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Particle moves in square, what is the expected distance before first return to edge?
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There is a unit square with a particle moving in it. After the particle collides with an edge, the angle of reflection is random and is drawn from the uniform distribution on $[-fracpi2, fracpi2].$
The question is to find the average distance the particle covers before it returns to the same edge next time (I guess that after a large number of collusions the starting point is not important).
To me, this sounds like a question about the stationary distribution of a Markov chain with a continuum of states. However, the problem actually is taken from a physics olympiad for high school students. It is claimed that the answer is $2sqrt2$.
If it is not a mistake, there probably is an intuitive non-rigorous argument why the answer is $2sqrt2$.
Update 1: I was asked to post the original text of the problem here. It is somewhat different from what I wrote above, but I believe that this is basically the same question:
Problem: In a computer model, movement of a particle inside of a square is
simulated. Square has sides of length L, the speed of the point is V.
After a collusion with an edge point bounces at a random angle
(equiprobable from -90 to 90 degrees) with the same speed. Estimate the
number of collusions with one of the sides after a large period of
time T.
Answer: $fracTV2sqrt2L$.
Update 2: There were attempts to do a simulation (see comments below), and the results tend to be somewhat smaller than $2sqrt2$.
Also, in my simulation the distribution of collision points is not uniform (points close to angles are more frequent) and distribution of distance from bounce to bounce is asymmetric and bimodal.
probability markov-chains
asked Jul 30 at 18:52
Hypsoline
1027
 |Â
show 16 more comments
up vote
11
down vote
favorite
There is a unit square with a particle moving in it. After the particle collides with an edge, the angle of reflection is random and is drawn from the uniform distribution on $[-fracpi2, fracpi2].$
The question is to find the average distance the particle covers before it returns to the same edge next time (I guess that after a large number of collusions the starting point is not important).
To me, this sounds like a question about the stationary distribution of a Markov chain with a continuum of states. However, the problem actually is taken from a physics olympiad for high school students. It is claimed that the answer is $2sqrt2$.
If it is not a mistake, there probably is an intuitive non-rigorous argument why the answer is $2sqrt2$.
Update 1: I was asked to post the original text of the problem here. It is somewhat different from what I wrote above, but I believe that this is basically the same question:
Problem: In a computer model, movement of a particle inside of a square is
simulated. Square has sides of length L, the speed of the point is V.
After a collusion with an edge point bounces at a random angle
(equiprobable from -90 to 90 degrees) with the same speed. Estimate the
number of collusions with one of the sides after a large period of
time T.
Answer: $fracTV2sqrt2L$.
Update 2: There were attempts to do a simulation (see comments below), and the results tend to be somewhat smaller than $2sqrt2$.
Also, in my simulation the distribution of collision points is not uniform (points close to angles are more frequent) and distribution of distance from bounce to bounce is asymmetric and bimodal.
probability markov-chains
asked Jul 30 at 18:52
Hypsoline
1027
1
@joriki I did it, thanks.
– Hypsoline
Jul 31 at 9:52
1
@joriki Thank you so much! I tried to do a simulation too, and the result seems to be correct! At least, in my simulation the average distance the point covers from bounce to bounce is close to sqrt2/2. This only makes the things more interesting.
– Hypsoline
Jul 31 at 11:21
2
@joriki Just a small update. I consistently get something like 2.76 as average distance of return, while 2*sqrt2 approximately equals 2.83. It might be some computational issue or a mistake in the code, but it also can be that the answer actually is wrong.
– Hypsoline
Jul 31 at 14:50
1
@Hypsoline: Interesting, thanks for the update! I'll write my own code to check, but I won't be getting around to this before tomorrow at the earliest. Note that the problem statement says "estimate", so there could be an argument why the value should be roughly but perhaps not exactly $2sqrt2$. (A very rough but I think ultimately unjustifiable argument could be made for it to be $pi$, which is also not too far off.)
– joriki
Jul 31 at 14:55
1
I also find something a little smaller than $2sqrt2$.
– Julian Rosen
Jul 31 at 14:59
 |Â
show 16 more comments
up vote
11
down vote
favorite
up vote
11
down vote
favorite
There is a unit square with a particle moving in it. After the particle collides with an edge, the angle of reflection is random and is drawn from the uniform distribution on $[-fracpi2, fracpi2].$
The question is to find the average distance the particle covers before it returns to the same edge next time (I guess that after a large number of collusions the starting point is not important).
