Classical 2-particle Mechanics - Noble Gases

[sethoflagos]

I've not really studied noble gases before, but on general principles, I understand that for 2 particles with mass ratio r, and initial scalar speeds u1,u2, the following equations should hold:

u1 + ru2 = v1 + rv2                   (conservation of momentum)

u1^2 + ru2^2 = v1^2 + rv2^2   (conservation of energy)

Leaving aside the trivial no collision solution (v1 = u1, v2 = u2), the quadratic formula gives the change in momenta during collision as:

v1 - u1 = 2r (u2 - u1) / (r+1) ; v2 - u2 = 2r (u1 - u2) / (r + 1) 

... which yields some rather puzzling consequences:

1) For a pure noble gas isotope (r = 1), v1 = u2, v2 = u1 - particle momenta are simply swapped.

2) If all particles have a uniform initial scalar speed (u2 = u1), they maintain their initial momenta indefinitely.

In both cases, there is no apparent trend from a disequilibrium particle velocity distribution toward equilibrium.

Is my reasoning correct so far? 

If so then where does the main mechanism for establishing an equilibrium particle velocity distribution come from:

a) rare 3-particle interactions?

b) quantum fluctuations?

c) something else?

 

Edited September 29, 2021 by sethoflagos
small clarification

[swansont]

3 hours ago, sethoflagos said:

) For a pure noble gas isotope (r = 1), v1 = u2, v2 = u1 - particle momenta are simply swapped.

No. If one particle is at rest, both particles will be moving after the collision.  

[sethoflagos]

1 hour ago, swansont said:

No. If one particle is at rest, both particles will be moving after the collision.  

Quoting from https://en.wikipedia.org/wiki/Elastic_collision

       "Collisions of atoms are elastic, for example Rutherford backscattering.

        A useful special case of elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta."

Is this assertion mistaken? It does at least preserve both conserved quantities (momentum & energy), which your response appears to contradict. 

[swansont]

2 hours ago, sethoflagos said:

Quoting from https://en.wikipedia.org/wiki/Elastic_collision

       "Collisions of atoms are elastic, for example Rutherford backscattering.

        A useful special case of elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta."

Is this assertion mistaken? It does at least preserve both conserved quantities (momentum & energy), which your response appears to contradict. 

Sorry, I missed the r=1 detail…but you are assuming a one-dimensional system, which is a special case. The speed distribution arises from having a 3-D system. Collisions don’t have to be head-on. Both particles will be moving after the collision, except for the head-on case.

[sethoflagos]

2 hours ago, swansont said:

Sorry, I missed the r=1 detail…but you are assuming a one-dimensional system, which is a special case. The speed distribution arises from having a 3-D system. Collisions don’t have to be head-on. Both particles will be moving after the collision, except for the head-on case.

Thanks! 

For a CoM reference frame with scattering into an arbitrary xy plane, I now have vx1 = +/- sqrt(1 - k^2) ux1; vy1 = k ux1; vx2 = -/+ sqrt(1 - k^2) ux1 / r; vy2 = - k ux1 / r for k = -1 ... +1 (0 = 'head-on')

which looks quite sufficient to fill in any gaps in the particle velocity distribution without recourse to anything exotic.