Elastic and Inelastic Collision in Three Dimensions
Firstly a note in order to avoid any misunderstandings: the exact kinematics of a particle collision is rarely of interest in plasma physics as it is impractical to track a large number of particles individually. I have derived the relationships below actually in a different context but could not find a better place to put them. Also, note that for the determination of the 'impact angle', I used here the hard sphere model which obviously would only be an approximation for charged particles (strictly speaking the Rutherford scattering formula would have to be used for this). Apart from this, the solution below is a completely general and exact description of a 3D collision event (and in any case it provides exact conservation of momentum and energy).
The collision in three dimensions can be treated analogously to the collision in two dimensions
. However, because of the additional dimension there are now two angles required to specify the velocity vector of ball 2 after the collision. It is convenient to choose these angles as polar coordinates, so that the x- and y- components of this vector can be expressed in terms of the z-component and these angles. With this, the conservation equations for momentum and energy can be written as:
(1) m1.vz,1 = m1.vz,1' + m2.Δvz,2'
(2) m1.vx,1 = m1.vx,1' + m2.Δvz,2'.tan(θ).cos(φ)
(3) m1.vy,1 = m1.vx,1' + m2.Δvz,2'.tan(θ).sin(φ)
(4) m1/2.(vx,12+vy,12+vz,12) =
As for the 2D case, it has been assumed here that mass 2 is initially resting i.e. vx,2
=0, and vz,2
' (this does not affect the general validity of the final result as the assumption will be dropped again later by referring the velocities to mass 2 explicitly).
Solving now Eq.(1) for vz,1
' , Eq.(2) for vx,1
' and Eq.(3) for vy,1
(5) vz,1' = vz,1 - m2/m1.Δvz,2'
(6) vx,1' = vx,1 - m2/m1.Δvz,2'.tan(θ).cos(φ)
(7) vy,1' = vy,1 - m2/m1.Δvz,2'.tan(θ).sin(φ)
Inserting Eqs.(5), (6) and (7) into Eq.(4) results in a quadratic equation for Δvz,2
' which (after some lengthy but straightforward algebraic manipulations) yields the solution:
Δvz,2' = 2[ vz,1 + tan(θ).( cos(φ). vx,1 + sin(φ). vy,1) ] / [(1+tan2(θ)).(1+m2 /m1 )] .
With Eqs.(1)-(3) and (5)-(7) the velocity components after the collision are then
(9) vz,2' = Δvz,2' ,
(10) vx,2' = tan(θ).cos(φ).Δvz,2' ,
(11) vy,2' = tan(θ).sin(φ).Δvz,2' ,
(12) vz,1' = vz,1 - m2/m1.Δvz,2' ,
(13) vx,1' = vx,1 - tan(θ).cos(φ). m2/m1.Δvz,2' ,
(14) vy,1' = vy,1 - tan(θ).sin(φ). m2/m1.Δvz,2' .
Note that in contrast to the corresponding 2D equations
, Eqs.(8)-(14) are still formulated in a coordinate system where the initial velocity of ball 2 is zero. This is because the angles θ and φ still have to be specified. The problem here is that the angles for the position and velocity vectors can not be simply added as in the 2D case as the vectors do in general not lie in a coordinate plane. One can therefore not characterise the angles by the corresponding arctan function but has to choose a different approach: the easiest way to define the scattering geometry for the 3D case is to choose the coordinate system such that the relative position vector coincides with the z-axis, i.e ball 1 is at the origin (moving with the relative velocity of both balls) and ball 2 rests on the z-axis at a point corresponding to the distance between both balls). This involves not only subtracting the velocity of ball 2 (as implicitly done already above in Eqs.(1)-(4) ) and subtracting the position vector of ball 1, but in addition a corresponding rotation of the coordinate system (in fact two successive rotations about the y and z axis). With these transformations, the required angles are then directly given by the polar coordinates of the velocity vector of ball 1. Applying the above equations and reversing the coordinate transformations gives then the velocities after the collision in the original coordinate system. The positions can then also be updated analogously to the 2D-case (for a better understanding of this, see the corresponding Fortran
Generalization to Inelastic Collisions
The generalization of the above formulae to inelastic collisions is ultimately simple: we just have to refer the velocity components (Eq.(9)-(14)) to the center of mass reference frame, apply the restitution coefficient to these, and add again the center of mass velocity to return to the lab frame, i.e. with
(15) vx,cm = ( m1.vx,1 + m2.vx,2 )/( m1+ m2)
(16) vy,cm = ( m1.vy,1 + m2.vy,2 )/( m1+ m2)
(17) vz,cm = ( m1.vz,1 + m2.vz,2 )/( m1+ m2)
(18) vx,1'' = (vx,1'-vx,cm).R + vx,cm
(19) vy,1'' = (vy,1'-vy,cm).R + vy,cm
(20) vz,1'' = (vz,1'-vz,cm).R + vz,cm
(21) vx,2'' = (vx,2'-vx,cm).R + vx,cm
(22) vy,2'' = (vy,2'-vy,cm).R + vy,cm
(23) vz,2'' = (vz,2'-vz,cm).R + vz,cm
where R is the restitution coefficient (=1 for a perfectly elastic collision; =0 for a perfectly inelastic collission (balls stick together after the collision)).