To me, this sounds like a question about the stationary distribution of a Markov chain with a continuum of states. However, the problem actually is taken from a physics olympiad for high school students. It is claimed that the answer is $2sqrt2$.
If it is not a mistake, there probably is an intuitive non-rigorous argument why the answer is $2sqrt2$.
Update 1: I was asked to post the original text of the problem here. It is somewhat different from what I wrote above, but I believe that this is basically the same question:
Problem: In a computer model, movement of a particle inside of a square is
simulated. Square has sides of length L, the speed of the point is V.
After a collusion with an edge point bounces at a random angle
(equiprobable from -90 to 90 degrees) with the same speed. Estimate the
number of collusions with one of the sides after a large period of
time T.
Answer: $fracTV2sqrt2L$.
Update 2: There were attempts to do a simulation (see comments below), and the results tend to be somewhat smaller than $2sqrt2$.
Also, in my simulation the distribution of collision points is not uniform (points close to angles are more frequent) and distribution of distance from bounce to bounce is asymmetric and bimodal.
probability markov-chains
asked Jul 30 at 18:52
Hypsoline
1027
There is a unit square with a particle moving in it. After the particle collides with an edge, the angle of reflection is random and is drawn from the uniform distribution on $[-fracpi2, fracpi2].$
The question is to find the average distance the particle covers before it returns to the same edge next time (I guess that after a large number of collusions the starting point is not important).
To me, this sounds like a question about the stationary distribution of a Markov chain with a continuum of states. However, the problem actually is taken from a physics olympiad for high school students. It is claimed that the answer is $2sqrt2$.
If it is not a mistake, there probably is an intuitive non-rigorous argument why the answer is $2sqrt2$.
Update 1: I was asked to post the original text of the problem here. It is somewhat different from what I wrote above, but I believe that this is basically the same question:
Problem: In a computer model, movement of a particle inside of a square is
simulated. Square has sides of length L, the speed of the point is V.
After a collusion with an edge point bounces at a random angle
(equiprobable from -90 to 90 degrees) with the same speed. Estimate the
number of collusions with one of the sides after a large period of
time T.
Answer: $fracTV2sqrt2L$.
Update 2: There were attempts to do a simulation (see comments below), and the results tend to be somewhat smaller than $2sqrt2$.
Also, in my simulation the distribution of collision points is not uniform (points close to angles are more frequent) and distribution of distance from bounce to bounce is asymmetric and bimodal.
probability markov-chains
asked Jul 30 at 18:52
Hypsoline
1027
edited Jul 31 at 15:17
asked Jul 30 at 18:52
Hypsoline
1027
asked Jul 30 at 18:52
Hypsoline
1027
asked Jul 30 at 18:52
Hypsoline
1027
1027
1
@joriki I did it, thanks.
– Hypsoline
Jul 31 at 9:52
1
@joriki Thank you so much! I tried to do a simulation too, and the result seems to be correct! At least, in my simulation the average distance the point covers from bounce to bounce is close to sqrt2/2. This only makes the things more interesting.
– Hypsoline
Jul 31 at 11:21
2
@joriki Just a small update. I consistently get something like 2.76 as average distance of return, while 2*sqrt2 approximately equals 2.83. It might be some computational issue or a mistake in the code, but it also can be that the answer actually is wrong.
– Hypsoline
Jul 31 at 14:50
1
@Hypsoline: Interesting, thanks for the update! I'll write my own code to check, but I won't be getting around to this before tomorrow at the earliest. Note that the problem statement says "estimate", so there could be an argument why the value should be roughly but perhaps not exactly $2sqrt2$. (A very rough but I think ultimately unjustifiable argument could be made for it to be $pi$, which is also not too far off.)
– joriki
Jul 31 at 14:55
1
I also find something a little smaller than $2sqrt2$.
– Julian Rosen
Jul 31 at 14:59
 |Â
show 16 more comments
1
@joriki I did it, thanks.
– Hypsoline
Jul 31 at 9:52
1
@joriki Thank you so much! I tried to do a simulation too, and the result seems to be correct! At least, in my simulation the average distance the point covers from bounce to bounce is close to sqrt2/2. This only makes the things more interesting.
