One of the basic processes of small-scale matter is collision - e.g. the molecues of a gas bouncing perpetually off one another, sustaining steady pressure. At least by the time we come to consider reactions between colliding particles, we need to allow for `inelastic' scattering. The easiest form of this to imagine is two molecules, colliding, each of which is capable of internal motions - oscillations of bond-lengths or of angles between bonds, for instance - hence allowing some of the energy of the collision to come from this internal reservoir and some to be left there after the collision. Consequently, although total energy is conserved, the externally visible kinetic energy of the molecules after the collision need not add up to the same total after the collision as before - the difference has come from or gone to internal intergy.

The archetypical collision involves two bodies, masses m and M moving with velocities v and u. Their centre of mass has momentum m.v+M.u and mass m+M so, relative to it, the two bodies' velocities are

- v - (m.v+M.u) / (m+M) = M.(v-u)/(m+M) and
- m.(u-v)/(m+M)

respectively; this gives total momentum 0, as expected, and kinetic energy

- (M.(v-u)/(m+M))**2 . m / 2 + (m.(u-v)/(m+M))**2 . M / 2
- = (v-u).(v-u).(m.M.M + M.m.m) / (m+M) / (m+M) / 2
- = m.M.(v-u).(v-u) / (m+M) / 2

That's how much energy is availabl in a collision as seen by an `outside' observer. In practice, I'll hereafter work in the rest frame of the centre of gravity of the colliding particles; again I'll call their masses m and M, again I'll call their velocities v and u; but the total momentum, m.v+M.u, is zero and the kinetic energies are just m.v.v/2 and M.u.u/2. Note that, with m.v+M.u = 0, we obtain

- (m+M).(m.v.v + M.u.u)
- = (m+M).(m.v.v + M.u.u) - (m.v+M.u)**2, as this just subtracts 0
- = m.m.v.v + M.m.v.v + m.M.u.u + M.M.u.u - (m.m.v.v + M.M.u.u + 2.M.m.v.u)
- = M.m.(v.v + u.u - 2.v.u)
- = M.m . (v-u)**2

so the general formula agrees (as it should) with the specific.

The available kinetic energy and the internal energies of the two molecules combine to give the total energy before or after the collision; and total energy is conserved. In the absence of any internal energy, or if transfer between it and (external) kinetic energy is impractical, we can expect the available kinetic energy to be conserved.

It seems reasonable to suppose that energy has more opportunity to pass between internal and external in a roughly head-on collision than in a glancing one. To quantify this, I'll need to have some description of `how far off head-on is the collision'. We can extrapolate their trajectories, when far from one another, forwards to where they meet; extrapolating without taking account of one another, we can observe how close their centres get to one another: if extrapolation puts their centres in one place together, that's a head-on collision; otherwise, the distance of closest extrapolated approach provides a measure of how far off head-on they were.

This closest-approach distance-between-centres is an offset inferred by extrapolating trajectories far from the collision into the vicinity of the collision; it can be done using the incoming or out-bound trajectories; it will be interesting to compare the offsets obtained from these two cases.

For each participant in the collision, in the simple case where movement far from the collision is in straight lines, we have in-bound and emergent trajectories which we can use as asymptotes for hyperbolae, though we'll need a length scale to chose which of these to use. The choice could be made by considering the two particles' hyperbolae in conjunction and solving for their length scales from the requirements of energy and momentum conservation as the particles traverse the trajectories (but possible transfers of energy between internal and external modes in the real collision may cause this to produce bizarre answers).

Back in the reference frame of the centre of mass, we have masses m and M inbound (roughly) towards the origin from opposite directions with equal (but opposite) momenta m.v and M.u. Chose x axis parallel to u and v, with v negative and u positive. The two trajectories are now parallel to the x axis, one coming in from positive x, the other from negative x. Chose the y axis parallel to some line from M's trajectory to the nearest point on m's trajectory; this is perpendicular to the trajectories, hence to the x axis. Since the centre of gravity is our origin, the two trajectories must lie on opposite sides of the x axis - by construction m's trajectory is the one with positive y co-ordinate, M's is negative - and there is some fixed h (with dimensions length/mass) for which m's y co-ordinate is M.h while M's is -m.h.

Thus we have mass m at (x,y)= (v.(t-Tc), M.h) and M at (u.(t-Tc), m.h) with m.v+M.u equal to zero (and u, v are now scalars) and Tc being the time of closest approach, as extrapolated from these prior trajectories. The distance of closest approach is then (m+M).h. Write r = u/m, so v = -M.r and write s(t) = (r.(t-Tc), h) so that m is at M.s(t), M is at m.s(t). All we really care about is s(t): the two bodies' trajectories are just scaled versions of it.

Total (externally visible) angular momentum is -m.v.M.h + M.u.m.h or r.h.m.M.(m+M), total kinetic energy is r.r.m.M.(m+M)/2. Any change in these must correspond to a matching change in the total internal angular momentum or energy, as appropriate, of the molecules.

A parabola with the x-axis as one asymptote and the other deflected through an angle q from it, so parallel to (-cos(q), sin(q)), satisfies the equation (sin(q).x + cos(q).y).y = const. We need to offset this from the origin to obtain (sin(q).x + cos(q).y -k).(y -h) = const where k, on the outward journey, corresponds to h in-bound: k is the distance between the origin and the nearest point to it on the asymptote of the emerging trajectory.

For the incoming and emerging asymptotes, look to where they meet. How close to the origin is this ? Compare to h and k. Asymptotes meet at y=h, sin(q).x= k-cos(q).y =k-h.cos(q), whereat |x,y|= |x.sin(q), y.sin(q)| / sin(q) = sqrt(h.h.sin(q)**2 + (k-h.cos(q))**2)/sin(q) = sqrt(h.h -2.k.h.cos(q) +k.k) / sin(q). Points of closest approach are (0, h) and (k.sin(q), k.cos(q)); difference is (-k.sin(q), h-k.cos(q)) whose length is sqrt(h.h -2.k.h.cos(q) +k.k), which is sin(q) times the distance from origin to asymptote-meeting.

For a particle deflected through an angle q, the collision mixes up the kinetic energy which was all in the x-mode but is now cos(q)**2 in x, sin(q)**2 in y. In some sense, (1-cos(q))**2 + sin(q)**2 = 2.(1-cos(q)) indicates the depth of the collision ?

Written by Eddy.$Id: collide.html,v 1.1 2001-10-17 18:00:21 eddy Exp $