# Particle Flux Derivation

1. Nov 3, 2012

### Unto

1. The problem statement, all variables and given/known data
Consider a cylindrical vessel with cross-sectional area 1m^2

Derive the particle flux (1/4n$\bar{v}$)

2. Relevant equations

I have the solid angle:

$\Omega$ = 2π(1-cosθ)

3. The attempt at a solution

I'm assuming that the solid angle represents the full area that the particles can possibly travel through when passing through the end of the cylinder. But to be honest I don't have a clue on anything else..

2. Nov 4, 2012

### Unto

I understand that about 1/2 the particles go in opposite directions, so for one end of the cylinder you already have 1/2nv.. But you have to take into account the angular distribution of particles with a velocity distribution [v, v + dv] coming out of the end.

And this is where the solid angle comes in, but every which way I try to compute the angle to account for the directions I don't get a value 1/4 for the final answer.

3. Nov 4, 2012

### Unto

I keep getting a factor of π/2 in my answer.

So I end up getting π/4nv which is wrong. Surely someone on this forum knows how to do this? I can't find any helpful sources on the internet, yet everyone quotes it religiously when the subject of particle flux comes up.

4. Nov 5, 2012

### Staff: Mentor

Is this for an infinite or semi-infinite cylinder? With or without absorption of the particles?

What is the particle source geometry: beam impinging upon diametral surface, distribued on diametral surface, line source on axis, distributed source through the cylinder?

5. Nov 5, 2012

### Unto

Hello,

It is a finite cylinder of lets say length vdt, with a cross-sectional area A of 1cm^2

The particle 'source' is a simple 3D gas with maxwellian velocity distribution (which I have already accounted for). The gas particles bounce elastically off of the walls until they exit through the cross-section.

This is actually a problem related to Plasma Physics.

6. Nov 6, 2012

### Staff: Mentor

Ah, so the gas behaves isotropically, except for perhaps some drift velocity, or low flow velocity. This is a similar problem in neutron diffusion in which the neutron current is the same in all directions, and one shows the the current is nv/4.

7. Nov 6, 2012

### Unto

Yeah but how do I show it?

I know the particles can exit the cross-section through a solid angle Ω = 2π(1-cosθ), but ingrating over this angle gives me π/4nv.

I think I am doing it completely wrong but I honestly don't know how to approach this problem any other way. Taking velocity components Vx, Vy and Vz in every direction would overcomplicate things because from when you know that at least half of the particles will LEFT and the other RIGHT in the cylinder, you already have 1/2nv from this.

But to get 1/4nv, you need to account for the solid angle Ω of which the particles can exit (flux) through the cross section A of the cylinder.

Any pointers?