Probability of Particle Collision on a Closed Surface

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SUMMARY

The probability of particle collision on a closed surface, such as a square, is influenced by the motion of the particles. If both particles are moving, the likelihood of collision increases compared to a scenario where one particle has a velocity of zero. The discussion emphasizes that independent motions and the average relative motion of the particles play critical roles in determining collision probabilities. For mathematical proof, the variance of independent variables is referenced, specifically the formula σX+Y² = σX² + σY².

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Fasso
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Is it more probable that particles will collide if both are moving or if one has velocity equal to 0?
Let's say we don't have any forces between them and they're on a closed surface (for example a square).
 
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Fasso said:
Is it more probable that particles will collide if both are moving or if one has velocity equal to 0?
Let's say we don't have any forces between them and they're on a closed surface (for example a square).
There is no such thing as "velocity equal to 0". EVERYTHING is moving in some frame of reference. Since you say "on a closed surface" I assume you mean relative to that surface, yes?

So, what do you think the answer is and why?
 
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Yes, relative to that surface, of course.
I don't know, I think it's more likely that they will collide if both particles are moving, but I don't have the mathematical proof for that opinion.
 
I assume you mean not only that there is no force between them, but also that no common force is affecting them both. So their motions are independent of each other. And I assume you are talking about the probability of colliding in a fixed amount of time. You can easily say something about the extreme cases: 1) no motion at all; 2) a lot of fast motion.
Draw your own conclusions.
To get a mathematical proof for the general case, you would have to be very specific about the situation.
 
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FactChecker said:
I assume you mean not only that there is no force between them, but also that no common force is affecting them both. So their motions are independent of each other. And I assume you are talking about the probability of colliding in a fixed amount of time. You can easily say something about the extreme cases: 1) no motion at all; 2) a lot of fast motion.
Draw your own conclusions.
To get a mathematical proof for the general case, you would have to be very specific about the situation.

Exactly. Well, I don't have any idea how to set up this problem. Maybe it's more probable that if one of the particles is standing still relative to the surface, they'll collide in the middle due to Gauss's law. But, in my opinion, it's more likely that they'll collide if both particles are moving.
 
Fasso said:
Exactly. Well, I don't have any idea how to set up this problem. Maybe it's more probable that if one of the particles is standing still relative to the surface, they'll collide in the middle due to Gauss's law. But, in my opinion, it's more likely that they'll collide if both particles are moving.
It probably depends on the average relative motion. If one point zipps around very fast then it obviously is more likely to collide. A rough conclusion can be obtained from the fact that for independent X and Y, the variance of X+Y is σX+Y2 = σX2 + σY2
So a non-zero separation is more likely to reach zero.
 
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FactChecker said:
It probably depends on the average relative motion. If one point zipps around very fast then it obviously is more likely to collide. A rough conclusion can be obtained from the fact that for independent X and Y, the variance of X+Y is σX+Y2 = σX2 + σY2
So a non-zero separation is more likely to reach zero.

Thank you for answer. What is the conclusion then? Which case is more likely to happen? And how did we get that formula? Is it some definition or?
 
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