– Hypsoline
Jul 31 at 11:21
2
@joriki Just a small update. I consistently get something like 2.76 as average distance of return, while 2*sqrt2 approximately equals 2.83. It might be some computational issue or a mistake in the code, but it also can be that the answer actually is wrong.
– Hypsoline
Jul 31 at 14:50
1
@Hypsoline: Interesting, thanks for the update! I'll write my own code to check, but I won't be getting around to this before tomorrow at the earliest. Note that the problem statement says "estimate", so there could be an argument why the value should be roughly but perhaps not exactly $2sqrt2$. (A very rough but I think ultimately unjustifiable argument could be made for it to be $pi$, which is also not too far off.)
– joriki
Jul 31 at 14:55
1
I also find something a little smaller than $2sqrt2$.
– Julian Rosen
Jul 31 at 14:59
1
1
@joriki I did it, thanks.
– Hypsoline
Jul 31 at 9:52
@joriki I did it, thanks.
– Hypsoline
Jul 31 at 9:52
1
1
@joriki Thank you so much! I tried to do a simulation too, and the result seems to be correct! At least, in my simulation the average distance the point covers from bounce to bounce is close to sqrt2/2. This only makes the things more interesting.
– Hypsoline
Jul 31 at 11:21
@joriki Thank you so much! I tried to do a simulation too, and the result seems to be correct! At least, in my simulation the average distance the point covers from bounce to bounce is close to sqrt2/2. This only makes the things more interesting.
– Hypsoline
Jul 31 at 11:21
2
2
@joriki Just a small update. I consistently get something like 2.76 as average distance of return, while 2*sqrt2 approximately equals 2.83. It might be some computational issue or a mistake in the code, but it also can be that the answer actually is wrong.
– Hypsoline
Jul 31 at 14:50
@joriki Just a small update. I consistently get something like 2.76 as average distance of return, while 2*sqrt2 approximately equals 2.83. It might be some computational issue or a mistake in the code, but it also can be that the answer actually is wrong.
– Hypsoline
Jul 31 at 14:50
1
1
@Hypsoline: Interesting, thanks for the update! I'll write my own code to check, but I won't be getting around to this before tomorrow at the earliest. Note that the problem statement says "estimate", so there could be an argument why the value should be roughly but perhaps not exactly $2sqrt2$. (A very rough but I think ultimately unjustifiable argument could be made for it to be $pi$, which is also not too far off.)
– joriki
Jul 31 at 14:55
@Hypsoline: Interesting, thanks for the update! I'll write my own code to check, but I won't be getting around to this before tomorrow at the earliest. Note that the problem statement says "estimate", so there could be an argument why the value should be roughly but perhaps not exactly $2sqrt2$. (A very rough but I think ultimately unjustifiable argument could be made for it to be $pi$, which is also not too far off.)
– joriki
Jul 31 at 14:55
1
1
I also find something a little smaller than $2sqrt2$.
– Julian Rosen
Jul 31 at 14:59
I also find something a little smaller than $2sqrt2$.
– Julian Rosen
Jul 31 at 14:59
 |Â
show 16 more comments
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1
@joriki I did it, thanks.
– Hypsoline
Jul 31 at 9:52
1
@joriki Thank you so much! I tried to do a simulation too, and the result seems to be correct! At least, in my simulation the average distance the point covers from bounce to bounce is close to sqrt2/2. This only makes the things more interesting.
– Hypsoline
Jul 31 at 11:21
2
@joriki Just a small update. I consistently get something like 2.76 as average distance of return, while 2*sqrt2 approximately equals 2.83. It might be some computational issue or a mistake in the code, but it also can be that the answer actually is wrong.
– Hypsoline
Jul 31 at 14:50
1
@Hypsoline: Interesting, thanks for the update! I'll write my own code to check, but I won't be getting around to this before tomorrow at the earliest. Note that the problem statement says "estimate", so there could be an argument why the value should be roughly but perhaps not exactly $2sqrt2$. (A very rough but I think ultimately unjustifiable argument could be made for it to be $pi$, which is also not too far off.)
– joriki
Jul 31 at 14:55
1
I also find something a little smaller than $2sqrt2$.
– Julian Rosen
Jul 31 at 14:59