# Computing expected number of particle collisions

1. Aug 3, 2009

### homomorphism

so i have a problem which can be formulated in the following way:

assume you have n particles and you know the initial position of each. There is some probability distribution P which contains the probability of having a particular velocity. Now, each particle has this same probability distribution of velocities. now choose some particle X and assume all other particles are moving toward X (for now we can assume X is still). In some time t, what is the expected number of collisions that X will have?

well, i know you can compute it by saying prob of s (s goes from 0 to n) particles hitting X and find expected value that way. but this requires a lot of computation. eg. if you want to compute prob of 1 particle hitting X, you need to compute probability of particle 1 hitting X and none others hitting X, or probability of particle 2 hitting and none others, or probabilty of particle 3, etc. and these all differ since they are at different places. is there some type of better way to formulate this?

2. Aug 3, 2009

### bpet

If the pairwise collision probabilities can be calculated, just add them up - the expectation of the number of events that occurs is the sum of the probabilities of the individual events, whether or not they are independent.

However, are the position/velocity distributions stationary or do they change as the number of collisions increases?

3. Aug 4, 2009

### homomorphism

i'm not sure i understand exactly what you meant by pairwise collisions.

if i have 100 particles, each with the same probability distribution of velocities going towards my particle X (assume none of these particles collide with each other but all start at different positions...each position is known before hand) it is kind of hard annoying to compute the probability of say...2 of those 100 colliding, and 98 not colliding, since there are a lot of ways this can happen...especially since there's a probability distribution of velocities. however, once i get this, the probabilities for 1 collision to 100 collisions, its easy to compute the expected value. while its definitely possible, and i know how to compute the probability of s collisions (1<= s <= 100), it is pretty tedious and not fun to implement. what i'd like to know is if there's a better way to reformulate this problem.

4. Aug 4, 2009

### bpet

Sorry I was thinking of the general problem of expected number of collisions between all particles not just X. Same principle applies though - just sum up the probabilities that X collides with the j'th particle.

Calculation of the individual collision probabilities will be easier if you can assume that the motion of X doesn't change significantly with the impacts and the particles don't interact significantly, e.g. earth-asteroid collisions.

5. Aug 5, 2009

### John Creighto

This seems more like a physics question then a math question. However, at any given time, all the particles take up a given amount of space. The velocity distribution will tell us in an instant how much everything moved. Consider a single particle with a velocity V. If the interval of time is small, the particle will occupy part of the space where it was and some new space.

Let:

v be the Velocity.
V be the volume.
N be the number of particles.
r be the radius of a particle.
P(v) be the probability of a given velocity

Each partial in an instant of time dt sweeps an an additional volume of:

$$2 \pi r^2 \ v \ dt$$

(*Note the 2 comes because the additional volume is half the surface area of a sphere multiplied by how much it moved)

The probability of this additional volume will occupy the space as another particle is:

$$2 \pi r^2 \ v \ dt ((4/3) \pi r^3N/V)$$

Therefor the expected number of collisions per unit time for a single particle is:

$$\frac{8}{3}\frac{\pi^2 \ N r^5}{V} \int_{-\infty}^{\infty} v \ P(v)dvdt$$

which simplifies to:

$$\frac{8}{3}\frac{ \pi^2 \ N r^5}{V} E[v]dt$$

To get the expected number of collisions, just multiply this by the time interval, then by the number of particles and divide by two since each collision involves two particles.

Disclaimer: The actually physics of whether a collision occurs my be more complicated. Also note, in typical derivations of a particles mean free path are based on a cylindrical being swept out. This is in contrast to the very thin half spherical shell used in the above derivation. Therefore, the above calculations should be close but isn't guaranteed to be exact.

Last edited: Aug 5, 2009