I recently stumbled upon the problem of three body collisions when reading The Emperors New Mind by Roger Penrose. In the book, the collision between three spherical bodies is used as an example of non-determinism in physics, if the resulting momenta are to predicted using classical conservation laws only. In particular, defining an algebra on bodies with collision as the group operation, then it turns out that it is not even associative! To clarify, let A, B and C be three bodies, then $$ A(BC) \neq (AB)C $$ such that the outcome depends on the specific order of operations. This does not mean that this holds for specific cases of collisions between bodies with certain masses and velocities. I googled the problem and got surprisingly little information back. The best answer is this thread on stack exchange, but I found the answers quite unsatisfactory. Some comments also mention non-associativity of the collision algebra as possible explanations. Other comments mention that there are not enough constraints to solve the problem. But no one explains why it is not associative or why additional constraints are needed, in a satisfactory way. Question is, instead of decomposing the three body problem into those between two, can we define a ternary operator that is physically motivated and correctly predicts outcomes?

My first attempt was to just proceed as when analyzing two bodies. In that case, one start by imposing conservation of linear momentum. Let $p_1$ and $p_2$ bet the momenta before the collection and $\tilde p_1$ and $\tilde p_2$ after, such that $$ p_1 + p_2 = \tilde p_1 + \tilde p_2 = p_1 + J_1 + p_2 + J_2 \Longleftrightarrow J_1 = -J_2 \equiv J \newline $$ where $J_1$ and $J_2$ are the impulses on each body, which necessarily must cancel out. Preservation of energy demands that (with some abuse of notation where I use multiplication of vectors to denote inner product) $$\begin{align} \frac{1}{m_1}p_1^2 + \frac{1}{m_2}p_2^2 &= \frac{1}{m_1}\tilde p_1^2 + \frac{1}{m_2}\tilde p_2^2 = \frac{1}{m_1}(p_1 + J)^2 + \frac{1}{m_2}(p_2 - J)^2 \newline \Longleftrightarrow 0 &= 2J^T\frac{p_1}{m_1} - 2J^T\frac{p_2}{m_2} + \frac{1}{m_1}J^2 + \frac{1}{m_2}J^2 \end{align}$$

This is a quadratic equation in two variables, which does not have a unique solution. Imposing that the there surfaces are frictionless, then the impulse must be along the surface normal $n$ ($J$ now a scalar) $$\begin{align} 0 &= 2\frac{Jn^Tp_1}{m_1} - 2\frac{Jn^Tp_2}{m_2} + \frac{1}{m_1}J^2 + \frac{1}{m_2}J^2 \newline \big( \frac{1}{m_1} + \frac{1}{m_2} \big) J &= 2 n^T \big( \frac{p_2}{m_2} - \frac{p_1}{m_1} \big) \newline \big( m_1 + m_2 \big) J &= 2 n^T \big( m_1p_2 - m_2p_1 \big) \newline J &= \frac{2}{m_1 + m_2} n^T \big( m_1p_2 - m_2p_1 \big) \newline &= \frac{2 m_1m_2 }{m_1 + m_2} n^T \big( \frac{p_2}{m_2} - \frac{p_1}{m_1} \big) \newline &= \frac{2 m_1m_2 }{m_1 + m_2} n^T \big( v_2 - v_1) \newline &= 2 m^* \Delta v_n \end{align}$$ where $m^*$ is the reduced mass of the system and $\Delta v_{ n}$ the difference in velocity along the surface normal $ n$ at the point of contact.

How does does this play out with three bodies? Linear momentum conservation alone is not sufficient due to deduce that the impulses between any two are opposite. Newtons third third law can be utilized here, which states that any force exerted by one body on another have an opposite force. Newtons laws and conservation laws (which can be derived from symmetry principles à la Noether’s theorem) are equivalent, so one can probably arrive at this principle by applying additional constraints, such as angular momentum. Anyhow, we can safely setup equations using three unknown impulses $J_{12}$, $J_{13}$ and $J_{13}$ only (instead of six two body impulses $J_{ij}$), each acting along the normal of the two body contact surfaces. One could also say that, if one have an “impulse tensor” $J_{ij}$, then it has to be skew symmetric. $$ J = \begin{pmatrix} 0 & J_{12} & J_{13} \newline -J_{12} & 0 & J_{23} \newline -J_{13} & -J_{23} & 0 \end{pmatrix} $$ The equation for conservation of energy then becomes $$\begin{align} 0 &= 2\frac{p_1^T n_{12}}{m_1}J_{12} + 2\frac{p_1^T n_{13}}{m_1}J_{13} - 2\frac{p_2^T n_{12}}{m_2}J_{12} + 2\frac{p_2^T n_{23}}{m_2}J_{23} - 2\frac{p_3^T n_{13}}{m_3}J_{13} - 2\frac{p_3^T n_{23}}{m_3}J_{23} \newline &+ \frac{1}{m_1}(J_{12}^Tn_{12} + J_{13}^Tn_{13})^2 + \frac{1}{m_2}(-J_{12}^Tn_{12} + J_{23}^Tn_{23})^2 + \frac{1}{m_3}(-J_{13}^Tn_{13} -J_{23}^Tn_{23})^2 \end{align}$$ Let $\alpha = J_{12}n_{12} + J_{13}n_{13}$, $\beta = -J_{12}n_{12} + J_{23}n_{23}$ and $\gamma = -J_{13}n_{13} - J_{23}n_{23}$, then $$ 0 = 2v_1^T \alpha + 2v_2^T\beta + 2v_3^T\gamma + \frac{1}{m_1}\alpha^2 + \frac{1}{m_2}\beta^2 + \frac{1}{m_3}\gamma^2 $$ which clearly does not have a unique solution.

I does feel reasonable that one could further assume that the two body impulses be consistent with two body collisions, such that they could be solved individually for each pair of bodies. However, this is inconsistent with taking the limit of a non-associative algebra. Physically, one simply has to recognize that the classical equations for collisions do have additional hidden assumptions baked in. In reality, bodies are not infinitely stiff, but they will deform and propagate forces through shock waves. This process happens over a finite timespan, which is idealized as a “delta event”. The classical equations assumes that collisions are between two bodies and that subsequent collisions does not happen faster than the relaxation time of the system, when treated as an elastic continuum. If they do happen more frequently than this, one has to setup the elastic partial differential equations for the system (e.g. by treating them as a spring ball system, including Hertz corrections, etc.) and solve it numerically. This is where the non-associativity of three body collisions come in!

I wonder, by setting up the timings of collisions and calculating relative “collision interaction times”, if that could be used together with the constraint on energy conservation to solve this in a simpler way…