball, there is always some measurement error. For many systems, the measurement error doesn't overly matter as the error does not grow with time. Hence if you are within some tolerance to start, you will remain within that tolerance.

For example, consider our ball again. Now, however, instead of knowing that the ball starts exactly at the origin, we only know that it starts somewhere nearby the origin, i.e. within a circle of radius ϵ \epsilon of the origin. However, it still has velocity (0,1) to start. The ball bounces off the wall N times. After 4N seconds, the ball is within a circle of radius L of the origin. What is L?

ϵ \epsilon ϵ N \epsilon^N ϵ N \epsilon^{-N} ϵ 2 N \epsilon^{2N}

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1 solution

Prasun Biswas
Apr 23, 2014

From the previous problem of the set, we can see that when we consider a bounded table and a ball placed at point ( 0 , 0 ) (0,0) to be rolled later and the vel. applied on the ball is ( 0 , 1 ) (0,1) m/s, then the movement is a periodic motion where the ball returns to the starting point ( 0 , 0 ) (0,0) after every 4 4 seconds.

So, we can say that the ball returns to the original position at origin O ( 0 , 0 ) O(0,0) after 4 X \Large 4X seconds where X X is an arbitrary positive integer, i.e., X = 1 , 2 , 3 , . . . . . X=1,2,3,..... . So, if the time is a multiple of 4 4 , then the ball is at the initial position again at O O .

Now, when we consider the no. of times ball bounces off the walls as N N and after 4 N 4N seconds, the ball is again at the initial position from where the ball was rolled off as 4 N 4N is a multiple of 4 4 and this states that the ball is at original position within a circle of radius L L as the ball is actually not to start rolling from O O but in a region close to O O in a radius L L , here according to the problem and in this case, it can be observed that L = ϵ \Large L=\epsilon . Since, it is itself said in the problem that the position is not accurately determined here, so we can say that the ball is at a circle of radius L = ϵ \Large L=\epsilon of the origin after the total time 4 N 4N rather than our consideration of being at O ( 0 , 0 ) O(0,0) as the ball is not starting to roll from O O in the first place. However, when a region is considered the ball will be in different positions after every 4 4 sec, but since original starting location was near origin, so we can say that the final position tends to be at O O and more accurately say that the region is within a radius of ϵ \large \epsilon from O O .

So, from all this we conclude that, L = ϵ \Large \boxed{L=\epsilon}

Is it true that no matter where you start, you will end up at the same position after 4s going at (0,1)m/s. I think that only happens for the case where (m,n)=(0,0)?

Jyotsna Singh - 7 years, 1 month ago

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Oh yes !! Sorry for that..I am editing the solution. Thanks for reading it !!

Prasun Biswas - 7 years, 1 month ago

I have edited the solution !! If you find any more mistakes, please be sure to notify me in the comments. Thanks again for pointing it out !!

Prasun Biswas - 7 years, 1 month ago

The previous problem has been taken down. Can you explain how the ball returns to the starting point every 4 seconds?

Shreyas Ravi - 1 year, 3 months ago

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